量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 9
Time-dependent
perturbation theory
Chapter 9
Time-dependent perturbation theory
9.1 Two
Two--level systems
9.2 Emission and absorption
of radiation
9.3 Spontaneous emission
9.1 TwoTwo-level systems
9.2 Emission and absorption
of radiation
9.3 Spontaneous emission
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:
H0ψ a = Eaψ a and
H0ψ b = Ebψ b
ψ a |ψ b = δab
Any state can be expressed as a linear combination of them
ψ (0) = caψ a + cbψ b
In the absence of any perturbation,each component evolves with its
characteristic exponential factor:
ψ (t ) = caψ a e
2
− iEa t / h
+ cbψ b e
2
| ca | + | cb | = 1
− iEa t / h
9.1.1 The perturbed system
Now suppose we turn on a timetime-dependent perturbation,
then the wave function can expressed as follows:
ψ (t ) = ca (t )ψ a e
− iEa t / h
+ cb (t )ψ b e
− iEa t / h
If the particle started out in the state ψ a We solve for
ca (t ) and cb (t ) by demanding that ψ (t) satisfy the tim-dependent
SchrÖdinger equation
∂ψ
Hψ = ih
, where
∂t
So, we find that
H = H 0 + H ' (t )
− iEa t / h
'
ca [ H 0ψ a ]e − iEa t / h + cb [ H 0ψ b ]e − iEb t / h + ca [ H ψ
]
e
a
'
+ cb [ H ψ b ]e
− iEb t / h
 iE a
+ caψ a  −
 h
− iE a t / h
− iEb t / h
&
&
= ih[ caψ a e
+ cbψ b e
 − iEa t / h
 iEb
+ cbψ b  −
e

 h
 − iEb t / h
]
e

The first two terms on the left cancel the last two terms on
the right, and hence
'
c a [ H ψ a ]e
− iE a t / h
= i h [ c& aψ a e
− iE a t / h
'
+ c b [ H ψ b ]e
+ c& bψ b e
− iE b t / h
− iE b t / h
]
Take the inner product with ψ a , and exploit the orthogonality
of ψ a and ψ b
For short, we define
'
ij
'
H ≡ ψ i | H |ψ
j
Note that the Hermiticity of H ' entails H 'ji = ( H 'ji ) * . M ultiplying
through by - (i / h ) e iE a t / h , we conclude that
i
&ca = − [ca Haa' + cb Hab' e−i ( Eb −Ea )t / h ]
h
Similarly, the inner product with ψ b picks out c&b :
i
c&b = − [cb Hbb' + ca Hba' e−i ( Eb −Ea )t / h ]
h
Typically, the diagonal matrix elements of H' vanish:
H
'
aa
= H
'
bb
=0
In that case the equations simplify
i
c& a = −
H
h
i
c& b = −
H
h
'
ab
e
− iω 0 t
cb
(1)
'
ba
e
− iω 0 t
cb
where
ω
0
Eb − E
≡
h
a
We'll assume that E b ≥ E a , so ω0 ≥ 0
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 H ' is "small", we can solve
Equation (1) by a process of successive approximations, as
follows. Suppose the particle starts out in the lower state:
ca (0) = 1,
cb (0) = 0
0..
If there were no perturbation at all , they would stay this way
foreever
Zeroth Order:
c
(0)
a
( t ) = 1,
c
(0)
b
= 0.
To calculate the first-order approximation, we insert these values
on the right side of Equation (1)
First Order:
dca
= 0 ⇒ ca(1) = 1;
dt
dcb
i ' iω0t
i t ' ' iω0t ' '
(1)
= − H ba e
⇒ cb = − ∫ H ba (t )e dt
dt
h
h 0
Now we insert these expressions on the right to obtain the
second-order approximation:
Second-Order:
dca
i ' −iω0t i t ' '' iω0t '' ''
= − H ab e ( h ) ∫ H ba (t )e dt ⇒
0
dt
h
'
t
t
'
1
− iω0t 
iω0t ''
(2)
'
'
'
''
'' 
'
ca (t ) = 1 − 2 ∫ H ab (t )e
H
(
t
)
e
dt
dt
ba
∫


0
0
h

Notes:
Notice that in my notation ca(2) (t ) includes the zeroth order
term; the second-order correction would be the integral term
alone.
In principle, we could continue this ritual indefinitely,
always inserting the nth − order approximation into the
right side of Equation (1) and solving for the (n +1)
th
−order. Notice that ca is modified in every even order,
and cb in every odd order.
9.1.3 Sinusoidal Perturbation
Suppose the perturbation has sinusoidal time dependence:
v
v
H ' ( r , t ) = V ( r ) cos(ω t )
so that
where
'
H ab
= Vab cos(ω t ),
Vab = ψ a | V | ψ b
To first order we have
t
i
iω0 t ' dt '
'
cb (t ) ≅ − Vba ∫ cos(ω t )e
0
h
iVba t  i (ω0 +ω ) t '
i ( ω 0 −ω ) t ' 
'
+
=e
e
dt

2h ∫0 
Vba  e i (ω0 +ω ) t − 1 e i (ω0 −ω ) t − 1 
+
=
2h  ω 0 + ω
ω 0 − ω 
This is the answer, but it is a little cumbersome to work with.
Dropping the first term:
We assume
ω0 +ω >>| ω0 −ω | . so
i(ω0 −ω)t /2
Vba e
i(ω0 −ω)t /2
−i(ω0 −ω)t /2


cb (t) ≅ −
e
e
−

2h ω0 −ω 
Vba sin[(ω0 −ω)t /2] i(ω0 −ω)t /2
=−i
e
h
ω0 −ω
The transition probability------the probability that a particle which
started out in the state ψa will be found, at time, in the state ψb :
2
2
| Vab | sin [(ω0 −ω)t /2]
Pa→b (t) =| cb (t)| ≅ 2
h
(ω0 −ω)2
2
(2)
As a function of time, the transition probability oscillates
sinusoidally.
P(t)
 | V ab | 


h
(
ω
−
ω
)
0


2π
| ω0 − ω |
4π
| ω0 − ω |
6π
| ω0 − ω |
Fig. 1 Transition probability as a function of time,
for a sinusoidal perturbation
2
t
The probability of a transition is greatest when the
driving frequency is close to the “natural”
frequency
P(t)
(ω0 − 2π t )
ω0
(ω0 + 2π t )
ω
Fig. 2 Transition probability as a function of driving frequency
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
exposed to a sinusoidally oscillating electric field
v
E = E cos(ω t ) kˆ
0
The perturbing Hamiltonian is
H ' = − qE0 z cos(ω t )
where q is the charge of the electron. Evidently
H ba' = −℘ E0 cos(ω t ), where ℘=q ψ b | z | ψ a
Chapter 9 TIME-DEPENDENT
PERTURBATION THEORY
9.1 TwoTwo-level systems
9.2 Emission and absorption
of radiation
9.3 Spontaneous emission
An electromagnetic wave consists of transverse
oscillating electric and magnetic fields.
Electric field
An electromagnetic
wave
v
E
v
u
x
v o
H
Magnetic field
Typically, ψ is an even or odd function of z; in
2
either case z|ψ | is odd , and integrates to zero.
This licenses our usual assumption that the
'
diagonal matrix elements of H vanish. Thus
the interaction of light with matter is governed
by precisely the kind of oscillatory perturbation
with
Vba = −℘E0
9.2.2 Absorption, Stimulated Emission,
and Spontaneous Emiss i on
If an atom starts out in the "lower" state ψ a , and you
shine a polarized monochromatic beam of light on it,
the probabilit y of a transition to the "upper" state ψ b
is given by Equation (2), which takes the form
2
2
|
℘
|
E
sin
[ (ω 0 − ω ) t / 2]

0 
Pa → b ( t ) = 

2
2
h
(
−
)
ω
ω


0
Of course, the probability of a transition down to
to the lo wer level:
2
2
|
℘
|
E
sin
[(ω 0 − ω )t / 2]

0 
Pb → a (t ) = 

(ω 0 − ω )
 h 
Three ways in which light interacts with atoms:
(a) Absorption
In this process, the atom absorbs energy
E b - E a = h ω 0 from the electromagnetic field.
we say that i t has "absorbed a photon"
(a) Absorption
(b) Stimulated emission
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 a transition is exactly
the same as for a transition upward from the lower state.
which process, which was first discovered by Einstein,
is called stimulated emission.
(b) Stimulated emission
(c) Spontaneous emission
If an atom in the excited state makes a transition
downward, with the release of a photon but without
any applied electromagnetic field to initiate the process.
This process is called spontaneous emission.
(c)
Spontaneous emission
9.2.3 Incoherent Perturbation
The energy density in an electromagnetic wave is
u=
ε0
E02
2
where E is the amplitude of the electric field. So the
transition probability
probability is proportional to the energy
density of the fields:
2
sin
[(ω0 − ω )t / 2]
2u
2
Pa →b (t ) =
|℘ |
2
2
ε 0h
(ω0 − ω )
This is for monochromatic perturbation, consisting of
a single frequency ω .
In many applications the system is exposed to electromagnetic
waves at a whole range of frequencies; in that case u → ρ (ω )dω,
and the net transition probablity takes the form of an integral:
2u
2
Pa→b (t ) =
|
℘
|
ε 0h 2
∫
∞
0
sin 2 [(ω0 − ω )t / 2]
dω
ρ (ω )
2
(ω0 − ω )
replace ρ (ω ) by ρ (ω0 ) and take it outside the integral, we get
2
∞ sin [(ω − ω )t / 2]
2 |℘|2
0
ρ
ω
Pa→b (t ) ≅
(
)
dω
0 ∫0
2
2
ε 0h
(ω0 − ω )
Changing variables to x ≡ (ω0 − ω ) / 2, extending the limits of
integration to x = ±∞, and looking up the define integral
sin 2 x
∫-∞ x2 dx = π
∞
we find
2 |℘|2
Pa→b (t ) ≅
ρ (ω0 )t
2
ε 0h
This time the transition probability is proportional ot t.
When we hit the system with an incoherent spread of
frequencies. The transition rate is now a constant :
Rb→a =
π
ε 0h
2
|℘|2 ρ (ω0 )
Note : the perturbing wave is coming in along the
x-direction and polarized in the z -direction.
But we shall be interested in the case of an atom
bathed in radiation coming from all direction,
and with all possible polarizations. What we
2
2
need, in place of |℘| ,is the average of | nˆ ⋅℘ | ,
where
v
℘ ≡ q ψ b | r |ψ a
and the average is over both polarizations (nˆ ) and
over all incident direction. This averaging can be
carried out as follows:
Polarization:
For propagation in the z-direction, the two possible
polarizations are iˆ and ˆj, so the polarization average
is
1 ˆ 2 ˆ 2
(nˆ ⋅℘) = [(i ⋅℘) + ( j ⋅℘) ]
2
1 2
1 2 2
2
= (℘x +℘y ) = ℘ sin θ
2
2
where θ is the angle between ℘ and the direction of
propagation.
2
p
Propagation direction:
Now let's set polar axis along ℘ and integrate over all
propagation directions to get the polarization-propagation
average:
1 1 2 2 
(nˆ ⋅℘) =
℘ sin θ  sin θ dθ dφ
∫

4π  2

2
2
π
℘
℘
3
=
sin θ dθ =
∫
4 0
3
So the transition rate for stimulated emission from state
b to state a, under the influence of incoherent, unpolarized
light incident from all directions, is
2
pp
So the transition rate for stimulated emission from
state b to state a, under the influence of incoherent,
unpolarized light incident from all directions, is
π
2
Rb→a =
|℘| ρ(ω0 )
2
3ε0h
Where ℘ is the matrix element of the electric
dipole moment between the two states and ρ(ω0 )
is the energy density in the fields, per unit frequency,
evaluated at ω0 = (Eb − Ea ) / h.
Chapter 9 TIME-DEPENDENT
PERTURBATION THEORY
9.1 TwoTwo-level systems
9.2 Emission and absorption
of radiation
9.3 Spontaneous emission
9.3 Spontaneous emission
9.3.1 Einstein's A and B coefficients
Picture a container of atoms, N a of them in the lower
state (ψ a ), and N b of them in the upper state (ψ b ).
Let A be the spontaneous emission rate, so th a t the
number of particle leaving the upper state by this
process, per unit time, is N b A . The transition rate for
stimulated emission is proportional to the energy
density of the electroma gnetic field---call it Bba ρ (ω 0 ).
The number of particles leaving the upper state by this
mechanism, per unit time, is N b Bba ρ (ω 0 ).
The absorption rate is likewise proportional to
ρ (ω0 )---call it Bab ρ (ω0 ); The number of particles
per unit joining the upper level is therefore
N a Bab ρ (ω0 ). All told, then,
dN b
= − N b A − N b Bba ρ (ω0 ) + N a Bab ρ (ω0 )
dt
Suppose that these atoms are in thermal equilibrium
with the ambient field, so that the number of particle
in each level is constant.
In that case
dN b / dt = 0, and it follows that
A
ρ (ω 0 ) =
( N a / N b ) Bab − Bba
The number of particles with energy E , in thermal
equilibrium at temperature T , is proportiona l to
t he Boltzmann fac to r , so
N a e − Ea / k B T
= − Ea / k B T = e hω 0 / k BT
Nb e
and hence
A
ρ (ω 0 ) = hω 0 / k B T
e
Bab − Bba
But Planck's blackbody formula
tells us the energy density of thermal radiation:
h
ω3
ρ (ω ) = 2 3 hω / kbT
π c e
− Bba
Comparing the two expressions, we conclude that
3
B ab = Bba
and
an d
and
ω h
A = 2 3 Bba
π c
π
2
|
℘
|
Bba =
3ε 0 h 2
and it follows that the spontaneous emission rateis
ω 3℘2
A=
3πε 0 h c 3
9.3.2 The Lifetime of an Excited S tate
Suppose, now, that you have a bottle full of
atoms, with N b (t ), of them in the excited state.
As a result of spontaneous emission, this number
w ill decrease
ecrease as time goes on; specifically, in a
time interval dt you will lose a fraction of Adt
them:
dN b = − AN b dt
Solving for N b (t ), we find
N b (t ) = N b (0)e
− At
evidently the number remaining in the excited
state decreases exponentially, with a time constant
1
τ=
A
We call this the lifetime of the state ---- technically,
it is the time it takes for N b (t ) to reach of its
1/e ≈ 0.368 initial value.
This spontaneous emission formula gives the
transition rate for ψ b → ψ a regardless of any
other allowed states.
evidently the number remaining in the excited
state decreases exponentially, with a time constant
1
τ=
A
We call this the lifetime of the state ---- technically,
it is the time it takes for N b (t ) to reach of its
1/e ≈ 0.368 initial value.
This spontaneous emission formula gives the
transition rate for ψ b → ψ a regardless of any
other allowed states.
Typically, an excited atom has many different decay
modes ( that is, ψ b can decay to a large number of
different lower-energy states, ψ a1 ,ψ a 2 ,ψ a3 L). In that
case the transition rates add , and the net lifetime is
1
τ=
A1 + A2 + A3 + L
While
h
( n' δ n,n' −1 + nδ n' ,n−1 )
n| x|n =
2mω
where ω is the natural frequency of the oscillator.
'
But we're talking about emission, so n '
must be low er than n; for our purpose, then,
nh
℘= q
δ n , n' −1lˆ
2 mω
Evidently transitions occur only to states one step
lower on the "ladder", and the frequency of the photon
emitted is
En − En'
( n + 1/ 2) hω − ( n ' + 1/ 2) hω
=
ω=
h
h
= ( n ' − n )ω = ω
Not surprisingly, the system radiates at the classical
oscillator frequency.
But the transition rate is
nq 2ω 2
A=
6πε 0 mc 3
th
and the lifetime of the n stationary state is
6πε 0 mc 3
τn =
2 2
nq ω
Meanwhile, each radiate photon carries an energy hω ,
so the power radiated is Ahω:
q 2ω 2
P=
( nh ω )
3
6πε 0 mc
or, since the energy of an oscillator
in the nth state is E = (n + 1)hω ,
q 2ω 2
1
P=
( E − hω )
3
6πε 0 mc
2
This is the average power radiated by a quntum
oscillator with (initial) energy E.
For comparison, let's determine the average power
radiated by a classical oscillator with the same energy
According to Larmor formula:
2
2
qa
P=
6πε 0 c 3
For a harmonic oscillator with amplitude
x0 ,x (t ) = x0 cos(ω t), and the acceleration is
a = − x0ω 2 co s(ω t). Averaging over a full cycle,
then
q 2 x 0ω 4
P=
12πε 0 c 3
But the energy of th e oscillator is E = (1 / 2) mω 2 x02 ,
so x02 = 2 E / mω 2 ,and herenc
q 2 x 0ω 4
P=
E
3
6πε 0 mc
This is the average power radiated by a classical
oscillator with (initial) energy E .
9.3.3 Selective Rules
The calculation of spontaneous emission rates has
been reduced to a matter of evaluating matrix elements
of the form
v
ψ b | r |ψ a
Suppose we are interested 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 m, and the matrix elements are
' ' ' v
n l m | r | nlm
'
a) Selection rules involving m and m :
Consider first the commutations of Lz with x, y, and z,
which we worked out in Chapter 4:
[Lz ,x] = ihy,
[Lz , y] = ihx,
[Lz ,z] = 0.
From the third of these it follows that
0 = n'l 'm' |[ Lz ,z]| nlm
= n'l 'm' | ( Lz z − zLz ) | nlm
' '
'
'
'
= n l m |[(m h) z − z(m h)) | nlm
'
' '
'
= (m − m)h n l m | z | nlm
Conclusion :
'
' '
'
Either m = m, or else n l m | z | nlm =0. (1)
'
So unless m = m, the matrix elements of z are
always zero.
Meanwhile, from the commutator of Lz with x
we get
' '
'
' '
'
n l m |[ Lz ,x]| nlm = n l m | ( Lz x − xLz ) | nlm
'
' '
'
' '
'
= (m − m)h n l m | x | nlm = ih n l m | y | nlm
Conclusio n :
( m ' − m ) n 'l ' m ' | x | nlm = i n 'l ' m ' | y | nlm
Finally, the commutator of Lz with y yields
n 'l ' m ' | [ Lz , y ] | nlm = n 'l ' m ' | ( Lz y − yLz ) | nlm
= ( m ' − m )h n 'l ' m ' | y | nlm = ih n 'l ' m ' | x | nlm
Conclusio n :
( m ' − m ) n 'l ' m ' | y | nlm = i n 'l ' m ' | x | nlm
and
( m ' − m ) 2 n 'l ' m ' | x | nlm = i ( m ' − m ) n 'l ' m ' | y | nlm
= n 'l ' m ' | x | nlm
and hence
'
2
either ( m − m ) = 1, or else
n 'l ' m ' | x | nlm = n 'l ' m ' | y | nlm = 0
( 2)
From Equations (1) and (2), we obtain the selection
rule for m:
No transitions occur unless ∆m = ±1 or 0
This is an easy result ot understand if you remember
that the photon carries spin 1, and henc its value of
m is 1, 0, or -1; conservation of angular momentum
requires that the atom give up whatever the photon
takes away.
b) Selection rules involving l and l' :
The commutation relation:
2
2 v
2 v 2
2v
[L ,[ L , r ]] = 2h ( rL + L r )
As before, we sandwich this commutation between
n 'l ' m ' and nlm to derive the selection rule:
v
' ' '
2
n l m | [ L ,[ Lz ,r ]] | nlm
v 2
2
' ' '
2v
= 2h n l m | ( rL − L r ) | nlm
4
' '
' ' ' v
= 2h [l (l + 1) + l (l + 1)] n l m | r | nlm
' ' '
2
2 v
2 v
= n l m | ( L [ L , r ] − [ L , r ]L2 ) | nlm
4 ' '
2
' ' ' v
= h [l (l + 1) + l (l + 1)] n l m | r | nlm
Conclusion :
Either 2[l (l + 1) + l ' (l ' + 1)] = [l ' (l ' + 1) − l (l + 1)]2
' ' ' v
or else n l m | r | nlm =0
But
'
'
'
'
'
[l (l + 1) − l (l + 1)]=(l + l + 1)(l + 1)
and
2[l (l + 1) + l ' (l ' + 1)]=(l ' + l ' + 1) 2 + (l ' − l ) 2 − 1
'
'
2
'
2
so (l + l + 1) + (l − l ) − 1=0
The first factor cannot be zero, so the condition
simplifies to l ' = l ± 1.
Thus we obtain the selection rule for l:
No transitions occur unless ∆l = ±1
Again, this result is easy to interpret: The
photon carries spin 1, so the rules for addition
'
'
of angular momentum would l = l + 1, l = l or
'
l = l − 1.
Allowed decays for the first four Bohr levels in
hydrogen