Quantum Dynamics – Quick View Concepts of primary interest ...

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, do 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, -2, … , 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

Contact: [email protected]

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 [ 0 < x < a], the

mixed state below exhibits time-dependence.

2222

21( , ) sin sin where ni ti t

max x

a aax t e e n

.

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 221( , ) sin sin i ti tx xa aax t e e

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.

6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 2

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

oppositely to the component of the acceleration2 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 density and, should not radiate in this semi-classical view.

Be warned: The flow of ideas for describing transitions between quantum states

rather than a 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.

22ˆ( , ) ( , ) ( ) ( , )2i r t H r t V r r tmt

[QMDyn.1]

Consider a state that is an eigenfunction of the Hamiltonian.

ˆ ( , ) ( , ) ( , )n n nH r t E r t i r tt

n

The wavefunction can be separated into temporal and spatial parts: ( , ) ( ) ( )n nr t u r T t

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. 2 The accerlation is evaluated at the retrarded time, the time at which the radiation was emitted to be observed here now.


leading to the equations ˆ ( ) ( ) and ( ) ( )n n n n n nH u r E u r E T t i T tt

and hence

( , ) ( )n n nni tr t c u r e

where n = -1 En and cn is a complex number usually of

magnitude 1. For any energy eigenstate, the probability density is stationary (time-

independent). 2

( ,r t) ( , ) ( ( ) ) ( ) ( )n n n n nn ni t i tr t u r e u r e u r

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 of 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 ( , ) ( ) ( ) mn mmixed

ni t i tr t a u r e bu r e with its

associated probability density 2 2 2 2 ) )*( (( , ) ( ) ( ) ( ) ( ) ( ( ) ( ) )mixed n m n m n m

m n m ni t ir t a u r b u r a bu r u r e ab u r u r e t

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 ( , ) ( ) ( ) mn mmixed

ni t i tr t a u r e bu r e is a mixed state of

( )nni tu r e and u r( )m e mi t 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 state-

to-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, dx/dt may not be zero for the state mixed.

A general mixed state 1

( , ) ( ) mm m

m

i tr t a u r e

does not describe state-to-state

transitions. Transitions correspond to the coefficients ak that depend on time.

Schrödinger’s equation shows that this is not the case as long as the are

eigenstates of the full hamiltonian.

( )mu r

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 that has eigenfunctions 0H

( ) mi tmu r e

0ˆ ( ) ( )m mi t i t

m m mH u r e E u r e .

(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, , ) ( ) mi tm m

mr t c u r e

. That is: the eigenfunctions form a complete

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:

*( ( ) ) ( )k mi t i tm mk kall space

u r e u r e dV mk

.

Begin with the system in a particular pure eigenstate, say the state |n


, ) ( ) ni tnr t u r e

. 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

states that: 0 0ˆ ˆ( , )H r t ( ) ( )n ni t i t

n n nddtH u r e E u r e i . It follows that:

0ˆ( , ) ( , ) ( , ) ( , ) ( ) ( )n ni t i t

n n nddt

i ir t t r t t r t H r t t u r e E u r e t .

0ˆ( , ) ( , ) ( ) ( ) ( )n ni t i t

n n nddt

i ir t t H r t t u r e e E t u r i t u r n n

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 rate:

1 ( )

( )

n

n

i tm n

ni tn

ddt E u r e

i u r e

. 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, ni te .

Exercise: Consider e-it. Show that ( ) ( )i t i t t i t i ti te e e e

Transitions between eigenstates of the base hamiltonian Ho are caused by interactions

which appear as perturbations, new terms added to the Hamiltonian to represent

external influences on the quantum system. The basic problem is described by the

unperturbed hamiltonian , and its eigen-solutions {… 0H ( ) mi tmu r e …} provide a

complete basis for expanding any well-behaved function defined over the same region

of space. The small interaction term or perturbation is added such that the full 1H


problem is described by , and the full time development equation is: 0ˆ ˆ ˆH H H 1

0 1ˆ ˆ ˆ( , ) ( , ) d

dtH r t H H r t i .

0 1ˆ ˆ ˆ( , )r t t ( , ) ( , ) ( , )i ir t H r t t H H r t t

With the assumption that the system starts at time t in state n of the unperturbed

Hamiltonian,

0 1ˆ ˆ( , )r t t ( ) ( )n ni t i t

n niu r e H H u r e t

0ˆ ( ) ni t

ni H u r e t

The piece is understood and not very interesting as it has nothing

to do with transitions; it is just the phase of n changing at the rate - n. Examine the

new small piece:

1ˆ ( ) ni t

ni H u r e t

What is 1ˆ ( ) ni t

nH u r e ? 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.

1ˆ ( ) , ) ( )n ki ti t

n new k kk

i H u r e t f r t b u r e

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 ( ) ji tju r e

, the

companion basis set function for bj, and the orthogonality relation is invoked.

( )

1 1

( ) ( )

ˆ ˆ( ) ( ) ( ) ( )

( ) ( )

j n jn

k j k j

i t i ti tj n j n

i t i t

k j k k jkk k

i iu r e H u r e t u r H u r e t

b u r u r e b e b

j

All the bj that are initially zero are proportional to t for short times; it follows that:

( ) ( ) *1 1

ˆ ˆ( ) ( ) ( ( )) ( )n j n ji t i tjj n j nall space

i idbu r H u r e e u r H u r dV

dt


The expression is a little more compact if a matrix element [H1]jn represents the Bra-

Ket 1ˆ

j nu H u .

( ) ( )

1 1ˆn j n ji t i tj

j n jni idb

e u H u e Hdt

[QMDyn.2]

1ˆ

nju H u is [H1]jn, the jn matrix element of the perturbation . 1H

Exercise: Rewrite the three lines of equations above replacing the Dirac Bra-kets with

the integrals that they represent. Rewrite [H1]jn = 1ˆ

ju H un in integral form. Which

forms of the equations are used when actual values are calculated?

State to State Transitions:

The system starts in the state un so at time to, cj = jn. All the cj = 0 for j n. Except for

small corrections ( )

1ˆn ji t

jj jdc db

dt dti e u H nu

)

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 ( n jie t means that the change oscillates rapidly averaging to 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.

What is H1? Consider a free space electromagnetic plane wave with frequency

washing over a charge.

*0 0 0

( ) ( )12( , ) Re cos( ) ,i k r t i k r t

0E r t E k r t E e E e E k

*0 0

( ) (12( , ) i k r t i k r tB r t k E e E e )


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. 1 .distr extH p E

A hydrogen atom consists of two equal, but opposite charges it is an electric dipole.

If is the electron’s position relative to the proton, then the dipole moment isr

p e r

and the interaction Hamiltonian is: 1(ˆ ( , ) i k r tH p E r t e r E e 0

) . The form of the

interaction 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

1H

p E 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 310k r

. The exponential 1 1ik re ik r .

To simplify the notation, is assumed to be real so that the interaction is: 0E

1 01

2ˆ ( ) i t i tH t e r E e

e . Substituting into the transition amplitude equation,

0( ) (

2n nj

njj i t idb

dtie u r u E e e )j t

. [QMDyn.3]

( )n nu u r so j nu r u

is time-independent

As we seek interesting behavior rather than the exact numerical predictions, the two

factors

[( ) (n nj ji t i

e e )t ] and 0j nu r u E 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 ( )n ji t

e or

( n ji te

) 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.

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 condition become: En Ej ]t < or [En -Ej

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 or the n - j oscillations.


]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.

0( ) (

2n nj j

njj i t idb

dtie u r u E e e )t

RULE ZERO: Our first selection rule is that = |n - j|. Energy must be

conserved. With energy conservation satisfied, we find:

02 njjdb

dtie u r u E

for = |n - j|.

The additional selection rules are those necessary for 0nju r u E

to be non-zero

or not excessively small. The selection rules derives 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: 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 0nju r u E

includes some

information about the polarization dependence of the interaction of the atom and the

electromagnetic field. To begin, the piece 1 nju r u

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 it cannot happen;

it only means that it is not caused in the lowest order approximation by that

perturbation 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: ( )1

ˆn ji tjn nj j

idbe u H u u r

dt u

that identifies r as the atomic parameter active in the perturbation.

The Character of the Perturbation:

Cartesian representation: ˆˆ ˆr xi y j z k

Spherical: ˆˆ ˆsin cos sin sin cosr r i j k

1,1 1, 1 1,1 1, 1 101

2 24

3ˆˆ ˆ( ( ( ( (ir r Y Y i Y Y j Y

k

The final form is useful when the angular momentum character of the perturbation is

discussed. The Ym 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.

The Cartesian representation reveals that the perturbation is an odd function.

*( ( )) ( )j n j nall spaceu r u u r r u r dV

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 ( ) and ( )n m n mr r

for 200,

210, 311 and 320. Conclude that: ( ) ( 1)n m n m ( )r 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.

The spherical representation of r

, the atomic factor in the perturbation, reveals its

angular momentum character. The spherical harmonics or Ym’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

of the final state must be the vector sum of the initial angular momentum jL

nL

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 .


A detailed study of the inner product in 0j nu r u E 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.

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 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.

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 1ik re ik r

1 . Keep the second term.

( ) nk r u

j n ju r u u r r i . The new term ( )ju r ik r u n

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 310k r

k r

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 110 .

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 )

Exercise: Given that has angular momentum character one, what angular r

momentum character would you predict for (r ik r ) ? Recall that is just a constant k

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, 1s21S 1s, 2s23S 1s, 2s21P 1s, 2p23S 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+1LJ

for the atom

S: the total spin angular momentum of the

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.


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 hyperphysics.phy-astr.gsu.edu/hbase/atomic/grotrian.html

= 0 and an electron spin flip (S 0) are needed. One expects millisecond decay

times1. As S = 0 is the allowed case selection rule, the para- and ortho- form almost

independent set 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

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. Metastable molecular states are used to store

energy in the chemical oxygen iodine laser (COIL). Excited state lifetimes can be minutes.


dynamics depend on trapping energy in the metastable helium. Desperately seeking to

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 metastable states when they are recaptured or to cause direct electron excitation. 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 (1g) 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 short high power pulse.



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 (Y00),

2s (Y10), 2p0 (Y10) and 2p1

u

0/10

r aAeu

0/ 220 (1 ) r aB b r eu

/ 21 sin riD r e e

0/ 220 cos r aC r eu

02,

a

(Y1,1) electron states of the hydrogen atom. The images

below are the basis for a discussion of mixed state probability densities.

001

4( , )Y

10

3cos

4( , )Y

1, 1

1 3sin

2 2( , ) ieY


All the orbital are viewed in along the y

axis except for the 2P+1 which is viewed

down the z axis. Relative phase is varying

form 0 to 2 by color varying from red to

magenta (for 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(,).


These images represent the wave amplitude; it is * that represents the probability

density.

The 210 to 100 Transition: Consider the mixed state:

10 1021 21( )1 1100 210 100 2102 2

, ) ( ) ( ) ( ) ( )i t i ti t i tA r t u r e u r e u r u r e e

.

where 100 1/ 2 3/ 2 3/ 2 3/ 20 0 0

0 0 000

/ / /1 2 1 2

4( , )( , , ) r a r a r a

a a au Yr e e e

and

210 1/ 2 1/ 23 / 2 3 / 20 00 0

0 010

/ 2 / 21 1

32 24( , )( , , ) cosr a r ar r

a aa au Yr e e

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.


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: 21 10( )1100 2102

, ) ( ) ( ) i t iA r t u r u r e e t ,

2 2 2

100 210 100 210 211 12 2, ) ( ) ( ) ( ) ( ) cos([ ] )A r t u r u r u r u r t

.

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.

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)

Study the 200 - 100 Mixed State: Consider the wavefunction:

10 20 20 10( )1 1100 200 100 2002 2

, ) ( ) ( ) ( ) ( )i t i t i t i tB r t u r e u r e u r u r e e .

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 2, )B r t .

Study the 211 - 100 Mixed State: Consider the wavefunction:

10 21 1021 ( )1 1100 211 100 2112 2

, ) ( ) ( ) ( ) ( )i t i t i ti tC r t u r e u r e u r u r e e .

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 2, )C r t .

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 210-

100 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 12

( )(i f i f ) has terms with factors

eimand e-im. Leading to there being probability lumps in proportional to

|e½imei½m

+ e-i½m 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

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.

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:

0( ) ( )

2 wherejn jnj n j

j i t i tdbdt

ie u r u E e e n j n

. Assume that the

perturbation is turned on at time t = 0 and is on until time , and find that:

0 0

( ) ( )

2( ) jn jnj j n

i t i tieb u r u E e e

dt


0

( ) ( )

21 1

( )jn jn

j j njn jn

i ie e e

b u r u E

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

is less than that of the initial atomic state n. With this restriction only jn + = 0 (or

equivalently: n - =j) is possible. The first term has a ‘zero’ in the denominator

and is large compared to the second term which can be neglected.

0

0

( )

( ) / 2

2

2

1( )

2 sin[( ) ]

nj

jn

j j njn

jnj n

jn

i

i

e

ie

eb u r u E

u r E u e

The probability amplitude is squared to find the probability:

2 2

22 22 2 2 20 0

1 1sin[( ) ] sin[( ) ]( ) cosjn jn

j j n jnjn jn

b u r E u H E

22 2 22 20

1( ) cos for shortj jnb H E

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 which is consistent

with the uncertainty principle. 2 22

( ) ( ) ( ) ( )n j jd h

c b E E b Edt


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 state

for the electromagnetic wave part of the system. The EM field has a rich continuum of

allowed states. The final states do not differ in the final atomic state and they differ

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 about 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 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. 2 2 2 20

1 cos ( )jnR H E E . At

this point, we cannot estimate the factors in this expression, but we observe that a

states decays at a constant 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 (atom state * field state) in the allowed E band

grows as 2 while the uncertainty band E shrinks as h/.


Return to:

2 2

22 22 2 2 20 0

1 1sin[( ) ] sin[( ) ]( ) cosjn jn

j j n jnjn jn

b u r E u H E

2

22 22 2 20

1 sin[( ) ]( ) ( ) cos ( )

E E jnn j jnE E

jn

dc b E dE H E

dt

E dE

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|.2

7 22

sin ( )( ) sin ( )n

n n xD x c nx

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},

PlotStyleThickness[0.006]]

Let n and x jn + .

2

222

0 22sin [( ) ]

( ) ([ ]

jnj j n

jn

eb u r E u

2

2 22 2 20 2

1 sin [( ) ]cos ( ( )

[ ]

E jnn jnE

jn

dc H E

dt

E dE

2

222

0 22sin [( ) ]

( ([ ]

E jnn j nE

jn

edc u r E u

dt

)E dE

222

02 ( ( ) ( )E

n j n njE

edc u r E u E E

dt

E dE

In the (nanosecond) long time limit, the Dirac delta enforces energy conservation.

22

0 22 (decay j n nje )eR u r E u E

and = |En – Ej|


CAUTION: These arguments have been qualitative. Many multiplicative factors have

been lost and a few concepts have been bruised.

REPEAT

222

01( ) ( ( )j j nb u r E u jn

In the short time limit, the probability to find the system in the state j grows linearly in

time.

222 2 20

1( ) ( ) cos ( ( )j j j n jnP b u e r u E

22 21( ) ( ) ( ( )j j jn jnP b H

The transition rate to the from the atomic state n to the atomic state j with the emission

of a single photon is the probability to be found in the state j divided by the time that

the perturbation has been acting: Rnj = -1 |bj|2 or

22

02 ( )nj j n njeR u r E u

END REPEAT

The delta function in frequency means that the transition1 is not likely unless the

energy of the initial excited state nis 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

1 A transition to a lower energy level of a system by emitting a photon is called a radiative decay.


the blackbody problem, (E), a density of states for photons can by found. (See your

modern physics course and you statistical mechanics course.)1 Call it (E) where E is

the energy. Sum or integrate over the final states for the photon and

22

0 02 ( ( ) )nj j n j ne u r E u E dE

R

22

02 ( (nj j n n je u r E u )

R where 2 321

( )( ) cE E

This final result is a form of Fermi’s Golden Rule2 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 ***

Fermi, Enrico (1901-1954) Italian-American physicist who was born in Rome. Fermi discovered the statistical laws, now called Fermi-Dirac 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, | 0ju r E u

n |2 is replaced by 2 2

0 cosj nu r u E . The angle

between the direction of j nu r u and the perturbing electric field is . The factor

cos2 has an average value of 1/3. Using this approximation, we can compute order of

magnitude estimates of transition rates without carefully evaluating 0j nu r E u |2.

1 The development is similar to the chapter 4 particle in a box. The details will be added in an appendix. 2 The Golden Rule was actually developed by Dirac 20 years before Fermi dubbed it one of two golden rules.


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 intensity of the applied electromagnetic wave.

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 could 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.

Einstein Coefficients: The form of the transition rate equation

2 2 222 202 cos ( );nj j n j n j n n j

eR u r u E u r u u r u

shows that the induced transition rate 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 wave. Spontaneous transitions happen. An

atom in an excited state can emit a photon and transition to a lower state in the

absence of any applied field (beyond the zero point field). Einstein postulated there

are three coefficients governing the transitions between an upper level u and a lower

level :

Au: the spontaneous transition rate from the upper to lower state

Bu: the induced transition rate per u from u to


Bu: the induced transition rate per u from to u

Note: u ) is a density w.r.t. frequency; (E) is w.r.t. energy

In equilibrium, the populations nu and n of the upper and lower states should be in

Boltzmann ratio: nu /n= )( uE E hkT kTe e

, and, as the relative population do not

change in equilibrium, the rate of transitions up should equal the rate down.

{ )} andu u u u u u u uR n A B R n B )

In thermal equilibrium: andh

kT

u un n e R R

u .

Substituting, rearranging and comparing with the Planck black-body radiation formula

for (vu) which must hold for systems in thermal equilibrium:

8 1

) )11

hhkTkT

u

uu Planck u

u

u

AB h

B c eeB

It follows that: Bu = Bu and that Au= 8hc-33 Bu . The equivalence of the

induced rates up and down B coefficients also follows from the rate formula. The

beauty is Au= 8hc-33 Bu which provides the value of the E-M wave radiation

density that is necessary to provide a transition rate equal to the spontaneous decay

rate. The quantum verification 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 â† |n = (n + 1)½|n + 1 remove a photon from a mode add a photon to a mode

The quantum state counting predicts that absorption of a photon from a mode is

proportional to the number of quanta n in that mode while emission of a quantum into


that mode is proportional to n + 1, the number of quanta 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. Spontaneous emission is properly

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: ˆˆ ˆx y zk k i k j k k

). 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

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.

Combining the factors discussed above (Kirkpatrick):

22

(3

absi f if fi

o

)R E p [QMDyn.4]

22

(3

stimi f if if

o

)R E p [QMDyn.5]

32

33ifspon

i f ifo

Rc

p [QMDyn.6]


where 2 22 2if e f x i f y i f z i p 2

[QMDyn.7]

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 |2po100| z |210.

|2po 0

03/2

0

2102

1 1

3( , )

ra

ar

aYe

10

1 3cos

2( , )Y

|1s 03/2

0 001 2 ( , )aa

rYe

= 0

3/2

0

1 2 14

aa

re

z = r cos = r 1032 ( ,Y

)

0 0

00

3 22

10 100 0 0

2 14

1 1100 210 sin

362

ra a

a

rr

az r r dr Y Ye e

d d

Note that the factor of (4)-½ is Y00, and the 2 (

/3)½ Y10 is cos.

0

0 0

4

0 0

321 1

100 21036

ar

r dra a

z a e

5

0 0

75

23

2100 210 4! 2

623

z a a

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.\ 3 3

2 203 3

1510

23

( )3 3

if ifsponi f if

o o

R ec c

p a

2

19 3 19 211 2

34 4 12 8 3

1511

23

(10.4 [1.602 10 ] ) [1.602 10 ](5.29 10 )

(1.054 10 ) (8.85 10 ) (3 10 )ifspon

i f J msC m

J CR m

Js


3 5 2 117

8 14 3 124

1511

23

(10.4 1.602 5.29 ) 106.63 10

(1.054 8.85 3 10 )sponi fR s

s

The lifetime is the inverse of the decay rate: = 1.51 ns.

Exercise: Compute the hydrogenic matrix elements 100| x |210 100| y |210

100| x |211 100| x |21-1100| y |211 and100| y |21-1

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.

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.

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 10-73) with the

widths in liquids and solids being potentially very large (E/E 10-1). Thus the energy

equation: = Eu – Especifies a range of frequencies for photons that can be

absorbed.


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

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.


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.

) )u u u u u u u u u uR R R n B n B n n B )

u

where the equality uB B 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 current developments were not rigorous, but rather

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)


http://upload.wikimedia.org/wikipedia/commons/a/ad/Boltzmann2.jpg�

Selection Rules: (pages 187-191.) Motivate equation (12.63).

Are the rules called selection rules when applied in the context of time independent

perturbations?

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.

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|nAn m| B|n which identifies the initial state and m| AB|nAm| B|n

Am m| B|nwhich 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|ncan be

expressed in terms of the eigenvalues associated with |m and |n and m|B|nthe

matrix element of interest. That is: m|C|n = fC(m,n) m| B|n

m| [A,B]|nm| AB-BA|n (Am – An) m| B|n= fC (m,n)m| B|n[QMDyn.8]

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.

In general, one should become familiar of the commutators of a new operator with

each of the current 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

ˆ ˆ ˆ[ , ] , [ , ] and [ , ]i j ijk k z zx L i x x L i y y L i x Let A Lz.

ˆ' [ , ] ( ) ' 'zm x L 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


Beginning with [y, Lz], one finds that ( ) ' 'm m m y m i m x m . Chaining

the results together, 2( ) ' 'm m 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.

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.8] 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:

2ˆ ˆ ˆ ˆ ˆ ˆ[ , ] 2 ( ) and 0y x x y zL z i xL L y xL yL zL .

It follows that:

2 2 2 2 2ˆ ˆ ˆ ˆ[ ,[ , ]] 2 ( )L L z L z z L [QMDyn.9]

Note that:

2 2 2 2 2 4ˆ ˆ ˆ ˆ' [ ,[ , ]] 2 ' ( ) 2 ( 1) ( 1) 'm L L z m m L z z L m m z m

Now the methods behind equation [QMDyn.8] can be applied to yield:

( 2)( )( 1)( 1) ' 0m z m [FQT.10]

The first factor can never vanish, so the possibilities for a non-zero outcome are

= = 0 and = 1. The = 0 states are spherically symmetric so

nm|z|nmn 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 2 ˆ ˆ ˆ ˆ ˆ[ , ] 2 ( ) and 0.y x x y z

L z i xL L y xL yL zL

Exercise: Evaluate 2 2ˆ ˆ' [ ,[ , ]]m L L z m and 2 2ˆ ˆ' ( )m L z z L m . Expand the relation

2 2 2 2 2ˆ ˆ ˆ ˆ' [ ,[ , ]] 2 ' ( )m L L z m m L z z L m to develop a result equivalent to

[FQT.10].One must first verify that 2 ˆ ˆ ˆ ˆ ˆ[ , ] 2 ( ) and 0.y x x y z

L z i xL L y xL yL zL

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 only

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.


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 o

34Coul

qrr

E

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

field given by 34BS

qv rr

B

. Finally, an accelerated charge has a radiation contribution

to the electric field that is approximated by the relation 24Rad

Ret

qac rE

(for v << c).

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 isto 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


distance while ECoul and BBS fall off as r -2. [There are corrections to RadE

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 and RadE

RadB

. 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.


ECoul

ERad

c t

v t

q

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


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 intervalt. 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

3 24Radqa r

c rB

Find an analogous expression for RadE

. It requires a double cross product! Show that

ˆ Rad Radr E c B

where is the direction of propagation. r

2.) Consider a charge q with acceleration a = - 2 d cos(t) in the z direction. Find an

expression for in terms of spherical components and unit vectors with the

coordinates centered on the charges location.

( )RadE r


3.) Compute Poynting's vector ( 10 [S 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 falls off as r -1 at

large distance if the radial component

( )RadE r

( )CoulE 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 0ˆcos cosqd t k t k ˆp p

ˆS n dA

. Compute the

average power radiated by the charge q by computing over a large sphere

concentric with the charge and averaging over one cycle. Express the result in terms of

po and plus other factors. (For the E-M transition problem, 22 2

0 e f r ip .)

Ans: Paverage = 20

c

p

Appendix III: The canonical interaction: q p A

One learns that the interaction energy of an electric dipole in an Electric field is: Udipole

= . (The symbol E Dip E p 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

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 = E Dip E p

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.

½ (i im x x q x y z, , ) L

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:

1( , , , )x y zA c A A A and ( , , , )x y zv c v v v

2 ½[1 ( ) ]vc

The lagrangian is a scalar so we try the inner product of these vectors as the potential

term. ( )A v v

½ (i im x x q

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, )v A L .

The previously 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 be

developed with Bi replaced by jijk kx A

, the component representation of B A


m mmag ijk j kmn n kij kmn j nx xF qv B q x A q x

A

mag, mi im jn in jm j n n mn i

mixAA

x xF q x A q x x

It is time to test our candidate for the lagrangian.

½ (k k m mm x x q x A ) L

i iix m x q A p L

i ; ( )mm

i i i

Ax x xq x L

kk

i mi m

i i i i

iA AAx x x t x

ddt m x q q x x

L L

x

Realizing that we are free to re-label dummy indices and noting that the electric field

is tE A

, it is confirmed. The proposed lagrangian works. The hamiltonian

follows as ½ (k k k k k k k k m m )p x m x x q x A m x x q x A H L .

½k k k kp x m x x q H L

The answer appears pretty simple, but we have yet to eliminate the coordinate

velocities in favor of the momenta as required. 1i i ix m p q A .

212 2k k k k

p p q qmm m 2mp q A p q A q p A A A q

H

Electromagnetic processes By Robert Joseph Gould

Google Books Electromagnetic interaction hamiltonian non-

relativistic.

Choosing the gauge 0A

and = 0,

2

2 2ˆ q q

m mH p A A p A A


The interaction hamiltonian above is correct 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.

ˆ ( )H 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.

Tools of the Trade:

Problems

1.) The harmonic oscillator wavefunctions are: 2 1 1

2 2 2( ) ( )1( , ) ( ) ( )

2 !n n n

n

i n t i n tzz t H z e e u z e

n

The lowest three harmonic oscillator wavefunctions are:

0

2 12 2( )1

( , )iz

z t e et

2 3

2 2( )2( , )n

i tzzz t e e

2

2

2 52 2( )2 1

( , )2

i tzzz t e e

Compute the time dependent probability densities for the mixed states:

01 0 1

1( , ) ( , ) ( , )

2M r t z t z t

02 0 2

1( , ) ( , ) ( , )

2M r t z t z 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

1 0 01

2ˆ ( , ) cos( ) i t i tH z t q z E t q z E e e

The spatial parts of harmonic oscillator wave functions satisfy the recursion relation:

1 11( ) ( ) ( )2 2n n

n nz u z u z u z n

Give the selection rules for allowed transitions caused by H1.

0

( ) (

2n j n j

j nj i t idb

dtiq E u z u e e )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

the matrix elements of the interaction.

2

12

1

1

1

n n

n n

n z nR

R n z n

. Compute this ratio. Read

the Einstein’s coefficients section. Recall that z = x = 12

ˆ ˆa a † . The real point is

that the answer is the same with or without the so just calculate it.

4.) Discuss the following statement. “By definition, the Hamiltonian acting on any

allowed state of the system yields the energy times that same state. ,

where E is the energy of the system.” Give examples to support your discussion.

H

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: 2( ) sinnn x

a au x . The allowed energies are

2 2

22n maE n . At time t = 0, the system is in the state (x,0) = 1 2

5 1213 13( ) ( )u x u 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

221.69ma eV

. 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 ( ). At time t = 0, a general solution to

the problem has the form

ˆ ( , ) ( , )n n nH x t E x t

0

( ,0) ( ,0)m mm

x c x

. Normalization:

*( , ) ( , )full range n mx t x t dx nm

a.) Give the form of (x,t).

b.) Give the probability to find the particle in the state j.

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 that Q


has a complete set of eigenfunctions ( ). 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.

ˆ ( , ) ( , )n n nQ x t q x t

0( , )

m0 0( , )m mx t c x t

Q

H Q

.

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 ? Q

The Hamiltonian and the Hermitian operator can have simultaneous

eigenfunctions if their commutator [ , ] = 0.

H

A 1-D quantum problem has the momentum operator ˆxp i that has a

complete set of eigenfunctions [ p ( , ) ( , )n n nx t p 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.

0 0 0 0, ) ikx

0

12

( , ) ( ) ( , )m mm

(x t c t x t

A k t e dk

.

a.) What is [ x , p ]?

b.) What condition must the Hamiltonian satisfy to ensure that [ , H p ] = 0

c.) Identify a problem that has solutions that are simultaneous eigenfunctions of

and H p .


9.) For the hydrogen atom, study 21 10( )1100 2102

, ) ( ) ( ) i t iA r t u r u r e e t

. Get the

explicit form for: 2 2 2

100 210 100 210 211 12 2, ) ( ) ( ) ( ) ( ) cos([ ] )A r t u r u r u r u r t

. The

probability density describes the electron which has charge – e. Compute the

oscillating electric dipole moment for the density 2, )A r t . 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.

10.) Estimate the oscillating electric dipole moment coupling between the states 210

and 100 of the hydrogen atom as: Re[ 210 100e r ] 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

that satisfy ( / )( , ) ( ) ni E tn nr t u r e ( / )

0ˆ ( , ) ( , ) ( ) ni E t

n n n n nH r t E r t E u r e . These

eigenstates form a complete set appropriate for expanding any function defined over

the same region of space. In particular, after adding a small perturbation 1H to the

original Hamiltonian, the full Hamiltonian becomes: 0 1ˆ ˆ ˆH H H 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.

( / )ji E te

1

, ) ( ) ( )j jj

r t c t u r

Use: 0 1ˆ ˆ

ti

and ( / ) ( / )0

ˆ ( ) ( )ni E t i E tn n nH u r e E u r e n to show that:


( / ) ( / )

1 1

ˆ( ) ( ) ( )ji E t i E t

j j jj j

i c t u r e u r e

j where ( )( )j

jdc tdtc t . Project out the relation

for by multiplying the relation by um* and invoking the orthogonality relation ( )mc t

*m n mnu u dV .

Show that you find: ( / )( / )

1

ˆ( ) jmi E ti E t

m m jj

i c t e u u e

. Assume 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 for very small positive times given these

initial conditions. Display explicitly, the integral that

( )mc t

ˆmu ju represents.

12.) A general mixed state 1

( , ) ( ) mm m

m

i tr t a u r e

where the ( )mu r are eigenstates of

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 ˆu Hk and the orthogonality relation k m kmu u .

13.) The spherical harmonics, the Ym(), are the angular factors in the hydrogenic

wavefunctions. The perturbation for electric dipole transitions has the form 0eE r

.

Express in terms of r and the Ym(). 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.

ˆˆ ˆr xi y j z k

001

4( , )Y

; 10

3cos

4( , )Y

; 1, 1

1 3sin

2 2( , ) ieY


Initial( , ) Final( , )

x

m y m

z

What selection rules are suggested by the equation above? Treat the cases of x, y and z

separately.

Exercise: Compute the hydrogenic matrix elements 100| x |210 100| y |210

100| x |211 100| x |21-1100| y |211 and100| y |21-1

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).

4. The Wolfram web site: mathworld.wolfram.com/

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