Overheads Lecture Chapter 7

8/7/2019 Overheads Lecture Chapter 7

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Chapter 7

Degenerate Perturbation Theory

Chapter 7 – p.1/ ?



Degenerate Perturbation Theory

The non-degenerate perturbation theory discussed in chapter 6 breaks down whenwe have degenerate energy levels E 01 = E 02 .

Note also that the results of 2nd order non-degenerate perturbation theory assume|V np | << E 0

p − E 0n so there is a problem if the spacing between levels is smallcompared to the perturbing matrix element.

| φ1

n = p p= n

φ0 p | V | φ0

n

E 0n − E 0 p | φ0

p (6 .15)

E 2n = − p

p= n

φ0

p |ˆV | φ

0

n

2

E 0 p − E 0n(6 .19)




The Degeneracy Problem

Suppose an eigenvalue E 0n of the unperturbed Hamiltonian H 0 is s -fold degenerate.






There are |u 0nα , (α = 1 , 2 · · · , s ) linearly independent orthonormal eigenfunctions

belonging to this eigenvalue,

u 0nα | u 0

nβ = δαβ , with α, β = 1 , · · · , s (7.1)

and any linear combination of these eigenfunctions is also an eigenfunction of H 0 .








u 0nα | u 0

nβ = δαβ , with α, β = 1 , · · · , s (7.1)


So the zero-order states that we want to use in the PT expansions are not uniquelydened.








u 0nα | u 0

nβ = δαβ , with α, β = 1 , · · · , s (7.1)


We must therefore nd s normed states

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)

where the ciα are constant coefcients which are the correct linear combinations forPT, i.e. they produce reasonable perturbation expansions.





| φ1n = p

p= n

φ0

p |ˆV | φ

0

nE 0n − E 0 p

| φ0 p (6 .15)

E 2n = − p

p= n

φ0 p | V | φ0n2

E 0 p − E 0n(6 .18)

For degenerate levels these expansions will only make sense if these terms also havevanishing numerators, i.e. we require for the degenerate states that

φ0ni | V | φ0

nj = φ0ni | V | φ0

ni δij for i, j = 1 , · · · , s (7.6)

That is, the s × s matrix φ0ni | V | φ0

nj i, j = 1 , 2, · · · , s is a diagonal matrix.





Givenφ0

ni | V | φ0ni δij for i, j = 1 , · · · , s (7.6)

That is, the s × s matrix is a diagonal matrix, we can carry out the same calculationsas in Chapter 6.





Givenφ0



For example, in order to determine the rst-order shift in the energy levels we can usethe equivalent of Eq. 6.7

E 1ni = φ

0ni | V | φ

0ni (7 .7)

c.f. equation 6.10.





Givenφ0



For example, in order to determine the rst-order shift in the energy levels we can usethe equivalent of Eq. 6.7

E 1ni = φ

0ni | V | φ

0ni (7 .7)

c.f. equation 6.10.

We now describe one method for nding the correct zero-order linear combinations ofeigenfunctions.




Finding the zeroth-order eigenfunctions for PT

− V − E 1n | φ

0ni = H o − E

0n | φ

1ni . (7 .8)

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)

Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .





− V − E 1

n | φ0

ni = H o − E 0

n | φ1

ni . (7 .8)

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)


Take the scalar product of eq. 7.8 with u 0

nβwe obtain

− u 0nβ | V − E 1n | φ0

ni = u 0nβ | H 0 − E 0n | φ1

ni





− V − E 1

n | φ0

ni = H o − E 0

n | φ1

ni . (7 .8)

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)



nβwe obtain

− u 0nβ | V − E 1n | φ0

ni = u 0nβ | H 0 − E 0n | φ1

ni

On the RHS

u 0nβ | H 0 − E 0n | φ1

ni = u 0nβ | H 0 | φ1

ni − u 0nβ | E 0n | φ1

ni





− V − E 1

n | φ0

ni = H o − E 0

n | φ1

ni . (7 .8)

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)



nβwe obtain

− u 0nβ | V − E 1n | φ0

ni = u 0nβ | H 0 − E 0n | φ1

ni

On the RHS

u 0nβ | H 0 − E 0n | φ1

ni = u 0nβ | H 0 | φ1

ni − u 0nβ | E 0n | φ1

ni

= E 0

n u0

nβ | φ1

ni − E 0

n u0

nβ | φ1

ni





− V − E 1

n | φ0

ni = H o − E 0

n | φ1

ni . (7 .8)

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)



nβwe obtain

− u 0nβ | V − E 1n | φ0

ni = u 0nβ | H 0 − E 0n | φ1

ni

and so

u 0nβ | V | φ0

ni = E 1ni u 0nβ | φ0

ni .





u 0nβ | V | φ0


ni .

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)


d h h d f f




u 0nβ | V | φ0


ni .

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)

Then substituting eq 7.2 into this equation we obtain

u 0nβ V

s

α =1ciα u 0nα = E 1ni u 0nβ

s

α =1ciα u 0nα


Fi di h h d i f i f PT




u 0nβ | V | φ0


ni .

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)


u 0nβ V

s

α =1ciα u 0nα = E 1ni u 0nβ

s

α =1ciα u 0nα

s

α =1

u 0nβ V u 0

nα ciα = E 1ni u 0nβ | u 0

n β ciβ


Fi di th th d i f ti f PT




u 0nβ | V | φ0


ni .

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)


u0nβ V

s

α =1ciα u

0nα = E

1ni u

0nβ

s

α =1ciα u

0nα

s

α =1

u 0nβ V u 0

nα ciα = E 1ni u 0nβ | u 0

nβ ciβ

= E 1ni ciβ


Fi di g th th d ig f ti f PT




u 0nβ | V | φ0


ni .

φ0ni =

s

α =1

ciα u 0nα , i = 1 , · · · , s (7 .2)

and so

s

α =1u

0nβ V u

0nα ciα = E

1ni ciβ

ors

α =1u

0nβ | V | u

0nα − E

1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)


Finding the zeroth order eigenfunctions for PT




s

α =1u 0nβ | V | u 0nα − E 1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)

This set of s equations 7.9 give the required values for ciα .


Finding the zeroth order eigenfunctions for PT




s

α =1u 0nβ | V | u 0nα − E 1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)

This set of s equations 7.9 give the required values for ciα .

The vanishing of the s × s determinant

det u 0nβ | V | u 0

nα − E 1ni δβα = 0

gives the s roots E 1ni , i = 1 , · · · , s that are the rst-order energy level shifts for the

states φ0ni with the (non-zero) coefcients ciα the solutions of eqs. 7.9.


A Simple Example



A Simple Example

The simplest illustration of degenerate PT is to assume that there is a 2-fold

degeneracy in the unperturbed state.

As a rst step we note that there is a degeneracy.

H 0 | u01 = E

01 | u

01

H 0 | u 02 = E 02 | u 0

2 = E 01 | u 02


A Simple Example



A Simple Example




H 0 | u01 = E

01 | u

01

H 0 | u 02 = E 02 | u 0

2 = E 01 | u 02

We write down a perturbing Hamiltonian H which "lifts" the degeneracy of the system

H = H 0 + β V

H | ψn = E n | ψn


A Simple Example



A Simple Example




H 0 | u01 = E

01 | u

01

H 0 | u 02 = E 02 | u 0

2 = E 01 | u 02

We write down a perturbing Hamiltonian H which "lifts" the degeneracy of the system

H = H 0 + β V

H | ψn = E n | ψn

So for our two-fold degenerate system

H | ψ 1 = E 1 | ψ 1

H | ψ 2 = E 2 | ψ 2 where E 1 > E2


A Simple Example



A Simple Example

We note that u01∗u 0

2 dτ = 0 as well as u0n

∗u 0m dτ = δnm for all other n, m .


A Simple Example



A Simple Example




If this is not the case we must rst choose a different set of eigenfunctions that areorthonormal.


A Simple Example



p p




If this is not the case we must rst choose a different set of eigenfunctions that areorthonormal.

Ideally we would like to obtain the correct eigenfunctions | ψn of the perturbed state.


A Simple Example



p p

Here we write down the eigenfunctions we require to rst order; these include some

linear combination of the degenerate wavefunctions.

| φ11 = a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 and E 1 = E01 + β E1

1


A Simple Example



p p



| φ11 = a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 and E 1 = E 01 + β E1

1

| φ12 = b1 | u 0

1 + b2 | u 02 + β | u 1

2 and E 2 = E02 + β E1

2 = E01 + β E1

2 .


A Simple Example



p p



| φ11 = a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 and E 1 = E 01 + β E1

1

| φ12 = b1 | u 0

1 + b2 | u 02 + β | u 1

2 and E 2 = E 02 + β E1

2 = E 01 + β E1

2 .

H | φ11 = H 0 + β V a 1 | u 0

1 + a 2 | u 02 + β | u 1

1

= E 01 + βE 11 a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 .


A Simple Example





| φ11 = a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 and E 1 = E 01 + β E1

1

| φ12 = b1 | u 0

1 + b2 | u 02 + β | u 1

2 and E 2 = E 02 + β E1

2 = E 01 + β E1

2 .

H | φ11 = H 0 + β V a 1 | u 0

1 + a 2 | u 02 + β | u 1

1

= E 01 + βE 11 a 1 | u 0

1 + a 2 | u 02 + β | u 1

1 .

H | φ12 = H 0 + β V b1 | u 01 + b2 | u 02 + β | u 12

= E 01 + βE 12 b1 | u 0

1 + b2 | u 02 + β | u 1

2 .


A Simple Example



Collect terms of order β

H 0 | u 11 + a 1 V | u 0

1 + a 2 V | u 02 = E 11 a 1 | u 0

1 + E 11 a 2 | u 02 + E 01 | u 1

1 (7 .10)


A Simple Example




H 0 | u 11 + a 1 V | u 0

1 + a 2 V | u 02 = E 11 a 1 | u 0

1 + E 11 a 2 | u 02 + E 01 | u 1

1 (7 .10)

ˆH 0 | u

1

2 + b1ˆV | u

0

1 + b2ˆV | u

0

2 = E 1

2 b1 | u0

1 + E 1

2 b2 | u0

2 + E 0

2 | u1

2 (7 .11)


A Simple Example




H 0 | u 11 + a 1 V | u 0

1 + a 2 V | u 02 = E 11 a 1 | u 0

1 + E 11 a 2 | u 02 + E 01 | u 1

1 (7 .10)

ˆH 0 | u

1

2 + b1ˆV | u

0

1 + b2ˆV | u

0

2 = E 1

2 b1 | u0

1 + E 1

2 b2 | u0

2 + E 0

2 | u1

2 (7 .11)multiply 7.10 by u 0

1

∗

and integrate

u0

1 |ˆ

H 0 | u1

1 + u0

1 | a 1ˆV | u

0

1 + u0

1 | a 2ˆV | u

0

2

= E 11 a 1 u 01 | u 0

1 + E 11 a 2 u 01 | u 0

2 + E 01 u 01 | u 1

1


A Simple Example




H 0 | u 11 + a 1 V | u 0

1 + a 2 V | u 02 = E 11 a 1 | u 0

1 + E 11 a 2 | u 02 + E 01 | u 1

1 (7 .10)

ˆH 0 | u

1

2 + b1ˆV | u

0

1 + b2ˆV | u

0

2 = E 1

2 b1 | u0

1 + E 1

2 b2 | u0

2 + E 0

2 | u1

2 (7 .11)multiply 7.10 by u 0

1

∗

and integrate

u0

1 |ˆ

H 0 | u1

1 + u0

1 | a 1ˆV | u

0

1 + u0

1 | a 2ˆV | u

0

2

= E 11 a 1 u 01 | u 0

1 + E 11 a 2 u 01 | u 0

2 + E 01 u 01 | u 1

1

giving

a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)

H 0 | u 11 + a 1 V | u 0

1 + a 2 V | u 02 = E 11 a 1 | u 0

1 + E 11 a 2 | u 02 + E 01 | u 1

1 (7 .10)

multiply 7.10 by u 02∗

and integrate to give

a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)

a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)

a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)

Write these two equations as a simultaneous equation in matrix form.

V 11 − E 11 V 12

V 21 V 22 − E 11

a 1

a 2= 0 (7 .14)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)

a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)


V 11 − E 11 V 12

V 21 V 22 − E 11

a 1

a 2= 0 (7 .14)

We can go through the same thing for E 2 giving

V 11 − E 12 V 12

V 21 V 22 − E 12

b1

b2= 0 (7 .15)


A Simple Example



a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)

a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)


V 11 − E 11 V 12

V 21 V 22 − E 11

a 1

a 2= 0 (7 .14)

We can go through the same thing for E 2 giving

V 11 − E 12 V 12

V 21 V 22 − E 12

b1

b2= 0 (7 .15)

We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if thedeterminant of the matrix on the LHS vanishes


A Simple Example



We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if the

determinant of the matrix on the LHS vanishes

V 11 − E 1 V 12

V 21 V 22 − E 1= 0 (7 .16)


A Simple Example





V 11 − E 1 V 12

V 21 V 22 − E 1= 0 (7 .16)

Note that the E 1 ’s (here E 11 and E 12 ) are the eigenvalues of the secular matrix

V 11 V 12

V 21 V 22


A Simple Example





V 11 − E 1 V 12

V 21 V 22 − E 1= 0 (7 .16)


V 11 V 12

V 21 V 22

There are two solutions to 7.16. One gives E 11 (and hence a 1 , a 2 )

E 11 = 12

(V 11 + V 22 ) + 12

(V 11 − V 22 )2 + 4 V 12 V 21

12

(7 .17)


A Simple Example





V 11 − E 1 V 12

V 21 V 22 − E 1= 0 (7 .16)


V 11 V 12

V 21 V 22

There are two solutions to 7.16. One gives E 11 (and hence a 1 , a 2 ), while the othergives E 12 (and hence b1 , b2 ).

E 11 =

12 (V 11 + V 22 ) +

12 (V 11 − V 22 )

2

+ 4 V 12 V 21

12

(7 .17)

E 12 =12

(V 11 + V 22 ) −12

(V 11 − V 22 )2 + 4 V 12 V 21

12

(7 .18)


A Simple Example



E 1

1 =12 (V 11 + V 22 ) +

12 (V 11 − V 22 )

2

+ 4 V 12 V 21

12

(7 .17)

E 12 =12

(V 11 + V 22 ) − 12

(V 11 − V 22 )2 + 4 V 12 V 21

12

(7 .18)

Frequently we nd from the symmetry of the physical problem that V 11 = V 22 = 0 .giving

E 11 = + |V 12 | and E 12 = − | V12 |


A Simple Example



E 1

1 =12 (V 11 + V 22 ) +

12 (V 11 − V 22 )

2

+ 4 V 12 V 21

12

(7 .17)

E 12 =12

(V 11 + V 22 ) − 12

(V 11 − V 22 )2 + 4 V 12 V 21

12

(7 .18)


E 11 = + |V 12 | and E 12 = − | V12 |

E 11 and E 12 can be substituted back into the simultaneous equations 7.14, 7.15 to nda 1 , a 2 , b1 , b2 .


A Simple Example



E 1

1 =12 (V 11 + V 22 ) +

12 (V 11 − V 22 )

2

+ 4 V 12 V 21

12

(7 .17)

E 12 =12

(V 11 + V 22 ) − 12

(V 11 − V 22 )2 + 4 V 12 V 21

12

(7 .18)


E 11 = + |V 12 | and E 12 = − | V12 |

E 11 and E 12 can be substituted back into the simultaneous equations 7.14, 7.15 to nda 1 , a 2 , b1 , b2 .

So we can write down the two zeroth-order eigenfunctions we were looking for which

are

| φ01 = a 1 | u 0

1 + a 2 | u 02 (7 .19)

| φ02 = b1 | u 0

1 + b2 | u 02 (7 .20)


A Simple Example




are

| φ01 = a 1 | u 0

1 + a 2 | u 02 (7 .19)

| φ02 = b1 | u 0

1 + b2 | u 02 (7 .20)


A Simple Example




are

| φ01 = a 1 | u 0

1 + a 2 | u 02 (7 .19)

| φ02 = b1 | u 0

1 + b2 | u 02 (7 .20)

It is usual to normalise i.e. |a 1 |2 + |a 2 |2 = 1 and |b1 |2 + |b2 |2 = 1 .

These eigenfunctions (| φ01 , | φ0

2 ) are the appropriate combinations of the

unperturbed states which split to give the energy levels E 11 and E

12 .


A Simple Example

d h h d f l k f h h




are| φ0

1 = a 1 | u 01 + a 2 | u 0

2 (7 .19)

| φ02 = b1 | u 0

1 + b2 | u 02 (7 .20)

It is usual to normalise i.e. |a 1 |2 + |a 2 |2 = 1 and |b1 |2 + |b2 |2 = 1 .

These eigenfunctions (| φ01 , | φ0

2 ) are the appropriate combinations of the

unperturbed states which split to give the energy levels E 11 and E

12 .

We can then proceed further to determine expressions for | φ11 , | φ1

2 and higher ordercorrections in a similar way to the non-degenerate case.


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