8. Spin and Adding Angular Momentum

1. 8. Spin and Adding Angular Momentum 8A. Rotations Revisited The Assumptions We Made • We assumed that |r formed a basis and R()|r =|r • From this we deduced R()(r) = (Tr) • Is this how other things work? • Consider electric field from a point particle • Can we rotate by R()E(r) = E(Tr)? • Let’s try it • This is not how electric fields rotate • It is a vector field, we must also rotate the field components • R()E(r) =  E(Tr) • Maybe we have to do somethingsimilar with ? +

2. Spin Matrices • How do the D()’s behave? • We want to find all matrices satisfying this relationship • Easy to show: when  = 1, D() = 1 • As before, Taylor expand D for small angles • In a manner similar to before, then show

3. We Already Know the Spin Matrices • We used to have identical expressions for the angular momentum L • From these we proved that L has the standard commutation relations • It follows that S has exactly the same commutation relations • The three S’s are generalized angular momentum • But in this case, they really are finite dimensional matrices • Logically, our wave functions would now be labeled • But s is a constant, so just label them • There are 2s + 1 of them total:

4. Restrictions on s? • Recall, for angular momentum, we had to restrict l to integers, not half-integers • Why? Because wave functions had to be continuous • Can we find a similar argument for spin? Consider s = ½ • Consider a rotation by 2: • This would imply if you rotate by 2, the state vector changes by |   – |  • But these states are indistinguishable, so this is okay! • Any value of s, integer or half-integer, is fine • The basic building blocks of matter are all s = ½ • Other particles have other spins part.s e-½ p+ ½ n0 ½ part.s 1 , 0 0 ’s 3/2

5. Basis States for Particles With Spin • Basis states used to be labeled by |r • But now we must label them also by whichcomponent we are talking about |r,ms • Comment: for spin ½, it is common to abbreviate the ms label: • The spin operators affect only the spin label: • Operators that concern position, like R, P, and L, only affect the position label • All these position operators must commute with spin operators

6. Sample Problem Define J = L + S. Find all commutators of J, J2, S2, and L2 • That’s 6 operators, so 65/2 = 15 possible commutators • I’ll just do five of them to give you the idea • Recall, for any angular momentum-like set of operators, [J2,J] = 0

7. Hydrogen Revisited • Recall our Hamiltonian: • Note that S commutes with the Hamiltonian • We can diagonalize simultaneously H, L2, Lz, S2, and Sz: • It is silly to label them by s, because s = ½ • Degeneracy: mstakes on two values,doubling the degeneracy Do all Hamiltonians commute with spin? • No! Magnetic interactions care about spin • Even hydrogen has small contributions (spin-orbit coupling) that depend on spin

8. 8B. Total Angular Momentum and Addition What Generates Rotations? • Recall that: • Rewrite this in ket notation • Define J: • J is what actually generates rotations • If a problem is rotationally invariant, we would expect J to commute with H • Not necessarily L or S

9. What are L,S and J? Consider the rotation of the Earth around the Sun: • It has orbital angular momentumfrom its orbit around the Sun: L • It has spin angular momentumfrom its rotation around the axis: S • The total angular momentum is • It is another set of angular momentum-like operators • It will have eigenvectors |j,m with eigenvalues: • Because L and S typically don’t commute with theHamiltonian, we might prefer to label our states byJeigenvalues, which do • To keep things as general as possible, imagine any two angular momentum operators adding up to yield a third:

10. Adding Angular Momentum • Commutation relations: • We could label states by their eigenvalues under the following four commuting operators: • Instead, we’d prefer to label them by the operators • These all commute with each other • These have the same j1 and j2 values, so we’ll abbreviate them: Two things we want to know: • Given j1 and j2, what will the states |j,m be? • How do we convert from one basis to another, i.e., what is: • Clebsch-Gordan coefficients

11. The procedure • It is easy to figure out what the eigenvalues of Jzare, because • For each basis vector |j1,j2;m1,m2, there will beexactly one basis vector |j,m with m = m1 + m2 • The ranges of m1 and m2are known • From this we can deduce exactly how many basisvectors in the new basis have a given value of m • By looking at the distribution of m values, wecan deduce what j values must be around • Easier illustrated by doing it than describing it

12. Sample Problem Suppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m. (a) What values of mwill appear in |j,m, and how many times? (b) What values of jwill appear in |j,m, and how many times? • First, find a list of all the m1 and m2 values that occur • I will do it graphically • Now, use the formula m = m1 + m2 to find the mvalue for each of these points • From these, deduce the m values and how manythere are • I will do it graphically • Note where the transitions are: m=3 m=2 m=-3 m=-1 m=1 m=-2 m=0

13. Sample Problem (2) Suppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m. (a) What values of mwill appear in |j,m, and how many times? (b) What values of jwill appear in |j,m, and how many times? • For any value of j, mwill run from –j to j • Clearly, there is no jbigger than 3 • But since m = 3 appears, there must be j= 3 • This must correspond to m’s from –3 to 3 • Now, there are still states with m up to 2 • It follows there must also be j = 2 • This covers another set of m’s from –2 to 2 • What remains has m up to 1 • It follows there must be j = 1 • And that’s it. Why did it run from j = 3 to j = 1? • Because it went from j1 + j2 down to j1 – j2

14. General Addition of Angular Momentum • The set of all (m1,m2) pairs forms a rectangle • The largest value of m is m = j1 + j2, which can only happen one way • As m decreases from the max value, there is one more way of making each m value for each decrease in m until you get to | j1 – j2 | • This implies that you get maximum jmax = j1 + j2 and minimum jmin = | j1 – j2 | • So, j runs from | j1 – j2 | to j1 + j2in steps of size 1

15. Check Dimensions • For fixed j1 and j2, the number of basis vectors |j1,j2;m1,m2 is How many basis vectors |j,m are there? • For each value of j, there are 2j + 1. • Therefore the total is • So dimensions work out

16. Sample Problem Suppose we have three electrons. Define the total spin as S = S1 + S2 + S3. What are the possible values of the total spin s, the corresponding eigenvalues of S2, and how many ways can each of them be made? • Electrons have spin s = ½, so • Combine the first two electrons: • Now add in the third: • If s1+2 = 0, this says: • If s1+2 = 1, this says • Final answer for s: • The repetition means there are two ways to combine to make s = ½ • For S2:

17. Hydrogen Re-Revisited • Recall hydrogen states labeled by • Because of relativistic corrections, these aren’t eigenstates • Closer to eigenstates are basis states • j = l ½ • States with different mj are related by rotation • Indeed, the value of mj will depend on choice of x, y, z axis • And they are guaranteed to have the same energy • Therefore, when labelling a state we need to specify n, l, j • We label l values by letters, in a not obvious way • Good to know the first four: s, p, d, f • We then denote j by a subscript, so example state could be 4d3/2 • Remember restrictions: l < n and j = l ½ • Often, we don’t care about j, so just label it 4d • Remember, number of states for given n,l is llet 0 s 1 p 2 d 3 f 4 g 5 h 6 i 7 k 8 l 9 m 10 n 11 o 12 q

18. 8C. Clebsch-Gordan Coefficients How do we change bases? • We wish to interchange bases |j1,j2;m1,m2  |j,m • These are complete orthonormal basis states in the same vector space • We can therefore use completeness either way • The coefficients are called Clebsch-Gordancoefficients, or CG coefficients for short • Our goal: Show that we can find them (almost) uniquely • Note that the states |j1,j2;m1,m2 are all related by J1and J2 • There are no arbitrary phases concerning how they are related • The |j,m states with the samej’s and different m’s are related by J • But there is no simple relation between |j,m’s different j’s – convention choice

19. Convention Confusion • If you ever have to look them up, be warned, different sources use different notations • Recall that the other states are alsoeigenstates of J12 and J22 • People also get lazy and drop some commas • In addition, the Clebsch-Gordan coefficients are defined only up to a phase • Everyone agrees on phase up to sign • As long as you use them consistently, itdoesn’t matter which convention you use. • They will turn out to be real, and therefore • Because of this ambiguity, people get lazy and often use what is logically the wrong one

20. Nonzero Clebsch-Gordan (C-G) Coefficients When are the coefficients meaningful and (probably) non-zero? (1) j range: (2) m range: (3) conservation of Jz: • Let’s prove the last one using • Act on the left with Jz and on the right with J1z and J2z: • Must be zero unless j1 + j2 – j is an integer j – m is an integer, etc.

21. Finding C-G Coefficients for m = j • Largest value for m is j, therefore • Recall in general • We therefore have • Recall: only if m1 + m2 = m (= j) are non-zero • This relates all the non-zero terms for m = j, all relative sizes determined • To get overall scale, use normalization • This determines everything up to a phase • We arbitrarily pick

22. Finding C-G Coefficients for m – 1 from m • We now have CG coefficients when m = j • I will now demonstrate that if we have them for m, we can get them for m – 1 • First note • Dagger this • So • So if we know them for m, we know them for m – 1 • Since we know them for m = j, we know them for m = j – 1, j – 2, etc. • Hence we have a (painful) procedure for finding all CG coefficients • Sane people don’t do it this way, they look them up or use computers

23. Properties of CG-coefficients • Adding j1 and j2 is thesame as adding j2and j1 • Corollary: if j1 = j2, then the combinations of spins is symmetric if j1 +j2 – j is even, anti-symmetric if it is odd • You can work your way up from m = –j in the same way we worked our way down from m = j: • Adding j1 = 0 or j2 = 0 is pretty trivial,because these imply J1 = 0 or J2 = 0 • If you ever look things up in tables, they will assume j1 j2 > 0, and assume you will use the first or third rule to get other CG coefficients • Or you can use computer programs to get them > clebsch(1,1/2,1,-1/2,3/2,1/2);

24. CG coefficients when j2 = ½ • For j2 small, we can find simple formulas for the CG coefficients • If j2 = ½, then j = j1 ½ • Example: • For one electron, J = L + S. Let j1 l, m  mj,drop j2 = s = ½, m2 = ½   • For adding two electron spins, drop s1and s2, abbreviate mi = ½  

25. Sample Problem Hydrogen has a single electron in one of the states |n,l,m,ms = |2,1,1,– or |2,1,0,+ , or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,1/2,1/2 . In all four cases, write explicitly the wave function • For s = ½, wave function looks like • Spin state mstells us which component exists • This lets us immediately write the wave functionfor the first two: • For the |j,mjstates we have:

26. Sample Problem (2) … or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,3/2,1/2 . In all four cases, write explicitly the wave function • You can also get the CG coefficients from Maple: > clebsch(1,1/2,1,-1/2,3/2,1/2); > clebsch(1,1/2,0,1/2,3/2,1/2); > clebsch(1,1/2,1,-1/2,1/2,1/2); > clebsch(1,1/2,0,1/2,1/2,1/2);

27. Sample Problem Hydrogen has a single electron in the state |n,l,j,mj = |2,1,3/2,1/2. If one of the following is measured, what would the outcomes and corresponding probabilities be, and what would the state afterwards look like: E, J2, Jz, L2, S2, Lz,Sz • For the first five choices, our state is an eigenstate of the operator • The eigenstate will be unchanged by this measurement • For the last two, we write it in terms ofeigenstates of Lz or Sz • Then we have • State afterwards is • Or we have • State afterwards is

28. 8D. Scalar, Vector, Tensor Definition and Commutation with J • A scalar operator S is anything that is unchanged under rotation • Examples: • Scalar operators commute with the generator of rotations J: • Vector operators V are operators that rotate like a vector: • Examples: • They have commutation relations with J given by • A rank 2 tensor Tij under rotation rotates as: • Can show that • Rank k tensor has k indicesand commutation relations: • Scalar = rank 0 tensor, Vector = rank 1 tensor • Rank 2 tensor is sometimes just called a “tensor”

29. How to Make a Tensor From Vectors • If V and W are any two vector operators,then we can define a rank-2 tensor operator: • One can similarly define higher rank tensor operators • This tensor has nine independent components • But it has pieces that aren’t very rank-2 tensor-like: • Dot product VW is a scalar operator • Cross product VW is a vector operator • The remaining five pieces are the truly rank-2 part • We want figure out how to extract the various pieces

30. Spherical Tensors • We start with a vector operator V • Define the three operators Vq by: • You can then show the following: • Proof by homework problem • Compare this with: • Another way to write it: Generalize this formula • Define a spherical tensor of rank k as 2k + 1 operators: • It must have commutation relations: • Trivial example: A scalar is a spherical tensor of rank 0

31. Combining Spherical Tensors (1) Theorem: Let V and W be spherical tensors of rank k1 and k2 respectively. Then we can build a new spherical tensor T of rank k defined by: • Those matrix elements are CG coefficients • Proof:

32. Combining Spherical Tensors (2) • We have complete set of states |k1,k2;q1,q2 • Now insert complete set of states |k’,q’: • J doesn’t change the k value, so k’ = k • So we have proven it

33. How it Comes Out • This sum only makes sense if CG coefficients are non-zero • Only non-zero terms are when q1 + q2 = q • So it’s really just a single sum • By combining two vectors, we can get k = 0, 1, 2 • k = 0: Scalar (dot product) • k = 1: Vector (cross product) • k = 2: Truly rank 2 tensor part • We can then combine rank 2 tensors with more vectors to make rank 3 spherical tensors

34. Sample Problem If we combine two copies of the position operator R, what are the resulting components of the rank-2 spherical tensor Tq(2)?

35. 8E. The Wigner-Eckart Theorem Why it should work • Suppose we have an atom or other rotationally invariant system • Eigenstates should be eigenstates of J2, Jz, probably other stuff • It is common to need matrix elementsof operators between these states: • We know how the ket and bra rotate • If we also know how the operator in the middle rotates, we should be able to find relations between these various quantities • Suppose the operator is a spherical tensor,or combinations thereof • Then we know how T rotates, and we should be able to find relations • This helps us because: • If the calculation is hard, we do it a few times and deduce the rest • If the calculation is impossible, we measure it a few times and deduce the rest

36. Similarities With CG coefficients (1) • I want to compare the matrix element above to the CG coefficient above • Recall relations for Jz: • Use commutationrelation: • Let Jzact on the braor the ket on the left: • Hence matrix elements are zero unless • Compare to the CG coefficient above: • This vanishes unless

37. Similarities With CG coefficients (2) • Recall relations for J: • For m = j, • Implies: • Our commutation relations tell us: • Compare to the CG coefficients:

38. Similarities With CG coefficients (3) • We have: • Equivalent to • Our commutation relations tell us:

39. These Matrix Elements are CG Coefficients • We have three relations that are identical for these two expressions: • These expressions were all that were used to find the CG coefficients • Plus, we had a normalization condition • Hence, these two expressions are identical • Up to normalization

40. The Wigner-Eckart Theorem What can the proportionality constant depend on? • Not m, m’, nor q • It can depend on , ’, j, j’, and of course T • The Wigner-Eckart Theorem: • The square root in the denominator is a choice, neither right nor wrong • That other thing is called a “reduced matrix element” • You don’t calculate it (directly) • You may be able to calculate left side for one value of m, m’, q • Or you may be able to measure left side for one value of m, m’, q • Then you deduce the reduced matrix element from this equation • Then you can use it for all the other values of m, m’, q

41. Why Is the Wigner-Eckart Theorem Useful? • The number of matrix elements is • For example, if j = 3, j’ =2, k = 1, this is 105 different matrix elements • Calculating them computationally may be difficult or impossible • Measuring them may be a great deal of work • By doing one (difficult) computation or one (difficult) measurement you can deduce a lot of others Comment: why is the factor of 2j + 1 there? • If T0(k) is Hermitian, then you can show

42. Sample Problem The magnetic dipole transition of hydrogen causing the 21 cm line is governed by the matrix element , where F is the total angular momentum quantum number and mFis the corresponding z-component, and S and L are spin and orbital angular quantum operators for the electron. Deduce as much as you can about these matrix elements for mF = +1, 0, or –1. We have no idea what most of this means, but it’s clear: • Fand mF are angular quantum number, effectively, j F and m  mF • S and L are vector operators • Call reduced matrix element A: • Non-vanishing only if q + mF = 0 • Get the CG coefficients from program • All other matrix elements vanish

43. Sample Problem Deduce as much as you can about these matrix elements for mF = +1, 0, or –1. • For mF = + 1, we also have • Vx and Vy: two equations, two unknowns • We therefore have: • Similarly:

44. 8F. Integrals of Spherical Harmonics Products of Spherical Harmonics • Consider the product of any two spherical harmonics: • By completeness, this can be written as a sum of spherical harmonics: • The coefficients clm can be found usingorthogonality: • Think of the expression as an operator acting on a wave function: • It is not hard to see that this operator is a spherical tensor operator • Think of clm then as a matrix element • By the Wigner-Eckart theorem: • All that remains is to findthereduced matrix elements

45. Working on the Reduced Matrix Element • Substitute the top equation in the bottom • Multiply this expression by and sum over m1, m2: • Rename l’ as l sum over complete states

46. Finishing the Computation • Must be true at all angles • Evaluate at  = 0 • Formula for the Y’s at  = 0 is simple • Now we solve for thereduced matrix element • We therefore have

47. When doesn’t it vanish? We want to know when this is non-zero, or likely to be non-zero: • We need: • We need: • Under parity, each of the spherical harmonics transforms to • So the whole integral satisfies • We need

48. Sample Problem Atoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f? • The initial state has n’ = 4 and l’ = 0, 1, 2, or 3 • Final state has unknown n and l, but • Must have n <4 because energy goes down • Must have l < n • The position operators can be written in terms of l = 1 spherical harmonics • So we have • To not vanish, we need • For 4s, l = 0: • Must have l < n < 4

49. Sample Problem (2) Atoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f? • For 4p: l’ = 1, so • Must have l < n < 4 • For 4d: l’ = 2, so • Must have l < n < 4 • For 4f: • Must have l < n < 4