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Algebraic Method of Solving the Linear Harmonic
Oscillator
Lecture by Gable Rhodes
PHYS 773: Quantum MechanicsFebruary 6th, 2012
Gable Rhodes, February 6th, 2012 2
Simple Harmonic Oscillator• Many physical problems can be modeled as small
oscillations around a stable equilibrium• Potential is described as parabolic around the
minimum energy
• The Hamiltonian is formulated in the usual way for canonical variables q and p.
22
21)( qmqV
)(2
2
qVmpH
Gable Rhodes, February 6th, 2012 3
Raising & Lowering Operators - Defined
• After substituting in our potential, we can rearrange the constants slightly to give the Hamiltonian a more suggestive appearance
• If we “factor” the operators, we get
• Recognizing that the last term is the commutator of the canonical variables, we get
22222222
21
21 pqm
mqmp
mH
qppqimipqmipqmm
H 21
pqimipqmipqmm
H ,21
Gable Rhodes, February 6th, 2012 4
pqimipqmipqmm
H ,21
Raising & Lowering Operators - Defined
• We can now define the operator a
• Upon substitution
• And Simplifying
mpiqma
2
pqimaamm
H ,221 †
mpiqma
2†
pqiaaH ,
2†
Gable Rhodes, February 6th, 2012 5
Raising & Lowering Operators - Defined
• If we reverse the order of the operators, a similar expression is obtained
• Subtracting the two forms yields the commutator
• Simplifying to
0,2
,2
††
pqiaapqiaaHH
0,†† pqiaaaa
pqiaa ,, †
pqiaaH ,
2†
pqiaaH ,
2†
Gable Rhodes, February 6th, 2012 6
Raising & Lowering Operators - Defined
• It is important to note that the a, a† operators are defined in such a way that as long as the state variables follow the canonical commutation relation, the a, a† commutator will be 1.
• And the Hamiltonian can be written as a linear function of a, a†
21†aaH
pqiaa ,, †
1, †
iiaa
21†aaH
Gable Rhodes, February 6th, 2012 7
Properties of a, a†
• If a wave function has the property that it is an eigenfunction of the Hamiltonian
• We can introduce the equivalent statement for the operator aa† with generic eigenvalue, λ
• We then apply the a† operator to a†ψ and test if the result is the same eigenvector.
)(2
2
qVmpH EH
aa†
†††††††† 111 aaaaaaaaaaa
Use commutator
Gable Rhodes, February 6th, 2012 8
Properties of a, a†
• And the equivalent method for a
• Using the relationship of Hamiltonian, we can then relate eigenvalues
aa†
aaaaaaaaaaa 1111 †††
EH aa†
21
21 †† aaaaH
21†aaEH
21
21† aaE
21E
Gable Rhodes, February 6th, 2012 9
Raising & Lowering Operators -Properties
• a† is called the raising (or creation) operator.
• And a is the lowering (or annihilating) operator.
• The rungs of the ladder are all evenly separated.
• No degeneracy.
ψ
a† ψ
aψ
aaψ
a† a† ψ
λ
λ+ 1
λ+2
λ-2
λ- 1W
ave
vect
ors
Eige
nval
ues
Gable Rhodes, February 6th, 2012 10
What is the Significance?• If we have any solution, we can find infinitely more
solutions by repeated application of the raising and lowering operators
• But, importantly, although there are infinite solutions we know that the are are no solutions with negative energy (both kinetic and potential components for the Hamiltonian are always positive)
• Therefore, a ground state must exist. (this is also a property of Sturm-Liouville PDE)
• Applying the lowering operator to the ground state will result in a null vector (trivial state).
• Eq.10-77 0a
Gable Rhodes, February 6th, 2012 11
Raising & Lowering Operators -Eigenvectors
• Once we have a ground state, repeated application of the raising operator will result in an infinite set of eigenvectors with distinct (non-degenerate) eigenvalues.
• And introducing an arbitrary starting point λ0.
• But for the special case of the ground state
• Must be zero on the right side (by our definition), so λ0 is exactly 0.
0† n
n a
0†
00†††
nnn anaaaaa
000† 0 aa
00 00 nn 0
...2,1,0n
Gable Rhodes, February 6th, 2012 12
Determine the Energy Levels• Using the previous equation relating the Hamiltonian
and aa†, we can relate energy to λ (n).
• And since we started at the ground state, we can relate the energy level to the eigenvectors
• If we use a normalization constant
• Where An can be found by direct integration at each step or by algebraic tricks (later)
21
21 nE
0† n
n a
21nEn
0† n
nn aA
Gable Rhodes, February 6th, 2012 13
What Are These Operators Good For Anyway?
• We found Energy exactly and wavevectors in abstract form.
• What else can we do with them?• What about expectation values? In chapter 5,
problem 2, we were asked to find <V>. This required direct integration with the (explicitly known) wavefunction.
• Can this be done without knowing the wavefunction?
22
21)( qmqV
22
21 qmV
2*2
21 qmV
Gable Rhodes, February 6th, 2012 14
Expectation value of Potential• If we use the definition of a†, a and rewrite q and
p operators in terms of a and a† we get
• Now these can be substituted into <V>
mpiqma
2†
mpiqma
2
22 †aam
q
22 †aaimm
p
2†*22*2
221
21 aa
mmqmV
)(22
1 ††††* aaaaaaaaV
Gable Rhodes, February 6th, 2012 15
Expectation value of Potential
• Of the four terms in the integral, we see that two of them vanish due to the orthogonality of the wavevectors.
• And the other two are known to us from the previous work.
2,2**
nnnnaa 02,2*††* nnnnaa
111 ,*†* nnnaa nnnn
nnnaa nnnn ,*†*
)(22
1 ††††* aaaaaaaaV
Gable Rhodes, February 6th, 2012 16
Expectation value of Potential• This makes the result pretty straightforward
• And the this result agrees with problem 5.2, and the virial theorem
• But, we did not need to know the explicit form of the wavefunction.
10022
1 nnV
nEnV
21
212
21
)(22
1 ††††* aaaaaaaaV
Gable Rhodes, February 6th, 2012 17
<q>, <p>, <q2>, <p2>• Other key expectation values can be easily
obtained.
22 †aam
q
22 †aaimm
p
02
†**
aam
02
†**
aaimm
pp
212 n
mq
2†*222*2
2aam
mpp
aaaaaaaamp ††††*2
2
21100
22 nmnnmp
Gable Rhodes, February 6th, 2012 18
Variance and Uncertainty?• The variance is therefore
• And this agrees with the uncertainty principle
21222 n
mqqq
21222 nmppp
221
npq
Gable Rhodes, February 6th, 2012 19
What is left then?• The Normalization constant, which can also be
determined algebraically.
• Square both sides and integrate
• After integration by parts and throwing out the boundary terms.
• So that gives us our wavefunction in terms of the ground state and raising operator
1†
nn Ba
1,12
1*
1†*†
nnnnnn BBBaa
2*† Baa nn nnnn nnB ,*2 11
1† 1 nn na
Gable Rhodes, February 6th, 2012 20
Normalization• Lets try ψ1.
• Next try ψ2.
• We can now write the general equation
0†
!1
nn a
n
1† 1 nn na
0†
1010 a 0†
1 a
1†
1111 a 0
2†2 !2
1 a
Gable Rhodes, February 6th, 2012 21
And what is ψ0? • Starting with our condition for the ground state
• And using the definition of the operator
• We get a first order ODE
00 a
mpiqma
2
02 00
mpiqma
00
mpiq
qip
00 qm
dqd
Gable Rhodes, February 6th, 2012 22
And what is ψ0?• The equation is separable and easily solved.
• After normalization we get
• Which is consistent with the solutions in chapter 5.
00 qm
dqd
2
0 2exp qmconst
2
41
0 2exp qmm
Gable Rhodes, February 6th, 2012 23
Finding ψ1. • Here we can use the raising operator to generate
further solutions
• Substitute in the operator
• After rearranging, we get the desired result.
0†
!1
nn a
n
01†
1 !11 a
01 2
mpiqm
01 22
qm
Gable Rhodes, February 6th, 2012 24
Finding ψn. • We find that continuing up the ladder is in fact a
generating algorithm for the Hermite polynomials, and the general equation then identical to eq. 5.39 0
†
!1
nn a
n
0!2
1
qmH
nnnn
Gable Rhodes, February 6th, 2012 25
Matrix representation of a, a†
• We can show that the lowering operator relates neighboring wave vectors with the normalization factor.
• Adding the bra.
• This gives the matrix elements of a as shown.
1 nn na
nna nnnn 111
00400
300200
10
a
Gable Rhodes, February 6th, 2012 26
Matrix representation of a, a†
• As an example, lowering n=2,nna nnnn 111
2020
010
100
00300
20010
010
221 a
Gable Rhodes, February 6th, 2012 27
Matrix representation of a, a†
• The equivalent representation of the raising operator is derived from the expression
• Adding the bra.
1† 1 nn na
03
002001
00
†a
11 11†
1 nna nnnn
Gable Rhodes, February 6th, 2012 28
Matrix representation of a, a†
• With the lowering and raising operators in matrix form, we can then solve for the q and p operators in terms of a, a†.
• For position we get,
22 †aam
q
03302
20110
222 †
maa
mq
Gable Rhodes, February 6th, 2012 29
Matrix representation of a, a†
• And the equivalent expression for momentum is,
• The complex constant insures that the anti-symmetric matrix is Hermitian.
22 †aaimm
p
03302
20110
222 †
miaaimm
p