degenerate time-independent perturbation theory
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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
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