Introduction to Differential Equations
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.
it (,)xt 1 sinxxeit sin2 e 2 where n2 . a aa n 2ma2 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.
it Exercise: Find the probability density for (,)xt 1 sinxxeit sin2 e 2 a aa 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.
2 i (,) rt Hˆ (,) rt 2 Vr () (,) rt [QMDyn.1] t 2m Consider a state that is an eigenfunction of the Hamiltonian. Hˆ (,)rt E (,) rt i (,) rt nn t n The wavefunction can be separated into temporal and spatial parts: nn(,)rt u () rTn () 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 3 leading to the equations Hˆ ur() Eur ()and ETt () i Tt () and hence nnnnnt n
itn -1 nnn(,)rt cu () re where n = En and cn is a complex number usually of magnitude 1. For any energy eigenstate, the probability density is stationary (time- independent).
2 itnn it nn(,rt) (,)rt ( u n () re ) u n () re u n () 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 itn itm mixed state might be of the form mixed (,)rt aurenm () bure () with its associated probability density
2222 it((mn)) * imnt mixed (,)rt a un () r b um () r aburun () m () re ( abun () ru m () re ) 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.
itn itm Exercise: Show that if mixed (,)rt aurenm () bure () is a mixed state of itn itm uren () and urm ()e 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 4 That is: this mixed (,)rt does not describe transitions or state evolution.
Note that expectation values of various operators in the state mixed can be time dx dependent. For example, /dt may not be zero for the state mixed.
itm A general mixed state (,)rt amm u () re does not describe state-to-state m1 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 urm () are eigenstates of the full hamiltonian.
State to State Transitions: Mixed states with time independent probability densities are less interesting than the cases in which an electron makes a transition from one quantum state to another. ˆ Consider a quantum problem described by the Hamiltonian H0 that has eigenfunctions
itm urem () ˆ itmmit Hu0 mm() re Eum () re . (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
itm eigenfunctions, rt,) cmm u () re . That is: the eigenfunctions form a complete m set and are an orthogonal basis for the space of all physical wavefunctions for the problem. The functions are assumed normalized and orthogonal and hence satisfy the relation: (()ureitkm )* ure ()it dV . all space kkmmmk Begin with the system in a particular pure eigenstate, say the state |n
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 5 itn rt,) un () re . 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 d states that: Hrtˆˆ(,)Hure()()itnn Eureit i . It follows that: 00nnn dt
d it it (,rt t ) (,) rt t (,) rt ii Hˆ (,) rtt ure ()nn Eure () t dt 0 nnn.
d itit (,)rt tii Hˆ (,) rt t u () renn e E tu () r i tur () dt 0 nn nnn 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: d 1 Eu() reitn dt mn . This result follows from the fact that the hamiltonian itn n iure n () 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, eitn .
Exercise: Consider e-it. Show that eeit i() t t e it ()it eit
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 ˆ itm unperturbed hamiltonian H0 , and its eigen-solutions {… urem () …} provide a complete basis for expanding any well-behaved function defined over the same region ˆ of space. The small interaction term H1 or perturbation is added such that the full
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 6 ˆˆ ˆ problem is described by HH0 H1, and the full time development equation is: Hrtˆˆˆ(,) H H (,) rti d 01 dt . (,)rt t (,)rtii Hˆˆ (,) rt t H Hˆ(,) rt t 01 With the assumption that the system starts at time t in state n of the unperturbed Hamiltonian, (,rt t )u() reitnni Hˆˆ H u() reit t nn 01 i Huˆ () reitn t The piece 0 n 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: i Hˆ ure()itn t 1 n ˆ itn What is Hu1 n () 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. i Hureˆ ()itnk t f rt ,) bu () reit 1 nnew kk k
Projection (the inner product with uj) is used to isolate the expansion coefficient bj.
it j That is: the expression is multiplied by the complex conjugate of ()urej , the companion basis set function for bj, and the orthogonality relation is invoked.
it i() t iiure()j Hureˆˆ ()itn t urHur () () enj t jn11jn
it()kjit () kj bururkj() k () e bkjk e bj kk
All the bj that are initially zero are proportional to t for short times; it follows that: db j iiurHur()ˆˆ () eit()nj e it()nj ( ur ())* HurdV () dt jn1 all space j1 n
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 7 The expression is a little more compact if a matrix element [H1]jn represents the Bra- ˆ Ket uHujn1 .
db j iieuHueHit()nj ˆ it()nj [QMDyn.2] dt jn11jn ˆ ˆ uHuj 1 n is [H1]jn, the jn matrix element of the perturbation H1 .
Exercise: Rewrite the three lines of equations above replacing the Dirac Bra-kets with ˆ the integrals that they represent. Rewrite [H1]jn = uHuj1 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 dc db jji euHit()nj ˆ u small corrections dt dt j1 n 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 15 -1 expected to be greater than 10 s 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. 1 ikr() t* ikr () t E(,)rt Re E000 cos( k r t ) Ee Ee, E0 k 2 1 ikrt()* ikrt ( ) Brt(,) k Ee00 Ee 2
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 8 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 -1 k = c E = / E). It follows that the magnitude of the magnetic force exerted on the v charge is less than or equal to /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.pEdistr ext
A hydrogen atom consists of two equal, but opposite charges it is an electric dipole. If r is the electron’s position relative to the proton, then the dipole moment is p er
ikr( t) ˆ and the interaction Hamiltonian is: HpErterEe1 (,) 0. The form of the ˆ interaction H1 to be considered as a given for now, but the identification is reviewed critically in Appendix II. We know that a point dipole has energy p E 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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 9 most complicated approximation that one can solve. The wavelength of light is of order 5000 times the size of an atom so kr 103 . The exponential eikrik r 11 . To simplify the notation, E0 is assumed to be real so that the interaction is:
it it ˆ 1 Ht10()2 erE e e . Substituting into the transition amplitude equation, db j ie iti()(nnj j)t uruj n E e e . [QMDyn.3] dt 2 0 uurnn () so urujn is time-independent As we seek interesting behavior rather than the exact numerical predictions, the two factors iti()(nnjj )t [ee ] and urujn E0 are to be investigated qualitatively.
Energy Considerations: The time derivative of bj is oscillatory, and the net change in
it()n j it(n j ) bj is close to zero on average unless either e or e is not 1 oscillatory. 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 10 ]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.
db j ie iti()(nnjj )t uruj n E e e dt 2 0
RULE ZERO: Our first selection rule is that = |n - j|. Energy must be conserved. With energy conservation satisfied, we find: db j ie uru E for = | - |. dt 2 j n 0 n j The additional selection rules are those necessary for uruj 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 formal matrix element rules following Griffiths (in Chp Guide 9)
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Selection Rules and Matrix Elements: The factor uruj n E0 includes some information about the polarization dependence of the interaction of the atom and the electromagnetic field. To begin, the piece uruj 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. db Selection rules are based on the relation: j i euHuurit()n j ˆ u dt jj1 nn that identifies r as the atomic parameter active in the perturbation.
The Character of the Perturbation: Cartesian representation: rxiyjzk ˆˆˆ Spherical: rrsin cos iˆˆ sin sin j cos kˆ rrYYiYY4 1 (( ˆˆ i (( jY ( kˆ 3 221,1 1,1 1,1 1,1 10 The final form is useful when the angular momentum character of the perturbation is
2 discussed. The Ym are eigenfunctions of L and Lz. It follows that the perturbation has
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 12 = 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.
* urujn (()) urj rurdV n () all space 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 nm()andrr nm ( )for 200,
210, 311 and 320. Conclude that: nm(rr ) ( 1) nm()
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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 13 that the interaction has angular momentum character = 1.1 The angular momentum Lj of the final state must be the vector sum of the initial angular momentum Ln and that of the perturbation. Using the vector model for the angular momenta, j = n + {-
1, 0, 1}. Including the required parity change (-1) = - 1,
Orbital angular momentum change: = ±1
Electric Dipole (E1) Transition Selection Rule #1: Transitions are allowed between states for which the orbital angular momentum of the electron differs by one unit.
An electrons has an intrinsic spin angular momentum of ½ that is combined with its orbital angular momentum to give its total angular momentum: JLS. 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 .
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 14 A detailed study of the inner product in urujn 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 15 A second method to skirt electric dipole (E1) rules is to extend the approximation for the interaction. Earlier eikrik r 1 1. Keep the second term. urujn ur j ri()kru n. The new term urikruj() 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 kr103 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 kr101 . 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.)
h Exercise: A photon has momentum p = / 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: Lrp ( .0014 ; 0.31 )
Exercise: Given that r has angular momentum character one, what angular momentum character would you predict for rikr( )? Recall that k is just a constant vector.
A Doubly Forbidden Example: The helium triplet 2S metastable state --- 23S.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 16 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).
S: the total spin angular momentum of the 2S+1 Notation: LJ electrons for the atom 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 17 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 18 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.
1 Metastable molecular oxygen ( g) can be used to store energy for times as long as 72 minutes, a feature exploited in the hybrid electro-chemical oxygen-iodine laser system being investigated for possible naval weapons applications. Energy stored at low power over a long time is extracted in a short time yielding a short high power pulse.
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The Electron Cloud View of Selection Rules: The images used in this discussion were extracted from the hydrogen atom applet posted at www.falstad.com/qmatom/.
ra/ 0 They represent amplitude and relative phase variation for the 1s u10 Ae (Y00),
ra/2 0 ra/2 0 2s u20 Bbre(1 ) (Y10), 2p0 u20 Crcos e (Y10) and 2p1
i r/2a0 u2,1 Drsin e e (Y1,1) electron states of the hydrogen atom. The images below are the basis for a discussion of mixed state probability densities.
1 3 13 Y (,) Y (,) cos Y (,) sin ei 00 4 10 4 1, 1 22
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 21 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(,).
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 22 These images represent the wave amplitude; it is * that represents the probability density.
The 210 to 100 Transition: Consider the mixed state: rt,)11 u () reit10 u () reit21 u() r u () rei() 21 t eit10 . A 22100 210 100 210
1212ra//00ra ra / 0 where uY100 (,reee , )1/2 3/2 3/2 3/2 00 (,) and aa004 a 0
11rrra/2 00ra/2 uY210 (,re , )1/2 cos 1/2 e10 (,) 3/2aa 3/2 32 aa0000 24 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 -1 frequencies. In a time ) , the relative phase between u100 and u210 changes by .
-1 Exercise: In which region is the electron most likely to be found at t = ) ?
Exercise: The hydrogen atom consists of a positive proton at the origin and an electron with a relative position described by A rt,)given above. What is the direction
-1 of the net electric dipole moment of the atom at times t = N ( ) for N = 0, 1, 2, 3 and 4? What is the oscillation frequency? Compare it to the frequency expected for the radiation emitted in a 210 to 100 transition.
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The 100-210 mixed state A rt,) 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: rt,)1 u () r u () reiti()21 e 10t, A 2 100 210
222 rt,)11 ur () ur () urur () ()cos([ ])t. A 22100 210 100 210 21
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: rt,)11 ure ()it10 ure () it20 urure() ()i()20 t e it10 . B 22100 200 100 200 -1 Describe the probability density at time t = 0 and at t = ) . Sketch the -1 direction of the net electric dipole moment at times t = N ( ) for N = 0, 1, 2, 3 and 4. Do you expect electric dipole transitions between the 200 and 100
2 states? Compute and discuss B rt,) .
Study the 211 - 100 Mixed State: Consider the wavefunction: rt,)11 ure ()it10 ure ()it21 urure() () i()21 t e it 10 . C 22100 211 100 211 -1 Describe the probability density at time t = 0 and at t = N ( ) for N = 0, 1, 2, 3 and 4. Sketch the direction of the net electric dipole moment at times t = N
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 24 -1 ( ) for N = 0, 1, 2, 3 and 4. Do you expect electric dipole transitions
2 between the 200 and 100 states? Compute and discuss C rt,) .
Exercise: Describe the apparent motion of the blob of enhanced electron probability density if the state C rt,) 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 = m – m . Show that 1 ()( ) has terms with factors f i 2 ifif
im -im e and e . Leading to there being probability lumps in proportional to ½im i½m -i½m |e e + e or equivalently cos[½ m ]. That is there are |m| lumps 2 spaced around in with separations of /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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 25 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 2 is |bj| , the probability to find the system in the state j that is of more direct interest. Use the perturbation result: db itit()() j ie uru E ejn e jn where . Assume that the dt 2 j nj0 njn perturbation is turned on at time t = 0 and is on until time , and find that:
itit()() bu() ie ru Eee jn jn dt jj2 n0 0
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 26 ii() () eejn 11jn buruE() e jj2 n0 jn jn As discussed earlier, the probability to find the system in state j is small unless either
jn + = 0 or jn - = 0. It is to be assumed that the system is to emit a photon adding energy to the E-M radiation. The energy (frequency) of the final atomic state j 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.
i() e nj 1 buruE() e jj2 n0 jn i()/2 2sin[(jn ) ] ie urEu e jn 2 jn0 jn The probability amplitude is squared to find the probability:
22 2 22 2 sin[( ) ] 2 sin[( ) ] bu()11rEujn HE 22cos jn jj00n jn jn jn
222 2 bHE() 1 22cos for short jj n 0 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.
dh2 22 cb() ( EEbE ) () ( ) dt nj j
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 27 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 h dictates that energy must be more and more precisely conserved. That is E /. 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
2 2 RHE 1 22cos (E ) equivalent to a constant decay rate for the state n. jn 0 . 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
2 h grows as while the uncertainty band E shrinks as /.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 28 Return to:
22 2 22 2 sin[( ) ] 2 sin[( ) ] bu()11rEujn HE 22cos jn jj00n jn jn jn
2 EE 22 2 sin[( ) ] d 2 1 22 jn cbEdEHEnj() ( ) jn0 cos (Ed ) E EE dt jn 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 72nnsin (x ) Dxn () sin( cnx ) 2 nx
sin(x) sinc(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 sin [( ) ] bu()e rEu ( jn jjn2 0 2 []jn
2 E 2 d 2 2 22 sin [(jn ) ] cHE1 cos ( (E ) dE njnE 0 2 dt []jn
2 2 2 d 2 E sin [(jn ) ] curEue (( E)dE njnE 2 0 2 dt []jn
2 2 d 2 E e curEuEEnjnnj 0 (( )() EdE dt E 2 In the (nanosecond) long time limit, the Dirac delta enforces energy conservation.
2 2 R eurEue( E) decay 2 j 0 n 2 nj and = |En – Ej|
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 29 CAUTION: These arguments have been qualitative. Many multiplicative factors have been lost and a few concepts have been bruised.
REPEAT
2 2 2 bu() 1 rEu(() jj 0 n jn In the short time limit, the probability to find the system in the state j grows linearly in time.
2 2 2 Pb() ()1 ueruE22cos ( ( ) jj jn0 jn
222 Pb() () ( 1 H () j j jn jn
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
-1 2 the perturbation has been acting: Rnj = |bj| or
2 2 RurEue () nj 2 j 0 n nj 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 30 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 R e urEu(() E )dE nj 2 j 0 n 0 j n
2 2 e urEu (( )where ()E 1 E 2 R nj 2 j 0 n n j 23()c 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
2 2 2 As an approximation, | urEuj 0 n| is replaced by urujn E0 cos . The angle between the direction of urujn and the perturbing electric field is . The factor
2 1 cos has an average value of /3. Using this approximation, we can compute order of 2 magnitude estimates of transition rates without carefully evaluating urEujn 0 | .
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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 31
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 2 energy must be conserved. The factor Eo 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 22 R e uru E22cos ( ); uru uru nj2 jn0 j n jn n j 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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 32 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
( EEu ) h kT kT nu Boltzmann ratio: /n= ee , and, as the relative population do not change in equilibrium, the rate of transitions up should equal the rate down.
RnABuuuu{) u}and RnB u uu )
h kT In thermal equilibrium: nneuuand R Ru. Substituting, rearranging and comparing with the Planck black-body radiation formula for (vu) which must hold for systems in thermal equilibrium:
Au Bu 81h u ))h h Planck u B kT c kT u e 1 e 1 Bu
-3 3 It follows that: Bu = Bu and that Au= 8hc Bu . The equivalence of the induced rates up and down B coefficients also follows from the rate formula. The
-3 3 beauty is Au= 8hc 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 ↠|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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 33 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 ˆˆˆ mode (propagation vector: kkikjkkx yz ). 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):
abs 2 Rif 2 p if(E fi) [QMDyn.4] 3o
stim 2 Ri f 2 p if(E if ) [QMDyn.5] 3o
3 spon if 2 Rif 3 p if [QMDyn.6] 3o c
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 34 222 where 22efxifyifzi [QMDyn.7] pif
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.
r 3/2 1 13 |2p 1 r 2a0 Y (,) Y (,) cos o 2a a e 10 10 0 3 0 2
r r 3/2 3/2 1 a0 1 a0 1 |1s a 2e Y00 (,) = a 2e 0 0 4
z = r cos = r 2(,Y10 ) 3
r r 3 1 2 1 aa00 r 2 2 1 100zr 210 a e e rdrY2 10Y 10 sin dd 0 000 a 4 6 0 3
-½ ½ Note that the factor of (4) is Y00, and the 2 ( /3) Y10 is cos.
3r 4 112a0 rdr 100za 210 e 0 0 aa 6 003
2 5 2 27 100za 21000 4! 2 a 3 6 35 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.\ 33 Respon if 22if ()a215 if 33pif 0 310 33oocc (10.4 [1.602 1019 ]JC ) 3 [1.602 1019 ] 2 Rmspon if (5.29 1011 ) 2 215 if 34 4 12 Jm8 3 311 (1.054 10Js ) (8.85 10Cm2 ) (3 10s )
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 35 3 5 2 117 (10.4 1.602 5.29 ) 10 15 Rspon 2 6.63 1081s if (1.0544 8.85 33 10124 )s 311
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
7 5 Exercise: Given 100| x |211 = 100| y |211 = ao (2 /3 ) 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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 36
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
E -73 important to remember that energy levels have finite widths ( /E 10 ) with the E -1 widths in liquids and solids being potentially very large ( /E 10 ). Thus the energy
equation: = Eu – Especifies a range of frequencies for photons that can be absorbed.
hyperphysics.phy-astr.gsu.edu/hbase/mod5.html
Stimulated Emission: A photon incident on an atom in an upper level u can induce or
stimulate the emission of a photon as the atom makes the transition to a lower level if
the photon energy is equal to the energy difference between the two levels: ( = Eu
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 37 – 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.
hyperphysics.phy-astr.gsu.edu/hbase/mod5.html
The character of a stimulated photon contrasts starkly with that of a spontaneously emitted photon. A spontaneously emitted photon is randomly emitted in most any direction with any phase. There is essentially no chance that the spontaneous photon will land in the same mode as the stimulated photons just discussed. A laser operates using the stimulated emission photons. The spontaneously emitted photons are uncorrelated and effectively contribute a noise field to the field of the good laser photons. For this reason, the spontaneous emission is omitted in the discussion of laser
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 38 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.
RR uuuuu R nB )) nB uuu nnB u u)
where the equality Bu B u has been used. The conclusion is that R > 0 if nu > n.
An active medium can provide photon gain if the upper state population exceeds the population in the lower state; that is: if there is a population inversion. Review the discussion of the helium neon laser dynamics.
Exercise: Why is the condition nu – n > 0 called a population inversion? What is the
expected value of nu/n for a system is in thermal equilibrium? Who was Boltzmann?
How did he die?
Many more wonders of quantum dynamics await you. Sweat the details when you next encounter the concepts. The 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)
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 39
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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 40
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 ˆˆ ˆ [,xijLi ] ijkk xxLiyyLi ,[,] z and[,z ]x Let A Lz. ˆ mxLm'[ ,z ] ( mm ) mxm ' m ' iym or ()'mm mxm i mym ' Note: That one need only compute the matrix element of x as the matrix element of y can be computed form it. mym'()' imm mxm
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 41 Beginning with [y, Lz], one finds that ()'mm mym i mxm '. Chaining the results together, ()'mm2 mxm mxm ' which means that mx' 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
2 interpret. That is: m| [L , 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: ˆˆˆˆˆˆ2 [,]2(Lz i xLLyyx )and xLyLzL x y z 0. It follows that: [,[,]]2(Lˆˆ22Lz 22 LzzL ˆ ˆ 2 ) [QMDyn.9] Note that: mLLz'[ˆˆ22 ,[ , ]] m 22 mLzzLm '(ˆ 2 ˆ 2 ) 24 ( 1) ( 1) mzm ' Now the methods behind equation [QMDyn.8] can be applied to yield: ( 2)( )( 1)( 1) mz ' m 0 [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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 42 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.
ˆˆˆˆˆˆ2 Exercise: Verify that [,]2(Lz i xLLyyx )and xLyLzL x y z 0.
Exercise: Evaluate mLLz'[ˆˆ22 ,[ , ]] mand mLzzLm'(ˆˆ22 ). Expand the relation
mLLz'[ˆˆ22 ,[ , ]] m 22 mLzzLm '(ˆ 2 ˆ 2 ) to develop a result equivalent to
ˆˆˆˆˆˆ2 [FQT.10].One must first verify that [,]2(Lz i xLLyyx )and xLyLzL x y z 0.
The Selection Rules for Electric Dipole Transitions: m = 0, 1 and = 1. The rules can be interpreted in terms of a photon being emitted or absorbed that carries one unit of angular momentum and that has odd parity. The atomic states have parity (-1) and well defined and m. The absorption (or emission) of a photon must lead to an atomic state of opposite parity which is consistent with changing by 1. This requirement is also consistent with the rule that = 0 to = 0 transitions are forbidden.
The rule that || < 2 is consistent with the photon carrying just one unit of angular momentum.1 Further, the photon is a fully relativistic particle and as such is 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 43 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 qr The static (Coulomb) electric field due to a charge at rest is ECoul 3 where r is the 4 o r displacement from the position S of the source charge q, to the field point P where the field is to be specified. In addition, a slowly moving charge has a Biot-Savart magnetic qv r field given by B . Finally, an accelerated charge has a radiation contribution BS 4 r3
qa to the electric field that is approximated by the relation ERad 2 (for v << c). 4 cr Ret
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 r at the retarded time t' = t - /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 -1 oppositely to the retarded value of a and that the radiation field falls off as r at large
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 44 -2 v distance while ECoul and BBS fall off as r . [There are corrections to ERad of order /c and r -2, etc.] The energy flow is described by the Poynting Vector S 1 EBwhich is proportional to the product of ERad and BRad . At large r the net energy flow out is
2 2 proportional to S 4πr = (1/µo) E B 4πr . 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 45 ERad v t
ECoul c 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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 46 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 qa r BRad 32 4 cr Find an analogous expression for ERad . It requires a double cross product! Show that rEˆRad cB Rad where rˆ is the direction of propagation.
2.) Consider a charge q with acceleration a = - 2 d cos(t) in the z direction. Find an expression for ErRad () in terms of spherical components and unit vectors with the coordinates centered on the charges location.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 47 1 3.) Compute Poynting's vector ( SE 0 [ 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.
-1 4.) Devise an argument that shows that the transverse component ErRad () falls off as r at
-2 large distance if the radial component ErCoul () falls off as r using the picture above these exercises. Motivate equation 3.
Vector Triple Product: A ,,BC 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 ˆˆ time dependent dipole moment is ppqdcos tk 0 cos tk . Compute the average power radiated by the charge q by computing SndA ˆ over a large sphere concentric with the charge and averaging over one cycle. Express the result in terms of
22 2 po and plus other factors. (For the E-M transition problem, p0 efri.)
2 p0 Ans: Paverage = c
Appendix III: The canonical interaction: q p A
One learns that the interaction energy of an electric dipole in an Electric field is: Udipole = pEDip E . (The symbol p is adopted for the electric dipole moment to avoid confusing it with the momentum.) This result is not directly applicable to the evaluation of the interaction energy of an electromagnetic wave and an atom. The result was derived by considering two equal, but opposite charges in an electrostatic field. E-M radiation is scarcely electrostatic. As this is a deep mystery, the resolution is rarely presented prior to graduate school. Here the path to the answer is to be sketched, but not presented in detail.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 48 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 = pEDip E The canonical method to identify a quantum operator for a quantity is to begin with its representation in classical Hamiltonian mechanics in terms of coordinates and momenta. Unfortunately, we need to develop the lagrangian as a step in identifying the canonical momenta and the hamiltonian itself. We could start with that T – U prescription.
L ½(mxxii q x,,) y z Summation notation is invoked so repeated indices are summed from 1 to 3. Relativity could guide me to the incorporation of magnetism. The electrostatic potential is the zero element of a four vector and magnetic interactions involve the velocity. The four- potential and the four-velocity are:
1 A (,,,cAAAx yz) and vcvvv (, x ,yz , )
v 2½ [1 (c ) ] The lagrangian is a scalar so we try the inner product of these vectors as the potential term. A vv ()AThis term obviously includes relativistic corrections that we are ignoring in this course and in the expression for the kinetic energy. In the 1 limit, L ½mxxii q (v A) .
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 vB is to be developed with Bi replaced by A , the component representation of B A ijkx j k
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 49 FqvBqx Aq xA mag ijk j kmnxxmm n kij kmn j n
AniA Fqmag,iimjninjmjnn xAqxxx m m xi xm It is time to test our candidate for the lagrangian.
L ½(mxkk x q x m A m)
L L Am mxii qA pi; qx() m xi xiixxi
d LL AimA Ai mxim q q x xk dt xii xx itx i xk Realizing that we are free to re-label dummy indices and noting that the electric field is E t A, it is confirmed. The proposed lagrangian works. The hamiltonian follows as HLpkkxmxxqxAmxxqxAkk kk ½( kk mm).
HL pkkxmxx½ kk q The answer appears pretty simple, but we have yet to eliminate the coordinate
1 velocities in favor of the momenta as required. xiimpqA i .
2 1 pp q q H 22mmpkkkkqA p qA q m p A 2m A A q
Electromagnetic processes By Robert Joseph Gould Google Books Electromagnetic interaction hamiltonian non- relativistic. Choosing the gauge A 0 and = 0,
2 ˆ qq H 22mm pA Ap AA
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 50 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. Hˆ BA () 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:
z2 11 1 22in() t in() 2 t nn(,)zt H () ze e u n () ze 2!n n The lowest three harmonic oscillator wavefunctions are:
z2 1 z2 3 1 22i()t 2 z 22it() 0 (,)zt e e n (,)zt e e
2 z2 5 21z 22it() 2 (,)zt e e 2 Compute the time dependent probability densities for the mixed states: 1 1 (,)rt (,) zt (,) zt (,)rt (,) zt (,) zt M 01 2 0 1 M 02 2 0 2 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
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 51 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
Hˆ (,) z t qzE cos( t ) qzE1 eit e it 10 0 2 The spatial parts of harmonic oscillator wave functions satisfy the recursion relation:
zu() znn u () z1 u () z nn2211n
Give the selection rules for allowed transitions caused by H1.
db iti()( )t j iq Euzue nj enj dt 2 0 jn
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
2 nzn1 Rnn1 the matrix elements of the interaction. 2 . Compute this ratio. Read Rnn1 nzn1 the Einstein’s coefficients section. Recall that z = x = 1 aaˆˆ† . The real point is 2 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. Hˆ , where E is the energy of the system.” Give examples to support your discussion.
5.) In the case of a particle in a 1-D box, an infinite well with the range [ 0 < x < a],
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 52 2 nx the normalized spatial states are: uxn () aa sin . The allowed energies are
2 En 2 . At time t = 0, the system is in the state (x,0) = 5 ux() 12 ux () where n 2ma2 1312 13 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.
2 c.) A detailed calculation shows that 1.69eV . Compute E, the expectation 2ma2 value of the energy for the state (x,t). d.) Compute | (x,t) |2. Discuss the result. 2 e.) Verify that un(x) is properly normalized. What is the average value of sin (x)
for 0 < x < 2 ?
6.) A general quantum problem with a time-independent Hamiltonian has a complete ˆ set of energy eigenfunctions ( HxtEnnn(,) (,) xt). At time t = 0, a general solution to
the problem has the form (,0)x cxmm (,0). Normalization: m0
*(,)xt (,) xt dx full range nm 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 Qˆ that
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 53 ˆ has a complete set of eigenfunctions (Qxtqnnn(,) (,) xt). Every Hermitian operator has a complete set of eigenfunction. As the set is complete, any allowed state
can be expanded in terms of them at time t = to. (,x tcx00 ) mm (,t ). m0
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 Qˆ ?
The Hamiltonian Hˆ and the Hermitian operator Qˆ can have simultaneous eigenfunctions if their commutator [ Hˆ , Qˆ ] = 0.
ˆ A 1-D quantum problem has the momentum operator pi x that has a complete set of eigenfunctions [ pˆ nnn(,xt) p (,) xt ]. 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.
(,x tctxt ) ( ) (, )1 Aktedk(,)ikx . 000 mm 2 0 m0 a.) What is [ xˆ , pˆ ]?
b.) What condition must the Hamiltonian satisfy to ensure that [ Hˆ , pˆ ] = 0 c.) Identify a problem that has solutions that are simultaneous eigenfunctions of
Hˆ and pˆ .
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 54 9.) For the hydrogen atom, study rt,)1 u () r u () reiti()21 e 10t. Get the A 2 100 210
222 explicit form for: rt,)11 ur () ur () urur () ()cos([ ])t. The A 22100 210 100 210 21 probability density describes the electron which has charge – e. Compute the
2 oscillating electric dipole moment for the density A rt,) . 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 er 100 ] where you need to include the time dependence of each state. Following problem 5 in Appendix I, compute the power P radiated by this oscillating dipole. Estimate the radiative lifetime of the 210 state of hydrogen as (E210 – E100)/ P. Compare your result with that of the previous problem. Do not sweat factors of 2 or 3 at this point.
11.) Consider a quantum mechanical problem with the Hamiltonian Ho and eigenstates
iE(/)n t ˆ iE(/)n t nn(,)rt u () re that satisfy HrtErtEure0 nnnnn(,) (,) () . These eigenstates form a complete set appropriate for expanding any function defined over ˆ the same region of space. In particular, after adding a small perturbation H1 to the ˆˆ ˆ original Hamiltonian, the full Hamiltonian becomes: HH01 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.
iE(/)j t rt,) cjj () tu () re j1 i ˆˆ Huˆ () reiE(/)n t Eu () reiE(/)n t Use: t 01 and 0 nnnto show that:
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 55 dc() t icture () ()iE(/)j t ˆ ure () iE(/)j t ct () j jj j where j dt . Project out the relation jj11 * for ctm () by multiplying the relation by um and invoking the orthogonality relation
* uudVmn mn.
iE(/) t iE(/)m t ˆ j Show that you find: ic mmj() te u u e . Assume that the system j1 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 ctm () for very small positive times given these ˆ initial conditions. Display explicitly, the integral that um uj represents.
itm 12.) A general mixed state (,)rt amm u () re where the urm () are eigenstates of m1 the full hamiltonian with energy eigenvalues Em = m. Use the Schrödinger equation
dak to show that /dt = 0 for all k. As always, one projects out a specific coefficient by ˆ using the inner product uHk and the orthogonality relation uukm km.
13.) The spherical harmonics, the Ym(), are the angular factors in the hydrogenic wavefunctions. The perturbation for electric dipole transitions has the form eE0 r . ˆˆˆ Express rxiyjzk 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.
1 3 13 Y (,) ; Y (,) cos ; Y (,) sin ei 00 4 10 4 1, 1 22
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 56 x Initial( ,my ) Final( , 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
7 5 Exercise: Given 100| x |211 = 100| y |211 = ao (2 /3 ) 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.
6/15/2010 Physics Handout Series.Tank: QDyn_QV QMDyn- 57 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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