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8/7/2019 Overheads Lecture Chapter 7
http://slidepdf.com/reader/full/overheads-lecture-chapter-7 1/57
Chapter 7
Degenerate Perturbation Theory
Chapter 7 – p.1/ ?
8/7/2019 Overheads Lecture Chapter 7
http://slidepdf.com/reader/full/overheads-lecture-chapter-7 2/57
Degenerate Perturbation Theory
The non-degenerate perturbation theory discussed in chapter 6 breaks down whenwe have degenerate energy levels E 01 = E 02 .
Note also that the results of 2nd order non-degenerate perturbation theory assume|V np | << E 0
p − E 0n so there is a problem if the spacing between levels is smallcompared to the perturbing matrix element.
| φ1
n = p p= n
φ0 p | V | φ0
n
E 0n − E 0 p | φ0
p (6 .15)
E 2n = − p
p= n
φ0
p |ˆV | φ
0
n
2
E 0 p − E 0n(6 .19)
Chapter 7 – p.2/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Suppose an eigenvalue E 0n of the unperturbed Hamiltonian H 0 is s -fold degenerate.
Chapter 7 – p.3/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Suppose an eigenvalue E 0n of the unperturbed Hamiltonian H 0 is s -fold degenerate.
There are |u 0nα , (α = 1 , 2 · · · , s ) linearly independent orthonormal eigenfunctions
belonging to this eigenvalue,
u 0nα | u 0
nβ = δαβ , with α, β = 1 , · · · , s (7.1)
and any linear combination of these eigenfunctions is also an eigenfunction of H 0 .
Chapter 7 – p.3/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Suppose an eigenvalue E 0n of the unperturbed Hamiltonian H 0 is s -fold degenerate.
There are |u 0nα , (α = 1 , 2 · · · , s ) linearly independent orthonormal eigenfunctions
belonging to this eigenvalue,
u 0nα | u 0
nβ = δαβ , with α, β = 1 , · · · , s (7.1)
and any linear combination of these eigenfunctions is also an eigenfunction of H 0 .
So the zero-order states that we want to use in the PT expansions are not uniquelydened.
Chapter 7 – p.3/ ?
8/7/2019 Overheads Lecture Chapter 7
http://slidepdf.com/reader/full/overheads-lecture-chapter-7 6/57
The Degeneracy Problem
Suppose an eigenvalue E 0n of the unperturbed Hamiltonian H 0 is s -fold degenerate.
There are |u 0nα , (α = 1 , 2 · · · , s ) linearly independent orthonormal eigenfunctions
belonging to this eigenvalue,
u 0nα | u 0
nβ = δαβ , with α, β = 1 , · · · , s (7.1)
and any linear combination of these eigenfunctions is also an eigenfunction of H 0 .
We must therefore nd s normed states
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
where the ciα are constant coefcients which are the correct linear combinations forPT, i.e. they produce reasonable perturbation expansions.
Chapter 7 – p.3/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
That is we look for
E ni = E 0n + βE 1ni + β 2 E 2ni + β 3 E 3ni + · · · (7 .3)
and
| ψni = | φ0ni + β | φ
1ni + β
2
| φ2ni + β
3
| φ3ni + · · · (7 .4)
Chapter 7 – p.4/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
That is we look for
E ni = E 0n + βE 1ni + β 2 E 2ni + β 3 E 3ni + · · · (7 .3)
and
| ψni = | φ0ni + β | φ
1ni + β
2
| φ2ni + β
3
| φ3ni + · · · (7 .4)
withH | ψni = E ni | ψni all for i = 1 , · · · , s (7.5)
The states | ψni with i = 1 , 2 · · · , s will ideally have different energies E ni .
Chapter 7 – p.4/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
That is we look for
E ni = E 0n + βE 1ni + β 2 E 2ni + β 3 E 3ni + · · · (7 .3)
and
| ψni = | φ0ni + β | φ
1ni + β
2
| φ2ni + β
3
| φ3ni + · · · (7 .4)
withH | ψni = E ni | ψni all for i = 1 , · · · , s (7.5)
The states | ψni with i = 1 , 2 · · · , s will ideally have different energies E ni .
i.e. the perturbation, V , will either lift the degeneracy of the s -fold degenerate levels or
in some cases V may just shift the level E 0
n .
Chapter 7 – p.4/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
| φ1n = p
p= n
φ0
p |ˆV | φ
0
nE 0n − E 0 p
| φ0 p (6 .15)
E 2n = − p
p= n
φ0 p | V | φ0n2
E 0 p − E 0n(6 .18)
For degenerate levels these expansions will only make sense if these terms also havevanishing numerators, i.e. we require for the degenerate states that
φ0ni | V | φ0
nj = φ0ni | V | φ0
ni δij for i, j = 1 , · · · , s (7.6)
Chapter 7 – p.5/ ?
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The Degeneracy Problem
| φ1n = p
p= n
φ0
p |ˆV | φ
0
nE 0n − E 0 p
| φ0 p (6 .15)
E 2n = − p
p= n
φ0 p | V | φ0n2
E 0 p − E 0n(6 .18)
For degenerate levels these expansions will only make sense if these terms also havevanishing numerators, i.e. we require for the degenerate states that
φ0ni | V | φ0
nj = φ0ni | V | φ0
ni δij for i, j = 1 , · · · , s (7.6)
That is, the s × s matrix φ0ni | V | φ0
nj i, j = 1 , 2, · · · , s is a diagonal matrix.
Chapter 7 – p.5/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Givenφ0
ni | V | φ0ni δij for i, j = 1 , · · · , s (7.6)
That is, the s × s matrix is a diagonal matrix, we can carry out the same calculationsas in Chapter 6.
Chapter 7 – p.6/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Givenφ0
ni | V | φ0ni δij for i, j = 1 , · · · , s (7.6)
That is, the s × s matrix is a diagonal matrix, we can carry out the same calculationsas in Chapter 6.
For example, in order to determine the rst-order shift in the energy levels we can usethe equivalent of Eq. 6.7
E 1ni = φ
0ni | V | φ
0ni (7 .7)
c.f. equation 6.10.
Chapter 7 – p.6/ ?
8/7/2019 Overheads Lecture Chapter 7
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The Degeneracy Problem
Givenφ0
ni | V | φ0ni δij for i, j = 1 , · · · , s (7.6)
That is, the s × s matrix is a diagonal matrix, we can carry out the same calculationsas in Chapter 6.
For example, in order to determine the rst-order shift in the energy levels we can usethe equivalent of Eq. 6.7
E 1ni = φ
0ni | V | φ
0ni (7 .7)
c.f. equation 6.10.
We now describe one method for nding the correct zero-order linear combinations ofeigenfunctions.
Chapter 7 – p.6/ ?
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Finding the zeroth-order eigenfunctions for PT
− V − E 1n | φ
0ni = H o − E
0n | φ
1ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
− V − E 1
n | φ0
ni = H o − E 0
n | φ1
ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Take the scalar product of eq. 7.8 with u 0
nβwe obtain
− u 0nβ | V − E 1n | φ0
ni = u 0nβ | H 0 − E 0n | φ1
ni
Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
− V − E 1
n | φ0
ni = H o − E 0
n | φ1
ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Take the scalar product of eq. 7.8 with u 0
nβwe obtain
− u 0nβ | V − E 1n | φ0
ni = u 0nβ | H 0 − E 0n | φ1
ni
On the RHS
u 0nβ | H 0 − E 0n | φ1
ni = u 0nβ | H 0 | φ1
ni − u 0nβ | E 0n | φ1
ni
Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
− V − E 1
n | φ0
ni = H o − E 0
n | φ1
ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Take the scalar product of eq. 7.8 with u 0
nβwe obtain
− u 0nβ | V − E 1n | φ0
ni = u 0nβ | H 0 − E 0n | φ1
ni
On the RHS
u 0nβ | H 0 − E 0n | φ1
ni = u 0nβ | H 0 | φ1
ni − u 0nβ | E 0n | φ1
ni
= E 0
n u0
nβ | φ1
ni − E 0
n u0
nβ | φ1
ni
Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
− V − E 1
n | φ0
ni = H o − E 0
n | φ1
ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Take the scalar product of eq. 7.8 with u 0
nβwe obtain
− u 0nβ | V − E 1n | φ0
ni = u 0nβ | H 0 − E 0n | φ1
ni
On the RHS
u 0nβ | H 0 − E 0n | φ1
ni = u 0nβ | H 0 | φ1
ni − u 0nβ | E 0n | φ1
ni
= E 0
n u0
nβ | φ1
ni − E 0
n u0
nβ | φ1
ni
= 0Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
− V − E 1
n | φ0
ni = H o − E 0
n | φ1
ni . (7 .8)
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Starting with the equivalent of equation 6.7 and eq. 7.2 we must determine thecoefcients ciα .
Take the scalar product of eq. 7.8 with u 0
nβwe obtain
− u 0nβ | V − E 1n | φ0
ni = u 0nβ | H 0 − E 0n | φ1
ni
and so
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
Chapter 7 – p.7/ ?
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Chapter 7 – p.8/ ?
d h h d f f
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Then substituting eq 7.2 into this equation we obtain
u 0nβ V
s
α =1ciα u 0nα = E 1ni u 0nβ
s
α =1ciα u 0nα
Chapter 7 – p.8/ ?
Fi di h h d i f i f PT
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Then substituting eq 7.2 into this equation we obtain
u 0nβ V
s
α =1ciα u 0nα = E 1ni u 0nβ
s
α =1ciα u 0nα
s
α =1
u 0nβ V u 0
nα ciα = E 1ni u 0nβ | u 0
n β ciβ
Chapter 7 – p.8/ ?
Fi di th th d i f ti f PT
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
Then substituting eq 7.2 into this equation we obtain
u0nβ V
s
α =1ciα u
0nα = E
1ni u
0nβ
s
α =1ciα u
0nα
s
α =1
u 0nβ V u 0
nα ciα = E 1ni u 0nβ | u 0
nβ ciβ
= E 1ni ciβ
Chapter 7 – p.8/ ?
Fi di g th th d ig f ti f PT
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
u 0nβ | V | φ0
ni = E 1ni u 0nβ | φ0
ni .
φ0ni =
s
α =1
ciα u 0nα , i = 1 , · · · , s (7 .2)
and so
s
α =1u
0nβ V u
0nα ciα = E
1ni ciβ
ors
α =1u
0nβ | V | u
0nα − E
1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)
Chapter 7 – p.8/ ?
Finding the zeroth order eigenfunctions for PT
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
s
α =1u 0nβ | V | u 0nα − E 1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)
This set of s equations 7.9 give the required values for ciα .
Chapter 7 – p.9/ ?
Finding the zeroth order eigenfunctions for PT
8/7/2019 Overheads Lecture Chapter 7
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Finding the zeroth-order eigenfunctions for PT
s
α =1u 0nβ | V | u 0nα − E 1ni δβα ciα = 0 , i = 1 , · · · , s (7 .9)
This set of s equations 7.9 give the required values for ciα .
The vanishing of the s × s determinant
det u 0nβ | V | u 0
nα − E 1ni δβα = 0
gives the s roots E 1ni , i = 1 , · · · , s that are the rst-order energy level shifts for the
states φ0ni with the (non-zero) coefcients ciα the solutions of eqs. 7.9.
Chapter 7 – p.9/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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A Simple Example
The simplest illustration of degenerate PT is to assume that there is a 2-fold
degeneracy in the unperturbed state.
As a rst step we note that there is a degeneracy.
H 0 | u01 = E
01 | u
01
H 0 | u 02 = E 02 | u 0
2 = E 01 | u 02
Chapter 7 – p.10/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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A Simple Example
The simplest illustration of degenerate PT is to assume that there is a 2-fold
degeneracy in the unperturbed state.
As a rst step we note that there is a degeneracy.
H 0 | u01 = E
01 | u
01
H 0 | u 02 = E 02 | u 0
2 = E 01 | u 02
We write down a perturbing Hamiltonian H which "lifts" the degeneracy of the system
H = H 0 + β V
H | ψn = E n | ψn
Chapter 7 – p.10/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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A Simple Example
The simplest illustration of degenerate PT is to assume that there is a 2-fold
degeneracy in the unperturbed state.
As a rst step we note that there is a degeneracy.
H 0 | u01 = E
01 | u
01
H 0 | u 02 = E 02 | u 0
2 = E 01 | u 02
We write down a perturbing Hamiltonian H which "lifts" the degeneracy of the system
H = H 0 + β V
H | ψn = E n | ψn
So for our two-fold degenerate system
H | ψ 1 = E 1 | ψ 1
H | ψ 2 = E 2 | ψ 2 where E 1 > E2
Chapter 7 – p.10/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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A Simple Example
We note that u01∗u 0
2 dτ = 0 as well as u0n
∗u 0m dτ = δnm for all other n, m .
Chapter 7 – p.11/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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A Simple Example
We note that u01∗u 0
2 dτ = 0 as well as u0n
∗u 0m dτ = δnm for all other n, m .
If this is not the case we must rst choose a different set of eigenfunctions that areorthonormal.
Chapter 7 – p.11/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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p p
We note that u01∗u 0
2 dτ = 0 as well as u0n
∗u 0m dτ = δnm for all other n, m .
If this is not the case we must rst choose a different set of eigenfunctions that areorthonormal.
Ideally we would like to obtain the correct eigenfunctions | ψn of the perturbed state.
Chapter 7 – p.11/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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p p
Here we write down the eigenfunctions we require to rst order; these include some
linear combination of the degenerate wavefunctions.
| φ11 = a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 and E 1 = E01 + β E1
1
Chapter 7 – p.12/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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p p
Here we write down the eigenfunctions we require to rst order; these include some
linear combination of the degenerate wavefunctions.
| φ11 = a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 and E 1 = E 01 + β E1
1
| φ12 = b1 | u 0
1 + b2 | u 02 + β | u 1
2 and E 2 = E02 + β E1
2 = E01 + β E1
2 .
Chapter 7 – p.12/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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p p
Here we write down the eigenfunctions we require to rst order; these include some
linear combination of the degenerate wavefunctions.
| φ11 = a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 and E 1 = E 01 + β E1
1
| φ12 = b1 | u 0
1 + b2 | u 02 + β | u 1
2 and E 2 = E 02 + β E1
2 = E 01 + β E1
2 .
H | φ11 = H 0 + β V a 1 | u 0
1 + a 2 | u 02 + β | u 1
1
= E 01 + βE 11 a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 .
Chapter 7 – p.12/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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Here we write down the eigenfunctions we require to rst order; these include some
linear combination of the degenerate wavefunctions.
| φ11 = a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 and E 1 = E 01 + β E1
1
| φ12 = b1 | u 0
1 + b2 | u 02 + β | u 1
2 and E 2 = E 02 + β E1
2 = E 01 + β E1
2 .
H | φ11 = H 0 + β V a 1 | u 0
1 + a 2 | u 02 + β | u 1
1
= E 01 + βE 11 a 1 | u 0
1 + a 2 | u 02 + β | u 1
1 .
H | φ12 = H 0 + β V b1 | u 01 + b2 | u 02 + β | u 12
= E 01 + βE 12 b1 | u 0
1 + b2 | u 02 + β | u 1
2 .
Chapter 7 – p.12/ ?
A Simple Example
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Collect terms of order β
H 0 | u 11 + a 1 V | u 0
1 + a 2 V | u 02 = E 11 a 1 | u 0
1 + E 11 a 2 | u 02 + E 01 | u 1
1 (7 .10)
Chapter 7 – p.13/ ?
A Simple Example
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Collect terms of order β
H 0 | u 11 + a 1 V | u 0
1 + a 2 V | u 02 = E 11 a 1 | u 0
1 + E 11 a 2 | u 02 + E 01 | u 1
1 (7 .10)
ˆH 0 | u
1
2 + b1ˆV | u
0
1 + b2ˆV | u
0
2 = E 1
2 b1 | u0
1 + E 1
2 b2 | u0
2 + E 0
2 | u1
2 (7 .11)
Chapter 7 – p.13/ ?
A Simple Example
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Collect terms of order β
H 0 | u 11 + a 1 V | u 0
1 + a 2 V | u 02 = E 11 a 1 | u 0
1 + E 11 a 2 | u 02 + E 01 | u 1
1 (7 .10)
ˆH 0 | u
1
2 + b1ˆV | u
0
1 + b2ˆV | u
0
2 = E 1
2 b1 | u0
1 + E 1
2 b2 | u0
2 + E 0
2 | u1
2 (7 .11)multiply 7.10 by u 0
1
∗
and integrate
u0
1 |ˆ
H 0 | u1
1 + u0
1 | a 1ˆV | u
0
1 + u0
1 | a 2ˆV | u
0
2
= E 11 a 1 u 01 | u 0
1 + E 11 a 2 u 01 | u 0
2 + E 01 u 01 | u 1
1
Chapter 7 – p.13/ ?
A Simple Example
8/7/2019 Overheads Lecture Chapter 7
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Collect terms of order β
H 0 | u 11 + a 1 V | u 0
1 + a 2 V | u 02 = E 11 a 1 | u 0
1 + E 11 a 2 | u 02 + E 01 | u 1
1 (7 .10)
ˆH 0 | u
1
2 + b1ˆV | u
0
1 + b2ˆV | u
0
2 = E 1
2 b1 | u0
1 + E 1
2 b2 | u0
2 + E 0
2 | u1
2 (7 .11)multiply 7.10 by u 0
1
∗
and integrate
u0
1 |ˆ
H 0 | u1
1 + u0
1 | a 1ˆV | u
0
1 + u0
1 | a 2ˆV | u
0
2
= E 11 a 1 u 01 | u 0
1 + E 11 a 2 u 01 | u 0
2 + E 01 u 01 | u 1
1
giving
a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
H 0 | u 11 + a 1 V | u 0
1 + a 2 V | u 02 = E 11 a 1 | u 0
1 + E 11 a 2 | u 02 + E 01 | u 1
1 (7 .10)
multiply 7.10 by u 02∗
and integrate to give
a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)
Write these two equations as a simultaneous equation in matrix form.
V 11 − E 11 V 12
V 21 V 22 − E 11
a 1
a 2= 0 (7 .14)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)
Write these two equations as a simultaneous equation in matrix form.
V 11 − E 11 V 12
V 21 V 22 − E 11
a 1
a 2= 0 (7 .14)
We can go through the same thing for E 2 giving
V 11 − E 12 V 12
V 21 V 22 − E 12
b1
b2= 0 (7 .15)
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a 1 V 11+ a 2 V 12 = a 1 E 11 (7 .12)
a 1 V 21+ a 2 V 22 = a 2 E 11 (7 .13)
Write these two equations as a simultaneous equation in matrix form.
V 11 − E 11 V 12
V 21 V 22 − E 11
a 1
a 2= 0 (7 .14)
We can go through the same thing for E 2 giving
V 11 − E 12 V 12
V 21 V 22 − E 12
b1
b2= 0 (7 .15)
We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if thedeterminant of the matrix on the LHS vanishes
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We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if the
determinant of the matrix on the LHS vanishes
V 11 − E 1 V 12
V 21 V 22 − E 1= 0 (7 .16)
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We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if the
determinant of the matrix on the LHS vanishes
V 11 − E 1 V 12
V 21 V 22 − E 1= 0 (7 .16)
Note that the E 1 ’s (here E 11 and E 12 ) are the eigenvalues of the secular matrix
V 11 V 12
V 21 V 22
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We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if the
determinant of the matrix on the LHS vanishes
V 11 − E 1 V 12
V 21 V 22 − E 1= 0 (7 .16)
Note that the E 1 ’s (here E 11 and E 12 ) are the eigenvalues of the secular matrix
V 11 V 12
V 21 V 22
There are two solutions to 7.16. One gives E 11 (and hence a 1 , a 2 )
E 11 = 12
(V 11 + V 22 ) + 12
(V 11 − V 22 )2 + 4 V 12 V 21
12
(7 .17)
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We obtain non-trivial solutions to 7.14 and 7.15 (i.e. a 1 , a 2 , b1 , b2 = 0) if and only if the
determinant of the matrix on the LHS vanishes
V 11 − E 1 V 12
V 21 V 22 − E 1= 0 (7 .16)
Note that the E 1 ’s (here E 11 and E 12 ) are the eigenvalues of the secular matrix
V 11 V 12
V 21 V 22
There are two solutions to 7.16. One gives E 11 (and hence a 1 , a 2 ), while the othergives E 12 (and hence b1 , b2 ).
E 11 =
12 (V 11 + V 22 ) +
12 (V 11 − V 22 )
2
+ 4 V 12 V 21
12
(7 .17)
E 12 =12
(V 11 + V 22 ) −12
(V 11 − V 22 )2 + 4 V 12 V 21
12
(7 .18)
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E 1
1 =12 (V 11 + V 22 ) +
12 (V 11 − V 22 )
2
+ 4 V 12 V 21
12
(7 .17)
E 12 =12
(V 11 + V 22 ) − 12
(V 11 − V 22 )2 + 4 V 12 V 21
12
(7 .18)
Frequently we nd from the symmetry of the physical problem that V 11 = V 22 = 0 .giving
E 11 = + |V 12 | and E 12 = − | V12 |
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E 1
1 =12 (V 11 + V 22 ) +
12 (V 11 − V 22 )
2
+ 4 V 12 V 21
12
(7 .17)
E 12 =12
(V 11 + V 22 ) − 12
(V 11 − V 22 )2 + 4 V 12 V 21
12
(7 .18)
Frequently we nd from the symmetry of the physical problem that V 11 = V 22 = 0 .giving
E 11 = + |V 12 | and E 12 = − | V12 |
E 11 and E 12 can be substituted back into the simultaneous equations 7.14, 7.15 to nda 1 , a 2 , b1 , b2 .
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E 1
1 =12 (V 11 + V 22 ) +
12 (V 11 − V 22 )
2
+ 4 V 12 V 21
12
(7 .17)
E 12 =12
(V 11 + V 22 ) − 12
(V 11 − V 22 )2 + 4 V 12 V 21
12
(7 .18)
Frequently we nd from the symmetry of the physical problem that V 11 = V 22 = 0 .giving
E 11 = + |V 12 | and E 12 = − | V12 |
E 11 and E 12 can be substituted back into the simultaneous equations 7.14, 7.15 to nda 1 , a 2 , b1 , b2 .
So we can write down the two zeroth-order eigenfunctions we were looking for which
are
| φ01 = a 1 | u 0
1 + a 2 | u 02 (7 .19)
| φ02 = b1 | u 0
1 + b2 | u 02 (7 .20)
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So we can write down the two zeroth-order eigenfunctions we were looking for which
are
| φ01 = a 1 | u 0
1 + a 2 | u 02 (7 .19)
| φ02 = b1 | u 0
1 + b2 | u 02 (7 .20)
Chapter 7 – p.16/ ?
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So we can write down the two zeroth-order eigenfunctions we were looking for which
are
| φ01 = a 1 | u 0
1 + a 2 | u 02 (7 .19)
| φ02 = b1 | u 0
1 + b2 | u 02 (7 .20)
It is usual to normalise i.e. |a 1 |2 + |a 2 |2 = 1 and |b1 |2 + |b2 |2 = 1 .
These eigenfunctions (| φ01 , | φ0
2 ) are the appropriate combinations of the
unperturbed states which split to give the energy levels E 11 and E
12 .
Chapter 7 – p.16/ ?
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So we can write down the two zeroth-order eigenfunctions we were looking for which
are| φ0
1 = a 1 | u 01 + a 2 | u 0
2 (7 .19)
| φ02 = b1 | u 0
1 + b2 | u 02 (7 .20)
It is usual to normalise i.e. |a 1 |2 + |a 2 |2 = 1 and |b1 |2 + |b2 |2 = 1 .
These eigenfunctions (| φ01 , | φ0
2 ) are the appropriate combinations of the
unperturbed states which split to give the energy levels E 11 and E
12 .
We can then proceed further to determine expressions for | φ11 , | φ1
2 and higher ordercorrections in a similar way to the non-degenerate case.
Chapter 7 – p.16/ ?