DEGENERATE Time-Independent Perturbation Theory

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


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


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


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


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


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


“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


“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


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


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).


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


Problem 6.7 (c) & (d)


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


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


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)


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


=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


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

DEGENERATE Time-Independent Perturbation Theory

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