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量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal UniversityShaanxi Normal University
Chapter 9
Time-dependent
perturbation theory
9.1 Two9.1 Two--level systemslevel systems
Chapter 9
Time-dependent perturbation theory
9.1 Two9.1 Two--level systemslevel systems
9.2 Emission and absorption9.2 Emission and absorption
of radiationof radiation
9.3 Spontaneous emission9.3 Spontaneous emission
9.1 Two9.1 Two--level systemslevel systems
9.2 Emission and absorption9.2 Emission and absorption
of radiationof radiationof radiationof radiation
9.3 Spontaneous emission9.3 Spontaneous emission
9.1 Two-level systemsSuppose that there are just two states of the system. They are
eigenstates of the unperturbed Hamiltonian,and they are eigenstates
orthonormal:
0 0and
|
a a a b b b
a b ab
H E H Eψ ψ ψ ψ
ψ ψ δ
= =
=
(0)a a b b
c cψ 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 aiE t iE t
a a b b
a b
t c e c e
c c
ψ ψ ψ− −
2 2
= +
| | + | | =
h h
9.1.1 The perturbed system
Now suppose we turn on a timeNow 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:
/ /( ) ( ) ( )a aiE t iE t
a a b bt c t e c t eψ ψ ψ− −= +h h
If the particle started out in the stateIf the particle started out in the statea
ψ We solve forWe solve fora We solve forWe solve for
( ) and ( ) by demanding that (t) satisfy the tim-dependent
Schr dinger equation
a bc t c t ψ
Ö
, H it
ψψ
∂=
∂h
'( )0H = H H t+where where
So, we find that
/ / /'
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 ta ba a b b
c H e c H e c H e
c H e i c e c e
iE iEc e c e
ψ ψ ψ
ψ ψ ψ
ψ ψ
− − −
− − −
− −
+ +
+ = +
+ − + −
h h h
h h h
h h
& &h
h h
The first two terms on the left cancel the last two terms on
the right, and hencethe 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 ψ
ψ ψ
− −
− −
+
= +
h h
h h& &h
a
a b
Take the inner product with , and exploit the orthogonality
of and
ψ
ψ ψ
For short, we define
' '| |ij i j
H Hψ ψ≡
( ) /' '[ ]b ai E E tic c H c H e
− −= − + h&
' ' *
/
Note that the Hermiticity of entails ( ) . Multiplying
through by ( / ) , we conclude thata
'
ji ji
iE t
H H H
- i h e
=
h
( ) /' '[ ]b ai E E t
a a aa b abc c H c H e− −= − + h
&h
bSimilarly, the inner product with picks out :bcψ &
( ) /' '[ ]b ai E E t
b b bb a ba
ic c H c H e
− −= − + h&
h'Typically, the diagonal matrix elements of H vanish:
' ' 0aa bbH H= =
In that case the equations simplify
0
0
'
'
i t
a a b b
i t
b b a b
ic H e c
ic H e c
ω
ω
−
−
= −
= −
&h
&h
(1)(1)
h
where
0b a
E Eω
−≡
h
b a 0We'll assume that E E , so 0ω≥ ≥
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:
0.a bc c= =
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= =
To calculate the first-order approximation, we insert these values
on the right side of Equation (1)
First Order:
'0 0
(1)
' (1) ' ' '
0
Now we insert expressions on the right to obtain the
0 1;
( )
aa
ti t i tb
ba b ba
dcc
dt
d
these
c i iH e c H t e dt
dt
ω ω
⇒= =
= − =⇒ − ∫h h
second-order approximation:
S
( )''
0 0
'' ''
0 0
' ' '' ''
0
(2) ' ' ' '' '' '
2 0 0
( )
1
econd-Order:
( ) 1 (
) )
(
ti t i ta i
ab ba
t ti t i t
a ab ba
dc iH e H t e dt
dt
c t H t e H t e dt dt
ω ω
ω ω
−
−
⇒= −
= −
∫
∫ ∫
hh
h
Notes:(2)Notice that in my notation ( ) includes the zeroth order
term; the second-order correction would be the integral term
alone.
ac t
In principle, we could continue this ritual indefinitely,
th
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 modifieda
n
n
c
−
+
− in every order,
and in every order.b
even
c odd
9.1.3 Sinusoidal Perturbation9.1.3 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
=
=
=
v v
' '
0
' '0 0
0 0
'
0
( ) ( ) '
0
( ) ( )
0 0
( ) cos( )
This is the ans
2
1 1
wer, but
=2
i
t
i
ti t dt
b ba
ti t i tba
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
+ −
+ −
≅ −
+
− −+
+ −
∫
∫
h
h
h
s a little cumbersome to work with.
Dropping the first term:
0
0 0
0
0 0
( ) /2( ) /2 ( ) /2
0
( ) /20
We assume | |
( )2
sin[( ) /2]
. s
o
i ti t i tba
b
i tba
V e c t e e
V ti e
ω ω
ω ω ω ω
ω ω
ω ω ω ω
ω ω
ω ω
−− − −
−
+ >> −
≅− − −
−=−
h
00
0
transition probaThe ------the proba
bility
b
bai eω ω
=−−h
2 22 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) ( )
aba b b
a b
V tP t c t
ψ
ω ω
ψ
ω ω→
−= ≅
−h
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
| |
( )
abV
ω ω
− h
t
Fig. 1 Transition probability as a function of time,
for a sinusoidal perturbation
The probability of a transition is greatest when the
driving frequency is close to the “natural”
frequency
P(t)
0ω0( 2 )tω π− 0( 2 )tω π+ ω
Fig. 2 Transition probability as a function of driving frequency
9.2 Emission and absorption
of radiation9.2.1 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
0
'
0
exposed to a sinusoidally oscillating electric field
The perturbing Hamiltonian is
where is the charge of th
ˆcos( )
cos(
e
electr
)
E = E t k
H q
q
E z
t
ω
ω= −
v
'
0 cos(
on. Ev
),
idently
where = | |ba b a
H E t q zω ψ ψ= −℘ ℘
�� 9.1 Two9.1 Two--level systemslevel systems
�� 9.2 Emission and absorption9.2 Emission and absorption
Chapter 9 TIME-DEPENDENT PERTURBATION THEORY
Chapter 9 TIME-DEPENDENT PERTURBATION THEORY
�� 9.2 Emission and absorption9.2 Emission and absorption
of radiationof radiation
�� 9.3 Spontaneous emission9.3 Spontaneous emission
An electromagnetic wave consists of transverse
oscillating electric and magnetic fields.
An electromagnetic
wave
vEv
An electromagnetic
wave
vEv
Electric field
uv
E
Hv x
o uv
E
Hv x
o
Magnetic field
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
ψ
0
the interaction of light with matter is governed
by precisely the kind of oscillatory perturbation
with
baV E= −℘
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
aψ
2
a transition to the "upper" state
is given by Equation (2), which takes the form
bψ
2 2
0 0
2 2
0
| | sin [
( ) / 2]
( )( )
a b
E tP t
ω ω
ω ω→
℘ − =
− h
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
tP t
ω ω
ω ω→
℘ − =
− h
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 aE - E ω= h
(a) Absorption
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,
stimuli ated es ca misslled ion.
(b) Stimulated emission
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
spontaneess is c ous emisalled sion.
(c) Spontaneous emission
200
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 = Eε
portional to the energy transition probability is pro
22 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
tuP t
monochromatic
ω ω
ε ω ω
ω
→
−= ℘
−h
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
2 200 0
0replace ( ) by ( ) and take it outside the integral, we
sin [( ) / 2]2| | ( )
( )
get
tud
ω ωρ ω ω
ε ω
ρ
ω
ωρω
∞ −℘
−∫h
22
002 20
0
0
0
0
replace ( ) by ( ) and take it outside the integral, we
sin [( ) / 2]2 | |( ) ( )
(
get
Ch
)
(anging variables to , exten) / 2
a b
tP t d
x
ρ
ω ωρ ω ω
ε
ω
ω
ρ
ω
ω
ω
ω
∞
→
−℘≅
−
≡ −
∫h
2
2-
ding the limits of
integration to , and looking up the define integral
sin
x
xdx
xπ
∞
∞
= ±∞
=∫
2
02
0
we find
This time the transition probability is proportional ot .
When we hit the system with an incoherent spread of
2
frequencies. The is
| | ( ) ( )
transition rat noe
a bP t t
t
ρ ωε
→
℘≅
h
w a constant :frequencies. The is transition rat noe
2
02
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 :
πρ ω
ε→ = ℘
h
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 f |o| e
all
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 aq r
n
ψ ψ℘≡v
2 2
For propagation in the z-direction, the two possible
polarizations are andˆ ˆ
1 ˆ ˆˆ (
Polarization:
, so
) [( ) ( ) ]
the polarization average
is
2
i j
n = i j⋅℘ ⋅℘ + ⋅℘2 2
2
1 ˆ ˆˆ ( ) [( ) ( ) ]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.
yθ
θ
= ℘
℘
2 2
Now let's set polar axis along and integrate over all
propagation directions to get the polarization-prop
1 1ˆ (
Propagation direction:
) sin sin4 2
agation
average:
2
ppn = d dθ θ θ φ
π
℘
⋅℘ ℘
∫4 2
ppπ ∫
2 23
0
So the transition rate for stimulated emission from state
to state , under the influence of incoherent, unpolarized
light i
= sin4 3
ncident from all directions, is
d
b a
π
θ θ℘ ℘
=∫
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
Rπ
ε→ = ℘
h
2
0( )ρ ω2
0
||3
b aR
ε→ = ℘
h0
0
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
( )
( )
( ) / .b a
E E
ρ ω
ρ ω
ω
℘
= − h
�� 9.1 Two9.1 Two--level systemslevel systems
�� 9.2 Emission and absorption9.2 Emission and absorption
Chapter 9 TIME-DEPENDENT PERTURBATION THEORY
Chapter 9 TIME-DEPENDENT PERTURBATION THEORY
�� 9.2 Emission and absorption9.2 Emission and absorption
of radiationof radiation
�� 9.3 Spontaneous emission9.3 Spontaneous emission
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
N
A
ψ 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
bN A
0
0
gnetic field---call it .
The number of particles leaving the upper state by this
mechanism, per unit time, i
s .
( )
( )
ba
b ba
B
N B
ρ ω
ρ ω
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 ρ ω
ρ ω0
a ab
bdN0 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 abN A N B N B
dt
constant
ρ ω ρ ω= − − +
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 Tk Ta
E k T
b
k T
ab ba
N ee
N e
A
B
t r
e
o
B
ω
ωρ ω
−
−= =
=−
h
h
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
bk T
bac e B
B B A
ω
ωρ ω
π=
−
=
h
h
3
Bω
=h
and ab ba
B B A=2 3
2
2
0
3 2
3
0
and
and it follows that the spontaneous emission rateis
|
3
|3
ba
ba
Bc
B
Ac
ω
ππ
ε
ω
πε
=
= ℘
℘=
h
h
h
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
bN t
crease as time goes on; specifically, in a w dill 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
N
= −
( ) (0) At
b bt N e
−=
evidently the number remaining in the excited
state decreases exponentially, with a time constant
We call this the lifetime of the state ---- technically,
1
Aτ =
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 ψ
≈
→
evidently the number remaining in the excited
state decreases exponentially, with a time constant
We call this the lifetime of the state ---- technically,
1
Aτ =
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 ψ
≈
→
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 L
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 A A
n x n n nm
τ
δ δω
ω
− −
=+ + +
= +
L
h
'
'
, 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 t
he "ladder",
2 n n
n
n
n
emission
low
q l
er
mδ
ω −℘ =
h
and the frequency of the photon
''
'
( 1/ 2) ( 1/ 2)
Not surprisingly, the sys
emitted is
tem radiates at the classical
oscillator frequen
cy.
( )
n n
E E n n
n n
ω ωω
ω ω
− + − += =
= − =
h h
h h
2 2
3
0
3
0
But the transition rate is
and the lifetime of the stationa
6
6
ry state is
th
n
nqA
mc
n
mc
ω
πε
πετ
=
=2 2
Meanwhile, each radiate photon carries an energy
nnq
τω
=
h
2 2
3
0
,
so the radiated is :
( )6
pow
qP
e A
mc
r
n
ω
ω
ωω
πε=
h
h
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
qP E
cm
ω
ωω
πε
= +
= −
h
h
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
Pπε
=
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 xP
c
ω
ω ω
ω
πε
=
= −
=0
But the energy of th 2 2
0
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
q xP E
mc
classical
E
ω
ω
ω
πε
=
=
=
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 a
rψ ψv
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 and , and the matrix elementsn,l, m
' ' '
are
| | nl m r nlmv
Consider first the commutations of with and
w Chaphich we worked out in :
,
[ ] , [ ]
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 :
= = =h h
of these it follows that
' ' '
' ' '
' ' ' ' '
' ' ' '
| [ ] |
| ( ) |
| [( ) ( )) |
( |
)
|
z
z z
0 nl m L ,z nlm
nl m L z zL nlm
nl m m z z m nlm
m m nl m z nlm
=
= −
= −
= −
h h
h
' ' '
So unless , the matrix elements of are
always zero.
Meanwhile, from the commutator of
(1
Either
with
, o | | =0. r else )'
'
m m nl m z nlm
m m z
L x
Conclusion :
=
=
' ' ' ' ' '
Meanwhile, from the commutator of with
we ge
|[ ] | | (
t
)
z
z z z
L x
nl m L ,x nlm nl m L x xL= −
' ' ' ' ' ' '
|
( ) | | | |
nlm
m m nl m x nlm i nl m y nlm= − =h h
' ' ' ' ' ' '
' ' ' ' ' '
' ' ' ' ' ' '
( ) | | | |
| [
Finally, the commutator of with yields
] | | ( ) |
(
) | | | |
z
z z z
m 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 m n l m y nlm i n l m
i
x
o
nlm
n :
− =
= −
= − =h h
' ' ' '( ) |
|m
Conc
m
lusio
n l m y nlm
n :
− ' ' '
' 2 ' ' ' ' ' ' '
' ' '
| |
( ) | | ( ) | |
an
| |
d
i n l m x nlm
m m n l m x nlm i m m n l m y nlm
n l m x nlm
=
− = −
=
' 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 m
n l m x nlm n l m y nlm
m
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
away.
2 2 2 2 2
' ' '
The commutation relation:
As before, we sandwich this commutation between
and to derive the sele
[ [ , ]] 2 ( )
cti o
'
L , L r rL L r
n l m nlm
b) Selection rules involving l and l :
= +v v v
h
' ' ' 2| [ ,[ ]] |
n rule:
n l m L L ,r nlmv' ' ' 2
2 ' ' ' 2 2
4 ' ' ' ' '
' ' ' 2 2 2 2
4 ' ' 2 ' ' '
| [ ,[ ]] |
2 | ( ) |
2 [ ( 1) ( 1)] | |
| ( [ , ] [ , ] ) |
[ ( 1) ( 1)] | |
zn 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
= −
= + + +
= −
= + + +
v
v vh
vh
v v
vh
' ' ' ' 2
' ' '
' ' ' ' '
2[ ( 1) ( 1)] [ ( 1) ( 1)]
| |
Either
or else
But
=0
[ ( 1) ( 1)]=( 1)( 1)
:
l l l l l l l l
n l m r nlm
l l l l l l l
Conclusion
+ + + = + − +
+ − + + + +
v
' ' ' ' 2 ' 2
' ' 2 ' 2
and
so
2[ ( 1) ( 1)]=( 1) ( ) 1
( 1) ( ) 1=0
l l l l l l l l
l l l l
+ + + + + + − −
+ + + − −
'
The first factor cannot be zero, so the condition
simplifies to 1.l l= ±
Thus we obtain the selection rule for :
Again, this result is easy to interpret: The
No transitions occur unless 1
l
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 l l l
l l
= + =
= −
Allowed decays for the first four Bohr levels in
hydrogen
