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量子力学

Quantum mechanics

School of Physics and Information Technology

Chapter 6

What

&

&

sligntly. potential the perturb we now potential, some for equation dinger o Schr t) independen -(time the solved have we Suppose

Problem

Difficulty & why

What

potential. mplicated -co more for this exactly, equation nger -ndi o Shr the solve to able be o unlikely t re we' lucky, very are we unless But , H : s eigenvalue and ions eigenfunct new the for solve to like d we' problem, For this n n = E Ψn

OUTLINE

•

Non-degenerate perturbation theory

• Degenerate perturbation theory

Non-degenerate Perturbation Theory

• General formulation

• First-order theory

• First-order theory

The unperturbed case

:

equation

dinger

o

Shr

t

independen

-Time

0 0

0 n 0

n n

E

H

Ψ

=

Ψ

Requirement:

nm m

n

Ψ

=

δ

Ψ

0 0

0 n

Ψ

0

n

E

Considering the perturbed case, we should solve for the new eigenvalues and eigenfunctions.

We can write the Hamiltonian of the new system into the

f o l l o w i n g t w o p a r t s Additional

Hamiltonian

H

H

H

=

0

+

′

Hamiltonian of the unperturbed system

of the perturbed

Rewrite the new Hamiltonian as the sum of two terms

H

H

H

=

₀

+

λ

′

Write the nth eigenfunction and eigenvalue as power series in λ as follows

A small number ;later we will crank it up to 1,and H will be the true,

exact Hamiltonian

+

+

+

=

+

Ψ

+

Ψ

+

Ψ

=

Ψ

2 2

1 0

n

2 2

1 0

n

n n

n

n n

n

E

E

E

E

λ

λ

λ

λ

The first-order correction

n

E

Ψ

=

Ψ

n

n

H

H,Plugging the above _n and E_n into the time-independent

equation and Collecting like

powers of λ

λ is a device to keep track of the different orders

To lowest order:

To first order:

0 0

0 n 0

n n

E

H

Ψ

=

Ψ

0 1

1 0

0 1

0

E

E

H

H

Ψ

+

′

Ψ

=

Ψ

+

Ψ

To second order：

And so on

n

H

n

E

n n

E

n n

H

Ψ

+

′

Ψ

=

Ψ

+

Ψ

0 2

1 1

2 0

1 2

n 0

n n

n n

n n

n

E

E

E

H

First-Order Theory

• The first-order correction to the

energy

The first-order

0 1 1 0 0 1 n 0 n n n n

n

E

E

H

H

Ψ

+

′

Ψ

=

Ψ

+

Ψ

g integratin and ) ( by g

Multiplyin Ψ_n0 ∗

0 0 1 1 0 0 0 0 1 n 0 0 n n n n n n n n

n H Ψ + Ψ H′Ψ = E Ψ Ψ + E Ψ Ψ

Ψ 0 0 = Ψ Ψ And, So, 1 0 0 1 n 0 0 0 0 1 n n n n n n E

H Ψ = Ψ Ψ

Ψ

= Ψ

Ψ _{1 0 0}

n n

n

H

E

=

Ψ

′

Ψ

The first-order correction

to the wave function

Rewrite equation 0 1 1 0 0 1 n 0 n n n n

n E E

H

H Ψ + ′Ψ = Ψ + Ψ

0 1 1 0 0 n n n

n H E

E

H − ）Ψ = −（ ′ − ）Ψ

≠ Ψ = Ψ n m m n m

C( ) 0

1 n The unperturbed wave

functi-ons cfuncti-onstitute a complete set

0 0 1 0 0 0 0 ) ( 0 0 )

( _{l n n l n}

n m m l n m n

m E C H E

E − Ψ Ψ = − Ψ ′ Ψ + Ψ Ψ

≠ 0 0 0 n 0 , Additional n n E

H Ψ = Ψ

0 1 0 ) ( 0 0 )

( _{n n}

n m m n m n

m E C H E

E − Ψ = − ′ − Ψ ≠ ） （ 0 th product wi inner the

.

get

we

n,

l

If

;

get

we

n,

l

If

) ( 1 n m n

C

E

≠

=

0 0 0 0 ) ( m n n m n m

E

E

H

C

−

Ψ

′

Ψ

=

Ψ

Ψ

′

Ψ

=

Ψ

_n m n _m

H

0 0 0 0 0 1 ≠

Ψ

−

=

Ψ

n m m m n n

E

E

0 0

Notice:

although perturbation theory often yields surprising by accurate energies, the

0 2 1 1 2 0 1 2 n 0 n n n n n n

n E E E

H

H Ψ + ′Ψ = Ψ + Ψ + Ψ

0 th product wi inner the

Take Ψ_n

0 0 2 1 0 1 2 0 0 1 0 2 n 0 0 n n n n n n n n n n n

n H Ψ + Ψ H′Ψ = E Ψ Ψ + E Ψ Ψ + E Ψ Ψ

Ψ

= Ψ

Ψ0 0 1 Meanwhile 2 0 0 2 n 0 n 0 2 n 0 0 n n n

n H Ψ = H Ψ Ψ = E Ψ Ψ

Ψ The Hermiticity of H0

We could proceed to

≠ Ψ Ψ = Ψ Ψ = Ψ Ψ n m m n m n n n n

C( ) _n0 0

1 0

0 0

1 Meanwhile， ；

≠ − Ψ ′ Ψ = n

m _{n m}

n m n E E H

E _{0 0}

2 0 0

2

We could proceed to

Results

( )

1

0 0 2 0 0 0 + − Ψ ′ Ψ + ′ + = ≠n

m _{n m}

n m nn n n E E H H E E

Correction to the eigenvalue:

( )

2

0 0 0 0 0 0 ≠ + Ψ − Ψ ′ Ψ + Ψ = Ψ n m m m n n m n n E E H

Degenerate Perturbation Theory

• Two-ford degeneracy

Non-degenerate perturbation theory fail?

• In many cases, where two (or more) distinct states share the same energy, non-degenerate p e r t u r b a t i o n t h e o r y f a i l ! • Sometimes, two energy levels exists so near • Sometimes, two energy levels exists so near that non-degenerate perturbation theory c a n n o t g i v e u s a s a t i s f i e d a n s w e r .

In order to see how the method generalizes, we begin with the twofold degeneracy

,

H

,

Suppose

0 0

0 a 0

Ψ

=

Ψ

E

_a

E

₀

.

0

,

H

0 b 0

a

0 0

0 b 0

=

Ψ

Ψ

Ψ

=

Ψ

E

_b

The perturbation will “break” the degeneracy

“Lifting of a degeneracy by a paerturbation

λ λ λ λ

0 b 0

a 0

Ψ

+

Ψ

=

Ψ

α

β

0 0

0 0

H

Ψ

=

E

Ψ

Any linear combination of the above states

is still an eigenstate of H0,with the same eigenvalue E0

Essential problem:

When we turn off the perturbation, the “upper” state reduces down to one perturbation, the “upper” state reduces down to one linear combination of and , and the “lower” state reduce to some other linear combination, but we don’t know a priori what these good linear combination will be. For this reason we can’t even calculate the first-order energy (equation 1) because we don’t know what unperturbed states to use.

0 a

+ Ψ + Ψ + Ψ = Ψ + + + = ′ + = 2 2 1 0 2 2 1 0 0 , , H H λ λ λ λ λ E E E E H

As before, we write H, E, in the following form

Plug into the stationary equation

0 0

0 n 0

H Ψ = E_n Ψ_n

0 1 1 0 0 1 0 Ψ + Ψ = Ψ ′ +

Ψ H E E

H

To lowest order,

0 1 1 0 0 1 0 Ψ + Ψ = Ψ ′ +

Ψ H E E

H 0 a th product wi inner the Take Ψ 0 0 1 1 0 0 0 0 1 0 0 Ψ Ψ + Ψ Ψ = Ψ ′ Ψ + Ψ

Ψ_a H _a H E _a E _a

0 0 0 0 0 0 0 a 0 , H , H Ψ = Ψ Ψ = Ψ _a E

E _{Meanwhile, H}0 _is

Hermitian, the first

1 0 b 0 a 0 a 0

a

H

β

H

α

E

α

Ψ

′

Ψ

+

Ψ

′

Ψ

=

0 b 0 a 0 0 b 0 a b , 0 , H Ψ + Ψ = Ψ = Ψ Ψ Ψ = Ψ

β

α

b

E Hermitian, the first

yields

th

product wi

inner

the

Similarly,

Ψ

_b0

)

,

,

(

,

Where,

W

_ij

≡

Ψ

_i0

H

′

Ψ

_j0

i

j

=

a

b

1

E

W

W

_aa

β

_ab

α

α

+

=

Results _{The “matrix}

elements” of H'

ab

W

β

Eliminate

1

E

W

W

_ba

β

_bb

β

α

+

=

b

[

W

_ab

W

_ba

−

(

E

1

−

W

_aa

)(

E

1

−

W

_bb

)

]

=

0

equation following

get the can

we zero,

not is

If α

( )

3 4

) (

2

1 ₂ 2

1

+ −

± +

=

± Waa Wbb Waa Wbb Wab

E

0 )

( )

( )

(E1 2 −E1 W_aa +W_bb + W_aaW_bb −W_abW_ba =

zero?

is

if

But what

α

2

The two roots

correspond to the two perturbed

How should we do in practice

Idea:

It would be greatly to our advantage if we could somehow guess the “good” states right from the start.

1. Look around for some Hermitian operator A that commutes with H' ;

M O

R A L

A that commutes with H' ;

2. Pick as your unperturbed states ones that are simultaneously eigenfunctions of H0

and A ;

3. Use ordinary first-order perturbation theory.

, , , s, eigenvalue distinct with of ions eigenfunct are and If . with commutes that operator Hermitian a be Let 0 0 0 0 0 0 b b a a b a A A A H A ≠ Ψ = Ψ Ψ = Ψ Ψ Ψ ′ ν µ ν µ

[

]

[

,

]

0

0 )

( ′_′ − 1 _′ =

k

k k k k

k E a

H δ

0 1

1 0

0 1

0

0 0

0 0

, a

Ψ +

Ψ =

Ψ ′ +

Ψ

Ψ =

Ψ =

E E

H H

E

E _n ， _{k nk} Matrix equation

0

det

H

_k

′

_′k

−

E

1

δ

_k′k

=

Secular equation

Get a_k , then get 0

Get E'

H

on

perturbati

the

es

Diagonaliz

′

Some Examples For Application

• The fine structure of hydrogen

• The zeeman effect

The Fine Structure of Hydrogen

• The relativistic correction

r e 1 4

2m H

0 2 2

2

πε

− ∇

− =

Correct for the motion of the nucleus

The simplest Hamiltonian of

the hydrogen

Replace m by the reduced mass

Relativistic correction

Fine structure

Lamb shift

Hyperfine structure

Relativistic correction

Spin-orbit coupling

Quantization of the Coulomb field

Hierarchy of correction to the Bohr

e n e r g i e s

o f

h y d r o g e n

2 5 2 4 2 2 order of : shift Lamb order of : structure Fine order of : energies Bohr mc mc mc α α α 2 4 order of : splitting Hyperfine order of : shift Lamb mc m m mc p α α

m

2

2

p

T

=

General discussion

r

e 1

4 2m

H

0 2 2

2

πε

− ∇

− =

Represent kinetic energy

Classical formula

2

The relativistic formula kinetic energy

( )

2 2

2

1

mc

c

mc

T

−

−

=

υ

The rest energy The total

Express T in terms of the relativistic momentum

( )

2

1 p momentum ic relativist The c m υ υ − = ：

( )

( )

(

)

2 2 2 2 4 2 2 2 2 4 2 2 2 1 mc T c c m c m c m c

p = +

− + = + υ υ υ mc p << limit istic nonrelativ The + − = − − +

= _{3 2}

4 2 4 2 2 8 2 1 8 1 2 1 1 c m p m p mc p mc p mc T

( )

2

(

)

1 c − υ 2 4 2 2 2 mc c m c p

+ −

= _{3 2}

4 2

8

2 m c

p m

p T

The classical result

The lowest-order

relativistic contribution to the Hamiltonian

In first-order perturbation theory, the correction to E_n is given by the expectation value of H' in the unperturbed state

ψ ψ

ψ

ψ 4 _{3 2} 2 2

2 3 1

r ˆ ˆ

8 1 ˆ 8 1 p p c m p c m

E = − = −

ψ ψ E V m p = + 2 ˆ And, 2

(

)

ψ

ψ m E V

pˆ 2 = 2 −

[

2 2

]

2 1 r 2 2 1 V V E E mc

In the case of hydrogen

+ + − = ₂ 2 0 2 0 2 2 n 2 1 r 1 4 1 4 2 2 1 r e r e E E mc E _n πε πε 1 1 1 1

Where E_n is the Bohr energy of the state in question

(

)

3 2 2 2 2 1 1 1 , 1 1 And, a n l r a n

r = = ₊

2 2 0

4

me

a

≡

πε

2

2 0 2 2 1 4 2 n e m

E_n = −

πε

(

+

)

+ + − = 2 3 2 2 2 2 2 n 2 1 r 1 1 4 1 4 2 2 1 a n l e a n e E E mc E _n πε πε

P l u g g i n g

(

+

)

3 2

0 2 0 n 2 r 2 1 4 4

2mc n πε n a πε _{l n a}

Spin-Orbit Coupling

The orbiting positive charge sets up a magnetic field B

in the electron frame, which exerts a to rque on th e spinning electron, tending t o a l i g n i t s m a g n e t i c

Β

Β

Β

Β

μ

μ

μ

μ

⋅

−

=

H

t o a l i g n i t s m a g n e t i c m o m e n t (µµµµµµµµ) a l o n g t h e d i r e c t i o n o f t h e f i e l d .

The Magnetic Field of the Proton

2r

: law Savart

-Biot the

to

According B = µ0 I

T e

I =

current effective

An The charge of proton

The period of the orbit

L

3 2

r mc

e

0000

4444 1111 Β

Β Β Β

πε

=

T mr rm

L

2

2

: electron the

of momentum angular

orbital

The = υ = π

direction same

in the point

and ,

The Magnetic Dipole Moment of the Electron

r

q

2

amics

electrodyn

Classical

π

µ

=

T

mr

S

T

r

q

2

2

π

π

µ

=

=

S

=

m

2

q

μ

μ

S

e

−

=

Classical

result

However, as it turns out, the electron’s

magnetic moment is

?

S

m

e

e

=

−

μ

μ

magnetic moment is twice the classical

answer

The “extra” factor of 2 was explained by Dirac in his relativistic theory of the electron. For a related

The Spin-Obit Interaction

L 3 2

r mc

e

0 00 0

4444 1111 Β

ΒΒ

Β

πε

=

S

m e

e = −

L S ⋅ = _{2 2 3}

2

1

r c m e

H

0 00 0

4444πε

FR A

We did the analysis in the rest frame of the electron, but that’s not an inertial

A U D

the electron, but that’s not an inertial system——it accelerates as the electron o r b i t s a r o u n d t h e n u c l e u s .

S O L U T I O N

The Spin-Obit Interaction

The modified

gyromagnetic ratio for the electron

the Thomas precession

factor

Exactly cancel one

another

D I S R E G A R D

L

S

⋅

=

′

3 2

2 2

1

r

c

m

e

H

_SO

0000

8888

πε

[

]

[

,

]

0

0

,

≠

′

≠

′

S

L

SO SO

H

H

Put it another way

momentum angular

total

The J ≡ L + S

The spin and orbital angular momenta are not conserved

on theory. perturbati in use to states good are quantities these of s eigenstate the and conserved, are , , , 2 2

2 z J J S L

S

L

J

≡

+

J

2

=

L

2

+

S

2

+

2

L

⋅

S

( ) ( ) ( )

[ 1 1 1 ]

2 of s eigenvalue The 2 + − + − + ⋅ s s l l j j : S L

(

)

( ) 3 3 3 3 2 2 2 1 2 1 1 1 1 Meanwhile, a n l l l r r c m e H_SO + + = ⋅ =

′ S L

0 00 0

8888πε

(

2 2 2

)

2 1

S L

J − −

= ⋅S L

2 l

(

l + 2

)

(l +1)n a

( ) ( )

[

]

(

)

( ) 3 3 2 2 2 2 1 1 2 1 4 3 1 1 2 1 a n l l l l l j j c m e E_SO + + − + − + = 0000 8888πε ( ) ( )

[

]

(

+

)

( + ) − + − + = 1 2 1 4 3 1 1 2 2 1 l l l l l j j n mc E E_SO n

− + − = 3 2 1 4 2 2 2 n 1 r l n mc E

E

[

( ) ( )

]

(

+

)

( + ) − + − + = 1 2 1 4 3 1 1 2 2 1 l l l l l j j n mc E E_SO n

The relativistic correction The spin-orbit coupling

+ − = 2 1 4 3 2 2 2 1 f j n mc E E _s n

The correction to the Bohr formula

Combine with the Bohr formula − + + − = 4 3 2 1 1 6 . 13 2 2 2 j n n n eV

E_nj α

The complete fine-structure formula

formula

•Fine structure breaks the degeneracy in

•The energies are determined by and

l

The Fine Structure of Hydrogen

The Zeeman Effect

• Weak-field zeeman effect

• Strong-field zeeman effect

Consider an atom placed in a

uniform external magnetic field

B

_ext

What

&

(

μ

μ

μ

μ

+

μ

μ

μ

μ

)

⋅

Β

Β

Β

Β

_extextextext

−

=

′

_{l s}

Z

H

For a single electron

&

How

(

+

)

⋅_{ΒΒΒΒextextextext} =

′ L 2S

2m e H_Z

Associated with orbital motion

Zeeman Effect

New phenomenon

The energy levels are shift when an atom is placed in a uniform external magnetic field

H

Depend on the strength of the external_{field in comparison with the internal field}

that gives rise to spin-orbit coupling

O

W

that gives rise to spin-orbit coupling

Weak-field Intermediate Strong-field

Comparable

Weak-field Zeeman Effect

B

_ext

<<B

_int

Fine structure

Fine structure

dominates

H

_Z

as a small

perturbation

[

]

[

,

]

0

0

,

≠

′

≠

′

S

L

SO SO

H

H

momentum angular total

The J ≡ L + S

Put it another way

The spin and orbital angular momenta are not conserved

on theory. perturbati in use to states good are quantities these of s eigenstate the and conserved, are , , , 2 2

2 z J J S L

numbers quantum

good" "

the are

and ,

,l j m _j

n

S L 2 2

1

+ ⋅

= ′

= _{ΒΒΒΒextextextext}

m e nljm

H nljm

E _{Z j Z j}

The zeeman correction to the energy

The expectation

S

?

(

)

J J

S S _{ave 2}

J

⋅ =

value of

S

?

In the presence of spin-orbit coupling,

L and S are not separately conserved; they precess about the fixed total

(

)

J J

S S_{ave 2}

Meanwhile, J ⋅ =

S

J

L

=

−

( ) [ ( 1) ( 1) ( 1)]

2 2

1 _{2 2 2} 2

+ − + + + = − + =

⋅J J S L j j s s l l

S

( )

( ) ( )

( ) J

J J S S L + + + − + + = ⋅ + = + 1 2 4 3 1 1 1 1 2 ₂ j j l l j j J The landé g-factor, g_J Choose the z-axis to lie along B

j ext

J B

Z

g

B

m

E

1

=

µ

Τ ×

=

≡ − e VVVV

m e B 5 10 788 . 5 2 Where,

µ

Choose the z-axis to lie along B_ext

Strong-field Zeeman Effect

Strong-field Zeeman Effect

B

_ext

>>B

_int

The zeeman

effect dominates

effect dominates

Fine structure

as the

perturbation

Strong-field Zeeman Effect

numbers quantum good" " the now are and ,

,l m_l m_s n

( _{z z})

ext

Z B L S

m e H 2 2 n Hamiltonia zeeman The + =

′ _{nm m B}B_ext (m_l m_s ) n eV E s l 2 6 . 13 energies d" unperturbe " The

2 + +

−

= µ

Choose the z-axis to lie along B_{ext Proof!}

( )

(

+

)

( + ) − + − = 1 2 1 1 4 3 6 . 13 2 3 1 l l l m m l l n n eV

E_fs α l s

m 2 n s l so r s l

fs nlm m H H nlm m

E 1 = ′ + ′

s l z z y y x

x L S L S L m m

S + + = 2 =

⋅J S

H'_r : The same as before

Intermediate-field Zeeman Effect

Neither H'

_Z

nor

H'

_fs

dominates

Treat them on

an equal footing

As perturbation

to Bohr

Intermediate-field Zeeman Effect

= ≡ Ψ = ≡ Ψ = −

− _{0 0 ,}

, 0 0 0 2 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 1 l j m j l n and , , by zed characteri states the : basis The 2 : case

The = ；

Matrix of H'

− − − + − β β γ β γ β γ β γ β γ 2

2 _{0 0}

0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5

( )

13 .6 and . 8

,

Where γ ≡ α 2 eV β ≡ µ _B B _ext

Energy levels for the n=2 states of hydrogen, with fine structure and zeeman splitting

2 2 5 5 2 4 2 3 2 2 2 1 β γ ε β γ ε β γ ε β γ ε − − = + − = − − = + − = E E E E

(

)

(

)

(

)

Brief introduction to proton

The quark structure of the proton. There are two up quark in it and one down quark. The s t r o n g f o r c e i s mediated by gluons

Made by Arpad Horvath.

commons.wikimedia.org/wiki/ Image:Quark_struct...

The proton itself constitutes a magnetic dipole, smaller than the electron’s

Brief introduction to proton

）

value

measured

(

59

.

5

;

2

=

=

g

m

ge

p p

p

S

magnetic field

p

(

)

[

μμμμ rrrr rrrr ----μμμμ

]

μμμμ

( )

rrrr 4

4 4 4 Β

ΒΒ

Β 0000

µ

0000

δ

3333

π

µ

3 2 3

3 ⋅ +

=

r

According to classical electrodynamics, a dipole

µ µµ

µ sets up a magnetic field.

The electron near proton

Β

Β

Β

Β

μ

μ

μ

μ

⋅

−

=

H

(

)

(

)

[

]

( )

rrrr

rrrr _{0000 3333}

0 00

0 µ _δ

µ ge p e p e ge

H′ = 3 S ⋅ S ⋅ − S ⋅S + S ⋅S

2 2

The Hamiltonian of the electron,due to the proton’s magnetic dipole moment

(

)

( )

( )2 2 3 2 1 3 3 0 Ψ e p e p e p e p e p hf m m ge r m m ge

E = 0000 S ⋅rrrr S ⋅rrrr − S ⋅S + 0000 S ⋅S

8888

µ π

µ

( )

_rrrr

8888π p e p e p eδ

hf m m r m m

H′ = + S ⋅S

3

3

For the ground state hydrogen

(

)

(

)

0 3

3 =

⋅ −

⋅ ⋅

r

e p

e

p S S S

S _{rrrr rrrr}

T h e w a v e f u n c t i o n i s

s p h e r i c a l l y s y m m e t r i c a l

( )

2 1₃

,

Meanwhile

a

π

=

0

Ψ₁₀₀

e p

e p

hf

a

m

m

ge

E

=

₃

S

⋅

S

2 1

3

π

( )

2 2 2 4 3 = = _p e S S

(

2 2 2

)

2

1

p e

e

p ⋅S = S − S − S

S

p

e

S

S

S

≡

+

Spin-spin coupling

The individual spin angular momenta are no longer conserved; the “good” states are eigenvectors of the total spin.

The electron and proton both have spin 1/2 .

− + = singlet). ( , 4 3 triplet); ( , 4 1 3 4 4 2 2 2 1 a c m m g E e p hf . 0 state, singlet In the 2 state, triplet In the 2 2 2 = = S S

( )

₄

= = _p

e S

S

Effect of spin-spin coupling

Hyperfine

splitting in

the ground

state of

•Lifting the triplet configuration

•Depressing the singlet configuration

eV a

c m m

g E

e p

6 4

2 2

2

10 88

. 5 3

4 ₋

× =

= ∆

z h

E

ΜΗ =

∆

= 1420

ν

The energy gap as a result of spin-spin coupling

The frequency of the photon emitted in a transition from the triplet to the singlet state

h

cm

21

c

:

h

wavelengt

ing

Correspond

=

ν

!" # This famous “21-centimeter line” is among

the most pervasive and ubiquitous forms of radiation in the universe.