Total Angular Momentum for Hydrogen

Physics 4 Lecture 7 Total Angular Momentum for Hydrogen Lecture 7 Physics 4 Quantum Mechanics I Friday, April th, 008 We have the Hydrogen Hamiltonian for central potential φ(r), we can write: H r = p p + φ(r). (7.) m We know that L and L z commute with the Hamiltonian, and, trivially, so do S and S z. Our current goal is to establish the eigenfunctions of the total angular momentum J = L + S, or, more precisely, J and J z, and look at the parity of the result. 7. Angular Momentum Addition We will need to be able to add two angular momenta to form a total angular momentum: J = L + S (7.) where L, S and hence J are angular momentum operators of some sort. We assume that we have in our possession a set of states: l l z and s s z such that L l l z = l (l + ) l l z, L z l l z = l z l l z, and similarly for s s z. The plan is to combine the eigenkets from the L and S subspaces our first question: How big is the space we need to span? Think of the operators L ± = L x ± i L y with L + l l z = α + l l z + for some normalization α +. Our requirement is that the maximum value of l z The normalization constant associated with L ± is: L ± l l z = p l (l + ) l z (l z ± ) l l z. (7.) of 9

7.. ANGULAR MOMENTUM ADDITION Lecture 7 is l, and the minimum value of l z is l, so that for a given l, we have l + eigenvectors for L z. So we need to span a combined space of size: ( l + ) ( s + ). We know, from the additive structure of J z = L z + S z that any eigenstate of L z and S z has eigenvalue j z = l z + s z. But then, this eigenket must lie in a state of total J > j z. For example, if we have l l z = and s s z =, then the combination has j z =, but j itself (the total momentum) can then be at least j =, and any number greater than this (integer or half-integer). Now we do not know what the total j corresponding to a particular j z = l z + s z is. We do know that once we have one value for j, we can construct j + total states with this angular momentum (the j + eigenstates of J z ). So we are really asking for the number of distinct series of j +, j z eigenkets associated with a particular j. We ll denote this number N(j), and it corresponds to the number of available combinations l l z s s z with total angular momentum j. In terms of the degeneracy n(j z ), the number of ways of making a total j z eigenket, we have n(j z ) = N(j). (7.4) j j z In other words, we can relate the total degeneracy of the j angular momentum to the degeneracy of the j z component. From our example, this says that j z = can be obtained by any allowed combination of total angular momenta with j (of which j z = could be a part). So we add the degeneracy of the total j for all possible j to get the degeneracy of the j z. Now consider the sums: n(j) = j j N(j ) n(j + ) = j j+ subtracting gives us a way to count N(j) using n(j): N(j ) (7.5) N(j) = n(j) n(j + ). (7.6) The advantage here is that we know n(j z ) that s just the number of ways of picking l z and s z so that j z = l z + s z. To count n(j z ), for example, we just add up all the choices of l z and s z that sum to j z. We can do this graphically, as shown in Figure 7. the points of 9

7.. ANGULAR MOMENTUM ADDITION Lecture 7 represent the z-component value of the spins (in this case, we have l =, s =, and the diagonals represent sums of constant j z (the three available sums for j z = are shown). So, from the figure, there are ways to get j z = 0, j z = ±, ways to have j z = ± and way to get j z = ±. j z = s z 0 l z 0 Figure 7.: Example of counting degeneracy for addition of l =, s = momenta. In general, we have (order so that l > s): 0 j z > l + s n(j z ) = l + s + ( j z ) l + s j z l s s + l s j z 0 (7.7) where the first line tells us that there are inaccessible values of j z for a given l and s, the third line covers the counting for the maximal orders (j z = ±, 0 in our example), and the middle line takes care of the rest of the terms. Using this form, we can compute (7.6) we learn, again assuming l > s, and looking at the form of the n(j) n(j + ), that only one of the terms depends at all on j the middle conditional will return: l + s + ( j ) (l + s + ( j + )) l + s j l s (7.8) or N(j) = j = l + s, l + s, l + s,..., l s. (7.9) So the recipe for success: In any angular momentum addition, there is a unique set (N(j) = ) of eigenvectors j j z with j = l + s l s in integer steps, and for each j, j z = j j in integer steps. of 9

7.. EXAMPLE Lecture 7 7. Example Take l =, and s =. Our basis kets for each of these are:, 0,,. (7.0) We know that there are combined states of total j =, and j =, so we ll pick a state with a j z = first: j =, (7.) and this must be a state with j = J = L + S + L S, and as well. We can check this assertion L S = 4 (L + + L ) (S + + S ) 4 (L + L ) (S + S ) + L z S z. Then the operator J acting on the composite state is: ( L + S + L S ) = ( + 4 ) + and this is just right for J = ( ) + (7.) (7.). (7.4) Now we can apply J = L + S to this eigenket in order to lower the j z. With normalization in place, we have and L l l z = (l + l z ) (l l z + ) l l z, (7.5) J We can normalize this: ( = 0 = ( 0 + + ). (7.6) ). (7.7) 4 of 9

7.. TOTAL ANGULAR MOMENTUM Lecture 7 The procedure continues, but is effectively covered by a Clebsch-Gordon table, this is basically how they are generated. What about the net j = state? We can generate this by taking the most general linear combination of states with j z = and projecting out the known state that will leave us with a pure j =, j z =, the top state of j =, and we can once again apply the lowering operator to get the other. Begin with then j = α + β 0 j, (7.8) = ( a + b ), (7.9) and to get this to be zero, and normalize the resulting state, we must have α =, β =, giving us the state: j = ( 0 This state has J j = 4 j as expected. ). (7.0) 7. Total Angular Momentum Since the Coulomb Hamiltonian H r has and, because there is no reference to spin at all, we have, trivially: [L, H r ] = [L z, H r ] = 0 (7.) [S, H r ] = [S z, H r ] = 0 (7.) [J, H r ] = [L + L S + S, H r ] = 0 (7.) and [J z, H r ] = 0. So the Hamiltonian has eigenfunctions that are simultaneously eigenfunctions of J and J z. What do these look like? Since the 5 of 9

7.. TOTAL ANGULAR MOMENTUM Lecture 7 Hydrogenic wave function is made up of an orbital part and a spin part, we can write Ψ(r) = ψ(r) (7.4) for a two-component spinor associated with the Pauli matrix representation of spin that is, S = σ has S z + = +, S z = and S ± = 4 ±. In addition, we know that the eigenfunctions of L are, in spherical representation: Yl m (θ, φ) = fl m e i m φ Pl m (cos θ) (7.5) for f m l a normalization factor introduced to ensure that (Yl m (θ, φ)) Yl m sin θ dθ dφ = δ ll δ mm (7.6) In this context, we have the operator eigenequation: and L z Y m l = m Y m l. L Y m l = l (l + ) Y m l, (7.7) The eigenfunction of J are combinations of the Yl m and + and, precisely the angular momentum addition we developed in the previous section. We know that there exist combinations, sometimes denoted YJ M that have: J Y M J = J (J + ) Y M J J z Y M J = M Y M J. (7.8) 7.. Computing Y M J Since we already know that the total angular momentum for an electron orbit and spin can come in only two combinations l = J +, l = J, it is reasonable to try to write down the actual expressions. For a state Y M J with definite M, we can have: Y M J = α Y M J+ + β Y M+ J+ + γ Y M J + + δ Y M+ J, (7.9) where each term is constructed so as to have J z component M (the total of S z + L z ). It is tedious at this point to actually compute the constraints on these terms that lead to the final form the line of reasoning is the same as in the above examples, you expand: J = L + S + L S (7.0) 6 of 9

7.4. PARITY FOR SPHERICAL HARMONICS Lecture 7 into ladder operators as in (7.), combine terms and set the whole thing equal to J (J + ) YJ M, then solve for (α, β, γ, δ). It is not surprising that the pairs (α, β) and (γ, δ) do not communicate (different l values). Nor will it come as a shock to learn that you have an overall normalization factor. In the end, the combination that solves the eigenvalue problem is: [ [ YJ M = α Y M + J + M J+ + + J M Y M+ J+ ]+γ Y M J M J + + J + M Y M+ J (7.) so we can divide this into two separate solutions (parity concerns, as we shall see below, suggest this) one with l = J + and one with l = J. In the end, after normalization, we get: Y M+ J = ( + J M Y M (J + ) J+ + + ) + J + M Y M+ J+ Y M J = ( J M + M Y J J + + ) J M Y M+ J (7.) ] 7.4 Parity for Spherical Harmonics The parity operator P in one dimension changes the sign of the coordinates: P f(x) = f( x). The eigenfunctions of P are even and odd functions with eigenvalue ±. When we think of the Hamiltonian operator with central potential: H = m + V (r), (7.) it is clear that [H, P ] = 0. So eigenfunctions of the Hamiltonian can be taken to be eigenfunctions of P. Since the spherical harmonics make up the angular solution, it is natural to ask if they are the desired simultaneous eigenfunction of P? Spherical harmonics are products of e i m φ and the associated Legendre polynomials Pl m (cos θ). The associated Legendre polynomials can be developed from the Legendre polynomials: P m l (z) =( z ) m / d m 7 of 9 dz m P l(z). (7.4)

7.5. PARITY FOR Y M J Lecture 7 What does x x (the three-dimensional parity operation) do to θ and φ? For θ, we see that reflecting the vector shown in Figure 7. has θ = π θ. θ φ θ Figure 7.: Relation of θ to θ under spatial reflection. Similarly, for the φ coordinate, we have φ = π + φ. polynomials have parity, from their definition: P l (x) = ( ) d l ( x l ) l l! dx Now the Legendre (7.5) of + when l is even, when l is odd (set y = x and rewrite the above). That means, looking at the associated Legendre polynomials, that the parity of Pl m is + for l + m even and for l + m odd. The parity of the phase e i mφ is e i m φ = e i m (π+φ) = ( ) m e i m φ, (7.6) so + for m even, for m odd. Suppose we have two functions f(x) and g(x), with P f(x) = ( ) p f(x) and P g(x) = ( ) q g(x), then the product has P (f(x) g(x)) = f( x) g( x) = ( ) (p+q) f(x) g(x). In this case, we have P Yl m = ( ) m ( ) l+ m = ( ) l+ m = ( ) l, (7.7) and we see that our spherical harmonics are eigenstates of P as well. 7.5 Parity for Y M J The parity of the total angular momentum states is determined by the parity of the underlying Yl m all we know for sure is that there are two opposite parities here since l = J + and l = J are separated by an integer, we have one combination even, the other odd, but until we know J, we cannot say more than this. For example, if J =, then there are l = and l = combinations in YJ M the l = form has parity, and l = has parity +. 8 of 9

7.5. PARITY FOR Y M J Lecture 7 Homework Reading: Griffiths, pp. 85 00. Problem 7. Some issues associated with the Stern-Gerlach wave function. a. Suppose you take the solution (6.), and treat it as a (potential) solution to Schrödinger s equation: (t) (t) γ B S (t) = i. (7.8) m z t Calculate the left and right-hand sides of this expression (set B = 0 for simplicity so no contribution, and take just the z-component of the magnetic field), and show that they differ by terms of order α. b. The magnetic field for the Stern-Gerlach experiment has an x component. Find the energies of the spin-portion of the Hamiltonian: H = γ B S if B = α x ˆx + (B 0 + α z) ẑ. Show that they reduce to the Stern-Gerlach energies when α is small (technically, we can compare αx to B 0, and the statement is B 0 α x Taylor expand your energies). Problem 7. If an electron is in the state = + (i.e. the spin up state w.r.t. S z ), what is the probability of obtaining a measurement of for S y? Problem 7. Form the eigenstates of J for J = L + S with l =, s = (think of an electron in some l = state of Hydrogen, while ignoring the proton spin) you will get four states of angular momentum and two states of angular momentum (that s j = + and j = ). Use the ladder approach to explicitly construct all states by finding the top state and working down to the bottom. Check your decompositions using the Clebsch-Gordon table (Table 4.8 on p. 88). 9 of 9