1 Central potentials and angular momentum

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1 1 Central potentials and angular momentum We are interested in systems governed by Hamiltonians of the form: H = m + V r) where V r) is in general a 3D potential with dependence on r, but not on θ, φ. Exercise Evaluate [H, L] = [H, ˆL x i + ˆL y j + ˆL z k] and [H, L ]. Solution The key is to rewrite H in spherical coordinates: H = r ψ ) + L mr r r mr + V r) Recall that the angular momentum operator L only acts on θ and φ. We know that [L, L ] = 0 and [L, L i ] = 0 for all i = x, y, z. The rest of the Hamiltonian doesn't act on θ or φ. Hence, [H, L x ] = [H, L y ] = [H, L z ] = 0 which altogether mean that [H, L] = 0, and [H, L ] = 0 too. Consider a set of commuting observables SCO) A, B, C,, which means that the commutator between any pair of such operators is zero. You can always nd a common set of eigenstates of all such operators. Note that {H, L, L i } for i = x, y, z is a SCO, but {H, L, L} is not, for instance. In particular, it is a convention to work with the SCO {H, L, L z }. So far, you know the eigenstates of L and L z : They are the spherical harmonics Y m l The common eigenstates of H, L and L z can be written in the form ψ nlm r, θ, φ) = Rr)Y m l Exercise Find the eective TISE for Rr) for xed l and m values. Solution Hψ nlm r, θ, φ) = mr r d mr dr r Rr) r ) r drr) dr As an aside, by writing Rr) = ur)/r, we get: [ d ur) ll + 1) + m dr mr which looks like a 1D TISE in the coordinate r! Hydrogenic atoms θ, φ). θ, φ). ) Yl m θ, φ) + ll + 1) Rr)Y m mr l θ, φ) + V r)rr)yl m θ, φ), [ ] ll + 1) + + V r) Rr) = ERr). 1) mr ] + V r) ur) = Eur), ) Rr) can be readily solved for a Coulombic potential V r) = e 4πɛ 0 to give a set of discrete r eigenenergies which only depend on n, E n = Z J) = Z 13.6eV ). n n

2 The probability density to nd the electron at distance r from the nucleus is Rr) r page 35 of McQuarrie). Note from Eq. 1), that Rr) = R nl r) but it has no dependence on m. No need to remember their functional forms they look like Laguerre polynomials). In Dirac notation we write r, θ, φ nlm = ψ nlm r, θ, φ). After solving the radial and angular parts, we have the following constraints: n = 1,, l = 0, 1, n 1 = S, P, D, F, m = l,, l 3) Exercise Prove that there are n eigenstates that share the quantum number n. Solution We are trying to prove that there is 1 1 = 1 state for n = 1 1s 1, = 4 states for n = s, p x, p y, p z ), and so on. We just need to show that n 1 l=0 l + 1) = n. For example, for n = 8, we need to evaluate S = = ) ) ) ) = 16 4 = 64. Not a general n, we need to add n n times, giving n. For ψ nlm r, θ, φ) = R nl r)yl m θ, φ) there are n 1 l radial nodes in R nl r) and l angular nodes in Yl m θ, φ). The latter are distributed as follows: m nodes along φ equatorial motion, yielding z projection to L), and l m nodes along θ. The number of radial nodes can be easily understood by looking at Eqs. 1) and ) at a xed l. First, note that the number of nodes in Rr) is the same as the number of nodes in ur). Eq. ) is just the 1D TISE we know and love. It has a ground state with no nodes, the rst excited state with one node, and so on. For concreteness take l = 1. The constraints in Eq. 3) tell us that the ground state for such equation must be R 1 r) with no nodes, and the successive excited states are R 31 r), R 41 r),, with one, two nodes, etc. But this precisely satises the rule of n l 1 nodes in R nl r). Exercise Based on Eqs. 1) and ), together with the constraints in Eq. 3), conclude that R nl r) has n l 1 nodes. Solution For xed l, R l+1,l r) is the ground state of Eq. 1), according to the constraint for the allowed values of l in Eq. 3), which has no nodes. R l+,l r) has one node, and following this pattern, R l+k,l r) has k 1 nodes, but n = l + k, hence, we get n l 1 nodes for R nl r).

3 Exercise Consider an electron prepared in the eigenstate ψ 30 r, θ, φ) = π Z Discuss the nodal struc- ) 3/ a 0 ρ e ρ/3 3 cos θ 1) of the hydrogenic atom. ture of the wavefunction and sketch it. Solution We rst identify ψ 30 as the 3p z orbital ψ 30 r, θ, φ) = R 3 r)y 0 θ, φ). The nodal structure must be analyzed indepedently for the real and imaginary parts, but in this case, the latter does not exist. We remember that Yl m has l nodal planes isolated points don't count), with m being along the equator. In this case, there are no equatorial nodes in φ, but there are l = nodes along the cones θ = arccos±1/3) = 1.3, 1.91, which are θ =70.5 o and o, respectively. This means that all the angular motion that ψ 30 encodes is along θ. Finally, R 3 has n l 1 = 0 nodes it is the ground state of Eq. 1) for l = ). A plain sketch of Y 0 θ, φ) is shown here, whereas one for Rr) is shown here, Warning: Note that the book normally sketches r Rr) instead of Rr).

4 Exercise Is the chemistry wavefunction 5d x y = ) an eigenstate of H for the hydrogenic atom? How about for L and L z? For the positive answers, write their eigenvalues too. Solution 5d x y is an eigenstate of H with energy E 5 = Z J) 5, it is also an eigenstate of L with eigenvalue ll + 1) = 6. But it is a superposition of m = ± eigenfunctions of L z..1 Magnetic eld B z in the z-direction Let H 0 = m e 4πɛ 0 be the hydrogenic Hamiltonian. Consider this hydrogenic atom in a r magnetic eld B z along the z direction: H = H 0 + e B z m e L z Note that the new H still commutes with L and L z, and also with H 0. Hence, can recycle the nlm eigenstates of H 0 : ) m ) m e H nlm = Z J) + e B z n = Z J) + β n B mb z nlm ) nlm Now, states with dierent m have dierent energies! Yet, the splittings OβB z ) 1 10 cm 1.. Spin Spin is postulated to be another angular momentum observable. By analogy to L and L, the following algebra is postulated: [S x, S y ] = i S z [S y, S z ] = i S x [S z, S x ] = i S y 4) From which it follows that [S, S i ] = 0. You might have not noticed this, but the determination of the eigenvalues and eigenvectors in bra-ket notation for L and L z only depends on the analogous relations to Eq. 4) see Appendix of Chapter 6, McQuarrie, if interested). Hence, I shall just write the following results right away: S s, m s = ss + 1) s, m s S z s, m s = m s s, m s and there are s + 1 m s values: m s = s, s + 1,, s 1, s. These eigenstates have no θ, φ representation, unlike the spherical harmonics, since the degrees of freedom are not the

5 angles θ and φ of a particle, but rather, some abstract spin degrees of freedom. An important dierence with orbital angular momentum is that particles have xed spin quantum number s. For example, electrons are spin 1 particles: s = 1. Exercise Consider the operator J = i L i) + S i) ), the sum of orbital and spin angular momentum operators for a series of particles i. Do the analog relations of Eq. 4) hold? Solution Yes, operators [L i) p, L j) q ] = [S p i), S q j) ] = 0 for i j, [L i) p, S q j) ] = 0 regardless of whether i = j or i j, and the only nonzero commutation relations are [L i) p, L i) q ] = r i ɛ pqrl i) r and [S p i), S q i) ] = r i ɛ pqrs r i). Hence, J is also an angular momentum operator, and it has eigenstates J, M such that: J J, M = JJ + 1) J, M J z J, M = M J, M.3 Spin-orbit interaction and term symbols A moving electron generates a magnetic eld which is proportional to L. This eld interacts with the spin to give rise to a spin-orbit interaction: H = H 0 + ξr) L S

6 Exercise Does H commute with L, S, J, L z, S z, J z, where J = L + S is the total angular momentum? Solution We know from our work before that H 0 commutes with L, L z. Since it is spin independent, H 0 also commutes with S and S z. Since H 0 commutes with L z and S z, it also commutes with J z : [H 0, J z ] = [H 0, L z ] + [H 0, S z ]. We just need to check that it commutes with J. Note that J = L + S) L + S) = L + S + L S, where we note that: L S = L x S x + L y S y + L z S z = i L i S i So we only need to check that [H 0, i L is i ] = 0. But that's easy because [H 0, L i ] = 0 and H 0 does not depend on S i, so [H 0, S i ] = 0. So H 0 commutes with all the list operators. This is not surprising. What are the common eigenstates of all those operators? Now we need to check if ξr) L S commutes with any of the operators. [ i L is i, L j ] = ik i ɛ ijkl k S i 0. Similarly, [ i L is i, S j ] = ik i ɛ ijkl i S k 0. However, [ i L i S i, L ] = i 0)S i = 0 [ i L i S i, S ] = i L i 0) = 0 [ i L i S i, J ] = [ i L i S i, L + S + i L i S i ] = 0 [ i L i S i, J j ] = [ i L i S i, L j + S j ] = ik = ik i ɛ ijk L k S i + i ɛ ijk L i S k ) i ɛ ijk L k S i i ɛ kji L k S i ) = 0 Given these results, it is only a few more straightforward steps to show that {H, L, S, J, J z } form a SCO. As an exercise, show that the remaining needed commutators are zero. For our time being, let us consider eigenstates only {L, S, J, J z }. We label them by JMls, such that the following are all true:

7 J JMls = JJ + 1) JMls J z JMls = M JMls L JMls = ll + 1) JMls S JMls = ss + 1) JMls 5) If I have an arbitrary wavefunction ψ of the angular and spin coordinates of the electron, I can always decompose it as: ψ = l,m,s,m s c l m;s m s l ms m s where c lm;sms is some complex coecient. I can always do this because the eigenstates of L, L z form a complete set for the angular degrees of freedom of the electron, and the eigenstates of S, S z fulll the analog requirement for the spin degrees of freedom. Notice that the sum over s could be ommited: As we mentioned before, the total spin of a particle cannot be changed. Nevertheless, we shall keep it for generality because, in fact, the results we are about to derive hold for sums of arbitrary angular momentum objects. Let us now focus on a more specic ψ, namely an eigenstate JMls of {L, S, J, J z }. Exercise Writing JMls = l,m,s,m s c l m;s m s l ms m s, what values of l and s enter the summation? Solution Acting L in both sides of the identity, L JMls = c l m;s m s l l + 1) l ms m s l,m,s,m s ll + 1) JMls = ll + 1) c l m;s m s l ms m s l,m,s,m s so l = l in the summation. This means that JMls is a combination of states with denite l = l. By acting S on the identity, we also conclude that JMls only contains states with denite s = s. Now, let's do some counting: There are D = l +1)s+1) states l ms m s with xed l = l and s = s. When I combine them to form a set of eigenstates of {L, S, J, J z }, I'd better get again D states, but of the form JMls. There's a theorem, which I won't prove, which says that for given values of l and s, the possible J values are l + s, l + s 1,, l s. For each of these J values, there are J + 1 values of M, from J to J, as expected from any angular momentum. As an example, consider l = 3 and s = 1, where D = 7)) = 14 The allowed J values are 7 and 5. So the JM31 states are: 7, 7, 3, 1, 7, 5, 3, 1, 7, 3, 3, 1, 7, 1, 3, 1, 7, 1, 3, 1, 7, 3, 3, 1, 7, 5, 3, 1, 7, 7, 3, 1

8 5, 5, 3, 1, 5, 3, 3, 1, 5, 1, 3, 1, 5, 1, 3, 1, 5, 3, 3, 1, 5, 5, 3, 1 Obviously, these states have an explicit form in terms of θ, φ and the spin degrees of freedom. We don't need to know this form for the time being. If you are mathematically inclined, you are welcome to show: l+s j= l s j + 1 = l + 1)s + 1) = D. Why do we care about all this? Turns out that if one makes very precise measurements of eigenenergies of the hydrogen atom, one will see that the original spectrum that we had predicted without the spin-orbit term is actually split into a ne structure. The eigenstates of H can be written as njm ls, where n is the quantum number associated to H. Now, the eigenenergies this time depend not only on n, but also on l, s, and J. Turns out they don't depend on M. You wouldn't know this if I don't tell it to you. In atomic physics, there is an additional nomenclature for these states, called "term symbols." For example, consider the 7 eigenstates 4f l = 3) of H 0. Upon inclusion of the spin orbit coupling, these l = 3 orbital states couple to the electron spin s = 1 and form eigenstates JM31 which we have listed above. Each group of 8 and 6 states carries a label: S+1 L J = F 7/, F 5/. Each of these terms is associated with two dierent energies, E4f F 5/ ) = cm 1 E4f F 7/ ) = cm 1 which are have tiny energy dierencesm, but yet, people can resolve these days.