Staff Site Universitas Negeri Yogyakarta

(1)

量子力学

Quantum mechanics

School of Physics and Information Technology

(2)

Chapter 9

Time-dependent

(3)

9.1 Two

9.1 Two--level systems

level systems

Chapter 9

Time-dependent perturbation theory

9.1 Two

9.1 Two--level systems

level systems

9.2 Emission and absorption

of radiation

(4)

9.1 Two

9.1 Two--level systems

level systems

of radiation

(5)

9.1 Two-level systems

Suppose that there are just two states of the system. They are

eigenstates of the unperturbed Hamiltonian,and they are eigenstates orthonormal:

0

and

0

|

a a a b b b

a b ab

H

ψ

E

ψ

H

ψ

E

ψ

ψ ψ

δ

=

(0)

c

_a _a

c

_b _b

ψ

=

ψ

+

ψ

Any state can be expressed as a linear combination of them

a b ab

In the absence of any perturbation,each component evolves with its characteristic exponential factor:

/ /

( )

1

a a

iE t iE t

a a b b

a b

t

c

e

c

e

c

ψ

−

ψ

−

2 2

=

+

(6)

9.1.1 The perturbed system

Now suppose we turn on a time

Now suppose we turn on a time--dependent perturbation, dependent perturbation, then the wave function can expressed as follows:

then the wave function can expressed as follows:

/ /

( )

iE ta

( )

iE ta

a a b b

t

c t

e

c t

e

ψ

=

ψ

−

+

ψ

−

If the particle started out in the state

If the particle started out in the state ψ_a_a _{We solve for}_{We solve for}_{We solve for}_{We solve for}

( ) and ( ) by demanding that (t) satisfy the tim-dependent Schr dinger equation

a b

c t c t ψ

H

i

,

t

ψ

=

∂

'

( )

0

H = H

+

H t

where where

(7)

/ / ' / 0 0 / / / ' / /

[

]

[

]

[

]

[

]

[

]

a b a

b a b

a b

iE t iE t iE t

a a b b a a

iE t iE t iE t

b b a a b b

iE t iE t

a b

a a b b

c H

e

c H

e

c H

e

c H

e

i c

e

c

e

iE

c

e

c

e

ψ

− − − − − − − −

+

=

+

−

+

−

The first two terms on the left cancel the last two terms on the right, and hence

the right, and hence

/ / ' ' / /

[

]

[

]

[

]

a b a b

iE t iE t

a a b b

iE t iE t

a a b b

c

H

e

c

H

e

i

c

e

c

e

ψ

− − − −

+

=

+

a a b

Take the inner product with

, and exploit the orthogonality

of

and

ψ

(8)

For short, we define

' '

|

ij i j

H

≡

ψ

H

ψ

( ) /

' '

[ i Eb E ta ]

i

c = − c H +c H e− −

' ' *

/

Note that the Hermiticity of entails ( ) . M ultiplying through by ( / ) a , we conclude that

'

ji ji iE t

H H H

- i h e

=

( ) /

' '

[ i Eb E ta ]

a a aa b ab

c = − c H +c H e− −

b

Similarly, the inner product with

ψ

picks out :c_b

( ) /

' '

[ i Eb E ta ]

b b bb a ba

i

c = − c H + c H e− −

'

Typically, the diagonal matrix elements of H vanish:

' '

0

aa bb

(9)

In that case the equations simplify

0

0 '

'

i t

a a b b

i t

b b a b

i

c

H

e

c

i

c

H

e

c

ω

−

= −

(1) (1)

where

0

b a

E

ω

≡

−

b a 0

(10)

9.1.2 Time-Dependent Perturbation Theory

So far, everything is exact: we have made no assumption about the size of the perturbation. But if is "small", we can solve Equation (1) by a process of successive approximations, as

follows. Sup

'

H

(0) 1, (0)

pose the particle starts out in the lower state: c_a = c_b = 0.

Zeroth Order:

(0) 1, (0)

If there were , they would stay this wa

0

y f

.

oreever

a b

no perturbation at all

c = c =

( 0 ) ( 0 )

( )

1,

0.

a b

c

t

=

c

=

(11)

First Order:

'

0 0

(1)

' (1) ' ' '

0

Now we insert

expressions on the right to obtain the

0

1;

( )

a a t

i t i t

b

ba b ba

dc

c

dt

d

these

c

i

H e

c

H

t e

dt

ω ω

=

= −

second-order approximation:

S

( )

'' 0 0 ' ' '' 0 0 ' ' '' '' 0

(2) ' ' ' '' '' '

2 ₀ ₀

( )

1

econd-Order:

( ) 1

(

)

(

)

t

i t i t

a i

ab ba

t t

i t i t

a ab ba

dc

i

H e

H

t e

dt

c

t

H

t e

H

t e

dt

(12)

Notes:

(2)

Notice that in my notation

( ) includes the zeroth order

term; the second-order correction would be the integral term

alone.

a

c

t

In principle, we could continue this ritual indefinitely,

th

In principle, we could continue this ritual indefinitely,

always inserting the

order approximation into the

right side of Equation (1) and solving for the (

1)

order. Notice that is modified

_a

n

c

−

+

−

in every

order,

and in every

_b

order.

even

(13)

9.1.3 Sinusoidal Perturbation 9.1.3 Sinusoidal Perturbation

'

( , ) ( ) cos(

Suppose the perturbation has sinusoidal time dependence:

so

that

where

To first o

) cos

rder we have

( ), | |

ab ab

ab a b

H r t V r t

H V t

V V ω ω ψ ψ = = =

0 ' '

' ' 0 0 0 0 ' 0 ( ) ( ) ' 0 ( ) ( ) 0 0

( ) cos( )

This is the ans

2 1 1 wer, but = 2 it i t

i t dt

b ba

t

i t i t

ba

i t i t

ba

i

c t V t e

iV

= - e e dt

V e e

-ω ω ω ω ω ω ω ω ω ω ω ω ω ω + − + − ≅ − + − − + + −

(14)

Dropping the first term:

0

0 0

0

0 0

( ) /2

( ) /2 ( ) /2 0

( ) /2 0

We assume

|

( )

2

sin[(

) / 2]

. s

o

i t

i t i t

ba b

i t

ba

V e

c t

e

V

t

i

e

ω ω ω ω ω ω ω ω

ω ω

− − − − −

+ >>

−

≅ −

−

=−

0 0

0

transition proba

The

---the proba

bility

b

i

e

ω ω

=−

−

2 2 2 ₀ 2 2 0

ility that a particle which

started out in the state

will be found, at time, in th

|

| sin [(

) / 2]

( ) |

e state

:

( )|

(2)

(

)

ab

a b b

a b

V

t

P

t

c t

(15)

As a function of time, the transition probability oscillates sinusoidally.

P(t)

0

2

| |

π

ω −ω ₀

4

| |

π

ω − ω ₀

6

| |

π ω − ω

2 0

| | ( )

ab

V

ω −ω

[image:15.842.61.789.54.532.2]

t

(16)

The probability of a transition is greatest when the driving frequency is close to the “natural”

frequency

P(t)

0

ω

0

[image:16.842.66.781.72.524.2]

(ω −2 )πt (ω₀ +2 )πt ω

(17)

9.2 Emission and absorption

of radiation

9.2.1 Electromagnetic Waves

An atom, in the presence of a passing light wave, responds primarily ot the electic component. If the wavelenght is long , we can ignore the spatial variation in the field; the atom is

0

'

0

exposed to a sinusoidally oscillating electric field

The perturbing Hamiltonian is

where is the charge of th

ˆ cos( )

cos(

e electr )

E = E t k

H

q

E z t

ω

= −

'

0 cos(

on. Ev

),

idently

(18)

9.1 Two

9.1 Two--level systems

level systems

Chapter 9 TIME-DEPENDENT

PERTURBATION THEORY

of radiation

(19)

An electromagnetic wave consists of transverse oscillating electric and magnetic fields.

An electromagnetic wave

E

An electromagnetic wave

E

Electric field

u

E

H

o

x

u

E

H

o

x

(20)

2

Typically,

is an even or odd function of ; in

either case

is

, and integrates to zero.

This licenses our usual assumption that the

matrix elements of

z|

vanish. Th

|

us

'

z

H

odd

diagonal

ψ

0

the interaction of light with matter is governed

by precisely the kind of oscillatory perturbation

with

(21)

If an atom starts out in the "lower" state , and

9.2.2 Absorption, Stimulated Emission,

you

shine a polarized monochromatic beam of light

and Spontaneous Emiss

on it, the probabilit i of on y a ψ

a transition to the "upper" state

is given by Equation (2), which takes the form

b ψ 2 ₂ 0 0 2 2 0

( ) | | sin [( ) / 2]

( )

a b

E t

P t ω ω

ω ω → ℘ − = − 2 ₂ 0 0 0

|

sin

Of course, the probability of a transition

to

[(

) / 2]

to the

lev l

( )

(

e

)

:

b a

down

lo

E

wer

t

P

t

ω

→

℘

−

=

(22)

0

Three ways in which light interacts with atoms:

In this process, the atom absorbs energy

from the electromagnetic field.

we say that

i

(a) Absorption

"absorbed a photon"

t has

b a

E - E

=

ω

(23)

If the particle is in the upper state, and you shine

light on it, it can make a transition ot the lower state,

and in fact the probability of such

(b) S

a tran

timulated emissio

sition is exa

n

ctly

the same as for a transition upward from the lower state.

which process, which was first discovered by Einstein,

stimul

is called ated emission.

(24)

If an atom in the excited state makes a transition

downward, with the release of a photon but

any applied electromagnetic fi

el

(c) Spontaneous emission

d to initiate the

process.

Thi p

s roc

without

spontane

ess is c

alled

ous emis

sion.

(25)

2 0

0

The energy density in an electromagnetic wave is

where E is the amplitude of the electric field.

9.2.3 Incoherent P

So the

transition

er

pr

2

obability is

tur

pro

bation

u =

ε

E

portional to the energy

transition probability is pro

2

2 ₀

2 2

0 0

sin [(

) / 2

portional to the energy

density of the fields:

This is for

perturbation,

]

2

( )

|

consisting of

(

)

a single frequency .

a b

t

u

P

t

monochromatic

ω

ε

ω

→

−

=

℘

(26)

In many applications the system is exposed to electromagnetic

waves at a whole range of frequencies; in that case

and the net transition probablity takes the form of an integral:

( ) , ( ) a b u d P t

ρ ω ω

→ → = 2 2 0

2 ₀ 2

0 0

0

replace ( ) by ( ) and take it outside the integral, we

sin [( ) / 2]

2 | | ( ) ( ) get t u d

ω

ρ ω

ω

ε

ω

ρ

ω

ρ

ω

∞ − ℘ − 2 2 0 0

2 ₀ 2

0

replace ( ) by ( ) and take it outside the integral, we

sin [( ) / 2]

2 | |

( ) ( ) ( get Ch ) (

anging variables to ) / 2, exten

a b

t

P t d

x

ρ

ω

ρ ω

ω

ε

ω

ρ

ω

∞ → − ℘ ≅ − ≡ − 2 2

-ding the limits of

integration to , and looking up the define integral

(27)

2

0 2

0

we find

This time the transition probability is proportional ot .

When we hit the system with an incoherent spread of

2 |

|

P

_a _b

( )

t

(

)

t

ρ ω

ε

→

℘

≅

frequencies. The

transition rat

e

is

no

2

0 2

0

the perturbing wave is coming in al

ong the

-direction and polarized in the

-dir

:

|

ec

w

ti

.

a

)

on

(

b a

Not

R

x

z

e

constant :

π

ρ ω

ε

(28)

2

But we shall be interested in the case of an atom

bathed in radiation coming from

direction,

and with

possible polarizations. What w

ˆ

|

e

need, in place of

|

, is th

e

o

f

|

all

average

n

℘

⋅

|

2

,

where

℘

| |

ˆ

(

where

and the average is over both polarizations

and

over all incident direction. This averaging can be

carried out as follows:

)

b a

q

r

n

ψ

(29)

2 2

For propagation in the z-direction, the two possible

polarizations are and

ˆ

1 ˆ

ˆ

(

Polarization:

, so

)

[(

)

(

) ]

the polarization average

is

2

i

j

n

⋅℘

=

i

⋅℘ +

2

j

⋅℘

2

1 ˆ

ˆ

(

)

[(

)

(

) ]

2

1

= (

2

2 p

x

n

⋅℘

=

i

⋅℘ +

j

⋅℘

℘ +℘

2 2 2

where is the angle between and the direction of

1

)

sin

2

propagation.

y

θ

= ℘

(30)

2 2

Now let's set polar axis along and integrate over all

propagation directions to get the polarization-prop

1 1 ˆ

(

Propagation direction:

) sin sin

4 2

agation

average:

2 pp

n =

θ

θ θ φ

d d

π

℘

⋅℘ ℘

4 2

pp

π

2 2

3

0

So the transition rate for stimulated emission from state

to state , under the influence of incoherent, unpolarized

light i

= sin

4 3

ncident from all directions, is

d

b a

π

θ θ

℘ ℘

(31)

2

So the transition rate for stimulated emission from

state to state , under the influence of incoherent,

unpolarized light incident from al

l direct

ions

|

s

|

3

, i

b a

b

a

R

π

ε

→

=

℘

2

0

(

)

ρ ω

2 0

|

3

b a

R

ε

→

=

℘

0

Where is the matrix element of the electric

dipole moment between the two states and

is the energy density in the fields, per unit frequency,

evaluated at

(

)

(

)

(

E

_b

E

_a

) / .

ρ ω

ω

℘

(32)

9.1 Two

9.1 Two--level systems

level systems

PERTURBATION THEORY

of radiation

(33)

9.3 Spontaneous emission

Picture a container of atoms, of them in the lower

state , and of them in the upper state .

Let be the spontaneous emission rate, so

9.3.1 Einstein's A and B coefficient

( )

th

)

t

s

a (

a

a b b

N

A

ψ

the number of particle leaving the upper state by this process, per unit time, is . The transition rate for

stimulated emission is proportional to the energy density of the electroma

b

N A

0

gnetic field---call it .

The number of particles leaving the upper state by this mechanism, per unit time, is .

( )

ba

b ba

B

N B

ρ ω

(34)

0 0

0

The absorption rate is likewise proportional to

---call it

The number of particles

per unit joining the upper level is therefore

All told, then,

(

)

(

);

(

).

ab

a ab

B

N B

ρ ω

₀

a ab

b

dN

0 0

Suppose that these atoms are in thermal equilibrium

with the ambient field, so that the number of particle

i

(

)

(

)

n each level is

.

b b ba a ab

N A

N B

dt

constant

ρ ω

(35)

0

In that case

and it follows that

The number of particles with energy , in thermal

equilibrium at temperature , is proportiona / 0, ( ) ( l to ) / b

a b ab ba

dN dt

A

N N B B

E T

ρ ω

= = − he

t Boltzmann fac

0 0 / / / 0 _/ , so and hence ( ) a B B a B B

E k T

k T a

(36)

3

/

2 3

(

But Planck's blackbody formula

tells us the energy density of thermal radiation:

Comparing the two expressions, we conclude that

an ) d b k T ba

c e B

B B A

ω

ρ ω

π

= − = 3 B

ω

=

B_ab = B_ba and A ₂ ₃

2 2 0 3 2 3 0 and

and it follows that the spontaneous emission rateis

(37)

Suppose, now, that you have a bottle full of

atoms, with

of them in the excited state.

As a resul

9.3.2 The Lifetime o

t of spontaneous em

( ),

issio

f an Excite

n, this num

d S

ber

tat

w d

e

ill e

b

N t

crease as time goes on; specifically, in a

w d

ill ecrease as time goes on; specifically, in a

time interval

you will lose a fraction of

them:

Solving for

, we find

( )

b b

b

dt

Adt

dN

A

N dt

t

N

= −

( )

(0)

At

b

t

N

b

e

−

(38)

evidently the number remaining in the excited

state decreases exponentially, with a time constant

We call this the lifetime of the state ---- technically,

1

A

τ

=

it s

i the time it takes for

to reach of its

initial valu

( )

e

1/

0.

368

.

b

b a

This spontaneous emission formula gives the

transition rate for

regardless of any

other allowed state

N t

e

s.

ψ

≈

(39)

evidently the number remaining in the excited

state decreases exponentially, with a time constant

1

A

τ

=

We call this the lifetime of the state ---- technically,

it s

i the time it takes for

to reach of its

initial valu

( )

e

1/

0.

368

.

b

b a

This spontaneous emission formula gives the

transition rate for

regardless of any

other allowed state

N t

e

s.

ψ

≈

(40)

1 2 3

Typically, an excited atom has many different decay

modes ( that is,

can decay to a large number of

different lower-energy states,

). In that

case the transition rates

, and the

,

net

,

b

a a a

add

ψ

ψ ψ ψ

1

lifetime is

τ

=

' '

1 2 3

' '

, 1 , 1

1

| |

(

While

where is the natural frequency of the oscillator.

)

2

n n n n

A

n x n

n

m

τ

δ

ω

− −

=

+

(41)

'

, 1

ˆ

But we're talking about , so

must be than ; for our purpose, then,

Evidently transitions occur only to states one step lower on the "ladder",

2 n n

n

emission

low

q l

er

m

ω

δ

−

℘ =

and the frequency of the photon

'

( 1 / 2) ( 1 / 2)

Not surprisingly, the sys

emitted is

tem radiates at the classical

oscillator frequency.

( )

En En n n

n n

ω

− ₊ ₋ ₊

= =

(42)

2 2

3 0

But the transition rate is

and the lifetime of the

stationa

6

ry state is

th

nq

A

mc

n

mc

ω

πε

τ

=

₂ ₂

Meanwhile, each radiate photon carries an energy

n

nq

τ

ω

=

2 2 3 0

,

so the

radiated is

:

(43)

2 2 3 0

(

1)

1

(

)

6

2

This

or, since the ene

is the average po

rgy of an oscillator

in the

st

wer radiated by a

ate is

,

th

quntum

n

E

n

q

P

E

c

m

ω

πε

=

+

=

−

oscillator with (initial) energy

.

For comp

E

2 2 3 0

arison, let's determine the average power

radiated by a

oscillator with the same energy

According to

:

6

classical

Larmor formu

a

c

a

q

l

P

πε

(44)

0 0 2 0 2 4 0 3 0

, ( ) cos( t)

co

For a harmonic oscillator with amplitude

, and the acceleration is

. Averaging over a full cycle,

then

s( t)

12

x x t x

a x q x P c

ω

πε

= = − = 0

But the energy of th 2 ₀2

2 2 0 2 4 0 3 0

This is t

(1

he average power radiated by a

oscillator with (initial)

e oscillator is ,

so ,and herenc

energy .

/ 2) 2 / 6

E m x

x E m

(45)

The calculation of spontaneous emission rates has

been reduced to a matter of evaluating matrix elements

of the form

9.3.3 Sele

ctive Rules

| |

b

r

a

ψ

Suppose we are i

b a

nterested in systems like hydrogen, for

which the Hamiltonian is spherically symmetrical. In that

case we may specify the states with the usual quantum

numbers

n,l,

and , and the matrix elements

m

' ' '

are

(46)

Consider first the commutations of with

and

w

hich we worked out in

Chap

:

,

[

]

, [

]

From the thi

,

rd

[

ter

]

0.

4

z

z z z

'

L

x, y,

z

L ,x

i y

L , y

i x

L ,z

a) Selection rules involving m and m :

=

of these it follows that

' ' '

' ' ' ' '

' ' ' '

| [

] |

| (

) |

| [(

)

(

)) |

(

|

)

|

_z

z z

0

n l m

L ,z nlm

n l m

L z

zL

nlm

n l m

m

z

z m

nlm

m

n l m z nlm

=

−

=

−

(47)

' ' '

So unless

, the matrix elements of are

always zero.

(1

Either

'

, o

r else

| |

=0.

)

'

m

n l m z nlm

m

z

Conclusion :

=

' ' ' ' ' '

Meanwhile, from the commutator of with

we ge

| [

] |

| (

t

)

z

z z z

L

x

n l m

L ,x nlm

=

n l m

L x xL

−

' ' ' ' ' ' '

|

(

)

| |

nlm

m

n l m x nlm

i

n l m y nlm

(48)

' ' ' ' ' ' '

' ' ' ' ' '

' ' ' ' ' ' '

(

)

|

| [

Finally, the commutator of

with yields

] |

| (

) |

(

)

|

z

z z z

m

n l m

x

Conclus

nlm

i n l m

y nlm

L

y

n l m

L , y

nlm

n l m

L y

yL

nlm

m

n l m

y nlm

i

n l m

i

x

o

nlm

n :

−

=

−

=

−

=

' ' ' '

(

)

|

m

|

Conc

m

lusio

n l m

y nlm

n :

−

' ' '

' 2 ' ' ' ' ' ' '

' ' '

|

(

)

|

(

)

|

an

|

d

i n l m

x nlm

m

n l m

x nlm

i m

m

n l m

y nlm

n l m

x nlm

=

−

=

−

(49)

' 2

' ' ' ' ' '

and hence

either

or else

From Equations (1) and (2), we obtain the selection

rule

(

)

1,

|

0

No transitions occur

for :

(

unless

1 or 0

2)

m

n l m

x nlm

n l m

y nlm

m

−

=

∆

= ±

No transitions occur

This is an easy

unless

1 or 0

r

m

∆

= ±

esult ot understand if you remember

that the photon carries spin 1, and henc its value of

is 1, 0, or -1; conservation of angular momentum

requires that the atom give up whatever the photon

takes

m

(50)

2 2 2 2 2

' ' '

The commutation relation:

As before, we sandwich this commutation between

and to derive the sele

[ [ , ]] 2 ( )

ctio

'

L , L r rL L r

n l m nlm

b) Selection rules involving l and l :

= +

' ' ' 2

| [ ,[ ]] |

n rule:

n l m' ' ' L2 L ,r nlm

2 ' ' ' 2 2

4 ' ' ' ' '

' ' ' 2 2 2 2

4 ' ' 2 ' ' '

| [ ,[ ]] |

2 | ( ) |

2 [ ( 1) ( 1)] | |

| ( [ , ] [ , ] ) |

[ ( 1) ( 1)] | |

z

n l m L L ,r nlm

n l m rL L r nlm

l l l l n l m r nlm

n l m L L r L r L nlm

l l l l n l m r nlm

= −

= + + +

= −

(51)

' ' ' ' 2

' ' '

' ' ' ' '

2[ (

1)

(

1)]

[ (

1)

(

1)]

|

Either

or else

But

=0

[ (

1)

(

1)]=(

1)(

1)

:

l l

n l m r nlm

l l

l

Conclusion

+

=

+

−

+

−

+

+ +

+

' ' ' ' 2 ' 2

' ' 2 ' 2

and

so

2[ (

1)

(

1)]=(

1)

(

)

1

(

1)

(

)

1=0

l l

l

+

+ +

+

−

+ +

+

−

'

The first factor cannot be zero, so the condition

(52)

Thus we obtain the selection rule for :

Again, this result is easy to interpret: The

No transitions occur unless

1

l

∆ = ±

Again, this result is easy to interpret: The

photon carries spin 1, so the rules for addition

of angular momen

' '

'

1,

tum woul

1

d

or

.

l

= +

=

(53)