Lecture 8 Dec: Selection, H and Line Intensities 

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|nAn m| B|n which identifies the initial state and m| AB|nAm| B|n Am m| B|nwhich identifies the final state. Additionally, it is assumed that the commutator [A, B] can be reduced to an equivalent operator C, and that m|C|ncan be expressed in terms of the eigenvalues associated with |m and |n and m|B|nthe matrix

element of interest. That is: m|C|n = fC(m,n) m| B|n m| [A,B]|nm| AB-BA|n (Am – An) m| B|n= fC (m,n)m| B|n[QMDyn.1]  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 new operators with each of the current (previously known) operators for a problem.

The hydrogenic states found in the Hydrogen Atom Math handout are eigenfunctions of energy, of the square of the orbital angular momentum and of the z component of angular momentum. Here the focus is on the orbital angular momentum so the initial state is represented as | m, and the final state as |m.

As the first example, the perturbation is –eEoz so B is effectively z and, as a first choice,

the (physics set of) hydrogenic states are eigenfunctions of Lz with eigenvalues m. Let A

 Lz.

 m| [Lz, z] | m =  m| 0| m = 0 = (m– m)  m| z |  m The matrix element  m| z |  m can only be non-zero if m= m. The perturbation z can only cause m = 0 transitions.

Page 1 Lecture 8 Dec: Selection, H and Line Intensities 

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  Beginning with [y, Lz], one finds that ()'mm   mym i  mxm '. Chaining the results together, ()'mm2  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.1] using [L2, z] fails to provide a result that is simple to interpret. 2 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.2] Note that: mLLz'[ˆˆ22 ,[ , ]] m 22 mLzzLm '(ˆ 2 ˆ 2 )  24  ( 1) ( 1) mzm '  Now the methods behind equation [QMDyn.1] can be applied to yield: ( 2)(   )(  1)(   1) mz '  m  0 [FQT.3] 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

nm|z|nmn 0m| z |n 0m as does n 0m| x |n 0m and n 0m| y |n 0m. In fact the  rules must be the same for x, y and z as they are just equivalent Cartesian components and reflect an arbitrary choice of a preferred direction, and the operator L2 is invariant with respect to permuting the labels x, y and z.

Page 2 Lecture 8 Dec: Selection, H and Line Intensities 

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 [FQT.3] ˆˆˆˆˆˆ2 .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 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 III: The canonical interaction: q p  A  One learns that the interaction energy of an electric dipole in an Electric field is: U    dipole =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

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.

Page 3 Lecture 8 Dec: Selection, H and Line Intensities  this is a deep mystery, the resolution is rarely presented prior to graduate school. Here the path to the answer is to be sketched, but not presented in detail. Just follow the flow of the ideas. By the time the dust settles, the conclusion is that the equation for the basic interaction energy does not change although the way that one regards it may change.   Matter-EM wave interaction: H1 =  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(mxxii 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 (,,,cAAA  x 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  ()AThis term obviously includes relativistic corrections that we are ignoring in this course and in the expression for the kinetic energy. In the   1 limit,   L ½(mxxii q v A).

The previous development seems suspect to validate the proposed function. The lagrangian is whatever it takes to reproduce the classical equations of motion. What do   we expect for the magnetic part of the Lorentz force? The equation for v B is to be     developed with Bi replaced by  A , the component representation of B  A ijkx j k  FqvBqx  Aq  xA mag ijk j kmnxxmm n kij kmn j n  AniA Fqmag,iimjninjmjnn   xAqxxx  m m  xi xm  It is time to test our candidate for the lagrangian.

L  ½(mxkk x q x m A m) L L  Am mxii qA pi; qx() m xi xiixxi d LL   AimA  Ai mxim  q q x xk dt  xii xx  itx   i xk 

Page 4 Lecture 8 Dec: Selection, H and Line Intensities 

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 HLpkkxmxxqxAmxxqxAkk kk ½( kk mm).

HL pkkxmxx½ kk q The answer is the total energy which appears pretty simple, but we have yet to eliminate 1 the coordinate velocities in favor of the momenta as required. xiimpqA i .    2  1 pp q q H 22mmpkkkkqA p  qA q m  p A 2m A A q   Choosing the gauge A 0 and  = 0,2  2  ˆ qq  H 22m pA   Ap  m AA   [QMDyn.4] The interaction hamiltonian above is appropriate for a point charge in an external field. For particles with intrinsic structure that leads to a magnetic moment, the magnetic dipole-field interaction must be added.      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. 

Examine the terms: q2  2m  A A operates on the field rather than the atom so we need not fret over it until we wish to quantize the electro-magnetic field. The operators for the particle momentum and the    vector potential act on different sets of coordinates so p  AAp2 pA . That is: the operators for p and A commute.

We are expanding in terms of eigenfunctions of the unperturbed hamiltonian and ˆˆ     Hˆ pp Vr(), Hˆ ,,xppx1 ˆˆ i pˆ and Hˆ ,rpi ˆ . o 2m ox2m m o m  Hˆˆ  qqpAm  H,, r  Aiq  Hˆ r  A mm   ioo   We are interested in matrix elements of the form:   Hˆˆ iq Hr,  A fi  foi 

2 Electromagnetic Processes By Robert Joseph Gould Google Books Search: Electromagnetic interaction hamiltonian non-relativistic.

Page 5 Lecture 8 Dec: Selection, H and Line Intensities 

Recall that i and f are eigenfunctions of the unperturbed hamiltonian with eigenvalues

Ei and Ef.   Hrˆˆˆ,  Hr  rH  Er  rE  f oifoifoiffifii Substituting     ˆ  iq  f HEErAiqrifififif ()   () Ai Recall: the vector potential operator does not act on the atomic coordinates. It acts as a scalar multiplier for the atomic wavefunctions.

Now we need to consider the vector potential for a plane wave given our choice of gauge   [0A  and  = 0].       ikr() t A A(,)rt Aeo so Et  i A      it As kr1, Art ( , )  Aeo .    A Energy conservation gives f - i = - . E  iA()fi  t   ˆ    f HiqrAqrE ififif() i   By comparison, we can represent the perturbation as Hˆ   qr Ep E where p is  the electric dipole moment qr er . The final effective form of the perturbation becomes:    Hˆ   qr Ep E er E [QMDyn.5]

Tools of the Trade: Note that an operator was re-expressed as a commutator of two operators. One of the operators in the commutator is an eigen-operator for the basis set of states so its introduction is reduced to eigenvalue multiplication rather than a general operator action.

Relative Intensities in the Zeeman Effect: 3 3 Consider the 5460.7 Å S1  P2 line in the spectrum of mercury. If the mercury is placed 3 3 in a strong uniform magnetic field, the S1 and P2 levels are each split into several levels. 3 Into how many levels is the S1 level split?

Page 6 Lecture 8 Dec: Selection, H and Line Intensities 

3 splits on mJ2J + 1 the P2 level? = 5. 3 3 3 Compute the g factors for the S1 {J=1; S=1; L=0; g = 2}. P2 {J=2; S=1; L=1; g = /2}

The selection rule for allowed level to level transitions is: m = 0, 1 The magnetically split levels as corresponding to energies: 33 3 ES01()kBext B E 01(SgBmES)(BextJ 01)2B BextmJ 33 33 E02()PBEPgBmEPBj B ext 02() B ext J  02() 2 B extm J

3 3 E01()2SB B ext S1 3 ES() g = 2 01 E ()23 SB  01 B ext mJ = -1, 0, 1

33 E ES01() EP 02 () B Bext   2to2by ½

E ()33 PB  3 02 B ext 3 3 3 P E02()PB 2 Bext 2 3 g = / ; EP02() 2 3 3 E02()PB 2 Bext 3 mJ = -2, …. , 2 E02()3PB B ext  

The relative intensities of the lines depend on the direction from which they are viewed. For that reason, the form of the transition rate formulae prior to averaging over directions must be used. 32223 ba  e ba  IJMeJMJMreab 33ff fp ˆˆ iii ff f iiiJM oocc  Note that the symbol ê in these equations represents the polarization of the perturbing electric field. The overall angular momentum quantum numbers for the initial and final

states are displayed explicitly. The symbols i and f represent all the other quantum numbers that remain fixed as the angular momentum options are investigated. For our

discussion, the only requirement is that i and f represent states of opposite parity. The assumption is that the matrix element factors. This conjecture is supported by the results for the n = 2 to n = 1 transition rates for atomic hydrogen that are computed in the appendix.

Page 7 Lecture 8 Dec: Selection, H and Line Intensities 

23 22 e ba IrJMreab 3  f i f fˆˆJ iM i  oc  Note that ê is the polarization/direction of the electric field.

2 3 3 The radial factor  fir  is assumed to be identical for all of the S1 to P2 transitions considered.3 One might look for an analogous situation for a hydrogen atom and establish the conjecture explicitly for that case. The relative intensities follow as they are just the 2 ˆˆ ratios of the angular momentum matrix elements squared JMff r eJM ii. Let’s recall the angular momentum nature of the position vector.  rxiyjzkrˆˆ ˆˆ (sin cos i ˆ  sin sin ˆj  cos k  rr22{ Y11 (,) Y (,)}  iiˆˆ  { Y11 (,)  Y (,)}  j 4 Y0 (,) kˆ  3311 11 31 2211ˆˆ11 40ˆ rYYiiYYjYˆ 33{11 (,)   (,)}  {11 (,)   (,)} 31 (,) k



3 The Wigner-Eckart theorem states that the matrix elements of the JfMf| Qkm|JiMi are proportional to some common value

Jf| Qk|JiC(JiMi,km;JfMf) provided the Qkm are an irreducible representation of a tensor of order k. The Ykm(,) are such representations. The theorem states that the angular integrals are proportional to Clebsch-Gordan coefficients. *** Motivate with H atom comparison.

Page 8 Lecture 8 Dec: Selection, H and Line Intensities 

y

z x

Consider the  lines which, when viewed from the transverse direction (in along the x axis) are polarized parallel to the magnetic field. That is: they are polarized in the z direction. 2 2 ˆ 4 0 JMffrˆ kJiMi JMff 3 Y1 (,)JiMi

In an attempt to simplify the calculation, the rations are first computed assuming that the addition of angular momenta is the only consideration. For example, 2 2 4 0 JMff 3 Y1 (,)JiMiJMff1, 0 JiMi

The three (m = 0)  lines have the following angular momentum characters: | 1 1  | 1 0 | 2 1| 1 0  | 1 0 | 2 0 and | 1 -1  | 1 0 | 2 -1. Using the 1  1 Clebsch-Gordan table:

Page 9 Lecture 8 Dec: Selection, H and Line Intensities 

1 2 1 The lines are predicted to be in the ratios /2 : /3 : /2 or 3:4:3.

For the + lines we only see the y projection of the polarization when viewed 2 11 transversely. Use iY3 {(,)11  Y (,)}. Note that the coefficients for the Y’s are less by a factor of the square root of two which is equivalent to a factor of 2 in the predicted relative intensity. Each emitted line has its characteristic frequency so as one computes the relative intensity of a line one needs only to consider the active factor coupling the two states. 2  JM 4 1 Y 1 (,) JMfor m 1 2  ff 321 ii JM rˆ ˆ jJM ff ii  2  JM 4 1 Y 1(,) JMfor m 1  ff 321 ii | 1 1  | 1 1 | 2 2| 1 0  | 1 1 | 2 1 and | 1 -1  | 1 1 | 2 0. Using the 1  1 Clebsch Gordan table, the matrix elements are 1, 2-½, 6-½ corresponding 1 1 1 1 1 to ratios of 1: /2 : /6 or, scaled to the  lines /2: /4 : /12. These are the relative intensities observed from the transverse direction with no 1 2 1 1 1 1 polarizer. ;   /2 : /3 : /2; /2: /4 : /12. or 6:8:6; 6:3:1. Multiplying by 2 to compare with the figure in the Chinese manual, we expect: {+; ;   {{2:6:12}; {12:16:12}; {12:6:2}}.

http://galileo.phys.virginia.edu/classes/317/zeeman/zeeman.html The Zeeman splitting of the 546.1 nm line is given by Melissinos, p. 294. You should understand the reasons for the form of the splitting. Because the splitting is not fully resolved with our apparatus, you will want to know the relative intensities of the components, which are given here (as viewed perpendicular to the field). Predicted intensities from the Chinese manual.

Page 10 Lecture 8 Dec: Selection, H and Line Intensities 

The preceding analysis generates the advertized intensities. Should we believe it? The wavefunction describes everything so it describes the electrons orbital and spin state. The perturbation acts only on the spatial part. In fact, some of our selection rules are based on the concept that the perturbation does not couple to the spins as that would require a magnetic field rather. Our perturbation is:   -p  EoabeIˆˆ  r e The perturbation is the position of the electron which has an angular momentum one character, but it combines with the spatial (orbital) character of the wavefunction and leaves the spin untouched. A proper treatment would involve a multi-electron wavefunction that is properly anti-symmetric and a tone of effort. We will just break the electronic wave function into spin and spatial components and then combine the angular moment on the perturbation with that of the spatial part in each term leaving the spin part alone. The results will then be assembled to predict the relative intensities.

3 Start with the Pi lines for which the perturbation is z or |10. The transitions are from S1 3 to P2. 3 3 S1  |00 |1 ms and P2 cs |1m |1 ms .

|00 |1 ms + |10 |NULL |10 |1 ms

-½ -½ |00 |1 1 + |10 |NULL |10 |1 1= 2 21 + 2 11  2 ½ 1 ½ |00 |1 0 + |10 |NULL |10 |1 0= ( /3) 20 + 0 10- ( /3) 11 -½ -½ |00 |1 -1 + |10 |NULL |10 |1 -1= 2 2 -1 + 2 1 -1 

Page 11 Lecture 8 Dec: Selection, H and Line Intensities 

For the Pi lines, we take the inner product of each line with Jf, mJf where Jf = 2 and mJf = mJi. 1 2 1 We find that the Pi lines are in the ratios /2: /3: /2 or 6:8:6. For the + lines: We will need 2-½ times our result from this part as: For the + lines we only see the y projection of the polarization when viewed transversely. Use 2 11 iY3 {(,)11  Y (,)}. Note that the coefficients for the Y’s are less by a factor of the square root of two which is equivalent to a factor of 2 in the predicted relative intensity.

|00 |1 ms + |11 |NULL |1, ms+1

|00 |1 1 + |11 |NULL |J, 2|2, 2 -½ -½ |00 |1 0 + |11 |NULL  |11|1 0  2 21 + 2 11  -½ -½ -½ |00 |1 -1 + |11 |NULL  |11|1 -1  6 20 + 2 10 + 3 00   -½ -½ We find that the + lines are in the ratios 1: 2 : 6 . Incorporating the factor of ½ and then multiplying by 12 as was done for the Pi line intensities, we find 6:3:1. Overall,  1:3:6:6:8:6: 6:3:1or{{2:6:12}; {12:16:12}; {12:6:2}} as before. The predicted ratios are the same. One should not extrapolate too far.

Discussion: Why are the answers the same? In the re-calculation of the relative intensities, the physical statement that the perturbation could not act on the spins, but only on the orbital angular momentum has manually inserted into the calculation. The results were unchanged. Realize that the initial and final states used in the first calculation have that character. The spin parts of the initial and final states are the same. Nature imposes that restriction. My manual injection of the requirement in the second calculation just restates what nature has declared. 

Page 12 Lecture 8 Dec: Selection, H and Line Intensities 

3 3 E01()2SB B ext S1 3 ES() g = 2 01 E ()23 SB  01 B ext mJ = -1, 0, 1

33 E ES01() EP 02 () B Bext   2to2by ½

E ()33 PB  3 02 B ext 3 3 3 P E02()PB 2 Bext 2 3 g = / ; EP02() 2 3 3 E02()PB 2 Bext 3 mJ = -2, …. , 2 E02()3PB B ext  

From Wikipedia, the free encyclopedia

http://en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem

The Wigner–Eckart theorem is a theorem of representation theory and quantum mechanics. It states that matrix elements of spherical tensor operators on the basis of angular momentum eigenstates can be expressed as the product of two factors, one of which is independent of angular momentum orientation,

Page 13 Lecture 8 Dec: Selection, H and Line Intensities 

and the other a Clebsch-Gordan coefficient. The name derives from physicists Eugene Wigner and Carl Eckart who developed the formalism as a link between the symmetry transformation groups of space (applied to the Schrödinger equations) and the laws of conservation of energy, momentum, and angular momentum.[1]

The Wigner-Eckart Theorem reads

where is a rank spherical tensor, and are eigenkets of total angular momentum and its z-component , has a value which is independent of and , and is the Clebsch-Gordan coefficient for adding and to get .

In effect, the Wigner–Eckart theorem says that operating with a spherical tensor operator of rank on an angular momentum eigenstate is like adding a state with angular momentum to the state. The matrix element one finds for the spherical tensor operator is proportional to a Clebsch-Gordan coefficient, which arises when considering adding two angular momenta. Example

Consider the position expectation value . This matrix element is the expectation value of a Cartesian operator in a spherically-symmetric hydrogen-atom-eigenstate basis, which is a nontrivial problem. However, using the Wigner–Eckart theorem simplifies the problem. (In fact, we could get the solution right away using parity, but we'll go a slightly longer way.)

We know that is one component of , which is a vector. Vectors are rank-1 tensors, so is some linear

combination of for . In fact, it can be shown that . Therefore

which is zero since both of the Clebsch-Gordan coefficients are zero.

The Wigner-Eckart theorem http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm

Why do we care about tensor operators?

The matrix elements of tensor operators with respect to angular momentum eigenstates satisfy

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The double bar matrix element is independent of m, m’, and q. This is the Wigner-Eckart theorem. (In the above equation (2j'+1)-1/2 is factored out of the double bar matrix element. This factoring is not unique. Different books may factor out a different factor.)

The matrix element is written as the product of two factors. The first factor is a Clebsch-Gordan coefficient which depends only on the way the system is oriented with respect to the z-axis. It does not depend on the physical nature of the particular tensor operator. The second term depends on the physical nature of the operator and the system, but it does not depend on the magnetic quantum numbers m, m’, and q. To evaluate for the various combinations of m, m’, and q we need to only evaluate one matrix element. The others are related via the Clebsch-Gordan coefficients.

Certain selection rules for tensor operators follow directly from the selection rule of the Clebsch-Gordan coefficients. The matrix element is zero unless m’=m+q and

Statment about the Wigner-Eckart theorem due to Tristan Hübsch at Howard University.

ˆ () th Consider Qq , the q element of the 2 + 1 elements of a tensor operator of rank . ˆ () Imagine a matrix element in which Qq links states with different j and m values.

;,j mj ; ,m ˆ () Qq :   ;,j mj  ;  ,m 

The symbols  and  represent all the quantum numbers exclusive of the angular momentum like j and m. The relative transition or mixing matrix elements are: ;,jmQ ˆ () ;, jm j,,;,mj mq q   ;,jm   Qˆ () ;, jm  j,,;,mj mq  q  Our irreducible tensor operator must represent r . Clearly, we need:  rxiyjzkrˆˆ ˆˆ (sin cos i ˆ  sin sin ˆj  cos k

Page 15 Lecture 8 Dec: Selection, H and Line Intensities 

 rr22{ Y11 (,) Y (,)}  iiˆˆ  { Y11 (,)  Y (,)}  j 4 Y0 (,) kˆ  3311 11 31 2211ˆˆ11 40ˆ rYYiiYYjYˆ 33{11 (,)   (,)}  {11 (,)   (,)} 31 (,) k We see that  = 1 and q assume the values -1, 0 and 1. The operator z involves only one element in the irreducible representation while x and y require using two of the elements.

49 Hübsch proposes the problem that the outer proton in 21 Sc (Scandium 49) has total angular momentum 3/ and spin angular momentum 1/ . The perturbing or mixing 2 2 operator is our friend r which has angular momentum character 1 so that the final sates 5 3 1 that can be reached have angular momentum /2, /2 and /2. If the proton initially has a z component projection of m, the final states can be mm +1, m and m - 1. Supposing that 3 the proton stays in the j = /2 state, the various projection of angular momentum are 22 2 33 33 33 proportional to: 22,1,1;,1;,mm 22 m ,1;;,0m 22 ,1,1;,1 mm Recall that the probability amplitudes are proportional to the matrix elements so that the probabilities are proportion to the absolute squares of the elements. There are four initial m values to consider. Let us evaluate the ratios for the case that m = ½.

333 12111113 3 3 3 8 2, 22 ,1; 2 ,15 ; 2 , 22 ,1; 2 , 0  152 ; ,  22 ,1; 2 ,  1  15 2 The three processes above occur in the ratios: 6:1:8. That is, /5’s of the time, the 3 1 8 projection will grow to /2; /15 ths of the time, the projection will stay the same; and /15’s 1 of the time, the projection will flip to - /2.

3 Exercise: Repeat the calculation for an initial projection of + /2.

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