Perturbation theory - Quantum mechanics 2 - Lecture 2fizika.unios.hr/.../qm2/Lecture_2_Perturbation_theory.pdf · Contents Time-independent nondegenerate perturbation theory Time-independent

ContentsTime-independent nondegenerate perturbation theory

Time-independent degenerate perturbation theoryTime-dependent perturbation theory

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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

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General formulationFirst-order theorySecond-order theory

Contents




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Do you remember this?




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Do you remember this?

H00n = E0n

0n

0n|0m = nm complete set




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




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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?




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AssumeH = H0 + H




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AssumeH = H0 + H

unperturbed Hamiltonian

perturbation Hamiltonian

small parameter




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Name Description Hamiltonian

L-S coupling Coupling between orbital and H = H0 + f (r)~L ~Sspin 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 + eE0zuniform 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 + K x4

oscilator force H = K x4

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




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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




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Assume




expand

n = 0n +

1n +

22n + . . . ,

En = E0n + E

1n +

2E 2n + . . .




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Assume




expand

n = 0n +

1n +

22n + . . . ,

En = E0n + E

1n +

2E 2n + . . .

and sort

(0) . . . H00n = E0n

0n ,

(1) . . . H01n + H0n = E

0n

1n + E

1n

0n ,

(2) . . . H02n + H1n = E

0n

2n + E

1n

1n + E

2n

0n ,

...




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Making 0n|/

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

First-order correction to the energy

E 1n = 0n|H |0n

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




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E 1n = 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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E 1n = 0n|H |0n

Example 1


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

2

asin(n

ax)




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E 1n = 0n|H |0n

Example 1



2

asin(n

ax)

Perturbation Hamiltonian: H = V0




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E 1n = 0n|H |0n

Example 1



2

asin(n

ax)


First-order correction:E 1n = 0n|V0|0n = V00n|0n = V0

corrected energy levels: En E 0n + V0




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Example 1



2

asin(n

ax)


First-order correction:E 1n = 0n|V0|0n = V00n|0n = V0

corrected energy levels: En E 0n + V0Compare this result with an exact solution for a constant perturbation all thehigher corrections vanish




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Example 1 (cont.)

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




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Example 1 (cont.)


E 1n =2V0a

a/20

sin2(n

ax)dx =

V02




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Example 1 (cont.)


E 1n =2V0a

a/20

sin2(n

ax)dx =

V02

HW. Compare this result with an exact one.




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

m 6=n

cmn0m give

First-order correction to the wavefunction

1n =m 6=n

0m|H |0n(E 0n E 0m)

0m





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First-order correction to the wavefunction

1n =m 6=n

0m|H |0n(E 0n E 0m)

0m


In conclusion

First-order perturbation theory gives:

often accurate energies

poor wave functions




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Example 2

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




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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3




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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3

H0 eigenenergies: E 0n =

(n +

1

2

)~




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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3


(n +

1

2

)~

H0 eigenfunctions:

0n(x) =

1

2nn!

e

x2

2 Hn(x)

Dont forget...

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




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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3


(n +

1

2

)~

H0 eigenfunctions:

0n(x) =

1

2nn!

e

x2

2 Hn(x)

E 1n = 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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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3


(n +

1

2

)~

H0 eigenfunctions:

0n(x) =

1

2nn!

e

x2

2 Hn(x)

E 1n = 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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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3


(n +

1

2

)~

H0 eigenfunctions:

0n(x) =

1

2nn!

e

x2

2 Hn(x)


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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Example 2


Hamiltonian: H = ~2

2m

d2

dx+

1

2kx2 + ax3


(n +

1

2

)~

H0 eigenfunctions:

0n(x) =

1

2nn!

e

x2

2 Hn(x)


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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Example 2


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

Dont forget...

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




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Example 2


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 3~n+1 ~n-1 ~n-3 3~




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Example 2

Compute the first-order corrections for a harmonic oscilatorwith applied small perturbation W = ax3.

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 3~n+1 ~n-1 ~n-3 3~

1n =a

2~

[1

3

n(n 1)(n 2)

20n3 + 3n

n

20n1

3(n + 1)

n + 1

20n+1

1

3

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

20n+3

]




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Making 0n|/

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

between 0n and 1n, we get

Second-order correction to theenergy

E 2n =m 6=n

|0m|H |0n|2

E 0n E 0m





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

m 6=n

cmn0m give

Second-order correction to the wave function

2n =m 6=n

[

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

(E 0n E 0m)2

+k 6=n

0m|H |0k0k |H |0n(E 0n E 0m)(E 0n E 0k )

]0m





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General formulationExample: Two-dimensional harmonic oscilator

Contents




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Symmetry Degeneracy Perturbation




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A question

Whats wrong with

E 2n =m 6=n

|0m|H |0n|2

E 0n E 0m, m, n q

if unperturbed eigenstates are degenerate E 01 = E02 = = E 0q ?




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H = H0 + H

H00n = E

0n

0n

E 01 q-fold degenerate

m1-

Construct

n =

qi=1

anm0m

which diagonalizes submatrixof H :n|H |k = E nk

m2

Nondegenerateperturbation theorywith basis B

q(H E )a = 0

m3- det

H E nI = 0m8 m6 m5 m46

? XXXXX

XXXXXX

Xy

?

{n}Gives new basis B

m7 {ani} E 1,E 2, . . . ,E q




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Matrix H in basis B

H =

H 11 H1,q+1 . . .

. . . 0H q/2,q/2

0 . . .H qq

H q+1,1...




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H = H0 + H

H00n = E

0n

0n


m1-

Construct

n =

qi=1

anm0m


m2


q(H E )a = 0

m3- det

H E nI = 0m8 m6 m5 m46

? XXXXX

XXXXXX

Xy

?


m7 {ani} E 1,E 2, . . . ,E q




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For n = 1,

qm=1

(H pm E npm)anm = 0 appear as

H 11 E 1 H 12 H 13 . . . H 1q

H 21 H22 E 1 H 23 . . . H 2q

......

......

...H q1

a11a12...

a1q

= 0




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H = H0 + H

H00n = E

0n

0n


m1-

Construct

n =

qi=1

anm0m


m2


q(H E )a = 0

m3- det

H E nI = 0m8 m6 m5 m46

? XXXXX

XXXXXX

Xy

?


m7 {ani} E 1,E 2, . . . ,E q




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

the submatrix of H .




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H = H0 + H

H00n = E

0n

0n


m1-

Construct

n =

qi=1

anm0m


m2


q(H E )a = 0

m3- det

H E nI = 0m8 m6 m5 m46

? XXXXX

XXXXXX

Xy

?


m7 {ani} E 1,E 2, . . . ,E q




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n = n +1n +

22n + . . . n qn =

0n +

1n +

22n + . . . n > qEn = E

0n +E

1n +

2E 2n + . . . n q (E 01 = = E 0q )

En = E0n + E

1n +

2E 2n + . . . n > qE n = n|H |n n qE 1n = 0n|H |0n n > q




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n = n +1n +

22n + . . . n qn =

0n +

1n +

22n + . . . n > qEn = E

0n +E

1n +

2E 2n + . . . n q (E 01 = = E 0q )

En = E0n + E

1n +

2E 2n + . . . n > qE n = n|H |n n qE 1n = 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 + p

2y

2m+

K

2(x2 + y 2)

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




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Two-dimensional harmonicoscilator Hamiltonian:

H0 =p2x + p

2y

2m+

K

2(x2 + y 2)

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

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




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Two-dimensional harmonicoscilator Hamiltonian:

H0 =p2x + p

2y

2m+

K

2(x2 + y 2)

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

Enp = ~(n + p + 1) is(n + p + 1)-fold degenerate.Whats the degeneracy of |01state?E10 = E01 = 2~0.




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Now, turn on the perturbation: H = K xy




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find w.f. which diagonalize H

1 = a10 + b01

2 = a10 + 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 = K

22, 2 =

m0~




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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 = K

22, 2 =

m0~

solve the secular equation

E EE E = 0 E = E E10

E+ = E10 + E

E = E10 E

@@@




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obtain the new w.f. from(E EE E

)(ab

)= 0

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




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obtain the new w.f. from(E EE E

)(ab

)= 0

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

HW

How does the threefold-degenerate energy

E = 3~0

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

H = K xy ?




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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?




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Problem


Let us assume:

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

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

H0n = E0nn

(~r , t) =n

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

i~t

= (H0 + H)




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Problem


Let us assume:

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

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

H0n = E0nn

(~r , t) =n

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

i~t

= (H0 + H)

Can you remember the meaning of these coefficients?




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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

e ikl tf (t)dt

2

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




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Contents




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Literature

Literature

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.


ContentsTime-independent nondegenerate perturbation theoryGeneral formulationFirst-order theorySecond-order theory

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

Time-dependent perturbation theoryLiterature

Documents

Perturbation theory - Quantum mechanics 2 - Lecture 2fizika.unios.hr/.../qm2/Lecture_2_Perturbation_theory.pdf · Contents Time-independent nondegenerate perturbation theory Time-independent