DEGENERATE Time-Independent Perturbation Theory
Transcript of 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
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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
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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
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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
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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
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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
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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
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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
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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
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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).
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Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 11
Problem 6.7 (b) & (c) (b)
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Physics 662 Spring 2013, Prof. S. Stone, Syracuse University 12
Problem 6.7 (c) & (d)
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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
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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
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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
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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
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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
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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