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
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.
2
2
2
2
2 1
( , )x t a sin ax e i t sin ax e i t where n n ma
.
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 ( , ) 1
sin x i t sin 2 x i t2
a a
a
x 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
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.
( , ) ˆ ( , ) 2 2 ( ) ( , ) 2 i r t H r t m V r r t t [QMDyn.1]
Consider a state that is an eigenfunction of the Hamiltonian. ˆ ( , ) n n( , ) n( , ) H r t E r t i r t t n The wavefunction can be separated into temporal and spatial parts: n( , )r t u r T tn( ) ( )
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.
leading to the equations ˆH u rn( ) E u rn n( ) and E T tn n( ) i T tn( ) t and hence ( , ) ( ) n r t c u r en n i tn
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 r t u r en i tn u r en i tn u rn
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 ( , ) ( ) ( ) m
n m
mixed r t a u r e i tn bu r e i t
with its
associated probability density
2 2 2 2 ( ) * ( )
( , ) ( ) ( ) ( ) ( ) ( ( ) ( ) )
mixed r t a u rn b u rm a bu r u r en m i m n t ab u r u r en m i m n
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 ( , ) ( ) ( ) m
n m
mixed r t a u r e i tn bu r e i t
is a mixed state of
( )
n i tn
u r e and u rm( ) ei tm 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 ( , ) ( ) m m m m i t r 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.
( )
m
u 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 Hˆ0 that has eigenfunctions
( ) i mt m u r e ˆ0 ( ) i mt ( ) i mt m m m H 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, , ) ( ) i mt
m m m
r t c u r e
. That is: the eigenfunctions form a completeset 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: * ( ( ) i tk ) ( ) i mt m m k k all space u r e u r e dV mk
. , ) ( ) i tn
n
r 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: Hˆ0 ( , )r t ˆ0 ( ) i tn ( ) i tn n n n ddt H u r e E u r e i . It follows that: 0 ˆ ( , ) ( , ) ( , ) ( , ) ( ) i tn ( ) i tn n n n d dt i i r t t r t t r t H r t t u r e E u r e t .
0 ˆ ( , ) ( , ) ( ) i tn i tn ( ) ( ) n n n d dt i i r t t H r t t u r e e E t u r i n t u rn 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 t m n n i t n d dt 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, ei tn .
Exercise: Consider e-it. Show that ei t ei(tt) ei t (i t)ei t
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 Hˆ0, and its eigen-solutions {… ( ) m
i t m
u r e …} provide a
complete basis for expanding any well-behaved function defined over the same region of space. The small interaction term Hˆ1 or perturbation is added such that the full
problem is described by Hˆ Hˆ0Hˆ1, and the full time development equation is: 0 1 ˆ ( , ) ˆ ˆ ( , ) d dt H r t H H r t i . 0 1 ˆ ˆ ˆ ( ,r t t) ( , )r t iH r t( , ) t iH H ( , )r t t
With the assumption that the system starts at time t in state n of the unperturbed Hamiltonian, 0 1 ˆ ˆ ( ,r t t) ( ) i tn ( ) i tn n i n u r e H H u r e t 0 ˆ ( ) i tn n i 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:
1 ˆ ( ) i tn n i H u r e t What is ˆ1 ( ) i tn n
H 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.
1 ˆ ( ) i tn , ) ( ) i tk n new k k k 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 ( ) i tj
j
u r e , the companion basis set function for bj, and the orthogonality relation is invoked.
( ) 1 1 ( ) ( ) ˆ ˆ ( ) ( ) ( ) ( ) ( ) ( ) j n n j k j k j i t i t i t j n j n i t i t k j k k jk k k i u r e H u r e t i 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 ( ) ( ) i n j t i n j t ( ( )) ( ) j j n all space j n i i db u 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 u H uj ˆ1 n .
( )
( )
1 1 ˆ n j n j i t i t j j n jn i i db e u H u e H dt [QMDyn.2] 1 ˆ n ju H u is [H1]jn,the jn matrix element of the perturbation Hˆ1.
Exercise: Rewrite the three lines of equations above replacing the Dirac Bra-kets with the integrals that they represent. Rewrite [H1]jn = u H uj ˆ1 n 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
( )
1 ˆ n j i t j j j dc db dt dt i e u H 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 ei( n j 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 1015s-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 12 0 ( ) 0 ( ) ( , ) Re cos( ) i k r t i k r t , 0 E r t E k r t E e E e E k
* 0 0 ( ) ( 1 2 ( , ) i k r t i k r t B 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. H1 pdistr.Eext
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: ˆ1 (
( , ) i k r t H 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
1
ˆ H
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 k r 103. The exponential eik r 1 ik r 1.
To simplify the notation, is assumed to be real so that the interaction is: E0
1 12 0
ˆ ( ) i t i t
H t e r E e e
. Substituting into the transition amplitude equation,
0
( ) ( 2 j n n j n j i t i db dt ie u r u E e e )j t . [QMDyn.3] ( ) n n u u r so u r uj n is time-independentAs we seek interesting behavior rather than the exact numerical predictions, the two factors
[ei( n j )t ei( n j )t] and
0 j n
u r u E are to be investigated qualitatively.
Energy Considerations: The time derivative of bjis oscillatory, and the net change in
bj is close to zero on average unless either
( n j )
i t
e or ei( 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.
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
]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
( ) ( 2 j n n j n j j i t i db dt ie 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:
2 j n 0
j db dt ie u r u E for = |n- j|.The additional selection rules are those necessary for u r uj n E0 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
Selection Rules and Matrix Elements: The factor u r uj n E0 includes some information about the polarization dependence of the interaction of the atom and the electromagnetic field. To begin, the piece u r uj 1 n 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: j
i( n j)t
ˆ1n n j j i db e 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: r xi y j z k ˆ ˆ ˆ Spherical: r rsin cos iˆ sin sin ˆj cos kˆ
1,1 1, 1
1,1 1, 1
10 1 2 2 4 3 ( ( ˆ i ( ( ˆ ( ˆ r r Y Y i Y Y j Y kThe 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 all space j n
u 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 n m ( ) andr n m (r)for 200,
210, 311 and 320. Conclude that: n m( ) ( 1)r 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.
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
j
L 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
A detailed study of the inner product in u r uj n 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.
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 eik r 1 ik r 1. Keep the second term.
(k r u ) n
j n j
u r u u r r i . The new term u r ik r uj ( ) 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 k r 103
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 101. 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.
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+1
L
Jfor 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
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 a Ae u 0 / 2 20 (1 ) r a B b r e u / 2 1 sin r i D r e e 0 / 2 20 cos r a C r e u 0 2, a
(Y1,1) electron states of the hydrogen atom. The images
below are the basis for a discussion of mixed state probability densities.
00 1 4 ( , ) Y 10 3 cos 4 ( , ) Y 1, 1 1 3 sin 2 2 ( , ) e i Y
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
These images represent the wave amplitude; it is * that represents the probability
density.
The 210 to 100 Transition: Consider the mixed state:
10 21 ( 21 ) 10 1 1 100 210 100 210 2 2 , ) ( ) i t ( ) i t ( ) ( ) i t i t A r t u r e u r e u r u r e e . where 100 1/ 2 3/ 2 3/ 2 3/ 2 0 0 0 0 0 0 00 / / / 1 2 1 2 4 ( , ) ( , , ) r a r a r a a a a u r e e e Y and 210 1/ 2 3 / 2 1/ 2 3 / 2 0 0 0 0 0 0 10 / 2 / 2 1 1 32 24 ( , ) ( , , ) r r a cos r r a a a a a u r e e Y
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 Ar 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 Ar 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: 1 ( 21 ) 10 100 210 2 , ) ( ) ( ) i t i A r t u r u r e e t , 2 2 2 100 210 100 210 21 1 1 2 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 1 100 200 100 200 2 2 , ) ( ) i t ( ) i t ( ) ( ) i t i t B 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 ( 21 ) 10 1 1 100 211 100 211 2 2 , ) ( ) i t ( ) i t ( ) ( ) i t i t C 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 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 Cr 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 directionwith 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 j n jn jn where j j i t i t db dt 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 jn j j n i t i t ie b u r u E e e dt
0
( ) ( ) 2 1 1 ( ) jn jn j j n jn jn i i e 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 n jn jn j n jn i i e ie e b u r u E u r E u e The probability amplitude is squared to find the probability:
2 2 2
2 2 2 2 2 2 0 0 1 sin[( ) ] 1 sin[( ) ] ( ) jn cos jn j j n jn jn jn b u r E u H E
2
2 2 2 2 2 0 1( ) cos for short
j jn
b 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 2 2 ( ) ( ) ( ) ( ) n j j d c b E E b E h dt
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 jn cos ( )
R 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 2
2 2 2 2 2 2 0 0 1 sin[( ) ] 1 sin[( ) ] ( ) jn cos jn j j n jn jn jn b u r E u H E
2 2 2 2 2 2 2 0 1 sin[( ) ] ( ) ( ) cos ( ) E E jn n E j E jn jn d c b E dE H E dt
E dEThe 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 2 2 sin ( ) ( ) sin ( ) n n n x D 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 2
2 2 0 2 2 sin [( ) ] ( ) ( [ ] jn j j n jn e b u r E u
2 2
2 2 2 2 0 2 1 cos ( sin [( ) ] ( ) [ ] E jn n E jn jn d c H E dt E dE
2 2
2 2 0 2 2 sin [( ) ] ( ( [ ] E jn n E j n jn e d c u r E u dt ) E dE
2 2 2 0 2 ( ( ) ( ) E n E e j n nj d c u r E u E E dt
E dEIn the (nanosecond) long time limit, the Dirac delta enforces energy conservation.
2 2
0 2
2 (
decay e j n e nj)
CAUTION: These arguments have been qualitative. Many multiplicative factors have been lost and a few concepts have been bruised.
REPEAT
2 2 2 0 1 ( ) ( ( ) j j n b u r E u jnIn the short time limit, the probability to find the system in the state j grows linearly in
time.
2 2 2 2 2 0 1 ( ) ( ) cos ( ( ) j j j n jn P b u e r u E
2 2 1 2 ( ) ( ) ( ( ) j j jn jn P b H The transition rate to the from the atomic state n to the atomic state j with the emission of a single photonis the probability to be found in the state j divided by the time that the perturbation has been acting: Rnj = -1|bj|2 or
2 2 0 2 ( ) nj e j n nj R 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
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
2 2 0 0 2 ( ( ) ) nj e u r E uj n E j n dE
R
2 2 0 2 ( ( nj e u r E uj n n j) R where 21 3 2 ( ) ( )E c 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, | u r E uj 0 n |2 is replaced by 2
2 0cos j nu r u E
. The anglebetween the direction of u r uj n 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 u r E uj 0 n |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 2 2 2 20
2 cos ( );
nj e j n j n j n n j
R 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 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=
)
(E Eu h
kT kT
e e , and, as the relative population do not
change in equilibrium, the rate of transitions up should equal the rate down.
{ )} and
u u u u u u u u
Rn AB R n B )
In thermal equilibrium: h kT and
u u
n n e R Ru .
Substituting, rearranging and comparing with the Planck black-body radiation formula for (vu) which must hold for systems in thermal equilibrium:
8 1 ) ) 1 1 h h kT kT u u u Planck u u u A B h B e c e B
It follows that: Bu = Bu and that Au= 8hc-33Bu . The equivalence of the induced rates up and down B coefficients also follows from the rate formula. The beauty is Au= 8hc-33Bu 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 ↠|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
