Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 1

DEGENERATE Time-Independent Perturbation Theory

  In 1st order non-degenerate perturbation theory, the wave function corrections are given by

  So if m=n, we are in trouble (also 2nd order energy correction

  Having m=n means we have two (or more) states with the same energy = Degenerate states

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 2

Case I: Two-Fold Degeneracy   Consider a two-state degenerate system

  Hoψao= Eoψa

o, Hoψbo= Eoψb

o, <ψao|ψb

o>=0,   Any linear combination ψo=αψa

o+βψbo is also an

eigenstate of Ho with eigenvalue Eo.

  What is the effect of adding a perturbation, on the degenerate states?   Here what we mean by a perturbation is a change to

the potential energy such that the changes in energy <O(10%)Eo.

  However the perturbation usually removes the degeneracy, so the levels split E

Eo

ΔV

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 3

2-fold Degeneracy II   Spilt states will have orthogonal

eigenfunctions of ψao & ψb

o. Lets proceed without knowing these

  As usual Hψ=Eψ, here H=Ho+λH’

  At order 1: so 1st terms cancel   At order λ:   Now take inner product with ψa

o

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 4

2-fold Degeneracy III   by Hermiticity  

  Rewrite as αWaa+βWab=αE1

  Inner product with ψbo: αWba+βWbb=βE1

  Since ψao & ψb

o are known & H’ is known Wij’s are known, but α & β are not known

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 5

Solving 2-fold degeneracy eqs   We have αWaa+βWab=αE1

  αWba+βWbb=βE1

  αWbaWab+βWabWbb=βWabE1   Now βWab=αE1-αWaa

  So αWbaWab+(αE1-αWaa)(Wbb-E1)=0   Or α[WabWba-(E1-Waa)(E1-Wbb)]=0

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 6

Solving 2-fold degeneracy eqs II

  Since Wab=Wba*   α[|Wba|2-(E1-Waa)(E1-Wbb)]=0, so providing α≠0, which would be non-degenerate

  (E1)2-(Waa+Wbb)E1+(Waa+Wbb-|Wba|2)=0   Using quadratic formula

  Note 2 solutions is just what we want for the perturbed energies

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 7

“Good” States   Back to α=0, then β=1 and Wab=0 from αWaa+βWab=αE1

  Then from αWba+βWbb=βE1, E1=Wbb

  Similarly if β=0, then α=1 and E1=Waa   Note that this implies that Wab=0, so the states a

& b were “good” states, i.e. could be treated as non-degenerate from the very beginning

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 8

“Good” States II   Theorem: Let A be a hermitian operator that

commutes with Ho & H’. If ψao & ψb

o, the degenerate eigenfunctions of Ho, are also eigenfunctions of A with distinct eigenvalues, then Wab=0, & ψa

o & ψbo are “good” states

  Proof: Let Aψao=µψa

o & Aψbo=νψb

o, with µ≠ν. Since [A,H’]=0,

Which can be 0, only if Wab=0

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 9

Problem 6.7(a) Particle of mass m free to move along a circular hoop of length L

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 10

Problem 6.7   (b) H’=-Voexp[-x2/a2], a<<L, find En

1

  (c) What are the good linear combinations of ψn & ψ-n? show that you get the 1st-order correction using Eq. 6.9

  (d) find a Hermitian operator A, that fits the requirements of the theorem and show that the simultaneous eigenstates of H0 & A are precisely the ones used in (c).

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 11

Problem 6.7 (b) & (c) (b)

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 12

Problem 6.7 (c) & (d)

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 13

Higher Order Degeneracies   In fact lets rewrite our equations as E1 represents the

eigenvalues & the “good” states are the eigenvectors   For higher order degeneracies just use nxn

matrix for Wij Wa=E1a.

  Example 6.2

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 14

Example 6.2   Cubical 3-d infinite well (3-d square well) with perturbation Vo.   Stationary states are

  Ground state 1,1,1 not degenerate   First excited state: 1,1,2 1,2,1 2,1,1 triply

degenerate with energy Eo1=3π2ħ2/ma2

V=Vo

a

a/2

V=∞ on walls

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 15

Example 6.2 II   Now calculate effect of Vo.   For ground state

  since

  What about for the 1st excited state. Need W   similar calculation gives Waa=Wbb=Wcc=V0/4

E01 = ψ 111 ′H ψ 111 =

2a

⎛⎝⎜

⎞⎠⎟

3

V0a4

⎛⎝⎜

⎞⎠⎟

2a2

⎛⎝⎜

⎞⎠⎟=14V0

sin2 t∫ dt = t / 2 − sin2t / 4 +C, sin2 πax

⎛⎝⎜

⎞⎠⎟0

a / 2

∫ dx =aπ

πa12a2−πa14sin 2 π

aa2

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =

a4− 0

E01 = ψ 111 ′H ψ 111 =

2a

⎛⎝⎜

⎞⎠⎟

3

V0 sin2 πax

⎛⎝⎜

⎞⎠⎟0

a / 2

∫ dx sin2 πay

⎛⎝⎜

⎞⎠⎟0

a / 2

∫ dy sin2 πaz

⎛⎝⎜

⎞⎠⎟0

a

∫ dz

Example 6.2 III   Off diagonal elements are more interesting

  z integral is 0. same for Wac.   Need Wbc. Find Wbc=(16/9π2)V0 (homework)

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 16

Wab = E11 = ψ 112 ′H ψ 121 =

2a

⎛⎝⎜

⎞⎠⎟

3

V0 sin2 πax

⎛⎝⎜

⎞⎠⎟0

a / 2

∫ dx sin πay

⎛⎝⎜

⎞⎠⎟0

a / 2

∫ sin 2πay

⎛⎝⎜

⎞⎠⎟dy sin π

az

⎛⎝⎜

⎞⎠⎟0

a

∫ sin 2πaz

⎛⎝⎜

⎞⎠⎟dz

Example 6.2 IV

  Matrix W Ignoring V0/4 &

finding eigenvalues   Solutions: 1-w=0, w1=1

  (1-w)2=κ2, w2=1+κ, w3=1-κ.

  Energies then are to order λ (E10 is unperturbed E)

    3 solutions breaks the degeneracy   Good unperturbed states are which are eigenvectiors of W

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 17

=V04

1 0 00 1 κ0 κ 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ κ =

83π

⎛⎝⎜

⎞⎠⎟

2

(1−w)3 −κ 2(1−w) = 0

E1 = E10 +V0 / 4,E1

0 + 1+κ( )V0 / 4,E10 + 1−κ( )V0 / 4( )

ψ 0 = αψ a + βψ b + γψ c

Example 6.2 V   Solve for each value of w

  w=1, gives α=1, β=γ=0   w=1+κ, gives α=(1+κ)α, so α=0, β+κγ=(1+κ)β,

so β=γ   w=1-κ, gives α=0, βκ+γ=(1-κ)γ, so β=-γ.   Thus the good states are

Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 18

1 0 00 1 κ0 κ 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

αβγ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟=w

αβγ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

ψ 0 = αψ a + βψ b + γψ c =ψ a, ψ b +ψ c( ) / 2, ψ b −ψ c( ) / 2