Time Independent Perturbation Theory, 1st order correction, 2nd order correction
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Time-Independent Perturbation Theory
Prepared by: James Salveo L. Olarve Graduate Student
January 27, 2010
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IntroductionThe presentation is about how to solve the approximate new energy levels and wave functions to the perturbed problems by building on the known exact solutions to the unperturbed case.
The intended reader of this presentation were physics students. The author already assumed that the reader knows Dirac braket notation.
This presentation was made to facilitate learning in quantum mechanics.
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The Perturb Hamiltonian
The Hamiltonian of a quantum mechanical system is written
Here, is a simple Hamiltonian whose eigenvalues and eigenstates are known exactly.
We shall deal only with nondegenerate systems; thus to each discrete eigenvalue there corresponds one and only one eigenfunction
And will be the additional term (can be due to external field)
EE
n
nE
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Task:
To find how these eigenkets and eigenenergies change if a small term (an external field, for example) is added to the Hamiltonian, so:
So on adding
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Assumption:
In perturbation theory we assume that is sufficiently small that the leading corrections are the same order of magnitude as itself, and the true energies can be better and better approximated by a successive series of corrections, each of order compared with the previous one.
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Strategy:
Expand the true wave function and corresponding eigenenergy as series in
It is more convenient to introduce dimensionless parameter λ
The series expansion
match the two sides term by term in powers of λ (taking λ=1).
Zeroth Term:
First Order Correction:
Second Order Correction: nEnEnEnHnH nnn2''2''2
Eq. 1
Eq. 2
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Matching the terms linear in on both λ sides (taking λ =1)
Taking the inner product of both sides with
Since it is normalized
The Hamiltonian is a Hermittian Operator
Eigenvalue Equation
1 oooo nn
*T
HH
nEnHn
First Order Correction
Eq. 1
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First Order CorrectionSo,
We find first order correction
for energy
1D continuous spectrum
nEnnEnnHnnEn nnn''''
nnEnnEnHnnnE nnn''''
dzHEn
'*'
'''' nnEnnEnnEnHn nnn
nHnEn''
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First Order Correction nEnEnHnH nn
''''''''
nHEnEH nn'''
mcn nm
nm'
nHEmcEH nnmn
nm
''
mEmH m
nHEmcEE nnmnm
nm
''
Solving for the 1st order change in the wave function
Since form a complete set thenn
The eigenvalue equation for unperturbed state m
nHlnlEmlcEE nnmnm
nm
''
Taking inner product with l
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First Order Correction
Cases: I.
)( nl
nHlnlEmlcEE nnmnm
nm
''
1 nlnl So,
0 mlml
nHnnHlEn'''
Now,
Cases: II.
)( nl So, 0 nlnl
1 mlml
nHlcEE nlnl
'
mn
nm
lnnl
nl EE
nHmc
EE
nHl
EE
nHlc
'''
Therefore the wave function correction to first order is:
mEE
nHmn
mnnm
'
'
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Second Order Correction
Taking the inner product with yieldsn
nEnnEnnEnnHnnHn nnn2''2''2
nEnEnEnHnH nnn2''2''2
nnEnnEnnEnHnnnE nnnn2''2''2
mnnm
n EE
nHmmHnnHnE
'
'''2
Now, '''2'2 nnEnHnE nn
But, 0)(' mn
nm
nm
nm
nmcmncnn
mnnm
n EE
nHmE
2'
2Second order correction for energy
Eq. 2
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Finally,The Eigenenergy
The Wave Function
...2'0 nnnn EEEE
...
2'
'0
mnnm
nn EE
nHmnHnEE
...' nnn
...'
m
EE
nHmnn
mnnm
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Twofold Degeneracy
When the unperturbed states are degenerate then two or more distinct states share the same energy. As a consequence of that the ordinary perturbation theory fails.
Suppose that:
Note that any linear combination of these states, ba 0
0 baba
is still an eigenstate of , with the same eigenvalue H E
aaH E bbH E
Typically the perturbation will break the degeneracy
We can’t even calculate the first-order energy because we don’t know what unperturbed states to use.
'H
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Twofold DegeneracyThe “good” unperturbed states in the general form EH
with'HHH
...22' EEEE
...22'
Plugging these and collecting like powers of ...... '''' EEEHHH
'''' EEHH
Taking the inner product with a
'''' EEHH aaaa
'''' aaaa EEEH
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Note: ba 0
Twofold Degeneracy
Then, baabaa EH ''
baaabaaa EEHH '''' ''' EHH baaa
0 baba
Let: aaaa HW ' baab HW '
'EWW abaa
where: jiij HW ' baji ,,
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Twofold DegeneracySimilarly, the inner product with yields
b
'EWW bbba
Multiplying at the right hand sideabW abbbba WEWW '
ababbbabba WEWWWW ' ababbbabba WEWWWW '
from'EWW abaa aaab WEW '
aaaabbabba WEEWEWWW '''
0''' 2
aaaabbbbabba WEEWWEWWW
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Twofold Degeneracy 0''' 2
aaaabbbbabba WEEWWEWWW
0''2 baabbbaabbaa WWWWWWEE
a) Suppose 0Then, baabbbaabbaa WWWWWWEE ''2
2
42' baabbbaabbaabbaa WWWWWWWWE
2
42' baabbbaabbaa WWWWWWE
here,
baabbbaabbbbaaaabaabbbaabbaa WWWWWWWWWWWWWW 424 222
*22 42 ababbbbbaaaa WWWWWW
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Twofold Degeneracyb) Suppose 0
ba 0 1
From:'EWW abaa 00 abW'EWW bbba '0 EWbb
Which is consistent with'E
aabbaabbaa WWWWWE 2'
2
1
bbbbaabbaa WWWWWE 2'
2
1
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The states and were already the “correct” linear combinations.
The answers for are precisely what we would have obtained using nondegenerate perturbation theory.
IMPLICATION: a
b
'E
In matrix form:
'E
WW
WW
aaba
abaa
Evidently the are nothing but the eigenvalues of the -matrix.
And the “good” linear combinations of the unperturbed states are the eigenvectors of W.
'E W
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Reference:
Retrieved from http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/Time_Ind_PT.htm, January 19 2010, Michael Fowler.
Introduction to Quantum Mechanics. David J. Griffiths. 1994