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Quantum Dynamics – Quick View
Concepts of primary interest:
The Time-Dependent Schrödinger Equation
Probability Density and Mixed States
Selection Rules
Transition Rates: The Golden Rule
Sample Problem Discussions:
Tools of the Trade
Appendix: Classical E-M Radiation
POSSIBLE ADDITIONS: After qualitative section, present the two state system, and then first and second order
transitions (follow Fitzpatrick). Chain together the dipole rules to get l = 2,0,-2 and m = -2, -1, … , 2. ??Where
do we get magnetic rules? Look at the canonical momentum and the Asquared term.
Schrödinger, Erwin (1887-1961) Austrian physicist who invented wave
mechanics in 1926. Wave mechanics was a formulation of quantum mechanics
independent of Heisenberg's matrix mechanics. Like matrix mechanics, wave
mechanics mathematically described the behavior of electrons and atoms. The
central equation of wave mechanics is now known as the Schrödinger
equation. Solutions to the equation provide probability densities and energy
levels of systems. The time-dependent form of the equation describes the
dynamics of quantum systems.
http://scienceworld.wolfram.com/biography/Schroedinger.html
© 1996-2006 Eric W. Weisstein
www-history.mcs.st-andrews.ac.uk/Biographies/Schrodinger.html
Quantum Dynamics: A Qualitative Introduction
Introductory quantum mechanics focuses on time-independent problems leaving
dynamics to be discussed in the second term. Energy eigenstates are characterized by
probability density distributions that are time-independent (static). There are
examples of time-dependent behavior that are by demonstrated by rather simple
introductory problems. In the case of a particle in an infinite well with the range
Contact: [email protected]
[ 0 < x < a], the mixed state below exhibits time-dependence.

i t
i t
 ( x, t )  1a sin  ax  e   sin  2a x  e 2


 ma  .


where n  n 2 
2
2
The probability density * for the function (x, t) has the form of a stationary piece
plus a piece that oscillates back and forth at the difference frequency 21 = 2 - 1.
This oscillation is perhaps the simplest example of quantum dynamics. According to
classical E&M, the system radiates light with the oscillation frequency if that
oscillating density is a charge density. More is to be said on this topic later.

Exercise: Find the probability density for  ( x, t )  1a sin  ax  eit  sin  2a x  ei2t

assuming that it applies for 0 < x < a and discuss its characteristics. Identify the
various time dependences.
Classically, Bohr’s orbiting electrons should radiate electromagnetic energy
continuously and spiral inward. Bohr postulated that electrons in his special orbits do
not radiate, but that they would radiate an electromagnetic energy chunk (a photon)
equal to the energy difference between allowed states when the electron in the
hydrogen atom made a transition between allowed orbits1. Before launching an attack
on quantum dynamics, the origin of classical electromagnetic radiation is to be
reviewed. A model for the radiation field can be found in Appendix I.
Classical Electromagnetic Radiation: Charges radiate when they are accelerated.
The radiation intensity varies as the square of the sine of the angle between the line of
sight direction and that of the acceleration. The radiation electric field is directed
1
Bohr postulated that, in the classical (large radius) limit, the radiated frequency would approach the orbital frequency.
This condition is consistent with the one given above, but its nature is not as quantum mechanical.
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oppositely to the component of the acceleration1 that is perpendicular to the line of
sight from the field point to the accelerated source charge so an analysis of the
acceleration provides information about the direction (polarization) of the electric
field in the radiation. The oscillating probability densities for charged particles in
mixed states correspond to charge moving and accelerating. Energy eigenstates
have static probability densities and, should not radiate in this semi-classical view.
Be warned: The flow of ideas for describing transitions between quantum states
rather than a careful, detailed development is to be presented. Normalizations, relative
sizes and numeric factors are omitted. Never use the equations in this handout to
calculate a value. That is: This entire handout should be regarded as a Meandering
Mind Segment.
Stationary and Mixed States: The governing equation for introductory quantum
dynamics is the time-dependent Schrödinger equation. 
i

t
 (r , t )  Hˆ  (r , t )   
2
 2m
 2  V (r )   (r , t )
[QMDyn.1]

Consider a state that is an eigenfunction of the Hamiltonian.

Hˆ  (r , t )  En  n (r , t )   i  n (r , t )
t
The wavefunction can be separated into temporal and spatial parts:  n (r , t )  un (r ) Tn (t )

Tn (t ) and hence
leading to the equations Hˆ un (r )  En un (r ) and En Tn (t )   i
t
 n (r , t )  cn un (r ) eint where n = -1 En and cn is a complex number usually of
magnitude 1. For any energy eigenstate, the probability density is stationary (timeindependent).
1
The accerlation is evaluated at the retrarded time, the time at which the radiation was emitted to be observed here now.
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 n (r , t )  n (r , t )  (un (r ) eint ) un (r ) eint  un (r )
2
The probability density is time independent so the particle described by the state is
not moving.
Exercise: What does it mean to say that a function is an eigenfunction of an operator?
What is true about the probability density for an eigenfunction of the hamiltonian?

A mixed state includes contributions from two or more energy eigenstates. A simple
mixed state might be of the form mixed (r , t )  a un (r ) eint  bum (r ) eimt with its
associated probability density
mixed (r , t )  a un (r )  b um (r )  ab un (r )um (r ) e i (m n )t  (ab un (r )um* (r ) e i (m n )t )
2
2
2
2
The probability density for mixed states has some time-independent components plus
components that oscillate at the difference frequencies,  |m - n|. An oscillating
probability density represents a particle that is accelerating.
Exercise: Show that if mixed (r , t )  a un (r ) eint  bum (r ) eimt is a mixed state of
un (r ) eint and um (r ) eimt which are eigenstates of the full Hamiltonian for the
problem, that the probabilities to find the particle with energies corresponding to the
states n or m are time-independent. We conclude that the system is not making stateto-state transitions.
That is: this mixed (r , t ) does not describe transitions or state evolution.
Note that expectation values of various operators in the state mixed can be time
dependent. For example, dx/dt may not be zero for the state mixed.
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
A general mixed state  (r , t )   am um (r ) ei t does not describe state-to-state
m
m 1
transitions. Transitions are described by coefficients ak that depend on time.
Schrödinger’s equation shows that this is not the case as long as the um (r ) are
eigenstates of the full hamiltonian.
State to State Transitions:
Mixed states with time independent probability densities are less interesting than the
cases in which an electron makes a transition from one quantum state to another.
Consider a quantum problem described by the Hamiltonian Ĥ 0 that has
eigenfunctions um (r ) ei t
m
 Hˆ 0 um (r ) eimt  Em um (r ) eimt .
(For definiteness, assume that the states describe the electron in the hydrogen atom.)
A general wavefunction for the problem is a mixed state expressed as a sum over the
eigenfunctions,   r , t )   cm um (r ) eimt . That is: the eigenfunctions form a complete
m
set and are an orthogonal basis for the space of all physical wavefunctions for the
problem. The functions are assumed normalized and orthogonal and hence satisfy the
relation:
all space ( uk (r ) e
ik t *
) um (r ) eimt dV  k m   mk .
Begin with the system in a particular pure eigenstate, say the state |n
 r , t )  un (r ) eint . The index m is a label representing anyone of the eigenstates.
At the initial time, cm = 0 except for cn = 1, or cm(t = to) = mn. At the initial time, there
is a 100% probability that a measurement will return a value consistent with the
system being in the state n. The time-dependent version of the Schrödinger equation
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states that: Hˆ 0  (r , t )  Hˆ 0 un (r ) eint  En un (r ) eint  i ddt . It follows that:
 (r , t  t )   (r , t )  ddt t   (r , t )  i Hˆ 0  (r , t )t  un (r ) e i t  i En un (r ) e i t t .
n

n

d
 (r , t )  dt t   i Hˆ 0  (r , t )t  un (r ) e int   i e int En t un (r )  in t un (r )
When the system is initially in a pure energy eigenstate n of the full Hamiltonian,
time development just adds in more of that same pure state, but ‘- i out of phase’. A
system in an eigenstate of the Hamiltonian just remains in that state. Continuously
adding a piece ‘-i out of phase’ just changes the complex phase of the function at a
d
rate:
dt
i

 1 Em un (r ) eint
un (r ) eint
  n . This result follows from the fact that the
hamiltonian operates on an eigenfunction to return a real constant times that same
function. This result means that the spatial form un remains fixed, and that every point
of the overall form is multiplied by the same complex phase variation, eint .
Exercise: Consider e-it. Show that eit  ei (t t )  eit  (i t ) eit 
Transitions between eigenstates of the base hamiltonian Ho are caused by interactions
which appear as perturbations, small terms added to the Hamiltonian to represent
external influences on the quantum system. The basic problem is described by the
unperturbed hamiltonian Ĥ 0 , and its eigen-solutions {… um (r ) ei t …} provide a
m
complete basis for expanding any well-behaved function defined over the same
region of space. The small interaction term Ĥ1 or perturbation is added such that the
full problem is described by Hˆ  Hˆ 0  Hˆ 1 , and the full time development equation is:
Hˆ  (r , t )   Hˆ 0  Hˆ 1   (r , t )  i ddt .


   (r , t  t )   (r , t )   i Hˆ  (r , t ) t   i  Hˆ 0  Hˆ 1   (r , t ) t

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
QMDyn- 6
With the assumption that the system starts at time t in state n of the unperturbed
Hamiltonian,
 (r , t  t )  un (r ) eint   i  Hˆ 0  Hˆ 1  un (r ) eint t


The piece  i Hˆ 0 un (r ) ei t t is understood and not very interesting as it has nothing
n
to do with transitions; it is just the phase of n changing at the rate - n. Examine the
new small piece:
 i Hˆ 1un (r ) eint t
What is Hˆ 1un (r ) eint ? It is an operator acting on a function so it is just another
function. As the original eigen-set is complete, this new function can be expanded in
terms of that orthogonal basis.
 i Hˆ 1un (r ) eint t  f new  r , t )   bk uk (r ) eik t
k
Projection (the inner product with uj) is used to isolate the expansion coefficient bj.
That is: the expression is multiplied by the complex conjugate of u j (r ) e
i j t
, the
companion basis set function for bj, and the orthogonality relation is invoked.
i
u j (r ) e
i j t
i (  )t
u j (r ) Hˆ 1un (r ) e n j t 
Hˆ 1un (r ) eint t   i
k bk
u j (r ) uk (r ) e
i (k  j )t
  bk  jk e
i (k  j )t
 bj
k
If the system is in the state n at time t = 0, then bn(0) = 1 and bj(0) = 0 for j  n. All
the bj that are initially zero are proportional to t for short times; it follows that:
db j
dt
 i
 i (  ) t
u j (r ) Hˆ 1un (r ) e n j   i
e
 i (n  j ) t

all space
(u j (r ))* Hˆ 1un (r ) dV
The expression is a little more compact if a matrix element [H1]jn represents the BraKet u j Hˆ 1 un .
db j
 i
dt
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e
i (n  j )t

u j Hˆ 1 un   i
e
i (n  j )t
Physics Handout Series.Tank: QDyn_QV
 H 
1 jn
[QMDyn.2]
QMDyn- 7
u j Hˆ 1 un is [H1]jn, the jn matrix element of the perturbation Ĥ1 .
Exercise: Rewrite the three lines of equations above replacing the Dirac Bra-kets
with the integrals that they represent. Rewrite [H1]jn = u j Hˆ 1 un in integral form.
Which forms of the equations are used when actual values are calculated?
State n to State j Transitions:
The system starts in the state un so at time to, bj = jn. All the bj = 0 for j  n. Except

db
for small corrections dtj   i ei (n  j )t

u j Hˆ 1 un so the amplitude to find the
system in a state other that n is growing (or at least changing in time). Taken at face
value, the factor e
i (n  j )t
means that the phase of the additions oscillates rapidly and
likely averages to near zero unless H1 has some special time dependence. For
electronic states in atoms (n - i) is expected to be greater than 1015 s -1 so averaging
to near a zero net value is very quick. As an example, the hydrogen atom is studied.
A Second Slice through the Material:
Let the state of the system  =

 b (t ) 
k 1
i t   i t

 b (t ) 
k 1
k
k
e
ik t
k
k
e
ik t
. The Schrödinger equation becomes:

  bk m (t ) m e imt  i btm (t ) m e imt
m 1


k 1
k 1
i t
 ( Hˆ o  Hˆ 1 )   bk (t ) ( Hˆ ok  Hˆ 1k )k e k   bm (t ) ( Em(0 )m  Hˆ 1m ) e imt
The terms highlighted in red have been shown to represent the uniform phase change
of the eigenstates rather than transition between states. Focusing on the other terms,
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

k 1
bk
t
k e
ik t
 i

b
m
m 1
(t ) Hˆ 1m eimt
Note that we have used distinct dummy summation indices. We will adopt Dirac
notation and take the inner product with j e-ijt to project out bj/t.


k 1
bk
t
 j k e
bj i
t 
ik t
 i
b
m 1

b
m 1
m

 
b j (t )  b j (0)  i

m1
m
(t )  j Hˆ 1m eimt
i (  )t
(t )  j Hˆ 1m e m j
t
0
i (  )t
bm (t )  j Hˆ 1 (t ) m e m j dt '
For short times, we assume that the bm have not changed very much.

 
b j (t )  b j (0)  i
m1
t
0
i (  )t
bm (0)  j Hˆ 1 (t ) m e m j dt '
First order time dependent perturbation result
We iterate this equation by substituting it for bm(t) in the relation above.

t
k 1
0
 
bm (t )  bm (0)  i

 
b j (t )  b j (0)  i

m 1
   b (0) 
 i
2
m ,k 1
k
t
0
t
t
0
0
i (  )t
bk (0) m Hˆ 1 (t ) k e k m dt 
i (  )t
{bm (0)  j Hˆ 1 (t ) m e m j dt '
i (  )t
i (  )t
 j Hˆ 1 (t )m e m j  m Hˆ 1 (t )k e k m dt dt '
Second order time dependent perturbation result
Transitions due to a time independent perturbation:
The system is assumed to be in the state n at t = 0 so bk(t = 0) = nk. We can find bk(t)
for short times (since t = 0) by ignoring states other than n and integrating.
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bk i
bn (t )
t 
k Hˆ 1n ei (n k )t  i bn (t ) k Hˆ 1n e i (n k )t
Assuming that H1 is time independent, we find:
bk (t )  0  i
ei (n k )t  1
i (n k )t
ˆ
i
ˆ
dt    (1) k H1n
 0 bn (t) k H1n e
i(n  k )
t
e
bk (t )   i (1) k Hˆ 1n t ei½(n k )t
i½(n k )t
 ei½(n k )t
(2i)(½)(n  k )t
bk (t )   i (1) k Hˆ 1n t sinc[(½)(n  k )t ] ei½(n k )t
Note that the sinc function has a delta function character with respect to  for large t.
So the only transitions caused by a time independent perturbation only connect
states with the same (frequency) energy.
Sinusoidal Time Dependence: Let use assume that H1(t) = H cos(t)
bk (t )   i 2 (1) k Hˆ n t
{e
i½(n k  )t
e
e
i½(n k  )t
i½(   )t
e n k
( 2i )(½)(n  k   )t
i½(n k  )t
e
i½(n k  )t
[QMDyn.3]
i½(   )t
e n k
( 2i )(½)(n  k   )t
}
The two terms correspond to conservation of energy with the emission or absorption
of a photon with energy .
The approximation above is a first order perturbation, and the selection rules are
buried in the time-independent matrix element k Hˆ  n .
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What is H1? Consider a free space electromagnetic plane wave with frequency 
washing over an atom.
E (r , t )  Re E0 cos(k  r   t )   12   E0 ei (k r  t )  E0* ei (k r  t )  , E0  k


B(r , t )   12   k   E0 ei (k r  t )  E0* ei (k r  t ) 


One half of the sum of a complex number and its complex conjugate is the real
part of that complex number. This cumbersome notation ensures that the
applied electric field is a real-valued plane wave. Re[z] = ½ (z + z*).
If you reviewed the properties of such a wave, you would find that the magnitude of
the magnetic field is the magnitude of the electric field divided by the speed of light
(B = c-1 E = k/ E). It follows that the magnitude of the magnetic force exerted on the
charge is less than or equal to v/c times the magnitude of the electrical force where v is
the speed of the charge. The dominance of the electric force means that the electric
field determines the directional character of the interaction of light (electromagnetic
radiation) with the charges in matter. The polarization of light describes the patterns
of the electric field direction in an EM wave. A full development includes electric
field and magnetic field interaction terms. We will focus on electric field term as it as
it is the dominate term causing radiative transition in atoms. An external electric field
interacts with the electric dipole moment of a charge distribution. H1   pdistr .  Eext
A hydrogen atom consists of two equal, but opposite charges  it is an electric
dipole. If r is the electron’s position relative to the proton, then the dipole moment
is p  e r and the interaction Hamiltonian is: Hˆ 1   p  E(r , t )   er  E0 ei (k r  t ) . The
form of the interaction Ĥ1 to be considered as a given for now, but the identification
is reviewed critically in Appendix II. We know that a point dipole has energy  p  E
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when place in an electric field so this form postulated is reasonable for an extended
charge distribution with a net dipole moment in an external electric field.
It’s too hard. Quantum mechanics is difficult, and even a slight complication can
lead to a problem that is impossible even to think about. One must always simplify to
the most complicated approximation that one can solve. The wavelength of light is of
order 5000 times the size of an atom so k  r  103 . The exponential eik r  1  ik  r  1.
To simplify the notation, E0 is assumed to be real so that the interaction is:


i t
i  t
Hˆ 1 (t )    12  er  E0 e   e  . Substituting into the transition amplitude equation,
 u
dbj
  ie 2
dt
j

i (   )t
i (   )t
r un  E0 e n j   e n j   .

[QMDyn.4]
un  un (r ) so u j r un is time-independent
As we seek interesting behavior rather than the exact numerical predictions, the two
factors [ e
i (n  j  )t
e
i (n  j  )t
] and u j r un  E0 are to be investigated
qualitatively.
Energy Considerations: The time derivative of bj is oscillatory, and the net change
in bj is close to zero on average unless either e
i (n  j  )t
or e
i (n  j  )t
is not
oscillatory.1 The time dependence of the perturbation H1(t) must have a time
dependence that cancels that due to the relative frequency between the initial and
final states. This occurs if: n j  if the energy of the final state Ej is equal to
the energy of the initial state plus or minus the photon energy . This equation is
the condition that energy is be conserved.
1
A typical duration of an atomic transition is 10 ns and the frequency of the emitted radiation is of order 600 THz. A
transition corresponds to averaging over 600,000 cycles of the n - j oscillations.
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Digression (Skip during your first reading.): Energy-Time Uncertainty: For
short times t, energy need not be conserved strictly. All that is needed is:
n j ]t < 1. Multiplying by , the conditions become: En Ej ]t < 
or [En -Ej ]t <. The energy defect is the amount by which energy
conservation is violated so E = | En Ej  |. The conclusion is that E t < .
For very short times the energy defect can be arbitrarily large for any state j although
the probability amplitude bj will surely be small. After longer and longer times, the
states with significant probability amplitude will include only those for which energy
conservation holds with one energy quantum absorbed from or emitted into the
perturbing field. All of this was expected.
 u
dbj
  ie 2
dt
j

i (   )t
i (   )t
r un  E0 e n j   e n j  

RULE ZERO: Our first selection rule is that  = |n - j|. Energy must be
conserved. With energy conservation satisfied, we find:
 u
db j
  ie 2
dt
j

r un  E0 for  = |n - j|.
Important Note: For j  n, the bj grow linearly with respect to time for short times.
Hence the probability to make a measurement identifying the system as have state j
character grows quadratically in time for small tomes. Stay alert for the explanation of
how this behavior is consistent with a linear radiative decay (constant decay rate) for
excited atomic states.

The additional selection rules are those necessary for u j r un  E0 to be non-zero
or not excessively small. The selection rules derived from this matrix element are
developed in chapter Guide 9: Time Dependent Perturbation Theory. The discussion
here will be more qualitative.
Atomic Dipole Selection Rules can be discussed in the context of:
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the matrix element of the perturbation
the character of the perturbation
the photon of odd parity, spin one and ms = 1 only. 
ms = 1  transverse; ms =   longitudinal not relevant
all massless particles have ms = s only
the oscillating probability densities
formal matrix element rules following Kirkpatrick
formal matrix element rules following Griffiths (in Chp Guide 9)
Selection Rules and Matrix Elements: The factor u j r un  E0 includes some
information about the polarization dependence of the interaction of the atom and the
electromagnetic field. To begin, the piece u j r1 un is to be studied to reveal
selection rules for transitions and to give some indication of the relative intensities of
various transitions. Selection rules identify transitions that are allowed which means
that they are caused by the perturbation being studied. Transitions that are not caused
are labeled as forbidden. The term forbidden as used does not mean that the transition
cannot occur; it only means that it is not caused by the particular perturbation in
the lowest order approximation so it does not happen with high probability. If a
transition violates energy conservation, angular momentum conservation or any other
fundamental principle, then it is absolutely forbidden. Review this paragraph after
you have studied selection rules.
Selection rules are based on the relation:
db j
 i
dt
e
i (n  j )t
u
j
Hˆ 1 un  u j r un
that identifies r as the atomic parameter active in the perturbation.
The Character of the Perturbation:
Cartesian representation: r  xiˆ  y ˆj  z kˆ
Spherical: r  r sin  cos iˆ  sin  sin  ˆj  cos kˆ 


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



r  4 3 r  1 2 Y1,1(    Y1,1(   iˆ  i 2 Y1,1(    Y1,1(   ˆj  Y10 (   kˆ 


The final form is useful when the angular momentum character of the perturbation is
discussed. The Ym are eigenfunctions of L2 and Lz. It follows that the perturbation has
 = 1 angular momentum character. The resulting rules can be understood by
considering the photon emitted or absorbed to be a particle with intrinsic spin angular
momentum 1 (and that the photon carries odd parity).
The Cartesian representation reveals that the perturbation is an odd function.
u j r un  
all space
(u j (r ))* r un (r ) dV ( photon has odd parity)
The photon links states that are odd (even) under coordinate inversion [ r  r ] to
those that are even (odd) under coordinate inversion. The initial and final atomic
states have opposite parity.
The eigenfunctions set used by physicists to describe the hydrogen atom
problem are all either even or odd. The even states are said to have even
parity and the odd states have odd parity. A state that is either even or odd is
said to be a state with good parity. The parity is given by P = (-1) where  is
the orbital angular momentum quantum number.
The overall integrand must be even so the states j and n must have opposite parity if
the perturbation is to link or couple the states (cause transitions between them).
Exercise: For hydrogen atom wavefunctions, compare  n m (r ) and  n m (r ) for 200,
210, 311 and 320. Conclude that:  n m (r )  (1)  n m (r ) 
Electric Dipole (E1) Transition Selection Rule #1 temp: Transitions between states
of the same parity (evenness or oddness) are forbidden. It is assumed that the states
are states of good parity.
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The spherical representation of r , the atomic factor in the perturbation, reveals its
angular momentum character. The spherical harmonics or Ym’s are eigenfunctions of
the square of the angular momentum and of its z component. The representation
shows that the interaction has angular momentum character  = 1.1 The angular
momentum L j of the final state must be the vector sum of the initial angular
momentum Ln and that of the perturbation. Using the vector model for the angular
momenta, j = n + {-1, 0, 1}. Including the required parity change (-1) = - 1,
Orbital angular momentum change: = ±1 

Electric Dipole (E1) Transition Selection Rule #1: Transitions are allowed
between states for which the orbital angular momentum of the electron differs by one
unit.
An electrons has an intrinsic spin angular momentum of ½  that is combined with its
orbital angular momentum to give its total angular momentum: J  L  S . The spin of
the electron has an associated magnetic moment that interacts more weakly than an
electric dipole does with an EM wave so the spin state is unlikely to change.
Considering this and other angular momentum tidbits, the
Electric Dipole (E1) Selection Rules for single electron transitions between atomic
states are:
 = s = 0; j = , 0 but not j = 0  j = 0; mj = , 0, but not mj = 0  mj =
0 if j = 0. These rules have not been derived or motivated; they have just been
stated. Check the index for a entry like selection rules for alkali atoms in a reference
such as Eisberg and Resnick, Quantum Physics or
http://en.wikipedia.org/wiki/Transition_rule.
1
The electromagnetic radiation is absorbed or emitted in quanta called photons. Photons carry intrinsic angular
momentum 1 .
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A detailed study of the inner product in u j r un  E0 would provide information about
the polarization of light that is most suited for causing or that would be emitted in a
particular transition. See Appendix I for a discussion of selection rules.
Forbidden Transitions: Forbidden transitions can occur, and it would be more
correct to use the phrase ‘not caused in (first) lowest order by the perturbation’
transitions rather than forbidden.
When we studied the helium spectrum, we noted that there were no (or only weak) transitions
between singlet and triplet spin states for the two electrons. This follows because the
perturbation does not couple to the spins (or magnetic moments) and hence does not flip them.
For very high orbital angular momentum states, the spin-orbit coupling does provide a channel
to flip spins and singlet to triplet transitions can be observed.
Circumventing the Rules:
Transitions that violate the electric dipole selection rules can occur, but they are
lower in probability by of order a factor of 1000 or more for each rule violated.
One possibility is that the EM wave causes a transition from n to j and then from j to
j. The states n and j would have the same parity and could have angular momenta
differing by 0 or 2 units. Second order transitions do occur; they are less probable
than first order transitions. In this transition, we might have two photons absorbed
leading to the energy requirement 2 = |En – Ej|. This double photon absorption
occurs when atoms are illuminated by intense laser beams. In these experiments, the
probability of the interaction improves if   |En – Ej|. The intermediate state is said
to be near resonant. The likelihood of double quantum transitions is small if the
number of photons per cubic wavelength is small as is the normal case in the visible.
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In the RF region, the number of photons per cubic wavelength can be made large, and
double quantum transition can be observed using the Teachspin Optical Pumping
Apparatus.
A second method to skirt electric dipole (E1) rules is to extend the approximation for
the interaction. Earlier eik r  1  ik  r  1. Keep the second term.
u j r un  u j r  r (ik  r ) un . The new term u j r (ik  r ) un will cause transitions
between states of the same parity and that differ in total angular momentum by up to
two units in first order. The rules for the newly included transitions would be called
electric quadrupole (E2) rules. They are of little importance because k  r  103 for
atoms and atomic transition photons. In nuclear physics, the photons are a million
times more energetic and the radius is only ten thousand times smaller so k  r  101 .
The electric quadrupole rules are of some importance to nuclear gamma ray
spectroscopy. The nucleus is a collection of positive charges that can have a
quadrupole moment, but not an electric dipole moment. Look for electric quadrupole
transitions (E2) between excited states of the same nucleus. If the magnetic
interactions are added to the hamiltonian, rules for magnetic dipole (M1) and
magnetic quadrupole (M2) transitions can be developed. The magnetic perturbations
are much smaller that the electric dipole term so atomic and molecular physicists
focus on the electric dipole selection rules. (You should consider (M1) and (M2)
transitions in nuclear physics.)
Exercise: A photon has momentum p = h/ Compute the maximum possible orbital
angular momentum relative to the hydrogen atom nucleus of a 2.5 eV optical photon
if it is emitted at one Bohr radius from the nucleus. Express the result in multiples of
. Repeat the calculation for an 8 MeV gamma ray emitted from the outer edge of a
lead nucleus. Recall: L  r  p (  .0014 ; 0.31 )
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Exercise: Given that r has angular momentum character one, what angular
momentum character would you predict for r (ik  r ) ? Recall that k is just a constant
vector.
A Doubly Forbidden Example: The helium triplet 2S metastable state --- 23S.
Upper case letters are full atom values; lower case letters are single electron state values.
Ground state helium has two electrons1 in the lowest allowed level, the 1S state. The
electrons have opposite spins, one up and one down, as required by the Pauli
exclusion principle. This relative spin state means that the net spin angular
momentum of the electrons is S = 0 leading to 2S + 1 or 1 possible mS values (a
singlet or parahelium state). The low lying excited states are formed by exciting one
of the 1s electrons into a higher state leading to the levels:
1S  1s, 1s21S  1s, 2s23S  1s, 2s21P  1s, 2p23S  1s, 2p
The spin parallel states, the ones with both spins UP, have total spin S = 1 and 2S + 1
or 3 possible mS values (a triplet or orthohelium state).
Notation: 2S+1L
for the atom
S: the total spin angular momentum of the
J
electrons
L: the total orbital angular momentum of the
electrons
J: the total (L+S) angular momentum of the
electrons
Notation: The choice of letters originates from a now-obsolete system of categorizing
spectral lines as "sharp", "principal", "diffuse" and "fine", based on their observed
fine structure: their modern usage indicates orbitals with an azimuthal quantum
number  of 0, 1, 2 or 3 respectively. After "f", the sequence continues alphabetically
"g", "h", "i"… (l = 4, 5, 6…), although orbitals with these high angular momentum
values are rarely required.
http://en.wikipedia.org/wiki/Electron_configuration
1
Helium: An atom with one electron too many.
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Helium Level Diagram with E1 Transitions
The energy levels for neutral helium are
illustrated with the allowed electric
dipole transitions shown as solid lines
linking the states. The atomic 2S states
have electrons in each of the single
electron 1s and 2s states. The triplet 23S
state lies 19.6 eV above the ground
state, and the singlet 21S lies at 20.4 eV.
A state at these energies would be
expected to decay electromagnetically
in nanosecond times. The decay from
21S to 11S is singly forbidden as  = 0
and one expects microsecond times. The
decay from 23S to 11S is doubly
forbidden as  = 0 and an electron spin
hyperphysics.phy-astr.gsu.edu/hbase/atomic/grotrian.html
flip (S  0) are needed. One expects millisecond decay times1. As S = 0 is the
allowed case selection rule, the para- and ortho-states form almost independent sets of
levels.
The absence of lines joining the 2S levels to the ground state indicates that energy
gets trapped in these levels for long times on the atomic time scale. The states are
metastable, and it is possible to collect small fractions of a percent of the helium
atoms in these states simply by running a low intensity RF discharge in the helium.
Energetic electrons in the discharge collide with helium atoms exciting electrons to
high levels or even ionizing the atoms leading to the electrons being recaptured into
high levels. The electrons then cascade down along the paths of allowed transitions
leading to the trapping of significant numbers in the metastable states from which
there are no allowed downward electromagnetic transitions. Helium-neon laser
dynamics depend on trapping energy in the metastable helium. Desperately seeking to
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return to the ground state, metastable atoms collide with neon atoms and resonantly
( the energy levels match to within kT so any deficit or surplus can appear as
change in the thermal kinetic energies of the atoms) transfer their excitation to the
neon atoms.
He-Ne Laser Level Diagram
Electron collisions are used to
ionize and excite helium
leading to electron pooling in
the 12S and 32S metastable
states when the electrons are
recaptured or transition from
higher levels. The energy is
transferred to the near resonant
neon 3s and 2s levels resulting
in these states having greater
population that the lower laser
levels 3s and 2p. These lower
levels decay rapidly to the 1s
levels preventing population
buildups that would destroy the
inversion required for laser
action.
Metastable molecular oxygen (1g) can be used to store energy for times
as long as 72 minutes, a feature exploited in the hybrid electro-chemical
oxygen-iodine laser system being investigated for possible naval
weapons applications. Energy stored at low power over a long time is
extracted in a short time yielding a high power pulse.
1
The lifetime of 12S state has been measured at 20 milliseconds and that of the 32S state at 8000 s. These are very long
and may reflect restrictions associated the J = 0 condition in the final state.
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Notice the notation key. The 2p is aligned above 2 p5 (n+1)p. Therefore the 2p
represents the configuration 2 p5 3p. The ground state neon is 1s22s22p6. The 2p  2
p5 3p 1s22s22p53p. Each closed subshell has net spin and net angular momentum
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zero. In 1s22s22p53p the 2p subshell has one vacancy and the 3p has occupancy one.
Thus 2p can have angular momentum 0, 1 or 2 and net spin 0 or 1.
The 2p levels lie between 18 and 19 eV above the ground state. The P3/2 corresponds
to a missing 2p electron with spin and orbital angular momentum parallel (J=1). The
P1/2 corresponds to a missing 2p electron with spin and orbital angular momentum
anti-parallel (J=0). NIST database: http://physics.nist.gov/PhysRefData/ASD/levels_form.html
Configuration
2s22p5(2P°3/2)3p
J
2
[1/2]
Energy (eV)
1 18.3816221
0 18.7113755
2s22p5(2P°3/2)3p 2[5/2]
3 18.5551069
2 18.5758348
2s22p5(2P°3/2)3p 2[3/2]
1 18.6127041
2 18.6367904
2s22p5(2P°1/2)3p 2[3/2]
1 18.6933582
2 18.7040700
2s22p5(2P°1/2)3p 2[1/2]
1 18.7263803
0 18.9659525
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The Electron Cloud View of Selection Rules: The images used in this discussion
were extracted from the hydrogen atom applet posted at www.falstad.com/qmatom/.
They represent amplitude and relative phase variation for the 1s  u10  Aer / a0 (Y00),
2s  u20  B (1  b r ) er / 2a0 (Y10), 2p0  u20  C r cos er / 2a0 (Y10) and 2p1
 u2,1  D r sin  ei e r / 2 a0 (Y1,1) electron states of the hydrogen atom. The images
below are the basis for a discussion of mixed state probability densities.
Y00 ( , ) 
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1
4
Y10 ( , ) 
3
cos 
4
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Y1,1 ( , ) 
1 3
sin  e i
2 2
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All the orbitals are viewed in along the y
axis except for the 2P+1 which is viewed
down the z axis. Relative phase is variation
from 0 to 2 is represented by color
varying from red to magenta (or violet).
Seen as viewed along z axis: A x-y plane
doughnut centered on the z axis. The
spectral fan represents the 0 to 2 phase
variation of the factor ei that appears in
Y1,1(,). 
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These images represent the wave amplitude ; it is the real value * that represents
the probability density.
The 210 to 100 Transition: Consider the mixed state:
A  r , t ) 
1
2
u (r ) ei10t  u (r ) ei21t  
210
 100

where u100 (r, , ) 
u210 (r , , ) 
1
2
u (r )  u (r ) ei (21  )t  e i10t .
210
 100

1
2
e r / a0  3/ 2 e r / a0
3/ 2
 a0
a0
1/ 2
1
 32  
1/ 2
1
2
 3/ 2 e r / a0 Y00 ( , ) and
a0
4
 r   r / 2 a0
 r   r / 2 a0
1
e
cos


Y10 ( ,  )


 e
1/
2
a03/ 2  a0 
 24  a03/ 2  a0 
At time t = 0, the two terms add in the region along the positive z axis and subtract in
the region along the negative z axis.
Exercise: In which region is the electron most likely to be found at t = 0?
Ans:
around + z-axis.
The 100 and 210 states have different energies so they time develop with different
frequencies. In a time )-1, the relative phase between u100 and u210 changes
by .
Exercise: In which region is the electron most likely to be found at t = )-1?
Exercise: The hydrogen atom consists of a positive proton at the origin and an
electron with a relative position described by  A  r , t ) given above. What is the
direction of the net electric dipole moment of the atom at times t = N ()-1
for N = 0, 1, 2, 3 and 4? What is the oscillation frequency? Compare it to the
frequency expected for the radiation emitted in a 210 to 100 transition.
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The 100-210 mixed state  A  r , t ) has a time dependent electric dipole moment that
corresponds to accelerating charge. Accelerated charge should radiate. We conclude
that the 210 to 100 transition should be allowed under electric dipole selection rules.
Given: A  r , t ) 
1
2
u100 (r )  u210 (r ) ei (21  )t  ei10t ,


 A  r , t )  12  u100 (r )  12 u210 (r )  u100 (r )u210 (r )cos([21   ]t )  .


2
2
2
Note that the result has a cosine term, but not a sine term or a complex-exponential. This
follows as u100 and u210 are real functions. The sine form can appear in other cases.
A  r , t ) 
1
2
u100 (r )  ei u210 (r ) ei (21  )t  ei10t


Exercise: The electric field in radiation emitted in a 210 to 100 transition should lie
in a plane that includes the line of sight and what other line. (Read the appendix.)
(Answer: the z – axis)
Exercise: Develop the probability density for:
1
2
u100 (r )  ei u210 (r ) ei (21  )t  ei10t


with  = /2. Compare it with that for A studies above.
Study the 200 - 100 Mixed State: Consider the wavefunction:
B  r , t ) 
1
2
u100 (r ) e i10t  u200 (r ) e i20t  
1
2
u100 (r )  u200 (r ) e i (20  )t  e i10t .
Describe the probability density at time t = 0 and at t = )-1. Sketch the
direction of the net electric dipole moment at times t = N ()-1 for N = 0,
1, 2, 3 and 4. Do you expect electric dipole transitions between the 200 and 100
states? Compute and discuss B  r , t ) .
2
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Study the 211 - 100 Mixed State: Consider the wavefunction:
C  r , t ) 
1
2
u100 (r ) ei10t  u211 (r ) e  i21t  
1
2
u100 (r )  u211 (r ) e i (21 t )  e i10t .
Describe the probability density at time t = 0 and at t = N ()-1 for N = 0,
1, 2, 3 and 4. Sketch the direction of the net electric dipole moment at times t = N
()-1 for N = 0, 1, 2, 3 and 4. Do you expect electric dipole transitions
between the 200 and 100 states? Compute and discuss C  r , t ) .
2
Exercise: Describe the apparent motion of the blob of enhanced electron probability
density if the state C  r , t ) is viewed from the positive z direction. Light emitted in the
z direction in a 211-100 transition is circularly polarized. The light emitted in a 210100 transition is linearly polarized. Describe the time dependence of the direction of
charge acceleration in a 210-100 mixed state and in a 211-100 mixed state.
The m Rule: Consider wavefunctions of the form  = R(r) f() eim for the initial
and final states. Let m = mf – mi. Show that
1
2
( i   f )( i   f ) has terms with
factors eimand e-im. Leading to there being probability lumps in  proportional to
|e½imei½m+ e-i½m or equivalently cos[½ m ]. That is there are |m| lumps
spaced around in  with separations of 2/m. If |m| > 1, then the vector sum of the
accelerations of the various lumps is zero and no net radiation is expected. For m =
0, there are no phi probability lumps, but the charge distribution may be oscillating
along the z axis. Look for such cases to radiate linearly polarized light when viewed
in the  = /2 plane.
Polarization and the Zeeman Effect: Nothing has been presented about the spin
angular momentum of the electron so only comments about the polarization of the
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light emitted during transitions is to be presented, not the details of the energy levels.
First, the z axis direction is set whenever a measurement is made or a direction is
made distinct. For the Zeeman problem, a magnetic field is applied to the atom
defining the field direction as the z direction with the result that the energy of the
states shift by m effective. In a 210 to 100 transition, the electron density oscillates
along the z direction and the emitted light is polarized along the projection of the z
direction onto a plane perpendicular to the plane of propagation.

3D to 2P Transitions with the levels split by an applied B field.
The three lines are the m = 0, 1 transitions.
Splittings are of order 60 (eV) or 14 GHz in a 1 Tesla field.
For a 211 to 100 transition, the electron density circles around the z direction at the
difference frequency 2 - 1. Viewing in along the z direction the light is circularly
polarized. View from a direction in the x-y plane, only the electron density projection
is observed – motion back and forth along a line – so the light is linearly polarized.
Again, consider the projection of the circular motion onto a plane perpendicular to the
propagation direction of the light. For light propagating at angles  relative to the z
direction, the projection on the motion onto the plane yields an elliptical path, and the
emitted light is elliptically polarized.
Begin by imagining the field line pattern due to a static point charge at the origin.
Next imagine the charge executing uniform circular motion about the origin in the
x-y plane. The information about the charges motion propagates out along the field
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lines at speed c. The transverse components are greatly exaggerated. Observer A
views inward along the x axis and sees horizontal electric field components. Observer
B views inward along the z axis and observes a transverse electric field with a
direction that rotates in the plane parallel to the x-y plane. If the intensity observed
along the z axis is normalized to 2, then the net intensity radiated along a direction at
angle  relative to the z axis is 1 + cos2. The light is circular polarized for lines of
sight at  = 0 or ; it is linearly polarized for  = ½; and it is elliptically polarized
for intermediate angles.
B is in the z direction. In the lab we view
in along the x axis, and the y axis is
vertical.

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y
z
x
x


Transition Rate: To this point the probability amplitude bj has been discussed, but it
is |bj|2, the probability to find the system in the state j that is of more direct interest.
Use the perturbation result:

dbj
  ie 2
dt
u
j

i (  )t
i (  )t
r un  E0 e jn   e jn   where  jn   j  n . Assume that the

perturbation is turned on at time t = 0 and is on until time , and find that:

b j ( )   ie 2

b j ( )   e 2
u

j
r un  E0
u j r un  E0



0
ei ( jn  )t   ei ( jn  )t   dt 


 ei ( jn  )  1 ei ( jn  )  1 



 jn   
  jn  
As discussed earlier, the probability to find the system in state j is small unless either
jn +  = 0 or jn -  = 0. It is to be assumed that the system is to emit a photon
adding energy to the E-M radiation. The energy (frequency) of the final atomic state j
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is less than that of the initial atomic state n. With this restriction only jn +  = 0 (or
equivalently: n -  =j) is considered. The first term has a ‘zero’ in the
denominator and is large compared to the second term which can be neglected.

b j ( )   e 2

u j r un  E0

 ie 2

 ei (nj  )  1 


  jn   
u
j

r  E0 un e
 2 sin[( jn   )  ] 


 jn  


i ( jn   ) / 2

The probability amplitude is squared to find the probability:
2
 
b j ( )  1
 sin[( jn   )  ] 

  1



jn



2
2
2
u j r  E0 un
 
2
b j ( )  1
2
H jn
2
E
2
0
 
2
H jn
2
 sin[( jn   )  ] 
E cos  

 jn  


2
0
2
2
2
cos2     for short elapsed times 
If the system starts in the state n, then the probability for the system to be found in the
state j grows quadratically in the elapsed time  (the time that perturbation has been
active).Note that this behavior is for the growth of probability for the single state j. 
The rate of decrease of the probability for the system to be found in state n is the sum
of the rates at which the sum of the probabilities for all the final states grows. All the
accessible final states must be in a band of energies of width E centered on (En - Ej)
which is consistent with the uncertainty principle. 
2
2
2
d
h
cn   b j ( )  ( E ) E  b j ( )  ( E )  
dt

[QMDyn.5]
In this model, it is assumed that there is a continuum of final states j that describe the
same state of the atomic part of the system coupled with a slightly different final
states for the electromagnetic wave part of the system.
Quantum State  (Atomic State)  (E-M Field State)
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The EM field has a rich continuum of allowed states. The final states do not differ in
the final atomic state, and they differ only a little in the state of the electromagnetic
field over the volume of the atom so that the matrix element for the transition form n
to all these allowed j is essentially equal for all the states in the energy band E. The
final states might have a photon of slightly different wavelength or some such. The
point is that the electric field strength does not vary much across the atom as  >>>
ao. As time marches on, the uncertainty requirement dictates that energy must be more
and more precisely conserved. That is E  h/. The probability to make the transition
to a singlet state j grows quadratically, but the allowed energy mismatch E shrinks
inversely as the time as required by the uncertainty principle. Over all, this leads to a
decrease in the probability in the state n probability that is proportional to time . The
probability that the atom makes a transition to a particular final atomic state grows
linearly in time for short times. This outcome is equivalent to a constant decay rate
 
for the state n. R  1
2
2
H jn E02 cos2   ( E ) . At this point, we cannot estimate the
factors in this expression, but we observe that a states decays at a constant relative
rate and hence exponentially when there is a continuum of closely related final states.
We should note that all of the final states may correspond to one (or to a few) final
states of the atom with the distinction being in the final states descriptions for the
electromagnetic field (which has a continuum of states that are essentially identical
over the atom’s volume). The result is that the probability to find the atom in a final
atomic state can grow linearly in  even though the probability for the probabilities
for each microstate ([atomic state]  [EM field state]) in the allowed E band grows
as 2 while the uncertainty band E shrinks as h/. 
A second approach is to use heavy math to avoid considering the mechanism by
which quadratically growing probabilities for each microstate leads to a linear decay.
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 
2
b j ( )  1
 sin[( jn   )  ] 

  1





jn

2
2
2
u j r  E0 un
 
E 
E 
2
d
2
1
cn   
b j ( )  ( E ) dE   
E

E

dt
 
2
H jn
2
 sin[( jn   )  ] 
E cos  

 jn  


2
0
2
2
 sin[( jn   )  ] 
E cos  
  ( E ) dE
 jn  


2
2
H jn
2
2
0
2
The result has a Dirac delta buried in it, and (E) is the density of final states for the a
final photon of energy E = |En – Ej|.
7
 sin 2 (n x) 
n
Dn ( x)    sin c 2 (nx)  
2 
 
 n x 
sinc(x) = sin(x)/x
See: mathworld.wolfram.com/
Plot[(Sin[10 x])^2/(10 Pi x^2),
{x,-1.5,1.5},PlotRange{0,3.5},
PlotStyleThickness[0.006]]

Let n  the elapsed time and x  jn + .

2

b j ( )  e 2
 
E 
d
2
1
cn   
E 
dt

E 
d
2
e
cn   
2
E 
dt

2
2

E 
d
2
e
cn   
2
E 
dt
 sin 2 [( jn   )  ] 

u j r  E0 un (   
2 
     [ jn   ] 
 sin 2 [( jn   )  ] 
2
2
2

H jn E0 cos  (   
 ( E ) dE
2 

     [ jn   ] 
2
2
u j r  E0 un


jn +  = 0 where jnj-n 
2
d
2
cn   e 2
dt
2
u j r  E0 un

2
 sin 2 [( jn   )  ] 

(   
 ( E ) dE
2 

     [ jn   ] 
2
(     ( En  [ E j  E ]) ( E ) dE
u j r  E0 un
2

(    ( En  E j ) 
The probability for the state n to have decayed grows linearly with respect to the
elapsed time  which corresponds to a constant decay rate. In the (nanosecond) long
time  limit, the Dirac delta enforces energy conservation.
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
Rdecay  e 2

2
u j r  E0 un
2
 e 2 (E
nj
)
and
 = |En – Ej|
CAUTION: These arguments have been qualitative. Many multiplicative factors have
been lost, and a few concepts have been bruised.
The delta function in frequency means that the transition1 is not likely unless the
energy of the initial excited state nis very close to the sum of the energy of the final
state of the atom j plus the energy of the emitted photon . The problem is
complicated because the photon can be emitted into a great many states (directions,
…. ). One must (average over the initial states and) sum over all the possible final
states for the photon that are more or less compatible with the energy . Reviewing
the blackbody problem, (E), a density of states for photons2 can by found. (See your
modern physics course or your statistical mechanics course.)3 Sum or integrate over
the final states for the photon and

R nj  e 2

R nj  e 2

2

2
u j r  E0 un
u j r  E0 un
2
2

(     ( E )   j    n ) dE
0
(   ( n   j ) where  ( E ) 

1
2(
c)3
E2
This final result is a form of Fermi’s Golden Rule4 for Transition Rates. You will
hear the phase ‘average over initial states and sum over final states’ chanted with
great reverence when you study quantum mechanics in graduate school.
*** Add
Fermi for EM radiation ***
1
A transition to a lower energy level of a system coupled with the emission of a photon is called a radiative decay.
2
See problem 8 in Chapter Guide 4, Quantum Mechanics in Three Dimensions.
3
The development is similar to the chapter 4 particle in a box. The details will be added in an appendix.
4
The Golden Rule was actually developed by Dirac 20 years before Fermi dubbed it one of two golden rules.
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Fermi, Enrico (1901-1954) Italian-American physicist who was born
in Rome. Fermi discovered the statistical laws, now called FermiDirac statistics that govern the particles subject to the Pauli exclusion
principle. Such particles are called fermions in Fermi's honor. Fermi
was appointed professor of theoretical physics at the University of
Rome, a post that he retained until 1938 when, immediately after
receiving the Nobel Prize in physics for his studies on the artificial
radioactivity produced by neutrons and for nuclear reactions of slow
neutrons, he escaped to United States to avoid Mussolini's fascism
and persecution of his Jewish wife. Fermi produced the first
controlled nuclear chain reaction in Chicago on December 2, 1942.
http://scienceworld.wolfram.com/biography/Fermi.html
© 1996-2006 Eric W. Weisstein
As an approximation, | u j r  E0 un |2 is replaced by u j r un  E0 cos  2 . The angle
2
between the direction of u j r un and the perturbing electric field is . When
averaged over all angles, cos2 has an average value of 1/3. Using this approximation,
we can compute order of magnitude estimates of transition rates without carefully
evaluating u j r  E0 un |2.
The decay rate is proportional to the square of the dipole matrix element linking the
states n and j. The radiation pattern has a cos2-dependence where  is the angle
between the line of sight and the direction of the electric dipole matrix element, the
direction of charge acceleration. The delta function factor (n -  - j) indicates that
energy must be conserved. The factor Eo2 indicates that photon absorption is
proportional to the time-averaged intensity of the applied electromagnetic wave.
I= c o (Erms)2  ½ c o Eo2
Again, recall that nothing has been developed rigorously. Hands were waved, and feet
were flapped. The goal was to present the general flow of the ideas so that you might
see how the concepts and mathematics fit together to describe quantum dynamics. Do
not expect to see rigorous developments of these concepts before the end of your first
year in graduate school.
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Einstein Coefficients: The form of the transition rate equation

Rnj  e 2

2
u j r un
2
u j r un
E02 cos2  (    j    n )
2
 un r u j
[QMDyn.6]
2
shows that the induced transition rate due to the applied field is the same forward and
backward, Rjn = Rnj and that the rate is proportional to the square of the electric
field and hence to the energy density  of the incident E-M wave1. Spontaneous
transitions happen. An atom in an excited state can emit a photon and make a
transition to a lower state in the absence of any applied field (beyond the zero point
field). Einstein postulated that there are three coefficients governing the transitions
between an upper level u and a lower level :
Note: u
Au: the spontaneous transition rate from the upper to lower state
Buu: the induced transition rate from u to 
Bu u:: the induced transition rate from to u
spectral energy density (w.r.t. frequency); (E) is a density of states w.r.t.
energy
In equilibrium, the populations nu and n of the upper and lower states should be in
 (E
Boltzmann distribution2 ratio: nu /n= e 
u E )
kT
  e h  kT  , and, as the relative population
does not change in equilibrium, the rate of transitions up is equal to the rate down.
Ru   nu { Au   Bu    u )} and Ru   n B u   u )
 h
In thermal equilibrium: nu  n e  kT  and Ru  Ru .
Substituting, rearranging and comparing with the Planck black-body radiation
formula for (vu) which must hold for systems in thermal equilibrium:
1
2
The various references adopt different radiation strength measures: energy density, intensity, … . Read carefully.
The Boltzmann distribution is the characteristic energy distribution of distinguishable (classical) systems in thermal equilibrium.
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  u ) 
Au 

h   B
Bu  e kT  u B

u

 8 h  
1

  Planck  u )
  h  kT 

  c e
1

It follows that: Bu = Bu and that Au= 8hc-33 Bu. The equivalence of the
induced rates up and down B coefficients also follows from the rate formula. The
beauty is Auis identified as 8hc-33 Bu which identifies the value of the E-M
wave spectral energy density that is necessary to cause a transition rate equal to the
spontaneous decay rate. The quantum verification of this identification comes only
when quantization is applied to the electromagnetic field. Each particle in a box mode
for the field becomes an independent harmonic oscillator mode and the excitation
quanta are the photons.
â |n = n½|n - 1
↠|n = (n + 1)½|n + 1
remove a photon from a mode
add a photon to a mode
The quantum state counting predicts that the probability to absorb a photon from a
mode is proportional to the number of quanta n in that mode while the probability to
emit a photon into that mode is proportional to n + 1, the number of quanta in the
mode after the emitted one is added.
That ‘+ 1’ is the spontaneous emission and the n part is the induced emission. A hint
of this can be seen in the outcome of problem 3. One does not need to add
spontaneous emission. Spontaneous emission is properly and automatically described
if the rules of quantum mechanics are applied to the electro-magnetic field as well as
to the atom. One imagines that the atom and fields are in a very large box with
perfectly conducting sides. The E-M fields have allowed quantum states that are the
particle-in-a-box spatial modes times two allowed polarizations for each spatial mode
(propagation vector: k  k x iˆ  k y ˆj  k z kˆ ). Each mode obeys the same equations as
does a simple 1D harmonic oscillator leading to allowed energies in each mode of (n
+ ½). Hence energy can be absorbed from of emitted into the modes only in
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chunks of . This harmonic oscillator link and the results of problem three suggest
that a n + 1 factor is expected for emission into a mode as compared to a factor of n
for absorption out of it. In the case that the field is in its lowest state (n = 0), emission
is proportional to 1 and absorption is proportional to 0. The proportional to 1 part is
atoms spontaneous emission which adds one unit of energy to the EM field. If the
field mode is initially in its lowest (n = 0) state, no absorption is possible.
Each mode of the EM field behaves as a QHO. The zero point energy depends on the density of
states and hence on the nature of the boundary of the region. An energy dependence on the boundary
leads to a pressure or force on the boundary which is called the Casimir force. This force must be
considered in the design of nano-machines. Electron state quantization becomes an issue as the traces
in microprocessors shrinks to sub-micron dimensions. Quantum mechanics is an issue for our
developing technologies. See Casimir Forces: Still surprising after 60 years, Physics Today,
February 2007, page 40.
Rework consistently and introduce new notation.
Combining the factors discussed above (Kirkpatrick):
Riabs
f 

3 o
Ristim
f 

3 o
spon
i f
R
where
2
pif2  ( E fi )
[QMDyn.7]
2
pif2  ( Eif )
[QMDyn.8]
if3

3  o c3
pif2  e2 
p
2
if
f xi
[QMDyn.9]
2
 f yi
2
 f zi
2

[QMDyn.10]
Sample Calculation:
Compute the spontaneous decay rate of the 2po state of hydrogen to the 1s state. For
this case, the only non-zero matrix element is 1s| z |2po100| z |210.
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 

0
|1s
a 
1
3/2
2
0
 
e
 
100 z 210  a1
0
3
r

a0 
Y00 ( , ) =



3
1

= r cos = r  2
z

1  r    r 2 a0 
Y10 ( , )

e
3  a0 
3/2
|2po 21a

6
a 
1
Y10 ( , ) 
3/2
2
0
e
 

r

a0 
1 3
cos  
2 
1
4

Y10 ( ,  ) 

r


e a r

0
0
 
r
a0
r


e 2 a  r dr 2

0
   2 1
Y10Y10 sin  d d

3   0  0 4


2
Note that the factor of (4)-½ is Y00, and that cos is 2 (/3)½ Y10.
100 z 210  a0
1    3r 2 a0  r
e
a0
6 0
 
100 z 210  a0  2 3 
5
4
dr
a0
 13 
 
2
7
4!  2 a0 25
3
6
It is easy to show that 100| x |210 = 100| y |210 = 0.
Compute the decay rate of the 210 state of hydrogen to the 100 state.
spon
i f
R
spon
i f
R

if3

3  o c3
p
2
if
if3
15

(ea0 )2 210
3
3
3  o c
(10.2  [1.602 1019 ]J )3if [1.602  1019 C ]2
(1.054  10
spon
i f
R
34
Js)  (8.85  10
4
12 J
C 2m
) (3  10
8m
(5.29  1011 m) 2 211
15
s
)
3
3
 
(10.23  1.6025  5.292 ) 10117 215
8 1

11  6.26  10 s
4
3
124
3
(1.054  8.85  3    10 ) s
The lifetime is the inverse of the decay rate:  = 1.60 ns.
Exercise: Compute the hydrogenic matrix elements 100| x |210 100| y |210
100| x |211 100| x |21-1100| y |211 and100| y |21-1
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Exercise: Given 100| x |211 = 100| y |211 = ao (27/35) and, 100| z |211 = 0,
compute the spontaneous decay rate of the 211 state of hydrogen. Compare it to the
decay rate for the 210 state.
Exercise: Compute the electric dipole decay rate of the 210 state of hydrogen to the
200 state. Only 200| z |210 is active. Assume that the 200 state lies 0.33 cm-1 below
the 210 state. (This is not exactly correct, but it puts us in the ballpark to make a
point.)
Note:  = 2 (3 x 1010) v where v = 0.33 cm-1; v = -1 with  expressed in cm.
Compare the sizes of the matrix elements 200| z |210 and 100| z |210. What factor
in the expression for R dominates the difference?
Exercise: Compute the electric dipole decay rate of the 200 state of hydrogen to the
100 state.
The decay rate of the 2s state is 420 s-1. The radiative decay rate must therefore be less than or
equal to this rate so the radiative lifetime is greater than 2.38 ms.

The fundamental radiative processes:
Spontaneous Emission: An atom in an upper level u can emit a photon as the atom
makes the transition to a lower level . The photon energy is equal to the energy
difference between the two levels: ( = Eu – E). Spontaneous emission occurs
randomly with the radiation spewing out into a broad spread of solid angles.
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hyperphysics.phy-astr.gsu.edu/hbase/mod5.html
Resonant Absorption: A photon incident on an atom in a lower level  can be
absorbed causing the atom to make a transition to an upper level u if the photon
energy is equal to the energy difference between the two levels: ( = Eu – E). It is
important to remember that energy levels have finite widths (E/E  10-73) with the
widths in liquids and solids being potentially very large (E/E  10-1). Thus the energy
equation:  = Eu – Especifies a range of frequencies for photons that can be
absorbed.
hyperphysics.phy-astr.gsu.edu/hbase/mod5.html
Stimulated Emission: A photon incident on an atom in an upper level u can induce
or stimulate the emission of a photon as the atom makes the transition to a lower level

if the photon energy is equal to the energy difference between the two levels: ( =
Eu – E). The magic of the event is that the emitted photon is a copy of the photon that
induced it – same direction, wavelength, polarization state, … . In turn, those two
identical photons can stimulate the emission of two more identical photons. As the
process continues the occupation n of that one photon state or mode becomes large
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leading to laser action as a emission into that mode is proportional to n + 1. Atomic
levels in gases are relative narrow leading to lasers that operate to emit on very
narrow frequency bands. Some liquid (dye) and solid state lasers have broad energy
levels leading to output frequencies that can be tuned.
hyperphysics.phy-astr.gsu.edu/hbase/mod5.html
The character of a stimulated photon contrasts starkly with that of a spontaneously
emitted photon. A spontaneously emitted photon is randomly emitted in most any
direction with any phase. There is essentially no chance that the spontaneous photon
will land in the same mode as the stimulated photons just discussed. A laser operates
using the stimulated emission photons. The spontaneously emitted photons are
uncorrelated and effectively contribute a noise field to the field of the good laser
photons. For this reason, the spontaneous emission is omitted in the discussion of
laser dynamics. The net rate of increase of photons in the beam is the rate of
downward transitions minus the rate of upward transitions in the lasing medium.
R  Ru  Ru  nu Bu   u )  n B u   u )   nu  n  Bu    u )
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where the equality Bu   B u has been used. The conclusion is that R > 0 if nu > n.
An active medium can provide photon gain if the upper state population exceeds the
population in the lower state; that is: if there is a population inversion. Review the
discussion of the helium neon laser dynamics.
Exercise: Why is the condition nu – n > 0 called a population inversion? What is the
expected value of nu/n for a system is in thermal equilibrium? Who was Boltzmann?
How did he die?
Many more wonders of quantum dynamics await you. Sweat the details when you
next encounter the concepts. The developments as presented are not rigorous, but
rather they are just a quick view of things to come.
Ludwig Eduard Boltzmann (1844 –1906) was subject to rapid alternation of
depressed moods with elevated, expansive or irritable moods, likely the symptoms
of undiagnosed bipolar disorder. He himself jestingly attributed his rapid swings in
temperament to the fact that he was born during the night between Mardi Gras and
Ash Wednesday. Meitner relates that those who were close to Boltzmann were
aware of his bouts of severe depression and his suicide attempts.
On September 5, 1906, while on a summer vacation in Duino, near Trieste,
Boltzmann hanged himself during an attack of depression. He is buried in the
Viennese Zentralfriedhof; his tombstone bears the inscription: S = k logW.
http://en.wikipedia.org/wiki/Ludwig_Boltzmann
***** Tidbits
Appendix I: Electric Dipole Selection Rules
(follows Fitzpatrick and Griffiths)
Selection Rules: (pages 187-191.) Motivate equation (12.63).
Are the rules called selection rules when applied in the context of time independent
perturbations?
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Several selection rules were motivated by discussions of the dipole moments of
mixed states. A second motivation of the rules focuses on the angular momentum 1
character of the photon. A third very formal development of the several selection
rules follows. Compare and contrast the approaches.
Tools of the Trade: Note that an operator was re-expressed as a commutator of two
operators. One of the operators in the commutator is an eigen-operator for the basis
set of states so its introduction is reduced to eigenvalue multiplication rather than a
general operator action.
Suppose we have a hermitian operator A with eigenstates |n and a perturbation B.
We want to know which pairs of the original eigenstates are linked by the
perturbation.
m| B |n  0
m| B A|nAn m| B|n which identifies the initial state and m| AB|nAm| B|n
Am m| B|nwhich identifies the final state. Additionally, it is assumed that the
commutator [A, B] can be reduced to an equivalent operator C, and that m|C|ncan
be expressed in terms of the eigenvalues associated with |m and |n and m|B|nthe
matrix element of interest. That is: m|C|n = fC(m,n) m| B|n
m| [A,B]|nm| AB-BA|n (Am – An) m| B|n= fC (m,n)m| B|n[QMDyn.11]

If A and B commute, then m|B|n must vanish for states with distinct eigenvalues for
the operator A. Similar conditions can be extracted for more general commutator
values.
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In general, one should become familiar of the commutators of new operators with
each of the current (previously known) operators for a problem.
The hydrogenic states found in the Hydrogen Atom Math handout are eigenfunctions
of energy, of the square of the orbital angular momentum and of the z component of
angular momentum. Here the focus is on the orbital angular momentum so the initial
state is represented as | m, and the final state as |m.
As the first example, the perturbation is –eEoz so B is effectively z and, as a first
choice, the (physics set of) hydrogenic states are eigenfunctions of Lz with
eigenvalues m. Let A  Lz.
 m| [Lz, z] | m =  m| 0| m = 0 = (m– m)  m| z |  m
The matrix element  m| z |  m can only be non-zero if m= m. The perturbation z
can only cause m = 0 transitions.
The case for x and y is not as simple, but the story begins the same way. The
perturbations –eEox x and –eEoy y are considered. Using
[ xi , Lˆ j ]  i  ijk xk , [ x, Lˆz ]  i y and [ y, Lˆz ]  i x Let A  Lz.
m ' [ x, Lˆz ] m  (m  m)
m ' x m  m '  i y m
or (m  m) m ' x m  i m ' y m
Note: That one need only compute the matrix element of x as the matrix element of y
can be computed form it.
m ' y m  i(m  m) m ' x m

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Beginning with [y, Lz], one finds that (m  m) m ' y m  i m ' x m . Chaining
the results together, (m  m) 2 m ' x m  m ' x m which means that
m ' x m
can be non-zero only if m = m  1; that is: if m =  1.
Tools of the Trade: Note that an operator was re-expressed as a commutator of two
operators. One of the operators in the commutator is an eigen-operator for the basis
set of states so its introduction is reduced to eigenvalue multiplication rather than a
general operator action.
Combining these results with those for z, the allowed changes to the magnetic
quantum number m are 1, 0 and -1. The rules for  require some perseverance. A
direct application of [QMDyn.11] using [L2, z] fails to provide a result that is simple
to interpret. That is:  m| [L2, z]|  m can not be reduced easily to the target form
fC(m,m) times a matrix element. The next step is to carry the process to the next
level and to try [L2, [L2, z]]. This compound commutator can be evaluated and
simplified using:
[ Lˆ2 , z ]  2i ( xLˆ y  Lˆx y ) and xLˆx  yLˆ y  zLˆz  0.
It follows that:
[ Lˆ2 ,[ Lˆ2 , z ]]  2 2 ( Lˆ2 z  z Lˆ2 )
[QMDyn.12]
Note that:
m ' [ Lˆ2 ,[ Lˆ2 , z ]] m  2
2
m ' ( Lˆ2 z  z Lˆ2 ) m  2
4

(   1)  (  1)  m ' z m
Now the methods behind equation [QMDyn.11] can be applied to yield:
(
   2)(  )(    1)(    1) m ' z m   0
[FQT.13]
The first factor can never vanish, so the possibilities for a non-zero outcome are
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 =  = 0 and  =  1. The  = 0 states are spherically symmetric so
nm|z|nmn 0m| z |n 0m as does n 0m| x |n 0m and n 0m| y |n 0m. In
fact the  rules must be the same for x, y and z as they are just equivalent Cartesian
components and reflect an arbitrary choice of a preferred direction, and the operator
L2 is invariant with respect to permuting the labels x, y and z.
Exercise: Verify that [ Lˆ2 , z ]  2i ( xLˆ y  Lˆ x y ) and xLˆ x  yLˆ y  zLˆ z  0.
Exercise: Evaluate
m ' [ Lˆ2 ,[ Lˆ2 , z ]] m and
m ' [ Lˆ2 ,[ Lˆ2 , z ]] m  2
2
m ' ( Lˆ2 z  z Lˆ2 ) m . Expand the relation
m ' ( Lˆ2 z  z Lˆ2 ) m to develop a result equivalent to
[FQT.13].One must first verify that [ Lˆ2 , z ]  2i ( xLˆ y  Lˆ x y ) and xLˆ x  yLˆ y  zLˆ z  0.
The Selection Rules for Electric Dipole Transitions: m = 0,  1 and  =  1. The
 rules can be interpreted in terms of a photon being emitted or absorbed that carries
one unit of angular momentum and that has odd parity. The atomic states have parity
(-1) and well defined  and m. The absorption (or emission) of a photon must lead to
an atomic state of opposite parity which is consistent with changing  by 1. This
requirement is also consistent with the rule that  = 0 to  = 0 transitions are
forbidden. The rule that || < 2 is consistent with the photon carrying just one unit of
angular momentum.1 Further, the photon is a fully relativistic particle and as such is
1
In atomic and molecular physics, it is unlikely that an emitted photon has orbital angular momentum. The photon
carries one unit of intrinsic (spin) angular momentum. In the case of nuclear physics, emitted photons can carry orbital
angular momentum as well as intrinsic angular momentum. Electric quadrupole and magnetic dipole must be considered.
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only permitted to have z components of angular momentum of m  . A photon
traveling in the z direction is associated with transverse electric fields (components in
the x and y direction) and hence with m = 1. A particular combination of the x and
y polarizations forms the +1 right-hand (-1 left-hand) circularly polarized light, and
the associated photons propagating in the z direction cause on m = +1 (-1)
transitions only. See the rubidium optical pumping experiment for example.
Appendix II: THE SOURCE OF E-M RADIATION: ACCELERATED CHARGE
The static (Coulomb) electric field due to a charge at rest is ECoul  4qr
 o r3
where r is the
displacement from the position S of the source charge q, to the field point P where the
field is to be specified. In addition, a slowly moving charge has a Biot-Savart magnetic
 qv  r
field given by BBS  4 r3 . Finally, an accelerated charge has a radiation contribution
to the electric field that is approximated by the relation ERad   4qac r  (for v << c).


 Ret

2
The subscript Ret directs that the source charge acceleration a, the component of the
acceleration perpendicular to the line of sight at the observation point, should be
evaluated at the retarded time t' = t - r/c where r is the distance between the observer now
and the source location at the time the radiation currently being detected was emitted by
the source. That is you want to evaluate a at the time that the light left the source S in
order to reach P at time t. Note that Erad isto the line of sight and that it is directed
oppositely to the retarded value of a and that the radiation field falls off as r -1 at large
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distance while ECoul and BBS fall off as r -2. [There are corrections to ERad of order v/c and
r -2, etc.] The energy flow is described by the Poynting Vector S  1 E  B which is
proportional to the product of ERad and BRad . At large r the net energy flow out is
proportional to S 4πr 2 = (1/µo) E B 4πr 2. As radiation must be capable of carrying energy
to infinity, the radiation contributions to the net E and B fields must fall off no faster than
r -1 with distance. We assume v << c which means that only non-relativistic sources are
to be considered.
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The form of the radiation field can be motivated using Gauss's Law. Electric field lines
begin and end only on charges. For a charge in uniform motion, the lines are directed
radially away from the instantaneous position of the charge. A short burst of acceleration
therefore puts kinks on the field lines that propagate outward at c. In the kinks, the field
has a component perpendicular to the line of sight in addition the Coulomb field along the
line of sight. If the charge accelerates to the right from time 0 to t, the field lines must
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adjust such that those inside a distance ct are directed away from the instantaneous
position of the source charge while the lines outside a distance c(t+t) are directed away
from the instantaneous position that the charge would have had if it had not been
accelerated. Between ct and c(t+t), the lines must be continuous (no starting or stopping
in the absence of charge). They join with straight line segments if the acceleration is
constant over the intervalt. Note that this behavior adds a transverse component to the
electric field that is directed oppositely to the projection of the acceleration perpendicular
to the line of sight at the retarded time t'=t-r/c. Please remember that the sketch is
exaggerated as v = a t << c. The two spheres (circles) should appear nearly concentric!
The center separation of centers is v t = t (a t). For the lines to be continuous, the ratio
of the transverse (Radiation) and longitudinal (Coulomb) field components must be the
distance that the source has traveled since acceleration divided by the distance that light
traveled during the time elapsed during the acceleration. ERad = ECoul v t/c t where t =
r/c and v = a t.
Radiation Exercises:
1.) The radiation contribution to the magnetic field can be represented as
 qa  r
BRad   3 2
4 c r
Find an analogous expression for ERad . It requires a double cross product! Show that
rˆ  ERad  c BRad where r̂ is the direction of propagation.
2.) Consider a charge q with acceleration a = - 2 d cos(t) in the z direction. Find an
expression for ERad (r ) in terms of spherical components and unit vectors with the
coordinates centered on the charges location.
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3.) Compute Poynting's vector ( S  01 [ E  B] ) for the charge discussed in #2. Describe
the dependence of the radiation on the angle measured relative to the direction of the
acceleration. Compute the average power radiated during one cycle.
4.) Devise an argument that shows that the transverse component ERad (r ) falls off as r -1 at
large distance if the radial component ECoul (r ) falls off as r -2 using the picture above these
exercises. Motivate equation 3.
Vector Triple Product:  A , B , C   A   B  C   B  C  A  C   A B 








5.) Consider a charge q with acceleration a = - 2 d cos(t) in the z direction. The
time dependent dipole moment is
p  qd cos t  kˆ  p0 cos t  kˆ .
average power radiated by the charge q by computing

Compute the
S  nˆ dA over a large sphere
concentric with the charge and averaging over one cycle. Express the result in terms
of po and  plus other factors.
Ans: Paverage =
(For the E-M transition problem, p02  e2 f r i .)
2
p02  
    c 
Appendix III: The canonical interaction: q p  A 
One learns that the interaction energy of an electric dipole in an Electric field is: U dipole
=  pE Dip  E . (The symbol
p
is adopted for the electric dipole moment to avoid
confusing it with the momentum.) This result is not directly applicable to the evaluation
of the interaction energy of an electromagnetic wave and an atom. The result was derived
by considering two equal, but opposite charges in an electrostatic field. E-M radiation is
scarcely electrostatic. As this is a deep mystery, the resolution is rarely presented prior to
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graduate school. Here the path to the answer is to be sketched, but not presented in detail.
Just follow the flow of the ideas. By the time the dust settles, the conclusion is that the
equation for the basic interaction energy does not change although the way that one
regards it may change.
Matter-EM wave interaction: H1 =  pE Dip  E
The canonical method to identify a quantum operator for a quantity is to begin with
its representation in classical Hamiltonian mechanics in terms of coordinates and
momenta. Unfortunately, we need to develop the lagrangian as a step in identifying
the canonical momenta and the hamiltonian itself. We could start with that T – U
prescription.
 L  ½ m xi xi  q ( x, y, z )
Summation notation is invoked so repeated indices are summed from 1 to 3.
Relativity could guide me to the incorporation of magnetism. The electrostatic
potential  is the zero element of a four vector and magnetic interactions involve the
velocity. The four-potential and the four-velocity are:
A  (c1, Ax , Ay , Az )  and  v   ( c,  vx ,  vy ,  vz ) 
  [1  (v c)2 ]½ 
The lagrangian is a scalar so we try the inner product of these vectors as the potential
term. A v   (  v  A) This term obviously includes relativistic corrections that
we are ignoring in this course and in the expression for the kinetic energy. In the  
1 limit, L  ½ m xi xi  q (  v  A) .
The previous development seems suspect to validate the proposed function. The
lagrangian is whatever it takes to reproduce the classical equations of motion. What
do we expect for the magnetic part of the Lorentz force? The equation for v  B is to
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be developed with Bi replaced by  ijk

x j
Ak , the component representation of
B    A 
Fmag  qv  B  q  ijk x j kmn xm An  q  kij kmn x j xm An

A
Fmag, i  q  im jn   in jm  x j xm An  q xn Axn  xm x i
m
i

It is time to test our candidate for the lagrangian.
L  ½ m xk xk  q (  xm Am )
L
xi
L
xi
 m xi  q Ai  pi ;
 
Am
 q ( 
x  xm x )
i
i
 

d L  L  m x  q    Ai  q x Am
i
m x
xi
xi
t
dt xi
i
x
k
Ai
xk

Realizing that we are free to re-label dummy indices and noting that the electric field
is E    t A , it is confirmed. The proposed lagrangian works. The hamiltonian
follows as H  pk xk  L  m xk xk  q xk Ak  ½ m xk xk  q (  xm Am )  .
H  pk xk  L  ½ m xk xk  q
The answer is the total energy which appears pretty simple, but we have yet to
eliminate the coordinate velocities in favor of the momenta as required.
xi  m1  pi  q Ai  .
H  21m  pk  q Ak  pk  q Ak   q 
p p
2m
q




q2


 m p  A  2m A  A  q
Choosing the gauge   A  0 and  = 0,1


q
q2
Hˆ    2m p  A  A  p  2m A  A
[QMDyn.14]
The interaction hamiltonian above is appropriate for a point charge in an external
field. For particles with intrinsic structure that leads to a magnetic moment, the
magnetic dipole-field interaction must be added.
1
Electromagnetic Processes By Robert Joseph Gould Google Books
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Search: Electromagnetic interaction hamiltonian non-relativistic.
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Hˆ      B     (  A)
The H term is active in the interaction electromagnetic transitions of an atom or
molecule, and the H term is active in the interaction of the field with an intrinsic
(say electron) spin. Terms describing the interaction of extended charge (and current)
distributions such as orbiting electrons can be found inside H. 
Examine the terms:
q2
2m
 A  A operates on the field rather than the atom so we need not fret over it until we
wish to quantize the electro-magnetic field. The operators for the particle momentum and
the vector potential act on different sets of coordinates so p  A  A  p  2 p  A . That is:
the operators for p and A commute.
We are expanding in terms of eigenfunctions of the unperturbed hamiltonian and
ˆ ˆ
Hˆ o  p p 2m  V (r ) ,  Hˆ o , x   12m  pˆ  pˆ , x    i m pˆ x and  Hˆ o , r    i m pˆ .




q
q
Hˆ    m p  A  m
 mi    Hˆ , r   A  iq   Hˆ , r   A
o
o
We are interested in matrix elements of the form:

 f Hˆ   i  iq  f  Hˆ o , r   i  A

Recall that i and f are eigenfunctions of the unperturbed hamiltonian with
eigenvalues Ei and Ef.
 f  Hˆ o , r   i   f Hˆ o r  i   f rHˆ o  i   f E f r  i   f rEi  i
Substituting


 f Hˆ   i  iq ( E f  Ei )  f r  i  A  iq( f  i )  f r  A  i
Recall: the vector potential operator does not act on the atomic coordinates. It acts as
a scalar multiplier for the atomic wavefunctions.
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Now we need to consider the vector potential for a plane wave given our choice of
gauge [   A  0 and  = 0].
A(r , t )  Aoe
i ( k r  t )
so E   At  i A
As k  r  1, A(r , t )  Aoe
i t
Energy conservation gives f - i = - . E   At  i( f  i ) A
 f Hˆ   i  iq( f  i )  f r  A  i  q  f r  E  i
By comparison, we can represent the perturbation as Ĥ   q r  E   p  E where p
is the electric dipole moment qr  er . The final effective form of the perturbation
becomes:
Ĥ   qr  E   p  E  er  E
[QMDyn.15]
Tools of the Trade: Note that an operator was re-expressed as a commutator of two
operators. One of the operators in the commutator is an eigen-operator for the basis
set of states so its introduction is reduced to eigenvalue multiplication rather than a
general operator action.
Problems
1.) The harmonic oscillator wavefunctions are:
1
 n ( z, t ) 
2n n ! 
H n ( z) e
2
z 2
e
i ( n 12 )t
 un ( z ) e
i ( n 12 )t
The lowest three harmonic oscillator wavefunctions are:
 0 ( z, t ) 
1

e
2
z 2
e
i ( 12 )t
 2 ( z, t ) 
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 n ( z, t ) 
2 z 2 1
2 
e
2
z 2
e
2 z

e
2
z 2
e
i ( 32 )t
i ( 5 2 )t
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Compute the time dependent probability densities for the mixed states:
M 01 (r , t ) 
1
 0 ( z, t )  1 ( z, t )
2
1
 0 ( z, t )  2 ( z, t ) 
2
M 02 (r , t ) 
Assume that the density is that of a charged particle. Find the oscillation frequencies
for the time dependent terms. Which state, M01 or would radiate
electromagnetically most strongly under the dipole model? Explain. Sketch the
probability densities at four equally spaced times during an oscillation period in each
case. Use the sketches to support your claim.
2.) The E-M electric dipole perturbation for a harmonic oscillator in one dimension is
i t
 i t
Hˆ 1 ( z, t )   q z E0 cos( t )   q z E0  12  e   e  
The spatial parts of harmonic oscillator wave functions satisfy the recursion relation:
z un ( z ) 
n u ( z) 
2 n1
n  1 u ( z)
2 n1
Give the selection rules for allowed transitions caused by H1.

dbj
 iq 2
dt
E  u
0
j
z un
 e
i (n  j  )t
e
i (n  j  )t


What should  be to cause the allowed transitions efficiently? What is the classical
oscillation frequency of the system? What frequency radiation should be emitted by a
charge oscillating at the classical frequency?
3.) Referring to the previous problem, the transition rates for a harmonic oscillator
from state n to n + 1 and from n to n – 1 are proportional to the square magnitudes of
n 1 z n
R
the matrix elements of the interaction. nn1 
Rnn 1
n 1 z n
the Einstein’s coefficients section. Recall that z = x =
2
2
. Compute this ratio. Read
1
2
 aˆ †  aˆ  . The real point is
that the answer is the same with or without the  so just calculate it.
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
4.) Discuss the following statement. “By definition, the Hamiltonian acting on any
allowed state of the system  yields the energy times that same state. Ĥ    ,
where E is the energy of the system.” Give examples to support your discussion.
5.) In the case of a particle in a 1-D box, an infinite well with the range [ 0 < x < a],
the normalized spatial states are: un ( x)  a2 sin  n a x  . The allowed energies are
 ma  . At time t = 0, the system is in the state (x,0) =

En  n 2 
2
2
2
5
12
13 u1 ( x)  13 u2 ( x)
where
u1(x) and u2(x) are the ground state and the first excited state .
a.) Give the form of (x,t).
b.) Give the probabilities to find the particle in the ground state and in the first
excited state.
c.) A detailed calculation shows that
 2
 1.69 eV
2 ma2
. Compute E, the
expectation value of the energy for the state (x,t).
d.) Compute | (x,t) |2. Discuss the result.
e.) Verify that un(x) is properly normalized. What is the average value of sin2(x)
for 0 < x < 2 ?
6.) A general quantum problem with a time-independent Hamiltonian has a complete
set of energy eigenfunctions ( Hˆ n ( x, t )  En n ( x, t ) ). At time t = 0, a general solution to

the problem has the form  ( x, 0)   cm m ( x, 0) . Normalization:
m0

full range
n* ( x, t ) m ( x, t ) dx   nm
a.) Give the form of (x,t).
b.) Give the probability to find the particle in the state j.
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c.) Compute E, the expectation value of the energy for the state (x,t) as a
sum.
d.) Compute | (x,t) |2. Discuss the result.
e.) Give the form of k(x,t).
f.) Under what conditions is (x,t) an eigenfunction of energy?
A general quantum problem has a (time-independent) Hermitian operator Q̂ that
has a complete set of eigenfunctions ( Qˆ  n ( x, t )  qn  n ( x, t ) ). Every Hermitian
operator has a complete set of eigenfunction. As the set is complete, any allowed state

can be expanded in terms of them at time t = to. ( x, t0 )   cm  m ( x, t0 ) .
m 0
a.) Give the probability for a measurement to yield a value qj consistent with the
system being in the state j.
b.) Compute Q, the expectation value of the operator at time to for the state
(x,to) as a sum.
c.) Can you give the form of k(x,t)?
d.) Under what conditions is (x,t) an eigenfunction of Q̂ ?
The Hamiltonian Ĥ and the Hermitian operator Q̂ can have simultaneous
eigenfunctions if their commutator [ Ĥ , Q̂ ] = 0.
A 1-D quantum problem has the momentum operator pˆ  i

x
that has a
complete set of eigenfunctions [ pˆ  n ( x, t )  pn  n ( x, t ) ]. Every Hermitian operator has a
complete set of eigenfunctions. If the spatial range for the problem is: - x < ,
then the expansion of an arbitrary function is a Fourier transform.
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
 ( x, t0 )   cm (t0 )  m ( x, t0 )  1
m 0

2


A(k , t0 ) eikx dk .
a.) What is [ x̂ , p̂ ]?
b.) What condition must the Hamiltonian satisfy to ensure that [ Ĥ , p̂ ] = 0
c.) Identify a problem that has solutions that are simultaneous eigenfunctions of
Ĥ and p̂ .
9.) For the hydrogen atom, study A  r , t ) 
1
2
u100 (r )  u210 (r ) e i (21  )t  e  i10t . Get the
explicit form for: A  r , t ) 2  12  u100 (r ) 2  12 u210 (r ) 2  u100 (r )u210 (r ) cos([21   ]t ) . The
probability density describes the electron which has charge – e. Compute the
oscillating electric dipole moment for the density  A  r , t ) . Following problem 5 in
2
Appendix I, compute the power P radiated by this oscillating dipole. Estimate the
radiative lifetime of the 210 state of hydrogen as (E210 – E100)/ P.
10.) Estimate the oscillating electric dipole moment coupling between the states 210
and 100 of the hydrogen atom as: Re[ 210  e r 100 ] where you need to include the
time dependence of each state. Following problem 5 in Appendix I, compute the
power P radiated by this oscillating dipole. Estimate the radiative lifetime of the 210
state of hydrogen as (E210 – E100)/ P. Compare your result with that of the previous
problem. Do not sweat factors of 2 or 3 at this point.
11.) Consider a quantum mechanical problem with the Hamiltonian Ho and
eigenstates n (r , t )  un (r ) ei ( E /
n
)t
that satisfy Hˆ 0 n (r , t )  En n (r , t )  En un (r ) ei ( E / )t .
n
These eigenstates form a complete set appropriate for expanding any function defined
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over the same region of space. In particular, after adding a small perturbation  Ĥ1 to
the original Hamiltonian, the full Hamiltonian becomes: Hˆ  Hˆ 0   Hˆ 1 where the
factor of  serves as an explicit reminder that the perturbation is small. A solution to
the full problem can be expanded in terms of the eigenfunctions of the original

problem.  r , t )   c j (t ) u j (r ) ei ( E /
j
)t
j 1
Use: i

t
  ˆ 0   ˆ 1   and Hˆ 0 un (r ) ei ( E / )t  En un (r ) ei ( E / )t to show that:
n

i
 c (t ) u (r ) e
j 1
j
j
i ( E j / )t

ˆ u (r ) e  i ( E j /
  
 j
j 1
)t
n
where c j (t ) 
dc j (t )
dt . Project out the relation
for cm (t ) by multiplying the relation by um* and invoking the orthogonality relation
*
i ( E /
 um un dV   mn . Show that you find: i cm (t ) e m

)t
ˆ u ei ( E j /
  um  

j
)t
. Assume
j 1
that the system was in the state n at time t = 0. This condition means that at t = 0, cn =
1 and that cj = 0 for j  n. Find the approximate form of  cm (t ) for very small positive
times given these initial conditions. Display explicitly, the integral that um  ˆ  u j
represents.

12.) A general mixed state  (r , t )   am um (r ) ei t where the um (r ) are eigenstates of
m
m 1
the full hamiltonian with energy eigenvalues Em = m. Use the Schrödinger equation
to show that dak/dt = 0 for all k. As always, one projects out a specific coefficient by
using the inner product uk Hˆ  and the orthogonality relation uk um   km .
13.) The spherical harmonics, the Ym(), are the angular factors in the hydrogenic
wavefunctions. The perturbation for electric dipole transitions has the form eE0  r .
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Express r  xiˆ  y ˆj  z kˆ in terms of r and the Ym(). The results reveal the  and
the m values (effectively, the angular momentum and z component of angular
momentum) that can be associated with x, y and z.
Y00 ( , ) 
1
;
4
Y10 ( , ) 
3
cos  ; Y1,1 ( , ) 
4
1 3
sin  e i
2 2
x
 
Initial( , m)   y   Final( , m)
z
 
What selection rules are suggested by the equation above? Treat the cases of x, y and
z separately.
14.) The spontaneous radiative decay rate of an atom from its initial (i) to final (f)
state can be calculated in the first order electric dipole (E1) approximation as:
spon
i f
R
if3

3  o c3
where
pif2  e2 
p
2
if
f xi
[QMDyn.16]
2
 f yi
2
 f zi
2

[QMDyn.17]
a.) Compute the radiative decay rate the hydrogen 210 state to the ground state.
b.) The lifetime of the hydrogen 210 state must be less than ??? ns.
c.) Compute the decay rate the hydrogen 200 state to the ground state in the electric
dipole approximation.
Exercise: Compute the hydrogenic matrix elements 100| x |210 100| y |210
100| x |211 100| x |21-1100| y |211 and100| y |21-1
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Exercise: Given 100| x |211 = 100| y |211 = ao (27/35) and, 100| z |211 = 0,
compute the spontaneous decay rate of the 211 state of hydrogen. Compare it to the
decay rate for the 210 state.
Exercise: Compute the decay rate of the 210 state of hydrogen to the 200 state. Only
200| z |210 is active. Assume that the 200 state lies 0.33 cm-1 below the 210 state.
(This is not quite correct, but it puts us in the ballpark to make a point.)
Note:  = 2 (3 x 1010) v where v = 0.33 cm-1; v = -1 with  expressed in cm.
Compare the sizes of the matrix elements 200| z |210 and 100| z |210. What factor
in the expression for R dominates the difference?
Exercise: Compute the decay rate of the 200 state of hydrogen to the 100 state.

References:
1. David J. Griffiths, Introduction to Quantum Mechanics, 2nd Edition, Pearson
Prentice Hall (2005).
2. Richard Fitzpatrick, Quantum Mechanics Note Set, University of Texas.
3. Robert Eisberg and Robert Resnick, Quantum Physics, 2nd Ed., John Wiley, New
York (1985).
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4. The Wolfram web site: mathworld.wolfram.com/
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