Lecture on Perturbation Theory

7/28/2019 Lecture on Perturbation Theory

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ContentsTime-independent nondegenerate perturbation theory

Time-independent degenerate perturbation theoryTime-dependent perturbation theory

Literature

Perturbation theoryQuantum mechanics 2 - Lecture 2

Igor Lukacevic

UJJS, Dept. of Physics, Osijek

17. listopada 2012.

Igor Lukacevic Perturbation theory
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1 Time-independent nondegenerate perturbation theoryGeneral formulationFirst-order theorySecond-order theory

2 Time-independent degenerate perturbation theoryGeneral formulationExample: Two-dimensional harmonic oscilator

3 Time-dependent perturbation theory

4 Literature

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General formulationFirst-order theorySecond-order theory

Contents




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Do you remember this?


C
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Do you remember this?

H0

0n = E

0n

0n

0n|0m = nm complete set


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Literature


Now, let us kick the potential bottom a little...


Contents


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Now, let us kick the potential bottom a little...

What wed like to solve now is...

Hn = Enn

A question

Does anyone have an idea how?


Contents
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Assume

H = H0

+ H


Contents
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AssumeH = H0 + H

unperturbed Hamiltonian

perturbation Hamiltonian

small parameter


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Time-independent nondegenerate perturbation theoryTime-independent degenerate perturbation theory

Time-dependent perturbation theoryLiterature


AssumeH = H0 + H



small parameter


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AssumeH = H0 + H



small parameter


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Name Description Hamiltonian

L-S coupling Coupling between orbital and H = H0 + f(r)L S

spin angular momentum in a H = f(r)L Sone-electron atom H0 = p2/2m Ze2/r

Stark effect One-electron atom in a constant H = H0 + eE0z

uniform electric field E = ezE0 H = eE0z

H0 = p2/2m

Ze2/rZeeman effect One-electron atom in a constant H = H0 + (e/2mc)J B

uniform magnetic field B H = (e/2mc)J BH0 = p2/2m Ze2/r

Anharmonic Spring with nonlinear restoring H = H0 + Kx4

oscilator force H = Kx4

H0 = p2

/2m + 1/2Kx2

Nearly free Electron in a periodic lattice H = H0 + V(x)electron V(x) =

n Vnexp[i(2nx/a)]

model H0 = p2/2m


ContentsTi i d d d b i h G l f l i
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Assume

H is small compared to H0

eigenstates and eigenvalues of H do not differ much from those of H0

eigenstates and eigenvalues of H0 are known


ContentsTime inde endent nondegenerate ert rbation theor General form lation
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Assume

H is small compared to H0

eigenstates and eigenvalues of H do not differ much from those of H0


expand

n = 0n +

1n +

2

2n + . . . ,

En = E0n + E

1n +

2E

2n + . . .


ContentsTime independent nondegenerate perturbation theory General formulation


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Assume

H is small compared to H0 eigenstates and eigenvalues of H do not differ much from those of H0


expand

n = 0n +

1n +

22n + . . . ,

En = E0n + E

1n +

2E

2n + . . .

and sort

(0) . . . H00n = E0n 0n ,(1) . . . H01n + H

0n = E0n

1n + E

1n

0n ,

(2) . . . H02n + H1n = E

0n

2n + E

1n

1n + E

2n

0n ,

...


ContentsTime-independent nondegenerate perturbation theory General formulation
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Making 0n|

(1) and using the normalization property of 0n, we get

First-order correction to the energy

E1n = 0n|H|0n

For calculationdetails, see Refs[2], [3] and [4].


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Time independent nondegenerate perturbation theoryTime-independent degenerate perturbation theory




E1n = 0n|H|0n

Example 1

Find the first-order corrections to the energy ofa particle in a infinite square well if the floorof the well is raised by an constant value V0.


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p g p yTime-independent degenerate perturbation theory


First-order theorySecond-order theory


E1n = 0n|H|0n

Example 1


Unperturbed w.f.: 0n(x) =

2a

sin

na

x


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gTime-independent degenerate perturbation theory




E1n = 0n|H|0n

Example 1


Unperturbed w.f.: 0

n(x) =2

a sinn

a x

Perturbation Hamiltonian: H = V0




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E1n = 0n|H|0n

Example 1



2

asin

n

ax

Perturbation Hamiltonian: H = V0First-order correction:E

1n = 0n|V0|0n = V00n|0n = V0

corrected energy levels: En E0n + V0



Ti i d d d b i hGeneral formulationFi d h


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Literature


Example 1



2a sin

na x


First-order correction:E

1n = 0n|V0|0n = V00n|0n = V0


Compare this result with an exact solution for a constant perturbation all thehigher corrections vanish



Ti i d d t d t t b ti thGeneral formulationFi t d th
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Literature


Example 1



2a sin

na x


First-order correction:E

1n = 0n|V0|0n = V00n|0n = V0


Compare this result with an exact solution for a constant perturbation all thehigher corrections vanish



Time independent degenerate perturbation theoryGeneral formulationFirst order theory
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Example 1 (cont.)

Now, cut the perturbation to only a half-wayacross the well



Time-independent degenerate perturbation theoryGeneral formulationFirst-order theory
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Example 1 (cont.)

Now, cut the perturbation to only a half-wayacross the well

E1n = 2V0a

a/20

sin2

n

ax

dx =V0

2



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Time independent degenerate perturbation theoryTime-dependent perturbation theory

Literature

First order theorySecond-order theory

Example 1 (cont.)

Now, cut the perturbation to only a half-way

across the well

E1n = 2V0a

a/20

sin2

n

ax

dx =V0

2

HW. Compare this result with an exact one.



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p g p yTime-dependent perturbation theory

Literature

ySecond-order theory

Now we seek the first-order correction to the wave function.(1) and 1n =

m=n

cmn0m give

First-order correction to the wavefunction

1n =

m=n

0m|H|0n(E0n E0m)

0m




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p g p yTime-dependent perturbation theory

Literature

ySecond-order theory

First-order correction to the wavefunction

1n = m=n

0m|H|0n(E0n

E0m)

0m


In conclusion

First-order perturbation theory gives:

often accurate energies

poor wave functions



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Second-order theory

Example 2

Compute the first-order corrections for aharmonic oscilator with applied smallperturbation W = ax3.



Time-independent degenerate perturbation theoryTi d d b i h

General formulationFirst-order theoryS d d h


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Second-order theory

Example 2

Compute the first-order corrections for a

harmonic oscilator with applied smallperturbation W = ax3.

Hamiltonian: H = 2

2m

d2

dx+

1

2kx

2 + ax3



Time-independent degenerate perturbation theoryTi d d t t b ti th

General formulationFirst-order theoryS d d th


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Second-order theory

Example 2

Compute the first-order corrections for aharmonic oscilator with applied small

perturbation W = ax3

.

Hamiltonian: H = 2

2m

d2

dx+

1

2kx

2 + ax3

H0 eigenenergies: E0n =

n +

1

2


ContentsTime-independent nondegenerate perturbation theoryTime-independent degenerate perturbation theory

Time dependent perturbation theory

General formulationFirst-order theorySecond order theory
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Second-order theory

Example 2


Hamiltonian: H =

2

2m

d2

dx +1

2 kx2

+ ax3


n +

1

2

H0 eigenfunctions:

0n(x) =

12nn!

e

x2

2 Hn(x)

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120



Time-dependent perturbation theory

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Second-order theory

Example 2Compute the first-order corrections for aharmonic oscilator with applied smallperturbation W = ax3.

Hamiltonian: H =

2

2m

d2

dx

+1

2

kx2 + ax3


n +

1

2

H0 eigenfunctions:

0n(x) = 1

2n

n!

ex2

2 Hn(x

)

E1n = 0n|ax3|0n = 0

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120






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Time dependent perturbation theoryLiterature

Second order theory

Example 2


Hamiltonian: H = 2

2m

d2

dx+

1

2kx

2 + ax3


n +

1

2

H0 eigenfunctions:

0n(x) =

1

2nn!

ex2

2 Hn(x

)

E1n = 0n|ax3|0n = 0For 1n we need expressions 0m|H|0n

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120




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p p yLiterature

y

Example 2


Hamiltonian: H = 2

2m

d2

dx+

1

2kx

2 + ax3


n + 12

H0 eigenfunctions:

0n(x) =

1

2nn!

ex2

2 Hn(x

)

E1n = 0n|ax3|0n = 0For 1n we need expressions 0m|H|0n for m = n 2k, k Z these are zero

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120




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p p yLiterature

y

Example 2

Compute the first-order corrections for a

harmonic oscilator with applied smallperturbation W = ax3.

Hamiltonian: H = 2

2m

d2

dx+

1

2kx

2 + ax3

H0 eigenenergies: E0n = n + 12

H0 eigenfunctions:

0n(x) =

1

2nn!

ex2

2 Hn(x

)

E1n =

0n

|ax3

|0n

= 0

For 1n we need expressions 0m|H|0n for m = n 2k, k Z these are zero so, well, for example, take only these:m = n + 3, n + 1, n

1, n

3

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120




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Literature

Example 2


n|ax3

|n + 3 = a (n + 1)(n + 2)(n + 3)(2)3n|ax3|n + 1 = 3a

(n + 1)3

(2)3

n|ax3

|n 1 = 3a n3

(2)3

n|ax3|n 3 = a

n(n 1)(n 2)(2)3

Dont forget...

n Hn()0 11 22 42 23 83 124 164 482 + 125 325 1603 + 120



Time-dependent perturbation theoryLi

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Literature

Example 2


n|ax3

|n + 3 = a (n + 1)(n + 2)(n + 3)(2)3n|ax3|n + 1 = 3a

(n + 1)3

(2)3

n|ax3

|n 1 = 3a n3

(2)3

n|ax3|n 3 = a

n(n 1)(n 2)(2)3

Energy differences

m En Emn+3 3n+1 n-1 n-3 3



Time-dependent perturbation theoryLit t

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Literature

Example 2

Compute the first-order corrections for a harmonic oscilator

with applied small perturbation W = ax

3

.

n|ax3|n + 3 = a

(n + 1)(n + 2)(n + 3)

(2)3

n|ax3|n + 1 = 3a

(n + 1)3

(2)3

n|ax3|n 1 = 3a

n3

(2)3

n|ax3|n 3 = a

n(n 1)(n 2)

(2)3

Energy differences

m En Emn+3 3n+1

n-1 n-3 3

1n = a2

1

3

n(n 1)(n 2)

2

0n3 + 3n

n

2

0n1

3(n + 1)

n + 1

2

0n+1 13

(n + 1)(n + 2)(n + 3)

2

0n+3




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Literature

Making 0n|

(2), using the normalization property of 0n and orthogonality

between 0n and 1n, we get

Second-order correction to theenergy

E2n =

m=n

|0m|H|0n|2E0n E0m





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Literature

Now we seek the second-order correction to the wave function.(2) and 1n =

m=n

cmn0m give

Second-order correction to the wave function

2n =

m=n

0n|H|0n0m|H|0n

(E0n E0m)2

+k=n

0m|H|0k0k|H|0n(E0n

E0m)(E0n

E0k)

0m





Literature

General formulationExample: Two-dimensional harmonic oscilator
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Symmetry Degeneracy Perturbation




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Symmetry Degeneracy Perturbation




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A question

Whats wrong with

E2n =m=n

|0m|H|

0n|

2

E0n E0m , m, n q

if unperturbed eigenstates are degenerate E01 = E02 = = E0q?




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H = H0 + HH0

0n = E

0n

0n

E01 q-fold degenerate

m1 E

Construct

n =

qi=1

anm

0

m

which diagonalizes submatrixof H:n|H|k = Enk

m2

%Nondegenerateperturbation theorywith basis B

q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

y

c

{n}Gives new basis B

m7' {ani} E1, E2, . . . , Eq




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Matrix H in basis B

H =

H11 H1,q+1 . . .

. . . 0

Hq/2,q/2

0. . .

HqqHq+1,1

...




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H = H0 + HH0

0n = E

0n

0n


m1 E

Construct

n =

qi=1 a

nm

0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

y

c


m7' {ani} E1, E2, . . . , Eq


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For n = 1,

qm=1

(Hpm Enpm)anm = 0 appear as

H11

E1 H

12 H

13 . . . H

1q

H21 H22 E1 H23 . . . H2q...

......

......

Hq1

a11

a12...

a1q

= 0


Contents



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H = H0 + HH0

0n = E

0n

0n


m1 E

Construct

n =

q

i=1 a

nm

0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

y

c


m7' {ani} E1, E2, . . . , Eq


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The q roots of secular equation detH EnI = 0 are the diagonal elements of

the submatrix of H.


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H = H0 + HH00n = E0n

0n


m1 E

Construct

n =

q

i=1

anm

0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

y

c


m7' {ani} E1, E2, . . . , Eq


Contents



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H = H0 + HH00n = E0n

0n


m1 E

Construct

n=

q

i=1

anm0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

yc


m7' {ani} E1, E2, . . . , Eq


Contents



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H = H0 + HH00n = E0n

0n


m1 E

Construct

n =

q

i=1

anm0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

yc


m7' {ani} E1, E2, . . . , Eq


Contents



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H = H0 + HH00n = E0n

0n


m1 E

Construct

n =

q

i=1

anm0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

yc


m7' {ani} E1, E2, . . . , Eq


Contents



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H = H0 + HH00n = E0n

0n


m1 E

Construct

n =

q

i=1

anm0

m


m2


q(H E)a = 0

m3E detH EnI

= 0

m8 m6 m5 m4T

c

yc


m7' {ani} E1, E2, . . . , Eq


Contents



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n = n +1n +

22n + . . . n qn =

0n +

1n +

22n + . . . n > qEn = E

0n +E

1n +

2E2n + . . . n

q (E01 =

= E0q)

En = E0n + E

1n +

2E2n + . . . n > qEn = n|H|n n qE1n = 0n|H|0n n > q


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n = n +1n +22n + . . . n qn =

0n +

1n +

22n + . . . n > qEn = E

0n +E

1n +

2E2n + . . . n q (E01 = = E0q)En = E

0n + E

1n +

2E2n + . . . n > qEn =

n

|H

|n

n

q

E1n = 0n|H|0n n > q

So, what do we get from degenerate perturbation theory:

1st-order energy corrections

corrected w.f. (with nondegenerate states they serve as a basis forhigher-order calculations)


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Two-dimensional harmonic oscilator Hamiltonian:

H0 =

p2x + p2y

2m +

K

2 (x

2

+ y

2

)

np = n(x)p(y) |np


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Two-dimensional harmonic

oscilator Hamiltonian:

H0 =p2x + p

2y

2m+

K

2(x2 + y2)

np = n(x)p(y) |np

Enp = (n + p+ 1) is(n + p+ 1)-fold degenerate.Whats the degeneracy of

|01

state?


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Two-dimensional harmonic

oscilator Hamiltonian:H0 =

p2x + p2y

2m+

K

2(x2 + y2)

np = n(x)p(y) |np

Enp = (n + p+ 1) is(n + p+ 1)-fold degenerate.Whats the degeneracy of |01state?E10 = E01 = 20.


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Now, turn on the perturbation: H = Kxy


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find w.f. which diagonalize H

1 =a

10 +b

012 = a

10 + b

01


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1 = a10 + b01

2 = a10 + b

01

calculate the elements of submatrix of H in the basis {10, 01}

H = K

10|xy|10 10|xy|0101|xy|10 01|xy|01

= E

0 11 0

E = K22 , 2 = m0


Contents




Now, turn on the perturbation: H = K xy
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Now, turn on the perturbation: H K xy


1 = a10 + b012 = a

10 + b

01

calculate the elements of submatrix of H in the basis {10, 01}

H = K

10|xy|10 10|xy|0101|xy|10 01|xy|01

= E

0 11 0

E =K

22, 2 =

m0

solve the secular equation

E EE E = 0 E = E E10

E+ = E10 + E

E = E10 E

ddd


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obtain the new w.f. from

E EE

E

a

b = 0= E

= +E 1 = 12 (10 + 01)E = E 2 = 12 (10 01)


Contents

Time-independent nondegenerate perturbation theoryTime-independent degenerate perturbation theoryTime-dependent perturbation theory

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obtain the new w.f. from E EE E

a

b

= 0

= E = +E 1 = 12 (10 + 01)

E = E 2 = 12 (10 01)

HW

How does the threefold-degenerate energy

E = 30

of the two-dimensional harmonic oscilator separate due to the perturbation

H = Kxy?


Contents


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4 Literature


Contents


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Problem

If the system is initially in H0, what is the probability that, after time t,transition to another state (of H0) occurs?


Contents


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ProblemIf the system is initially in H0, what is the probability that, after time t,transition to another state (of H0) occurs?

Let us assume:

H(r, t) = H0(r) + H(r, t)

n(r, t) = n(r)eit

H0n = E0n n

(r, t) =

ncn(t)n(r, t) , t > 0

it

= (H0 + H)


Contents


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Problem

If the system is initially in H0, what is the probability that, after time t,transition to another state (of H0) occurs?

Let us assume:

H(r, t) = H0(r) + H(r, t)n(r, t) = n(r)e

it

H0n = E0n n

(r, t) =n

cn(t)n(r, t) , t > 0

i

t = (H0 + H)

Can you remember the meaning of these coefficients?


Contents


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Inserting (r, t) and cn(t) = c0n + c

1n(t) +

2c2n (t) + . . . into time-dependentS.E. and factorizing the perturbation Hamiltonian as H(r, t) = H(r)f(t) gives

Probability that the system has undergone a transition from state l to statek at time t

Plk = Plk = |cn|2 =Hkl

2

t

eiklt

f(t)dt2

For calculation details, see Refs [2] and [3].


Contents


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4 Literature


Contents


Literature

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1 I. Supek, Teorijska fizika i struktura materije, II. dio, Skolska knjiga,Zagreb, 1989.

2 D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., PearsonEducation, Inc., Upper Saddle River, NJ, 2005.

3 R. L. Liboff, Introductory Quantum Mechanics, Addison Wesley, SanFrancisco, 2003.

4 A. Szabo, N. Ostlund, Modern Quantum Chemistry, Introduction toAdvanced Electronic Structure theory, Dover Publications, New York,1996.

5

Y. Peleg, R. Pnini, E. Zaarur, Shaums Outline of Theory and Problems ofQuantum Mechanics, McGraw-Hill, 1998.

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Documents

Lecture on Perturbation Theory