Quantisation schemes of quantum mechanics

Advanced Quantum Mechanics 2lecture 2

Schemes of Quantum Mechanics

Yazid Delenda

Departement des Sciences de la matiereFaculte des Sciences - UHLB

http://theorique05.wordpress.com/f411/

Batna, 30 October 2014

1/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

An operator is a function that acts on a vector to produce another:

χ(x) = Qψ(x), |χ〉 = Q|ψ〉

A linear operator is one that satisfies:

Q(α|χ1〉+ β|χ2〉) = α|ψ1〉+ β|ψ2〉,

where|ψ1〉 = Q|χ1〉, |ψ2〉 = Q|χ2〉,

2/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

An operator is a function that acts on a vector to produce another:

χ(x) = Qψ(x), |χ〉 = Q|ψ〉

A linear operator is one that satisfies:

Q(α|χ1〉+ β|χ2〉) = α|ψ1〉+ β|ψ2〉,

where|ψ1〉 = Q|χ1〉, |ψ2〉 = Q|χ2〉,

2/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

An operator is a function that acts on a vector to produce another:

χ(x) = Qψ(x), |χ〉 = Q|ψ〉

A linear operator is one that satisfies:

Q(α|χ1〉+ β|χ2〉) = α|ψ1〉+ β|ψ2〉,

where|ψ1〉 = Q|χ1〉, |ψ2〉 = Q|χ2〉,

2/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Suppose that a state / wavefunction |ψ〉 of the system isdecomposed into the eigenstates / eigenfunctions |n〉 of someother operator:

|ψ〉 =∑n

an|n〉, 〈x|ψ〉 =∑n

an〈x|ψ〉, ψ(x) =∑n

ann(x)

and suppose that the operator Q’s action on the basis ofeigenstates |n〉 is known:

Q|n〉 =∑

Qmn|m〉, Qmn = 〈m|Q|n〉

with Qmn the (matrix) representation of the operator Q in thebasis |n〉 (or the operator Q in |n〉 representation).

3/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Suppose that a state / wavefunction |ψ〉 of the system isdecomposed into the eigenstates / eigenfunctions |n〉 of someother operator:

|ψ〉 =∑n

an|n〉, 〈x|ψ〉 =∑n

an〈x|ψ〉, ψ(x) =∑n

ann(x)

and suppose that the operator Q’s action on the basis ofeigenstates |n〉 is known:

Q|n〉 =∑

Qmn|m〉, Qmn = 〈m|Q|n〉

with Qmn the (matrix) representation of the operator Q in thebasis |n〉 (or the operator Q in |n〉 representation).

3/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Suppose that a state / wavefunction |ψ〉 of the system isdecomposed into the eigenstates / eigenfunctions |n〉 of someother operator:

|ψ〉 =∑n

an|n〉, 〈x|ψ〉 =∑n

an〈x|ψ〉, ψ(x) =∑n

ann(x)

and suppose that the operator Q’s action on the basis ofeigenstates |n〉 is known:

Q|n〉 =∑

Qmn|m〉, Qmn = 〈m|Q|n〉

with Qmn the (matrix) representation of the operator Q in thebasis |n〉 (or the operator Q in |n〉 representation).

3/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Suppose that a state / wavefunction |ψ〉 of the system isdecomposed into the eigenstates / eigenfunctions |n〉 of someother operator:

|ψ〉 =∑n

an|n〉, 〈x|ψ〉 =∑n

an〈x|ψ〉, ψ(x) =∑n

ann(x)

and suppose that the operator Q’s action on the basis ofeigenstates |n〉 is known:

Q|n〉 =∑

Qmn|m〉, Qmn = 〈m|Q|n〉

with Qmn the (matrix) representation of the operator Q in thebasis |n〉 (or the operator Q in |n〉 representation).

3/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Therefore the action of the operator Q on the state vector |ψ〉 is:

Q|ψ〉 =∑n

anQ|n〉 =∑mn

anQmn|m〉 =∑m

bm|m〉, bm =∑m

Qmnan

Given that the |n〉 representation Q, Qmn of an operator isknown,we want to express this operator in terms of these matrixelements.We have:

Q =1Q1

=∑mn

|m〉〈m|Q|n〉〈n|

=∑mn

Qmn|m〉〈n| (1)

4/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Therefore the action of the operator Q on the state vector |ψ〉 is:

Q|ψ〉 =∑n

anQ|n〉 =∑mn

anQmn|m〉 =∑m

bm|m〉, bm =∑m

Qmnan

Given that the |n〉 representation Q, Qmn of an operator isknown,we want to express this operator in terms of these matrixelements.We have:

Q =1Q1

=∑mn

|m〉〈m|Q|n〉〈n|

=∑mn

Qmn|m〉〈n| (1)

4/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Therefore the action of the operator Q on the state vector |ψ〉 is:

Q|ψ〉 =∑n

anQ|n〉 =∑mn

anQmn|m〉 =∑m

bm|m〉, bm =∑m

Qmnan

Given that the |n〉 representation Q, Qmn of an operator isknown,we want to express this operator in terms of these matrixelements.We have:

Q =1Q1

=∑mn

|m〉〈m|Q|n〉〈n|

=∑mn

Qmn|m〉〈n| (1)

4/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Therefore the action of the operator Q on the state vector |ψ〉 is:

Q|ψ〉 =∑n

anQ|n〉 =∑mn

anQmn|m〉 =∑m

bm|m〉, bm =∑m

Qmnan

Given that the |n〉 representation Q, Qmn of an operator isknown,we want to express this operator in terms of these matrixelements.We have:

Q =1Q1

=∑mn

|m〉〈m|Q|n〉〈n|

=∑mn

Qmn|m〉〈n| (1)

4/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Hence we deduce that an operator maybe expressed as a sum ofouter-products of the basis |n〉 with coefficients being thecorresponding matrix-elements:

Q =∑mn

Qmn|m〉〈n| (2)

In the particular case that the states |n〉 are eigenstates of theoperator Q itself with eigenvalues qn,Q|n〉 = qn|n〉, thenQmn = 〈m|Q|n〉 = qnδmn

5/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Hence we deduce that an operator maybe expressed as a sum ofouter-products of the basis |n〉 with coefficients being thecorresponding matrix-elements:

Q =∑mn

Qmn|m〉〈n| (2)

In the particular case that the states |n〉 are eigenstates of theoperator Q itself with eigenvalues qn,Q|n〉 = qn|n〉, thenQmn = 〈m|Q|n〉 = qnδmn

5/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Hence we deduce that an operator maybe expressed as a sum ofouter-products of the basis |n〉 with coefficients being thecorresponding matrix-elements:

Q =∑mn

Qmn|m〉〈n| (2)

In the particular case that the states |n〉 are eigenstates of theoperator Q itself with eigenvalues qn,Q|n〉 = qn|n〉, thenQmn = 〈m|Q|n〉 = qnδmn

5/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Hence we deduce that an operator maybe expressed as a sum ofouter-products of the basis |n〉 with coefficients being thecorresponding matrix-elements:

Q =∑mn

Qmn|m〉〈n| (2)

In the particular case that the states |n〉 are eigenstates of theoperator Q itself with eigenvalues qn,Q|n〉 = qn|n〉, thenQmn = 〈m|Q|n〉 = qnδmn

5/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

Hence we deduce that an operator maybe expressed as a sum ofouter-products of the basis |n〉 with coefficients being thecorresponding matrix-elements:

Q =∑mn

Qmn|m〉〈n| (2)

In the particular case that the states |n〉 are eigenstates of theoperator Q itself with eigenvalues qn,Q|n〉 = qn|n〉, thenQmn = 〈m|Q|n〉 = qnδmn

5/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

so the operator Q is expressed as a diagonal matrix in its owneigenstates basis:For example:

H =∑n

En|φn〉〈φn|

p =

∫p|p〉〈p|dp

We have the following properties of operators in quantummechanics:

6/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

so the operator Q is expressed as a diagonal matrix in its owneigenstates basis:For example:

H =∑n

En|φn〉〈φn|

p =

∫p|p〉〈p|dp

We have the following properties of operators in quantummechanics:

6/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

so the operator Q is expressed as a diagonal matrix in its owneigenstates basis:For example:

H =∑n

En|φn〉〈φn|

p =

∫p|p〉〈p|dp

We have the following properties of operators in quantummechanics:

6/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators

so the operator Q is expressed as a diagonal matrix in its owneigenstates basis:For example:

H =∑n

En|φn〉〈φn|

p =

∫p|p〉〈p|dp

We have the following properties of operators in quantummechanics:

6/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsCorrespondence principle:

For every classical observable there is a corresponding quantummechanical (hermitian) operator:

f(p, q)→ f

(−i~ ∂

∂x, x

)however the opposite is not true.Some quantum mechanicaloperators do not have a corresponding observable in classicalphysics, such as spin, parity, identity operator.

7/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsCorrespondence principle:

For every classical observable there is a corresponding quantummechanical (hermitian) operator:

f(p, q)→ f

(−i~ ∂

∂x, x

)however the opposite is not true.Some quantum mechanicaloperators do not have a corresponding observable in classicalphysics, such as spin, parity, identity operator.

7/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsCorrespondence principle:

For every classical observable there is a corresponding quantummechanical (hermitian) operator:

f(p, q)→ f

(−i~ ∂

∂x, x

)however the opposite is not true.Some quantum mechanicaloperators do not have a corresponding observable in classicalphysics, such as spin, parity, identity operator.

7/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsCorrespondence principle:

For every classical observable there is a corresponding quantummechanical (hermitian) operator:

f(p, q)→ f

(−i~ ∂

∂x, x

)however the opposite is not true.Some quantum mechanicaloperators do not have a corresponding observable in classicalphysics, such as spin, parity, identity operator.

7/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsCorrespondence principle:

For every classical observable there is a corresponding quantummechanical (hermitian) operator:

f(p, q)→ f

(−i~ ∂

∂x, x

)however the opposite is not true.Some quantum mechanicaloperators do not have a corresponding observable in classicalphysics, such as spin, parity, identity operator.

7/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

We want to prove the relation for the expectation value of a givenobservable Q, for a system in the state |ψ〉, which is:

〈Q〉 = Q = 〈ψ|Q|ψ〉

using the representation of the operator Q in its eigenstates basis:Q =

∑n qn|n〉〈n|, with qn the eigenvalue of the operator Q

corresponding to the eigenstate |n〉.Assuming the system is writtenin this basis |n〉 as a linear superposition:

|ψ〉 =∑

an|n〉

8/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

We want to prove the relation for the expectation value of a givenobservable Q, for a system in the state |ψ〉, which is:

〈Q〉 = Q = 〈ψ|Q|ψ〉

using the representation of the operator Q in its eigenstates basis:Q =

∑n qn|n〉〈n|, with qn the eigenvalue of the operator Q

corresponding to the eigenstate |n〉.Assuming the system is writtenin this basis |n〉 as a linear superposition:

|ψ〉 =∑

an|n〉

8/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

We want to prove the relation for the expectation value of a givenobservable Q, for a system in the state |ψ〉, which is:

〈Q〉 = Q = 〈ψ|Q|ψ〉

using the representation of the operator Q in its eigenstates basis:Q =

∑n qn|n〉〈n|, with qn the eigenvalue of the operator Q

corresponding to the eigenstate |n〉.Assuming the system is writtenin this basis |n〉 as a linear superposition:

|ψ〉 =∑

an|n〉

8/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

where we from the postulates we know that the probability thatthe measurement of the observable Q yielding the result qn is|an|2.Thus the average value, which by definition is the sum overall measured values of the observable weighed with probability foreach value, is:

〈Q〉 ≡∑n

qn|an|2

=∑n

qn|〈n|ψ〉|2

=∑n

qn〈ψ|n〉〈n|ψ〉

=〈ψ|(∑

n

qn|n〉〈n|)|ψ〉

=〈ψ|Q|ψ〉 (3)

9/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

where we from the postulates we know that the probability thatthe measurement of the observable Q yielding the result qn is|an|2.Thus the average value, which by definition is the sum overall measured values of the observable weighed with probability foreach value, is:

〈Q〉 ≡∑n

qn|an|2

=∑n

qn|〈n|ψ〉|2

=∑n

qn〈ψ|n〉〈n|ψ〉

=〈ψ|(∑

n

qn|n〉〈n|)|ψ〉

=〈ψ|Q|ψ〉 (3)

9/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

where we from the postulates we know that the probability thatthe measurement of the observable Q yielding the result qn is|an|2.Thus the average value, which by definition is the sum overall measured values of the observable weighed with probability foreach value, is:

〈Q〉 ≡∑n

qn|an|2

=∑n

qn|〈n|ψ〉|2

=∑n

qn〈ψ|n〉〈n|ψ〉

=〈ψ|(∑

n

qn|n〉〈n|)|ψ〉

=〈ψ|Q|ψ〉 (3)

9/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

where we used in the second equality the fact that an = 〈n|ψ〉,and in the last equality we used the relation (2)

10/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian conjugate

if the operation of the operator Q on the vector |ψ〉 yields thevector |χ〉, i.e |χ〉 = Q|ψ〉,then in the dual space the operator thatacts in the bra 〈ψ| to produce the bra 〈ψ| is the hermitian adjoint〈χ| = 〈ψ|Q†.Suppose that operator Q is represented into the eigenstates |n〉 ofsome other operator, Q =

∑mnQmn|m〉〈n|,so that

|χ〉 = Q|ψ〉 =∑mn

Qmn|m〉〈n|ψ〉

11/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian conjugate

if the operation of the operator Q on the vector |ψ〉 yields thevector |χ〉, i.e |χ〉 = Q|ψ〉,then in the dual space the operator thatacts in the bra 〈ψ| to produce the bra 〈ψ| is the hermitian adjoint〈χ| = 〈ψ|Q†.Suppose that operator Q is represented into the eigenstates |n〉 ofsome other operator, Q =

∑mnQmn|m〉〈n|,so that

|χ〉 = Q|ψ〉 =∑mn

Qmn|m〉〈n|ψ〉

11/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian conjugate

if the operation of the operator Q on the vector |ψ〉 yields thevector |χ〉, i.e |χ〉 = Q|ψ〉,then in the dual space the operator thatacts in the bra 〈ψ| to produce the bra 〈ψ| is the hermitian adjoint〈χ| = 〈ψ|Q†.Suppose that operator Q is represented into the eigenstates |n〉 ofsome other operator, Q =

∑mnQmn|m〉〈n|,so that

|χ〉 = Q|ψ〉 =∑mn

Qmn|m〉〈n|ψ〉

11/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian conjugate

if the operation of the operator Q on the vector |ψ〉 yields thevector |χ〉, i.e |χ〉 = Q|ψ〉,then in the dual space the operator thatacts in the bra 〈ψ| to produce the bra 〈ψ| is the hermitian adjoint〈χ| = 〈ψ|Q†.Suppose that operator Q is represented into the eigenstates |n〉 ofsome other operator, Q =

∑mnQmn|m〉〈n|,so that

|χ〉 = Q|ψ〉 =∑mn

Qmn|m〉〈n|ψ〉

11/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

so〈χ| = (|χ〉)† =

∑mn

Q∗mn〈ψ|n〉〈m|

but since 〈χ| = 〈ψ|Q† then:

〈ψ|Q† = 〈ψ|∑mn

Q∗mn|n〉〈m|

meaning that

Q† =∑mn

Q∗mn|n〉〈m| =∑nm

Q∗nm|m〉〈n|

12/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

so〈χ| = (|χ〉)† =

∑mn

Q∗mn〈ψ|n〉〈m|

but since 〈χ| = 〈ψ|Q† then:

〈ψ|Q† = 〈ψ|∑mn

Q∗mn|n〉〈m|

meaning that

Q† =∑mn

Q∗mn|n〉〈m| =∑nm

Q∗nm|m〉〈n|

12/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

so〈χ| = (|χ〉)† =

∑mn

Q∗mn〈ψ|n〉〈m|

but since 〈χ| = 〈ψ|Q† then:

〈ψ|Q† = 〈ψ|∑mn

Q∗mn|n〉〈m|

meaning that

Q† =∑mn

Q∗mn|n〉〈m| =∑nm

Q∗nm|m〉〈n|

12/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

so〈χ| = (|χ〉)† =

∑mn

Q∗mn〈ψ|n〉〈m|

but since 〈χ| = 〈ψ|Q† then:

〈ψ|Q† = 〈ψ|∑mn

Q∗mn|n〉〈m|

meaning that

Q† =∑mn

Q∗mn|n〉〈m| =∑nm

Q∗nm|m〉〈n|

12/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

The operator Q† on the other hand is expressed as (just line anyother linear operator A =

∑mnAmn|m〉〈n| with Amn = 〈m|A|n〉):

Q† =∑mn

Q†mn|m〉〈n|

thus by comparison we deduce that:

Q†mn = Q∗nm

so the matrix elements of Q† are constructed from those of Q bytaking the complex conjugate of the transpose (or simply thehermitian conjugate).

13/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

The operator Q† on the other hand is expressed as (just line anyother linear operator A =

∑mnAmn|m〉〈n| with Amn = 〈m|A|n〉):

Q† =∑mn

Q†mn|m〉〈n|

thus by comparison we deduce that:

Q†mn = Q∗nm

so the matrix elements of Q† are constructed from those of Q bytaking the complex conjugate of the transpose (or simply thehermitian conjugate).

13/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsExpectation values:

The operator Q† on the other hand is expressed as (just line anyother linear operator A =

∑mnAmn|m〉〈n| with Amn = 〈m|A|n〉):

Q† =∑mn

Q†mn|m〉〈n|

thus by comparison we deduce that:

Q†mn = Q∗nm

so the matrix elements of Q† are constructed from those of Q bytaking the complex conjugate of the transpose (or simply thehermitian conjugate).

13/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operators[Combination of operators

it holds true that:

Q = ABC ⇔ Q† = C†B†A†

and|χ〉 = ABC|ψ〉 ⇔ 〈χ| = 〈ψ|C†B†A†

The proof is easy. Consider |χ〉 = B|ψ〉 (thus 〈χ| = 〈ψ|B),and |ζ〉 = A|χ〉 (thus 〈ζ| = 〈χ|A†),so that |ζ〉 = AB|ψ〉 = Q|ψ〉,with Q = AB, and 〈ζ| = 〈ψ|Q†, then we have:

〈ζ| = 〈χ|A† = 〈ψ|B†A†

hence we clearly see that (AB)† = B†A†. The proof can easily begeneralised to the case of the product of three operators.

14/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorscommutators

Generally:[A, B] = AB − BA 6= 0

for example [x, px] = i~. We have the following useful relation:

[A2, B] = A[A, B] + [A, B]A

15/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorscommutators

Generally:[A, B] = AB − BA 6= 0

for example [x, px] = i~. We have the following useful relation:

[A2, B] = A[A, B] + [A, B]A

15/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorscommutators

Generally:[A, B] = AB − BA 6= 0

for example [x, px] = i~. We have the following useful relation:

[A2, B] = A[A, B] + [A, B]A

15/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsHermitian operators

A hermitian operator is one for which

Q† = Q, Q∗mn = Qnm

Physical observables correspond to hermitian operators,whichfollows from the requirements that the eigenvalues be real.To seethis consider the average value of a physical observable which mustbe a real number:

Q = 〈ψ|Q|ψ〉 = real

then taking the complex conjugate:

Q∗

=(〈ψ|Q|ψ〉

)†= 〈ψ|Q†|ψ〉 = Q

which must equal the mean value (since it is real). Thus we seethat Q = Q†.

16/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsUnitary operators

A unitary operator is one that satisfies:

U †U = 1

Unitary operatrs conserve the norm of wavefunctions. To see thisconsider operating on a state by a unitary operator, e.g. theevolution operator:|ψt〉 = U |ψ0〉, such that |ψ0〉 is normalised,〈ψ0|ψ0〉 = 1.We have 〈ψt| = 〈ψ0|U †. Then the norm of theresulting is:

〈ψt|ψt = 〈ψ0|U †U |ψ0〉 = 〈ψ0|1|ψ0〉 = 〈ψ0|ψ0〉 = 1

so the norm of the state is conserved.

17/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsEigenvalues and eigenstates

We say that λ is an eigenvalue of the operator Q corresponding tothe eigenstate |ψ〉 if:

Q|ψ〉 = λ|ψ〉and in quantum mechanics we only deal with normalisable states,〈ψ|ψ〉 <∞.Eigenvalues may be quantised (leading to a discreetspectrum of eigenvalues)or continuous (leading to a continuousspectrum).If |ψ〉 is an eigenstate of the operator Q, then theexpectation value is of Q is just the corresponding eigenvalue:〈ψ|Q|ψ〉 = λ.If the operator Q is hermitian then its eigenvectors |q〉 form a basisin Hilbert space.

18/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

Hermitian operators have real eigenvalues and their eigenvectorsform a complete orthogonal set of basis, which can also be chosento be normalised, 〈n|m〉 = δmn. So a Hermitian operator may beexpressed in terms of its eigenstates and eigenvalues, as before:

Q = 1Q1 =∑n

qn|n〉〈n|

and the expectation value of the observable Q if the state is aneigenstate of it is just the corresponding eigenvalues:

〈Q〉 = 〈n|Q|n〉 = qn

19/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

Hermitian operators have real eigenvalues and their eigenvectorsform a complete orthogonal set of basis, which can also be chosento be normalised, 〈n|m〉 = δmn. So a Hermitian operator may beexpressed in terms of its eigenstates and eigenvalues, as before:

Q = 1Q1 =∑n

qn|n〉〈n|

and the expectation value of the observable Q if the state is aneigenstate of it is just the corresponding eigenvalues:

〈Q〉 = 〈n|Q|n〉 = qn

19/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

Hermitian operators have real eigenvalues and their eigenvectorsform a complete orthogonal set of basis, which can also be chosento be normalised, 〈n|m〉 = δmn. So a Hermitian operator may beexpressed in terms of its eigenstates and eigenvalues, as before:

Q = 1Q1 =∑n

qn|n〉〈n|

and the expectation value of the observable Q if the state is aneigenstate of it is just the corresponding eigenvalues:

〈Q〉 = 〈n|Q|n〉 = qn

19/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

Hermitian operators have real eigenvalues and their eigenvectorsform a complete orthogonal set of basis, which can also be chosento be normalised, 〈n|m〉 = δmn. So a Hermitian operator may beexpressed in terms of its eigenstates and eigenvalues, as before:

Q = 1Q1 =∑n

qn|n〉〈n|

and the expectation value of the observable Q if the state is aneigenstate of it is just the corresponding eigenvalues:

〈Q〉 = 〈n|Q|n〉 = qn

19/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

Hermitian operators have real eigenvalues and their eigenvectorsform a complete orthogonal set of basis, which can also be chosento be normalised, 〈n|m〉 = δmn. So a Hermitian operator may beexpressed in terms of its eigenstates and eigenvalues, as before:

Q = 1Q1 =∑n

qn|n〉〈n|

and the expectation value of the observable Q if the state is aneigenstate of it is just the corresponding eigenvalues:

〈Q〉 = 〈n|Q|n〉 = qn

19/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The proof of the orthogonality of the eigenstates is as follows.Consider the eigenvalue equation for the Hermitian operator Q fortwo non-degenerate eigenvalues:

Q|n〉 = qn|n〉, Q|m〉 = qm|m〉

Taking inner products with the bras 〈m| and 〈n| respectively leadsto:

〈m|Q|n〉 = qn〈m|n〉, 〈n|Q|m〉 = qm〈n|m〉Taking the hermitian conjugation of the second equation, andusing Q† = Q and that q∗m = qm, we find

〈m|Q†|n〉 = 〈m|Q|n〉 = q∗m〈m|n〉 = qm〈m|n〉

20/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The proof of the orthogonality of the eigenstates is as follows.Consider the eigenvalue equation for the Hermitian operator Q fortwo non-degenerate eigenvalues:

Q|n〉 = qn|n〉, Q|m〉 = qm|m〉

Taking inner products with the bras 〈m| and 〈n| respectively leadsto:

〈m|Q|n〉 = qn〈m|n〉, 〈n|Q|m〉 = qm〈n|m〉Taking the hermitian conjugation of the second equation, andusing Q† = Q and that q∗m = qm, we find

〈m|Q†|n〉 = 〈m|Q|n〉 = q∗m〈m|n〉 = qm〈m|n〉

20/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The proof of the orthogonality of the eigenstates is as follows.Consider the eigenvalue equation for the Hermitian operator Q fortwo non-degenerate eigenvalues:

Q|n〉 = qn|n〉, Q|m〉 = qm|m〉

Taking inner products with the bras 〈m| and 〈n| respectively leadsto:

〈m|Q|n〉 = qn〈m|n〉, 〈n|Q|m〉 = qm〈n|m〉Taking the hermitian conjugation of the second equation, andusing Q† = Q and that q∗m = qm, we find

〈m|Q†|n〉 = 〈m|Q|n〉 = q∗m〈m|n〉 = qm〈m|n〉

20/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The proof of the orthogonality of the eigenstates is as follows.Consider the eigenvalue equation for the Hermitian operator Q fortwo non-degenerate eigenvalues:

Q|n〉 = qn|n〉, Q|m〉 = qm|m〉

Taking inner products with the bras 〈m| and 〈n| respectively leadsto:

〈m|Q|n〉 = qn〈m|n〉, 〈n|Q|m〉 = qm〈n|m〉Taking the hermitian conjugation of the second equation, andusing Q† = Q and that q∗m = qm, we find

〈m|Q†|n〉 = 〈m|Q|n〉 = q∗m〈m|n〉 = qm〈m|n〉

20/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The proof of the orthogonality of the eigenstates is as follows.Consider the eigenvalue equation for the Hermitian operator Q fortwo non-degenerate eigenvalues:

Q|n〉 = qn|n〉, Q|m〉 = qm|m〉

Taking inner products with the bras 〈m| and 〈n| respectively leadsto:

〈m|Q|n〉 = qn〈m|n〉, 〈n|Q|m〉 = qm〈n|m〉Taking the hermitian conjugation of the second equation, andusing Q† = Q and that q∗m = qm, we find

〈m|Q†|n〉 = 〈m|Q|n〉 = q∗m〈m|n〉 = qm〈m|n〉

20/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The left-hand-side of these two equations are now identical, thusthe right-hand-sides are also equal:

qn〈m|n〉 = qm〈m|n〉 ⇒ (qm − qn)〈m|n〉 = 0

so either qm = qn, or 〈m|n〉 = 0, which means that the eigenstatesfor different eigenvalues are orthogonal.Now if the operators A, B, · · · are all hermitian and commutepairwise,then there exists a complete basis |n〉 in which alloperators are diagonal,i.e. a basis formed by common eigenstatesof all these operators:

A =∑n

an|n〉〈n|, B =∑n

bn|n〉〈n|, · · ·

21/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The left-hand-side of these two equations are now identical, thusthe right-hand-sides are also equal:

qn〈m|n〉 = qm〈m|n〉 ⇒ (qm − qn)〈m|n〉 = 0

so either qm = qn, or 〈m|n〉 = 0, which means that the eigenstatesfor different eigenvalues are orthogonal.Now if the operators A, B, · · · are all hermitian and commutepairwise,then there exists a complete basis |n〉 in which alloperators are diagonal,i.e. a basis formed by common eigenstatesof all these operators:

A =∑n

an|n〉〈n|, B =∑n

bn|n〉〈n|, · · ·

21/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The left-hand-side of these two equations are now identical, thusthe right-hand-sides are also equal:

qn〈m|n〉 = qm〈m|n〉 ⇒ (qm − qn)〈m|n〉 = 0

so either qm = qn, or 〈m|n〉 = 0, which means that the eigenstatesfor different eigenvalues are orthogonal.Now if the operators A, B, · · · are all hermitian and commutepairwise,then there exists a complete basis |n〉 in which alloperators are diagonal,i.e. a basis formed by common eigenstatesof all these operators:

A =∑n

an|n〉〈n|, B =∑n

bn|n〉〈n|, · · ·

21/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The left-hand-side of these two equations are now identical, thusthe right-hand-sides are also equal:

qn〈m|n〉 = qm〈m|n〉 ⇒ (qm − qn)〈m|n〉 = 0

so either qm = qn, or 〈m|n〉 = 0, which means that the eigenstatesfor different eigenvalues are orthogonal.Now if the operators A, B, · · · are all hermitian and commutepairwise,then there exists a complete basis |n〉 in which alloperators are diagonal,i.e. a basis formed by common eigenstatesof all these operators:

A =∑n

an|n〉〈n|, B =∑n

bn|n〉〈n|, · · ·

21/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum operatorsMore on hermitian operators

The left-hand-side of these two equations are now identical, thusthe right-hand-sides are also equal:

qn〈m|n〉 = qm〈m|n〉 ⇒ (qm − qn)〈m|n〉 = 0

so either qm = qn, or 〈m|n〉 = 0, which means that the eigenstatesfor different eigenvalues are orthogonal.Now if the operators A, B, · · · are all hermitian and commutepairwise,then there exists a complete basis |n〉 in which alloperators are diagonal,i.e. a basis formed by common eigenstatesof all these operators:

A =∑n

an|n〉〈n|, B =∑n

bn|n〉〈n|, · · ·

21/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Consider the Schrodinger equation,

H|ψ〉 = i~∂

∂t|ψ〉

Now since any operator can be written in any orthonormal basisQ =

∑nmQmn|n〉〈m|,with (generally non-diagonal) matrix

elements Qmn = 〈m|Q|n〉, consider doing this for the Hamiltonianin the x-representation:

H =∑xx′

Hxx′ |x〉〈x′|, Hxx′ = 〈x|H|x′〉

The matrix elements are explicitly evaluated in thex-representation as follows:

Hxx′ = 〈x|H|x′〉 =1

2m〈x|p2|x′〉+ 〈x|V (x)〉|x′〉

Hxx′ =

(− ~2

2m∇2 + V (x)

)δxx′

22/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Thus

H =∑xx′

(− ~2

2m∇2 + V (x)

)δxx′ |x〉〈x′|

H =∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|

Note that any general classical observable f(p, q) can be written:

f =∑xx′

f

(−i~ ∂

∂x, x

)δxx′ |x〉〈x′|

In the Schrodinger equation we get:∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|ψ〉 = i~

∂

∂t|ψ〉

23/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Thus

H =∑xx′

(− ~2

2m∇2 + V (x)

)δxx′ |x〉〈x′|

H =∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|

Note that any general classical observable f(p, q) can be written:

f =∑xx′

f

(−i~ ∂

∂x, x

)δxx′ |x〉〈x′|

In the Schrodinger equation we get:∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|ψ〉 = i~

∂

∂t|ψ〉

23/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Thus

H =∑xx′

(− ~2

2m∇2 + V (x)

)δxx′ |x〉〈x′|

H =∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|

Note that any general classical observable f(p, q) can be written:

f =∑xx′

f

(−i~ ∂

∂x, x

)δxx′ |x〉〈x′|

In the Schrodinger equation we get:∑x

(− ~2

2m∇2 + V (x)

)|x〉〈x|ψ〉 = i~

∂

∂t|ψ〉

23/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Multiplying by the bra 〈x′| we obtain

∑x

(− ~2

2m∇2 + V (x)

)〈x′|x〉〈x|ψ〉 = i~

∂

∂t〈x′|ψ〉

so using 〈x′|x〉 = δxx′ the sum over x is reduced(− ~2

2m∇2x′ + V (x′)

)〈x′|ψ〉 = i~

∂

∂t〈x′|ψ〉

or (− ~2

2m∇2 + V (x)

)ψ(x) = i~

∂

∂tψ(x)

i.e. the Schrodinger equation in the x-representation.Problem: Find the Schrodinger equation in p representation for theHarmonic oscillator potential.

24/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Multiplying by the bra 〈x′| we obtain

∑x

(− ~2

2m∇2 + V (x)

)〈x′|x〉〈x|ψ〉 = i~

∂

∂t〈x′|ψ〉

so using 〈x′|x〉 = δxx′ the sum over x is reduced(− ~2

2m∇2x′ + V (x′)

)〈x′|ψ〉 = i~

∂

∂t〈x′|ψ〉

or (− ~2

2m∇2 + V (x)

)ψ(x) = i~

∂

∂tψ(x)

i.e. the Schrodinger equation in the x-representation.Problem: Find the Schrodinger equation in p representation for theHarmonic oscillator potential.

24/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Multiplying by the bra 〈x′| we obtain

∑x

(− ~2

2m∇2 + V (x)

)〈x′|x〉〈x|ψ〉 = i~

∂

∂t〈x′|ψ〉

so using 〈x′|x〉 = δxx′ the sum over x is reduced(− ~2

2m∇2x′ + V (x′)

)〈x′|ψ〉 = i~

∂

∂t〈x′|ψ〉

or (− ~2

2m∇2 + V (x)

)ψ(x) = i~

∂

∂tψ(x)

i.e. the Schrodinger equation in the x-representation.Problem: Find the Schrodinger equation in p representation for theHarmonic oscillator potential.

24/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Multiplying by the bra 〈x′| we obtain

∑x

(− ~2

2m∇2 + V (x)

)〈x′|x〉〈x|ψ〉 = i~

∂

∂t〈x′|ψ〉

so using 〈x′|x〉 = δxx′ the sum over x is reduced(− ~2

2m∇2x′ + V (x′)

)〈x′|ψ〉 = i~

∂

∂t〈x′|ψ〉

or (− ~2

2m∇2 + V (x)

)ψ(x) = i~

∂

∂tψ(x)

i.e. the Schrodinger equation in the x-representation.Problem: Find the Schrodinger equation in p representation for theHarmonic oscillator potential.

24/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equation

Multiplying by the bra 〈x′| we obtain

∑x

(− ~2

2m∇2 + V (x)

)〈x′|x〉〈x|ψ〉 = i~

∂

∂t〈x′|ψ〉

so using 〈x′|x〉 = δxx′ the sum over x is reduced(− ~2

2m∇2x′ + V (x′)

)〈x′|ψ〉 = i~

∂

∂t〈x′|ψ〉

or (− ~2

2m∇2 + V (x)

)ψ(x) = i~

∂

∂tψ(x)

i.e. the Schrodinger equation in the x-representation.Problem: Find the Schrodinger equation in p representation for theHarmonic oscillator potential.

24/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equationNote that the Hamiltonian is not diagonal in the x representation(is it orthogonal in the p representation?).The Hamiltonian is a hermitian operator, H† = H, and itseigenvalues are real. The eigenvalue equation is the TISE:

H|ψn〉 = En|ψn〉where the eigenstates are orthonormal 〈ψn|ψm〉 = δnm,Note alsothat multiplying the time-independent eigenstates by thetime-evolution e−iEnt/~ yields the solutions to the time-dependentSchrodinger equation:

|ψn(t)〉 = e−iEnt/~|n〉so

i~∂

∂t|ψn(t)〉 = i~(−iEn/~)e−iEnt/~|n〉 = Ene

−iEnt/~|n〉 = e−iEnt/~H|n〉

i~∂

∂t|ψn(t)〉 = H|ψn〉

25/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equationNote that the Hamiltonian is not diagonal in the x representation(is it orthogonal in the p representation?).The Hamiltonian is a hermitian operator, H† = H, and itseigenvalues are real. The eigenvalue equation is the TISE:

H|ψn〉 = En|ψn〉where the eigenstates are orthonormal 〈ψn|ψm〉 = δnm,Note alsothat multiplying the time-independent eigenstates by thetime-evolution e−iEnt/~ yields the solutions to the time-dependentSchrodinger equation:

|ψn(t)〉 = e−iEnt/~|n〉so

i~∂

∂t|ψn(t)〉 = i~(−iEn/~)e−iEnt/~|n〉 = Ene

−iEnt/~|n〉 = e−iEnt/~H|n〉

i~∂

∂t|ψn(t)〉 = H|ψn〉

25/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equationNote that the Hamiltonian is not diagonal in the x representation(is it orthogonal in the p representation?).The Hamiltonian is a hermitian operator, H† = H, and itseigenvalues are real. The eigenvalue equation is the TISE:

H|ψn〉 = En|ψn〉where the eigenstates are orthonormal 〈ψn|ψm〉 = δnm,Note alsothat multiplying the time-independent eigenstates by thetime-evolution e−iEnt/~ yields the solutions to the time-dependentSchrodinger equation:

|ψn(t)〉 = e−iEnt/~|n〉so

i~∂

∂t|ψn(t)〉 = i~(−iEn/~)e−iEnt/~|n〉 = Ene

−iEnt/~|n〉 = e−iEnt/~H|n〉

i~∂

∂t|ψn(t)〉 = H|ψn〉

25/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equationNote that the Hamiltonian is not diagonal in the x representation(is it orthogonal in the p representation?).The Hamiltonian is a hermitian operator, H† = H, and itseigenvalues are real. The eigenvalue equation is the TISE:

H|ψn〉 = En|ψn〉where the eigenstates are orthonormal 〈ψn|ψm〉 = δnm,Note alsothat multiplying the time-independent eigenstates by thetime-evolution e−iEnt/~ yields the solutions to the time-dependentSchrodinger equation:

|ψn(t)〉 = e−iEnt/~|n〉so

i~∂

∂t|ψn(t)〉 = i~(−iEn/~)e−iEnt/~|n〉 = Ene

−iEnt/~|n〉 = e−iEnt/~H|n〉

i~∂

∂t|ψn(t)〉 = H|ψn〉

25/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger equationNote that the Hamiltonian is not diagonal in the x representation(is it orthogonal in the p representation?).The Hamiltonian is a hermitian operator, H† = H, and itseigenvalues are real. The eigenvalue equation is the TISE:

H|ψn〉 = En|ψn〉where the eigenstates are orthonormal 〈ψn|ψm〉 = δnm,Note alsothat multiplying the time-independent eigenstates by thetime-evolution e−iEnt/~ yields the solutions to the time-dependentSchrodinger equation:

|ψn(t)〉 = e−iEnt/~|n〉so

i~∂

∂t|ψn(t)〉 = i~(−iEn/~)e−iEnt/~|n〉 = Ene

−iEnt/~|n〉 = e−iEnt/~H|n〉

i~∂

∂t|ψn(t)〉 = H|ψn〉

25/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Properties of the eigenstates of the Hamiltonian

orthonotmality of the eigenstates 〈n|m〉 = δmn. The proof oforthogonality of the eigenstates follows from theproof of orthogonality of Hermitian operators (seebefore).

Completness the eigenstates |n〉 form a complete basis:∑n |n〉〈n| = 1

Representation in its own eigenstates: Since the Hamiltonian isHermitian, then:

H = 1H 1 =∑mn

|m〉〈m|H|n〉〈n| =∑mn

Hmn|m〉〈n|

with the matrix elements:

Hmn = 〈m|H|n〉 = En〈m|n〉 = Enδmn

which means H is diagonal in this basis, so:

H =∑mn

Enδmn|m〉〈n| =∑n

En|n〉〈n|26/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Properties of the eigenstates of the Hamiltonian

orthonotmality of the eigenstates 〈n|m〉 = δmn. The proof oforthogonality of the eigenstates follows from theproof of orthogonality of Hermitian operators (seebefore).

Completness the eigenstates |n〉 form a complete basis:∑n |n〉〈n| = 1

Representation in its own eigenstates: Since the Hamiltonian isHermitian, then:

H = 1H 1 =∑mn

|m〉〈m|H|n〉〈n| =∑mn

Hmn|m〉〈n|

with the matrix elements:

Hmn = 〈m|H|n〉 = En〈m|n〉 = Enδmn

which means H is diagonal in this basis, so:

H =∑mn

Enδmn|m〉〈n| =∑n

En|n〉〈n|26/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Properties of the eigenstates of the Hamiltonian

orthonotmality of the eigenstates 〈n|m〉 = δmn. The proof oforthogonality of the eigenstates follows from theproof of orthogonality of Hermitian operators (seebefore).

Completness the eigenstates |n〉 form a complete basis:∑n |n〉〈n| = 1

Representation in its own eigenstates: Since the Hamiltonian isHermitian, then:

H = 1H 1 =∑mn

|m〉〈m|H|n〉〈n| =∑mn

Hmn|m〉〈n|

with the matrix elements:

Hmn = 〈m|H|n〉 = En〈m|n〉 = Enδmn

which means H is diagonal in this basis, so:

H =∑mn

Enδmn|m〉〈n| =∑n

En|n〉〈n|26/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Properties of the eigenstates of the Hamiltonian

orthonotmality of the eigenstates 〈n|m〉 = δmn. The proof oforthogonality of the eigenstates follows from theproof of orthogonality of Hermitian operators (seebefore).

Completness the eigenstates |n〉 form a complete basis:∑n |n〉〈n| = 1

Representation in its own eigenstates: Since the Hamiltonian isHermitian, then:

H = 1H 1 =∑mn

|m〉〈m|H|n〉〈n| =∑mn

Hmn|m〉〈n|

with the matrix elements:

Hmn = 〈m|H|n〉 = En〈m|n〉 = Enδmn

which means H is diagonal in this basis, so:

H =∑mn

Enδmn|m〉〈n| =∑n

En|n〉〈n|26/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Properties of the eigenstates of the Hamiltonian

orthonotmality of the eigenstates 〈n|m〉 = δmn. The proof oforthogonality of the eigenstates follows from theproof of orthogonality of Hermitian operators (seebefore).

Completness the eigenstates |n〉 form a complete basis:∑n |n〉〈n| = 1

Representation in its own eigenstates: Since the Hamiltonian isHermitian, then:

H = 1H 1 =∑mn

|m〉〈m|H|n〉〈n| =∑mn

Hmn|m〉〈n|

with the matrix elements:

Hmn = 〈m|H|n〉 = En〈m|n〉 = Enδmn

which means H is diagonal in this basis, so:

H =∑mn

Enδmn|m〉〈n| =∑n

En|n〉〈n|26/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have the evolution equation for a quantum state:

|ψ(t)〉 = U(t, t0)|ψ(t0)〉

with the evolution operator:

U(t, t0) = T exp

(− i~

∫ t

t0

H(τ)dτ

),

In the case that the Hamiltonian is time-independent, H 6= H(t),then [Ht1 , Ht2 ] = 0, so:

U(t, t0) = exp

(− i(t− t0)

~H

),

27/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have the evolution equation for a quantum state:

|ψ(t)〉 = U(t, t0)|ψ(t0)〉

with the evolution operator:

U(t, t0) = T exp

(− i~

∫ t

t0

H(τ)dτ

),

In the case that the Hamiltonian is time-independent, H 6= H(t),then [Ht1 , Ht2 ] = 0, so:

U(t, t0) = exp

(− i(t− t0)

~H

),

27/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have the evolution equation for a quantum state:

|ψ(t)〉 = U(t, t0)|ψ(t0)〉

with the evolution operator:

U(t, t0) = T exp

(− i~

∫ t

t0

H(τ)dτ

),

In the case that the Hamiltonian is time-independent, H 6= H(t),then [Ht1 , Ht2 ] = 0, so:

U(t, t0) = exp

(− i(t− t0)

~H

),

27/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have that the evolution operator is unitary: U †U = 1, whichmeans that the norm of the wavefunction is conserved.To see thiswe have |ψ(t)〉 = U |ψ(t0)〉 ⇒ 〈ψ(t)| = 〈ψ(t0)|U †, so the norm ofthe wavefunction at time t is

〈ψ(t)|ψ(t)〉 = 〈ψ(t0)|U †U |ψ(t0)〉 = 〈ψ(t0)|1|ψ(t0)〉 = 〈ψ(t0)|ψ(t0)〉

The adjoint of the evolution operator is:

U †(t, t0) = T † exp

(+i

~

∫ t

t0

H(τ)dτ

)= T exp

(− i~

∫ t0

tH(τ)dτ

),

where the time limits in the integration over τ have been swappedhere.

28/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have that the evolution operator is unitary: U †U = 1, whichmeans that the norm of the wavefunction is conserved.To see thiswe have |ψ(t)〉 = U |ψ(t0)〉 ⇒ 〈ψ(t)| = 〈ψ(t0)|U †, so the norm ofthe wavefunction at time t is

〈ψ(t)|ψ(t)〉 = 〈ψ(t0)|U †U |ψ(t0)〉 = 〈ψ(t0)|1|ψ(t0)〉 = 〈ψ(t0)|ψ(t0)〉

The adjoint of the evolution operator is:

U †(t, t0) = T † exp

(+i

~

∫ t

t0

H(τ)dτ

)= T exp

(− i~

∫ t0

tH(τ)dτ

),

where the time limits in the integration over τ have been swappedhere.

28/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

Quantum Evolution

We have that the evolution operator is unitary: U †U = 1, whichmeans that the norm of the wavefunction is conserved.To see thiswe have |ψ(t)〉 = U |ψ(t0)〉 ⇒ 〈ψ(t)| = 〈ψ(t0)|U †, so the norm ofthe wavefunction at time t is

〈ψ(t)|ψ(t)〉 = 〈ψ(t0)|U †U |ψ(t0)〉 = 〈ψ(t0)|1|ψ(t0)〉 = 〈ψ(t0)|ψ(t0)〉

The adjoint of the evolution operator is:

U †(t, t0) = T † exp

(+i

~

∫ t

t0

H(τ)dτ

)= T exp

(− i~

∫ t0

tH(τ)dτ

),

where the time limits in the integration over τ have been swappedhere.

28/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Here we define the states such that they carry all the timedependence (|Ψ〉 = |Ψ(t)〉)and operators Os aretime-independent.The states obey the Schrodinger equation:

H|Ψ(t)〉 = i~∂|Ψ(t)〉∂t

,

whose solution can be expressed as:

|Ψ(t)〉 = e−iHt/~|Ψ(0)〉. (4)

The expectation value for an observable Os is:

〈Os〉 = 〈Ψ(t)|Os|Ψ(t)〉 (5)

29/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Here we define the states such that they carry all the timedependence (|Ψ〉 = |Ψ(t)〉)and operators Os aretime-independent.The states obey the Schrodinger equation:

H|Ψ(t)〉 = i~∂|Ψ(t)〉∂t

,

whose solution can be expressed as:

|Ψ(t)〉 = e−iHt/~|Ψ(0)〉. (4)

The expectation value for an observable Os is:

〈Os〉 = 〈Ψ(t)|Os|Ψ(t)〉 (5)

29/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Here we define the states such that they carry all the timedependence (|Ψ〉 = |Ψ(t)〉)and operators Os aretime-independent.The states obey the Schrodinger equation:

H|Ψ(t)〉 = i~∂|Ψ(t)〉∂t

,

whose solution can be expressed as:

|Ψ(t)〉 = e−iHt/~|Ψ(0)〉. (4)

The expectation value for an observable Os is:

〈Os〉 = 〈Ψ(t)|Os|Ψ(t)〉 (5)

29/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Here we define the states such that they carry all the timedependence (|Ψ〉 = |Ψ(t)〉)and operators Os aretime-independent.The states obey the Schrodinger equation:

H|Ψ(t)〉 = i~∂|Ψ(t)〉∂t

,

whose solution can be expressed as:

|Ψ(t)〉 = e−iHt/~|Ψ(0)〉. (4)

The expectation value for an observable Os is:

〈Os〉 = 〈Ψ(t)|Os|Ψ(t)〉 (5)

29/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Here we define the states such that they carry all the timedependence (|Ψ〉 = |Ψ(t)〉)and operators Os aretime-independent.The states obey the Schrodinger equation:

H|Ψ(t)〉 = i~∂|Ψ(t)〉∂t

,

whose solution can be expressed as:

|Ψ(t)〉 = e−iHt/~|Ψ(0)〉. (4)

The expectation value for an observable Os is:

〈Os〉 = 〈Ψ(t)|Os|Ψ(t)〉 (5)

29/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Thus the time-evolution of the expectation value is:

d〈Os〉dt

=∂〈Ψ(t)|∂t

Os|Ψ(t)〉+ 〈Ψ(t)|Os∂|Ψ(t)〉∂t

=i

~

(〈Ψ(t)|HOs|Ψ〉 − 〈Ψ(t)|OsH|Ψ〉

)d〈Os〉dt

=i

~

(〈Ψ(t)|

[H, Os

]|Ψ(t)〉

), (6)

where we used the fact that the Hamiltonian is Hermitian(H = H†) and that

∂|Ψ(t)〉∂t

= − i~H|Ψ(t)〉 ⇒ ∂〈Ψ(t)|

∂t= +

i

~〈Ψ(t)|H

30/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Schrodinger picture of Quantum Mechanics

Thus the time-evolution of the expectation value is:

d〈Os〉dt

=∂〈Ψ(t)|∂t

Os|Ψ(t)〉+ 〈Ψ(t)|Os∂|Ψ(t)〉∂t

=i

~

(〈Ψ(t)|HOs|Ψ〉 − 〈Ψ(t)|OsH|Ψ〉

)d〈Os〉dt

=i

~

(〈Ψ(t)|

[H, Os

]|Ψ(t)〉

), (6)

where we used the fact that the Hamiltonian is Hermitian(H = H†) and that

∂|Ψ(t)〉∂t

= − i~H|Ψ(t)〉 ⇒ ∂〈Ψ(t)|

∂t= +

i

~〈Ψ(t)|H

30/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Here the states 1 |ψ〉 is time-independent and the observablesOH(t) carry all the time dependence.Let us choose both pictures to coincide at time t = 0, such that:

|Ψ(0)〉s = |ψ〉H ,Os = OH(t = 0),

The expectation value for the observable OH at time t is:

〈OH〉(t) = 〈ψ|OH(t)|ψ〉and since physics is independent of the picture we write using eq.(4):

〈OH〉(t) = 〈Os〉(t) = 〈Ψ(t)|Os|Ψ(t)〉= 〈Ψ(0)|e+itH/~Ose−itH/~|Ψ(0)〉= 〈ψ|e+itH/~Ose−itH/~|ψ〉

1We use uppercase |Ψ〉 for time-dependent wavefunction, while lowercase|ψ〉 for time-independent wavefunction

31/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Here the states 1 |ψ〉 is time-independent and the observablesOH(t) carry all the time dependence.Let us choose both pictures to coincide at time t = 0, such that:

|Ψ(0)〉s = |ψ〉H ,Os = OH(t = 0),

The expectation value for the observable OH at time t is:

〈OH〉(t) = 〈ψ|OH(t)|ψ〉and since physics is independent of the picture we write using eq.(4):

〈OH〉(t) = 〈Os〉(t) = 〈Ψ(t)|Os|Ψ(t)〉= 〈Ψ(0)|e+itH/~Ose−itH/~|Ψ(0)〉= 〈ψ|e+itH/~Ose−itH/~|ψ〉

1We use uppercase |Ψ〉 for time-dependent wavefunction, while lowercase|ψ〉 for time-independent wavefunction

31/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Here the states 1 |ψ〉 is time-independent and the observablesOH(t) carry all the time dependence.Let us choose both pictures to coincide at time t = 0, such that:

|Ψ(0)〉s = |ψ〉H ,Os = OH(t = 0),

The expectation value for the observable OH at time t is:

〈OH〉(t) = 〈ψ|OH(t)|ψ〉and since physics is independent of the picture we write using eq.(4):

〈OH〉(t) = 〈Os〉(t) = 〈Ψ(t)|Os|Ψ(t)〉= 〈Ψ(0)|e+itH/~Ose−itH/~|Ψ(0)〉= 〈ψ|e+itH/~Ose−itH/~|ψ〉

1We use uppercase |Ψ〉 for time-dependent wavefunction, while lowercase|ψ〉 for time-independent wavefunction

31/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Here the states 1 |ψ〉 is time-independent and the observablesOH(t) carry all the time dependence.Let us choose both pictures to coincide at time t = 0, such that:

|Ψ(0)〉s = |ψ〉H ,Os = OH(t = 0),

The expectation value for the observable OH at time t is:

〈OH〉(t) = 〈ψ|OH(t)|ψ〉and since physics is independent of the picture we write using eq.(4):

〈OH〉(t) = 〈Os〉(t) = 〈Ψ(t)|Os|Ψ(t)〉= 〈Ψ(0)|e+itH/~Ose−itH/~|Ψ(0)〉= 〈ψ|e+itH/~Ose−itH/~|ψ〉

1We use uppercase |Ψ〉 for time-dependent wavefunction, while lowercase|ψ〉 for time-independent wavefunction

31/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

and therefore:OH(t) = eiHt/~Ose−iHt/~, (7)

or equivalently:

Os = e−iHt/~OHeiHt/~. (8)

Furthermore the time-evolution of the expectation value is:

d

dt〈OH〉 =

d

dt〈ψ|OH(t)|ψ〉 = 〈ψ|dO

H(t)

dt|ψ〉 =

⟨dOHdt

⟩.

since the states are time-independent.

32/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

and therefore:OH(t) = eiHt/~Ose−iHt/~, (7)

or equivalently:

Os = e−iHt/~OHeiHt/~. (8)

Furthermore the time-evolution of the expectation value is:

d

dt〈OH〉 =

d

dt〈ψ|OH(t)|ψ〉 = 〈ψ|dO

H(t)

dt|ψ〉 =

⟨dOHdt

⟩.

since the states are time-independent.

32/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

and therefore:OH(t) = eiHt/~Ose−iHt/~, (7)

or equivalently:

Os = e−iHt/~OHeiHt/~. (8)

Furthermore the time-evolution of the expectation value is:

d

dt〈OH〉 =

d

dt〈ψ|OH(t)|ψ〉 = 〈ψ|dO

H(t)

dt|ψ〉 =

⟨dOHdt

⟩.

since the states are time-independent.

32/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

and therefore:OH(t) = eiHt/~Ose−iHt/~, (7)

or equivalently:

Os = e−iHt/~OHeiHt/~. (8)

Furthermore the time-evolution of the expectation value is:

d

dt〈OH〉 =

d

dt〈ψ|OH(t)|ψ〉 = 〈ψ|dO

H(t)

dt|ψ〉 =

⟨dOHdt

⟩.

since the states are time-independent.

32/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

and therefore:OH(t) = eiHt/~Ose−iHt/~, (7)

or equivalently:

Os = e−iHt/~OHeiHt/~. (8)

Furthermore the time-evolution of the expectation value is:

d

dt〈OH〉 =

d

dt〈ψ|OH(t)|ψ〉 = 〈ψ|dO

H(t)

dt|ψ〉 =

⟨dOHdt

⟩.

since the states are time-independent.

32/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Since physics is independent of the picture we write using eq. (6)and (4):

d〈OH〉dt

=d〈Os〉dt

=i

~

(〈Ψ(t)|

[H, Os

]|Ψ(t)〉

)=

i

~

(〈Ψ(0)|e+iHt/~

[H, Os

]e−iHt/~|Ψ(0)〉

)=

i

~

(〈ψ|[H, e+iHt/~Ose−iHt/~

]|ψ〉)

=i

~

(〈ψ|[H, OH

]|ψ〉),

where we used the fact that the evolution operator commutes withthe Hamiltonian. But since on the other hand we have:

d

dt〈OH〉 = 〈ψ|dO

H(t)

dt|ψ〉,

33/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

Since physics is independent of the picture we write using eq. (6)and (4):

d〈OH〉dt

=d〈Os〉dt

=i

~

(〈Ψ(t)|

[H, Os

]|Ψ(t)〉

)=

i

~

(〈Ψ(0)|e+iHt/~

[H, Os

]e−iHt/~|Ψ(0)〉

)=

i

~

(〈ψ|[H, e+iHt/~Ose−iHt/~

]|ψ〉)

=i

~

(〈ψ|[H, OH

]|ψ〉),

where we used the fact that the evolution operator commutes withthe Hamiltonian. But since on the other hand we have:

d

dt〈OH〉 = 〈ψ|dO

H(t)

dt|ψ〉,

33/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

we deduce that:dOH

dt=i

~

[H, OH

], (9)

which is known as the Heisenberg equation of motion.ExerciseShow that

dx

dt=pxm.

34/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

we deduce that:dOH

dt=i

~

[H, OH

], (9)

which is known as the Heisenberg equation of motion.ExerciseShow that

dx

dt=pxm.

34/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The Heisenberg picture of Quantum Mechanics

we deduce that:dOH

dt=i

~

[H, OH

], (9)

which is known as the Heisenberg equation of motion.ExerciseShow that

dx

dt=pxm.

34/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

In the case that the Hamiltonian is a sum of a “free” HamiltonianH0 and an interaction term V (t):

H = H0 + V (t)

and letting the “free” evolution operator be:

U0(t, t0) = exp(−i(t− t0)H0/~)

so the expectation value (in the Schrodinger picture) of someoperator Q (which may depend on time) is given by:

〈Qs〉(t) =〈ψ(t)|Q|ψ(t)〉=〈ψ(t)|1Q1|ψ(t)〉=〈ψ(t)|U0U

†0QU0U

†|ψ(t)〉

35/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

In the case that the Hamiltonian is a sum of a “free” HamiltonianH0 and an interaction term V (t):

H = H0 + V (t)

and letting the “free” evolution operator be:

U0(t, t0) = exp(−i(t− t0)H0/~)

so the expectation value (in the Schrodinger picture) of someoperator Q (which may depend on time) is given by:

〈Qs〉(t) =〈ψ(t)|Q|ψ(t)〉=〈ψ(t)|1Q1|ψ(t)〉=〈ψ(t)|U0U

†0QU0U

†|ψ(t)〉

35/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

In the case that the Hamiltonian is a sum of a “free” HamiltonianH0 and an interaction term V (t):

H = H0 + V (t)

and letting the “free” evolution operator be:

U0(t, t0) = exp(−i(t− t0)H0/~)

so the expectation value (in the Schrodinger picture) of someoperator Q (which may depend on time) is given by:

〈Qs〉(t) =〈ψ(t)|Q|ψ(t)〉=〈ψ(t)|1Q1|ψ(t)〉=〈ψ(t)|U0U

†0QU0U

†|ψ(t)〉

35/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

In the case that the Hamiltonian is a sum of a “free” HamiltonianH0 and an interaction term V (t):

H = H0 + V (t)

and letting the “free” evolution operator be:

U0(t, t0) = exp(−i(t− t0)H0/~)

so the expectation value (in the Schrodinger picture) of someoperator Q (which may depend on time) is given by:

〈Qs〉(t) =〈ψ(t)|Q|ψ(t)〉=〈ψ(t)|1Q1|ψ(t)〉=〈ψ(t)|U0U

†0QU0U

†|ψ(t)〉

35/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

In the case that the Hamiltonian is a sum of a “free” HamiltonianH0 and an interaction term V (t):

H = H0 + V (t)

and letting the “free” evolution operator be:

U0(t, t0) = exp(−i(t− t0)H0/~)

so the expectation value (in the Schrodinger picture) of someoperator Q (which may depend on time) is given by:

〈Qs〉(t) =〈ψ(t)|Q|ψ(t)〉=〈ψ(t)|1Q1|ψ(t)〉=〈ψ(t)|U0U

†0QU0U

†|ψ(t)〉

35/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us define the state of the system |ψ(t)〉 and the operatorˆQ in

the interaction picture to be:

|ψ(t)〉 = U †0(t, t0)|ψ(t)〉, ˆQ = U †0(t, t0)QU0(t, t0)

such that:〈QS〉(t) = 〈ψ(t)| ˆQ|ψ(t)〉 = 〈Q〉(t)

Note that the state of the system in the interaction picture |ψ(t)〉would just be a constant in time when the interaction term V iszero,so the interaction and the Heisenberg pictures coincide in thiscase. However in the presence of the interaction term the state of

the system evolves in time as well as the operatorˆQ.

How is the time evolution of these goverened?

36/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us take the time-derivative of the state of the system in theinteraction picture |ψ(t)〉 = U †0(t, t0)|ψ(t)〉:

i~∂

∂t|ψ(t)〉 =i~

∂

∂t(U †0(t, t0)|ψ(t)〉)

=i~∂

∂t(U †0(t, t0))|ψ(t)〉+ U †0(t, t0)i~

∂

∂t|ψ(t)〉

Now using the Schrodinger equation for the state of the system inthe Schrodinger picture, i~ ∂

∂t |ψ(t)〉 = (H0 + V (t))|ψ(t)〉,and usingthe time derivative of the hermitian conjugate of the evolutionoperator:

i~∂

∂tU †0(t, t0) =i~

∂

∂texp

(+i(t− t0)H0/~

)=i~(+iH0/~) exp

(+i(t− t0)H0/~

)= −H0U

†0 = −U †0H0

37/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us take the time-derivative of the state of the system in theinteraction picture |ψ(t)〉 = U †0(t, t0)|ψ(t)〉:

i~∂

∂t|ψ(t)〉 =i~

∂

∂t(U †0(t, t0)|ψ(t)〉)

=i~∂

∂t(U †0(t, t0))|ψ(t)〉+ U †0(t, t0)i~

∂

∂t|ψ(t)〉

Now using the Schrodinger equation for the state of the system inthe Schrodinger picture, i~ ∂

∂t |ψ(t)〉 = (H0 + V (t))|ψ(t)〉,and usingthe time derivative of the hermitian conjugate of the evolutionoperator:

i~∂

∂tU †0(t, t0) =i~

∂

∂texp

(+i(t− t0)H0/~

)=i~(+iH0/~) exp

(+i(t− t0)H0/~

)= −H0U

†0 = −U †0H0

37/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us take the time-derivative of the state of the system in theinteraction picture |ψ(t)〉 = U †0(t, t0)|ψ(t)〉:

i~∂

∂t|ψ(t)〉 =i~

∂

∂t(U †0(t, t0)|ψ(t)〉)

=i~∂

∂t(U †0(t, t0))|ψ(t)〉+ U †0(t, t0)i~

∂

∂t|ψ(t)〉

Now using the Schrodinger equation for the state of the system inthe Schrodinger picture, i~ ∂

∂t |ψ(t)〉 = (H0 + V (t))|ψ(t)〉,and usingthe time derivative of the hermitian conjugate of the evolutionoperator:

i~∂

∂tU †0(t, t0) =i~

∂

∂texp

(+i(t− t0)H0/~

)=i~(+iH0/~) exp

(+i(t− t0)H0/~

)= −H0U

†0 = −U †0H0

37/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

Let us take the time-derivative of the state of the system in theinteraction picture |ψ(t)〉 = U †0(t, t0)|ψ(t)〉:

i~∂

∂t|ψ(t)〉 =i~

∂

∂t(U †0(t, t0)|ψ(t)〉)

=i~∂

∂t(U †0(t, t0))|ψ(t)〉+ U †0(t, t0)i~

∂

∂t|ψ(t)〉

Now using the Schrodinger equation for the state of the system inthe Schrodinger picture, i~ ∂

∂t |ψ(t)〉 = (H0 + V (t))|ψ(t)〉,and usingthe time derivative of the hermitian conjugate of the evolutionoperator:

i~∂

∂tU †0(t, t0) =i~

∂

∂texp

(+i(t− t0)H0/~

)=i~(+iH0/~) exp

(+i(t− t0)H0/~

)= −H0U

†0 = −U †0H0

37/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

where we used the fact that the Hamiltonian commutes with itsevolution operator. Hence

i~∂

∂t|ψ(t)〉 =− U †0(t, t0)H0|ψ(t)〉+ U †0(t, t0)(H0 + V (t))|ψ(t)〉

=U †0(t, t0)V (t)1|ψ(t)〉=U †0(t, t0)V0(t)U0(t, t0)U

†0(t, t0)|ψ(t)〉

=ˆV (t)|ψ(t)〉

where the interaction term in the interaction picture is:

ˆV (t) = U †0(t, t0)V0(t)U0(t, t0)

38/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

and so the time-evolution of the states in the interaction picture isjust due to the interaction hamiltonian, which obeys theSchrodinger equation with the Hamiltonian being just theinteraction term:

i~∂

∂t|ψ(t)〉 =

ˆV (t)|ψ(t)〉

Furthermore the operators in the interaction picture obey theequation:

d

dtˆQ =

i

~

[H0,

ˆQ]

+

ˆ(dQ

dt

)(prove this).In the case that the interaction term istime-independent this reduces to an the Heisenberg equation butwith the Hamiltonian being the free-Hamiltonian.

39/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

and so the time-evolution of the states in the interaction picture isjust due to the interaction hamiltonian, which obeys theSchrodinger equation with the Hamiltonian being just theinteraction term:

i~∂

∂t|ψ(t)〉 =

ˆV (t)|ψ(t)〉

Furthermore the operators in the interaction picture obey theequation:

d

dtˆQ =

i

~

[H0,

ˆQ]

+

ˆ(dQ

dt

)(prove this).In the case that the interaction term istime-independent this reduces to an the Heisenberg equation butwith the Hamiltonian being the free-Hamiltonian.

39/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

and so the time-evolution of the states in the interaction picture isjust due to the interaction hamiltonian, which obeys theSchrodinger equation with the Hamiltonian being just theinteraction term:

i~∂

∂t|ψ(t)〉 =

ˆV (t)|ψ(t)〉

Furthermore the operators in the interaction picture obey theequation:

d

dtˆQ =

i

~

[H0,

ˆQ]

+

ˆ(dQ

dt

)(prove this).In the case that the interaction term istime-independent this reduces to an the Heisenberg equation butwith the Hamiltonian being the free-Hamiltonian.

39/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

The interaction (Dirac) picture of Quantum Mechanics

and so the time-evolution of the states in the interaction picture isjust due to the interaction hamiltonian, which obeys theSchrodinger equation with the Hamiltonian being just theinteraction term:

i~∂

∂t|ψ(t)〉 =

ˆV (t)|ψ(t)〉

Furthermore the operators in the interaction picture obey theequation:

d

dtˆQ =

i

~

[H0,

ˆQ]

+

ˆ(dQ

dt

)(prove this).In the case that the interaction term istime-independent this reduces to an the Heisenberg equation butwith the Hamiltonian being the free-Hamiltonian.

39/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

as any operator maybe expressed as Q =∑

mnQmn|m〉〈n|,wehave:

p =∑xx′

pxx′ |x〉〈x′|, pxx′ = 〈x|p|x′〉 = −δxx′i~∂

∂x

hence

p = −i~∑x

∂

∂x|x〉〈x|

To see the action of the operator p on a certain wavefunction〈x|ψ〉 let us consider its action on a the corresponding statep|ψ〉 = |χ〉, so:

|χ〉 = −i~∑x

∂

∂x|x〉〈x|ψ〉

projecting on the 〈x′| representation:40/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

〈x′|χ〉 = −i~∑x

∂

∂x〈x′|x〉〈x|ψ〉 = −i~

∑x

δxx′∂

∂x〈x|ψ〉 = −i~ ∂

∂x′〈x′|ψ〉

so in position representation:

〈x|p|ψ〉 = −i~ ∂∂x〈x|ψ〉

Working in the p-representation now we have:

p =∑p

p|p〉〈p|

where the operator p is diagonal in this representation, since thekets |p〉 are eigenkets of the momentum operator, p|p〉 = p|p〉.

41/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

〈x′|χ〉 = −i~∑x

∂

∂x〈x′|x〉〈x|ψ〉 = −i~

∑x

δxx′∂

∂x〈x|ψ〉 = −i~ ∂

∂x′〈x′|ψ〉

so in position representation:

〈x|p|ψ〉 = −i~ ∂∂x〈x|ψ〉

Working in the p-representation now we have:

p =∑p

p|p〉〈p|

where the operator p is diagonal in this representation, since thekets |p〉 are eigenkets of the momentum operator, p|p〉 = p|p〉.

41/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

〈x′|χ〉 = −i~∑x

∂

∂x〈x′|x〉〈x|ψ〉 = −i~

∑x

δxx′∂

∂x〈x|ψ〉 = −i~ ∂

∂x′〈x′|ψ〉

so in position representation:

〈x|p|ψ〉 = −i~ ∂∂x〈x|ψ〉

Working in the p-representation now we have:

p =∑p

p|p〉〈p|

where the operator p is diagonal in this representation, since thekets |p〉 are eigenkets of the momentum operator, p|p〉 = p|p〉.

41/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

〈x′|χ〉 = −i~∑x

∂

∂x〈x′|x〉〈x|ψ〉 = −i~

∑x

δxx′∂

∂x〈x|ψ〉 = −i~ ∂

∂x′〈x′|ψ〉

so in position representation:

〈x|p|ψ〉 = −i~ ∂∂x〈x|ψ〉

Working in the p-representation now we have:

p =∑p

p|p〉〈p|

where the operator p is diagonal in this representation, since thekets |p〉 are eigenkets of the momentum operator, p|p〉 = p|p〉.

41/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

So now what is the action of the operator p on the wavefuction inmomentum representation, ψ(p)? We have that:

p|ψ〉 =∑p

p|p〉〈p|ψ〉

so projecting onto the bra 〈p′|, we find:

〈p′|p|ψ〉 =∑p

p〈p′|p〉〈p|ψ〉 =∑p

pδpp′〈p|ψ〉 = p′〈p′|ψ〉

hence the action of the momentum operator on the wavefunctionin momentum space is:

〈p|p|ψ〉 = p〈p|ψ〉Note that the wavefunction in momentum space 〈p|ψ〉 ≡ ψ(p) isjust the Fourier transform of the wavefunction 〈x|ψ〉 ≡ ψ(p)(prove this).

42/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

So now what is the action of the operator p on the wavefuction inmomentum representation, ψ(p)? We have that:

p|ψ〉 =∑p

p|p〉〈p|ψ〉

so projecting onto the bra 〈p′|, we find:

〈p′|p|ψ〉 =∑p

p〈p′|p〉〈p|ψ〉 =∑p

pδpp′〈p|ψ〉 = p′〈p′|ψ〉

hence the action of the momentum operator on the wavefunctionin momentum space is:

〈p|p|ψ〉 = p〈p|ψ〉Note that the wavefunction in momentum space 〈p|ψ〉 ≡ ψ(p) isjust the Fourier transform of the wavefunction 〈x|ψ〉 ≡ ψ(p)(prove this).

42/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

So now what is the action of the operator p on the wavefuction inmomentum representation, ψ(p)? We have that:

p|ψ〉 =∑p

p|p〉〈p|ψ〉

so projecting onto the bra 〈p′|, we find:

〈p′|p|ψ〉 =∑p

p〈p′|p〉〈p|ψ〉 =∑p

pδpp′〈p|ψ〉 = p′〈p′|ψ〉

hence the action of the momentum operator on the wavefunctionin momentum space is:

〈p|p|ψ〉 = p〈p|ψ〉Note that the wavefunction in momentum space 〈p|ψ〉 ≡ ψ(p) isjust the Fourier transform of the wavefunction 〈x|ψ〉 ≡ ψ(p)(prove this).

42/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

So now what is the action of the operator p on the wavefuction inmomentum representation, ψ(p)? We have that:

p|ψ〉 =∑p

p|p〉〈p|ψ〉

so projecting onto the bra 〈p′|, we find:

〈p′|p|ψ〉 =∑p

p〈p′|p〉〈p|ψ〉 =∑p

pδpp′〈p|ψ〉 = p′〈p′|ψ〉

hence the action of the momentum operator on the wavefunctionin momentum space is:

〈p|p|ψ〉 = p〈p|ψ〉Note that the wavefunction in momentum space 〈p|ψ〉 ≡ ψ(p) isjust the Fourier transform of the wavefunction 〈x|ψ〉 ≡ ψ(p)(prove this).

42/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesMomentum operator

So now what is the action of the operator p on the wavefuction inmomentum representation, ψ(p)? We have that:

p|ψ〉 =∑p

p|p〉〈p|ψ〉

so projecting onto the bra 〈p′|, we find:

〈p′|p|ψ〉 =∑p

p〈p′|p〉〈p|ψ〉 =∑p

pδpp′〈p|ψ〉 = p′〈p′|ψ〉

hence the action of the momentum operator on the wavefunctionin momentum space is:

〈p|p|ψ〉 = p〈p|ψ〉Note that the wavefunction in momentum space 〈p|ψ〉 ≡ ψ(p) isjust the Fourier transform of the wavefunction 〈x|ψ〉 ≡ ψ(p)(prove this).

42/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

The operator for kinetic energy is

T =p2

2m

In the momentum representation we have that p2 is diagonal:

p2 =∑p

p2|p〉〈p|

as |p〉 are eigenstates of p2, p2|p〉 = p2|p〉, and therefore thekinetic energy operator is also diagonal:

T =∑p

p2

2m|p〉〈p|

43/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

The operator for kinetic energy is

T =p2

2m

In the momentum representation we have that p2 is diagonal:

p2 =∑p

p2|p〉〈p|

as |p〉 are eigenstates of p2, p2|p〉 = p2|p〉, and therefore thekinetic energy operator is also diagonal:

T =∑p

p2

2m|p〉〈p|

43/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

The operator for kinetic energy is

T =p2

2m

In the momentum representation we have that p2 is diagonal:

p2 =∑p

p2|p〉〈p|

as |p〉 are eigenstates of p2, p2|p〉 = p2|p〉, and therefore thekinetic energy operator is also diagonal:

T =∑p

p2

2m|p〉〈p|

43/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

The operator for kinetic energy is

T =p2

2m

In the momentum representation we have that p2 is diagonal:

p2 =∑p

p2|p〉〈p|

as |p〉 are eigenstates of p2, p2|p〉 = p2|p〉, and therefore thekinetic energy operator is also diagonal:

T =∑p

p2

2m|p〉〈p|

43/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

thus the action of the kinetic energy operator on the wavefunctionin momentum space is:

〈p|T |ψ〉 =p2

2m〈p|ψ〉

What is the kinetic energy operator in position representation?

44/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesKinetic energy in momentum representation:

thus the action of the kinetic energy operator on the wavefunctionin momentum space is:

〈p|T |ψ〉 =p2

2m〈p|ψ〉

What is the kinetic energy operator in position representation?

44/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We want to express the potential energy in momentumrepresentation:

V (x) =∑pp′

Vpp′ |p〉〈p′|, Vpp′ = 〈p|V (x)|p′〉

Since potential energy is usually only known in positionrepresentation we introduce other unities as follows:

V (x) =11V (x)11

=∑pp′xx′

|p〉〈p|x〉〈x|V (x)|x′〉〈x′|p′〉〈p′|

where we have that 〈x|V (x)|x′〉 = V (x)δxx′ , and〈x|p〉 = Ne−ixp/~, hence:

45/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We want to express the potential energy in momentumrepresentation:

V (x) =∑pp′

Vpp′ |p〉〈p′|, Vpp′ = 〈p|V (x)|p′〉

Since potential energy is usually only known in positionrepresentation we introduce other unities as follows:

V (x) =11V (x)11

=∑pp′xx′

|p〉〈p|x〉〈x|V (x)|x′〉〈x′|p′〉〈p′|

where we have that 〈x|V (x)|x′〉 = V (x)δxx′ , and〈x|p〉 = Ne−ixp/~, hence:

45/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We want to express the potential energy in momentumrepresentation:

V (x) =∑pp′

Vpp′ |p〉〈p′|, Vpp′ = 〈p|V (x)|p′〉

Since potential energy is usually only known in positionrepresentation we introduce other unities as follows:

V (x) =11V (x)11

=∑pp′xx′

|p〉〈p|x〉〈x|V (x)|x′〉〈x′|p′〉〈p′|

where we have that 〈x|V (x)|x′〉 = V (x)δxx′ , and〈x|p〉 = Ne−ixp/~, hence:

45/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We want to express the potential energy in momentumrepresentation:

V (x) =∑pp′

Vpp′ |p〉〈p′|, Vpp′ = 〈p|V (x)|p′〉

Since potential energy is usually only known in positionrepresentation we introduce other unities as follows:

V (x) =11V (x)11

=∑pp′xx′

|p〉〈p|x〉〈x|V (x)|x′〉〈x′|p′〉〈p′|

where we have that 〈x|V (x)|x′〉 = V (x)δxx′ , and〈x|p〉 = Ne−ixp/~, hence:

45/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

V (x) =∑pp′xx′

N∗e+ixp/~V (x)δxx′Ne−ix′p′/~|p〉〈p′|

V (x) =∑pp′xx′

|N |2V (x)e−i(x′p′−xp)/~δxx′ |p〉〈p′|

=|N |2∑pp′x

e−ix(p′−p)/~V (x)|p〉〈p′|

Let us define the Fourier transform of the potential energy:

V (p− p′) = |N |2∑x

e−ix(p′−p)/~V (x)→ |N |2

∫e−ix(p

′−p)V (x)dx

such that the potential energy operator is written as:

V (x) =∑pp′

V (p− p′)|p〉〈p′|

46/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

V (x) =∑pp′xx′

N∗e+ixp/~V (x)δxx′Ne−ix′p′/~|p〉〈p′|

V (x) =∑pp′xx′

|N |2V (x)e−i(x′p′−xp)/~δxx′ |p〉〈p′|

=|N |2∑pp′x

e−ix(p′−p)/~V (x)|p〉〈p′|

Let us define the Fourier transform of the potential energy:

V (p− p′) = |N |2∑x

e−ix(p′−p)/~V (x)→ |N |2

∫e−ix(p

′−p)V (x)dx

such that the potential energy operator is written as:

V (x) =∑pp′

V (p− p′)|p〉〈p′|

46/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

V (x) =∑pp′xx′

N∗e+ixp/~V (x)δxx′Ne−ix′p′/~|p〉〈p′|

V (x) =∑pp′xx′

|N |2V (x)e−i(x′p′−xp)/~δxx′ |p〉〈p′|

=|N |2∑pp′x

e−ix(p′−p)/~V (x)|p〉〈p′|

Let us define the Fourier transform of the potential energy:

V (p− p′) = |N |2∑x

e−ix(p′−p)/~V (x)→ |N |2

∫e−ix(p

′−p)V (x)dx

such that the potential energy operator is written as:

V (x) =∑pp′

V (p− p′)|p〉〈p′|

46/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

V (x) =∑pp′xx′

N∗e+ixp/~V (x)δxx′Ne−ix′p′/~|p〉〈p′|

V (x) =∑pp′xx′

|N |2V (x)e−i(x′p′−xp)/~δxx′ |p〉〈p′|

=|N |2∑pp′x

e−ix(p′−p)/~V (x)|p〉〈p′|

Let us define the Fourier transform of the potential energy:

V (p− p′) = |N |2∑x

e−ix(p′−p)/~V (x)→ |N |2

∫e−ix(p

′−p)V (x)dx

such that the potential energy operator is written as:

V (x) =∑pp′

V (p− p′)|p〉〈p′|

46/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

V (x) =∑pp′xx′

N∗e+ixp/~V (x)δxx′Ne−ix′p′/~|p〉〈p′|

V (x) =∑pp′xx′

|N |2V (x)e−i(x′p′−xp)/~δxx′ |p〉〈p′|

=|N |2∑pp′x

e−ix(p′−p)/~V (x)|p〉〈p′|

Let us define the Fourier transform of the potential energy:

V (p− p′) = |N |2∑x

e−ix(p′−p)/~V (x)→ |N |2

∫e−ix(p

′−p)V (x)dx

such that the potential energy operator is written as:

V (x) =∑pp′

V (p− p′)|p〉〈p′|

46/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

which is not a diagonal operator in the p-representation. Inmomentum space we can represent the “interaction” of a particlewith the external potential by a “vertex”:

47/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

which is not a diagonal operator in the p-representation. Inmomentum space we can represent the “interaction” of a particlewith the external potential by a “vertex”:

47/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

which is not a diagonal operator in the p-representation. Inmomentum space we can represent the “interaction” of a particlewith the external potential by a “vertex”:

p p′

q = p′ − p

47/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Now we can see the action of the potential energy operator inmomentum representation. We have:

V (x)|ψ〉 =∑pp′

V (p− p′)|p〉〈p′|ψ〉

and projecting this onto the bra 〈p′′| we find:

〈p′′|V (x)|ψ〉 =∑pp′

V (p− p′)〈p′′|p〉〈p′|ψ〉

=∑pp′

V (p− p′)δpp′′〈p′|ψ〉

=∑p′

V (p′′ − p′)〈p′|ψ〉

48/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Now we can see the action of the potential energy operator inmomentum representation. We have:

V (x)|ψ〉 =∑pp′

V (p− p′)|p〉〈p′|ψ〉

and projecting this onto the bra 〈p′′| we find:

〈p′′|V (x)|ψ〉 =∑pp′

V (p− p′)〈p′′|p〉〈p′|ψ〉

=∑pp′

V (p− p′)δpp′′〈p′|ψ〉

=∑p′

V (p′′ − p′)〈p′|ψ〉

48/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Now we can see the action of the potential energy operator inmomentum representation. We have:

V (x)|ψ〉 =∑pp′

V (p− p′)|p〉〈p′|ψ〉

and projecting this onto the bra 〈p′′| we find:

〈p′′|V (x)|ψ〉 =∑pp′

V (p− p′)〈p′′|p〉〈p′|ψ〉

=∑pp′

V (p− p′)δpp′′〈p′|ψ〉

=∑p′

V (p′′ − p′)〈p′|ψ〉

48/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus:〈p|V (x)|ψ〉 =

∑p′

V (p− p′)〈p′|ψ〉

Let us take the example that we dealt with before:

V (x) = u0δ(x)

with u0 < 0. We solved this problem before in the positionrepresentation.In the momentum representation we first need tofind the Fourier Transform of the potential:

V (q) = |N |2∫e−iqx/~V (x)dx

49/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus:〈p|V (x)|ψ〉 =

∑p′

V (p− p′)〈p′|ψ〉

Let us take the example that we dealt with before:

V (x) = u0δ(x)

with u0 < 0. We solved this problem before in the positionrepresentation.In the momentum representation we first need tofind the Fourier Transform of the potential:

V (q) = |N |2∫e−iqx/~V (x)dx

49/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus:〈p|V (x)|ψ〉 =

∑p′

V (p− p′)〈p′|ψ〉

Let us take the example that we dealt with before:

V (x) = u0δ(x)

with u0 < 0. We solved this problem before in the positionrepresentation.In the momentum representation we first need tofind the Fourier Transform of the potential:

V (q) = |N |2∫e−iqx/~V (x)dx

49/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with q = p′ − p the momentum transfer.Substituting we simplyfind:

V (q) = |N |2∫e−iqx/~u0δ(x)dx = |N |2u0

so the potential in Fourier space is a constant. So it is expressedas:

V = |N |2u0∑pp′

|p〉〈p′|

and

〈p|V |ψ〉 = |N |2u0∑p′

〈p′|ψ〉 = |N |2u0∫p space

ψ(p′)dp′

50/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with q = p′ − p the momentum transfer.Substituting we simplyfind:

V (q) = |N |2∫e−iqx/~u0δ(x)dx = |N |2u0

so the potential in Fourier space is a constant. So it is expressedas:

V = |N |2u0∑pp′

|p〉〈p′|

and

〈p|V |ψ〉 = |N |2u0∑p′

〈p′|ψ〉 = |N |2u0∫p space

ψ(p′)dp′

50/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with q = p′ − p the momentum transfer.Substituting we simplyfind:

V (q) = |N |2∫e−iqx/~u0δ(x)dx = |N |2u0

so the potential in Fourier space is a constant. So it is expressedas:

V = |N |2u0∑pp′

|p〉〈p′|

and

〈p|V |ψ〉 = |N |2u0∑p′

〈p′|ψ〉 = |N |2u0∫p space

ψ(p′)dp′

50/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with q = p′ − p the momentum transfer.Substituting we simplyfind:

V (q) = |N |2∫e−iqx/~u0δ(x)dx = |N |2u0

so the potential in Fourier space is a constant. So it is expressedas:

V = |N |2u0∑pp′

|p〉〈p′|

and

〈p|V |ψ〉 = |N |2u0∑p′

〈p′|ψ〉 = |N |2u0∫p space

ψ(p′)dp′

50/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We can now write the total hamiltonian in the momentumrepresentation:

H =p2

2m+ u0δ(x)

as:

H =∑p

p2

2m|p〉〈p|+

∑pp′

V (p− p′)|p〉〈p′|

In our example we have:

H =∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

So we can now solve the same problem we solved before but inmomentum space, i.e. solve the TISE:

H|ψ〉 = E|ψ〉51/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We can now write the total hamiltonian in the momentumrepresentation:

H =p2

2m+ u0δ(x)

as:

H =∑p

p2

2m|p〉〈p|+

∑pp′

V (p− p′)|p〉〈p′|

In our example we have:

H =∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

So we can now solve the same problem we solved before but inmomentum space, i.e. solve the TISE:

H|ψ〉 = E|ψ〉51/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We can now write the total hamiltonian in the momentumrepresentation:

H =p2

2m+ u0δ(x)

as:

H =∑p

p2

2m|p〉〈p|+

∑pp′

V (p− p′)|p〉〈p′|

In our example we have:

H =∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

So we can now solve the same problem we solved before but inmomentum space, i.e. solve the TISE:

H|ψ〉 = E|ψ〉51/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

We can now write the total hamiltonian in the momentumrepresentation:

H =p2

2m+ u0δ(x)

as:

H =∑p

p2

2m|p〉〈p|+

∑pp′

V (p− p′)|p〉〈p′|

In our example we have:

H =∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

So we can now solve the same problem we solved before but inmomentum space, i.e. solve the TISE:

H|ψ〉 = E|ψ〉51/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

with normalised states 〈ψ|ψ〉 = 1.In momentum representation wewrite this as:

〈p′′|H|ψ〉 = E〈p′′|ψ〉substituting:

〈p′′|

∑p

p2

2m|p〉〈p|+ |N |2u0

∑pp′

|p〉〈p′|

|ψ〉 =E〈p′′|ψ〉

∑p

p2

2m〈p′′|p〉〈p|ψ〉+ |N |2u0

∑pp′

〈p′′|p〉〈p′|ψ〉 =E〈p′′|ψ〉

∑p

p2

2mδpp′′〈p|ψ〉+ |N |2u0

∑pp′

δpp′′〈p′|ψ〉 =E〈p′′|ψ〉

p′′2

2m〈p′′|ψ〉+ |N |2u0

∑p′

〈p′|ψ〉 =E〈p′′|ψ〉52/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

Thus we obtain the equation

p2

2mψ(p) + |N |2u0

∫p′ space

ψ(p′)dp′ = Eψ(p)

where we note that the result of the integral∫p′ space ψ(p′)dp′ is in

fact just a number because one integrates over all momentumspace(it is also independent of p), so we treat it just like anumber.Thus solving for ψ(p) we find that:(

p2

2m− E

)ψ(p) = −|N |2u0

∫p′ space

ψ(p′)dp′ ⇒

ψ(p) =−|N |2u0p2

2m − E

∫p′ space

ψ(p′)dp′ (10)

Hence we see that

ψ(p) =A

p2 − 2mE53/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and the constant A is just a normalisation constant. Obviously werequire that E < 0 so that the wavefunction be finite andsquare-integrable.We can find the normalisation constant (inone-dimensional space) to be:

A =

√2

π(−2mE)3/4

, so:

ψ(p) =

√2

π(−2mE)3/4

1

p2 − 2mE

In order to find the energy E we integrate both sides of equation(10) over p, then we arrive at:∫

p spaceψ(p)dp =

∫space

−|N |2u0p2

2m − Edp×

∫p′ space

ψ(p′)dp′

54/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and the constant A is just a normalisation constant. Obviously werequire that E < 0 so that the wavefunction be finite andsquare-integrable.We can find the normalisation constant (inone-dimensional space) to be:

A =

√2

π(−2mE)3/4

, so:

ψ(p) =

√2

π(−2mE)3/4

1

p2 − 2mE

In order to find the energy E we integrate both sides of equation(10) over p, then we arrive at:∫

p spaceψ(p)dp =

∫space

−|N |2u0p2

2m − Edp×

∫p′ space

ψ(p′)dp′

54/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and the constant A is just a normalisation constant. Obviously werequire that E < 0 so that the wavefunction be finite andsquare-integrable.We can find the normalisation constant (inone-dimensional space) to be:

A =

√2

π(−2mE)3/4

, so:

ψ(p) =

√2

π(−2mE)3/4

1

p2 − 2mE

In order to find the energy E we integrate both sides of equation(10) over p, then we arrive at:∫

p spaceψ(p)dp =

∫space

−|N |2u0p2

2m − Edp×

∫p′ space

ψ(p′)dp′

54/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and the constant A is just a normalisation constant. Obviously werequire that E < 0 so that the wavefunction be finite andsquare-integrable.We can find the normalisation constant (inone-dimensional space) to be:

A =

√2

π(−2mE)3/4

, so:

ψ(p) =

√2

π(−2mE)3/4

1

p2 − 2mE

In order to find the energy E we integrate both sides of equation(10) over p, then we arrive at:∫

p spaceψ(p)dp =

∫space

−|N |2u0p2

2m − Edp×

∫p′ space

ψ(p′)dp′

54/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and the constant A is just a normalisation constant. Obviously werequire that E < 0 so that the wavefunction be finite andsquare-integrable.We can find the normalisation constant (inone-dimensional space) to be:

A =

√2

π(−2mE)3/4

, so:

ψ(p) =

√2

π(−2mE)3/4

1

p2 − 2mE

In order to find the energy E we integrate both sides of equation(10) over p, then we arrive at:∫

p spaceψ(p)dp =

∫space

−|N |2u0p2

2m − Edp×

∫p′ space

ψ(p′)dp′

54/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and since∫p space ψ(p)dp =

∫p′ space ψ(p′)dp′ we deduce that the

energy eigenvalue E must satisfy:∫space

−|N |2u0p2

2m − Edp = 1

Let us solve this equation for E in the following cases:

55/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

and since∫p space ψ(p)dp =

∫p′ space ψ(p′)dp′ we deduce that the

energy eigenvalue E must satisfy:∫space

−|N |2u0p2

2m − Edp = 1

Let us solve this equation for E in the following cases:

55/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

One-dimensional space Solving the integral we find:∫space

−|N |2u0p2

2m − Edp =

∫ +∞

−∞

−|N |2u02mp2 − 2mE

dp = −|N |2u02mπ√−2mE

= 1

where we used the fact that∫ ∞−∞

1

x2 + adx =

π√a, a > 0

which gives (squaring both sides):

E = −|N |4u20m2π2

Now using the normalisation N = 1√2π~

we obtain:

E = − 1

4π2~2u20m2π2 = −u

20m

2~2which is exactly the result we found before.

56/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

One-dimensional space Solving the integral we find:∫space

−|N |2u0p2

2m − Edp =

∫ +∞

−∞

−|N |2u02mp2 − 2mE

dp = −|N |2u02mπ√−2mE

= 1

where we used the fact that∫ ∞−∞

1

x2 + adx =

π√a, a > 0

which gives (squaring both sides):

E = −|N |4u20m2π2

Now using the normalisation N = 1√2π~

we obtain:

E = − 1

4π2~2u20m2π2 = −u

20m

2~2which is exactly the result we found before.

56/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

One-dimensional space Solving the integral we find:∫space

−|N |2u0p2

2m − Edp =

∫ +∞

−∞

−|N |2u02mp2 − 2mE

dp = −|N |2u02mπ√−2mE

= 1

where we used the fact that∫ ∞−∞

1

x2 + adx =

π√a, a > 0

which gives (squaring both sides):

E = −|N |4u20m2π2

Now using the normalisation N = 1√2π~

we obtain:

E = − 1

4π2~2u20m2π2 = −u

20m

2~2which is exactly the result we found before.

56/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

One-dimensional space Solving the integral we find:∫space

−|N |2u0p2

2m − Edp =

∫ +∞

−∞

−|N |2u02mp2 − 2mE

dp = −|N |2u02mπ√−2mE

= 1

where we used the fact that∫ ∞−∞

1

x2 + adx =

π√a, a > 0

which gives (squaring both sides):

E = −|N |4u20m2π2

Now using the normalisation N = 1√2π~

we obtain:

E = − 1

4π2~2u20m2π2 = −u

20m

2~2which is exactly the result we found before.

56/56 Advanced Quantum Mechanics 2 - lecture 2

Quantisation schemes of quantum mechanics

ExamplesPotential energy in momentum space:

One-dimensional space Solving the integral we find:∫space

−|N |2u0p2

2m − Edp =

∫ +∞

−∞

−|N |2u02mp2 − 2mE

dp = −|N |2u02mπ√−2mE

= 1

where we used the fact that∫ ∞−∞

1

x2 + adx =

π√a, a > 0

which gives (squaring both sides):

E = −|N |4u20m2π2

Now using the normalisation N = 1√2π~

we obtain:

E = − 1

4π2~2u20m2π2 = −u

20m

2~2which is exactly the result we found before.

56/56 Advanced Quantum Mechanics 2 - lecture 2