PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 8: Solutions. Topics covered: hydrogen fine structure

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1 PHYS85 Quantum Mechanics II, Spring HOMEWORK ASSIGNMENT 8: Solutions Topics covered: hydrogen fine structure. [ pts] Let the Hamiltonian H depend on the parameter λ, so that H = H(λ). The eigenstates and eigenvalues of H are then also functions of λ, i.e. E n = E n (λ) and n = n(λ). Use the property H n = E n n to prove the Feynman-Hellmann theorem: E n (λ) λ = n(λ) H(λ) λ n(λ) Start from the definition Differentiation gives E n = n H n ( ) ( ) λ E n = λ n H n + n λ H n + n H λ n Since H n = E n n and n H = n E n, we can rewrite this as ( ) ( ) λ E n = E n λ n n + n λ n E n + n H λ n Since n n =, it follows that = E n n n + n H λ λ n E n (λ) λ = n(λ) H(λ) λ n(λ)

2 . [5 pts] The effective Hamiltonian which governs the radial wave equation is The exact eigenvalues in terms of e and l are H = M r + l(l + ) Mr e 4πɛ r. Me 4 E n = 3π ɛ n(l), where n(l) = n r + l +, with n r being the highest power in the series expansion or R nl (r). Apply the Feynman-Hellman theorem with λ = e to derive: nl () R nl () = n a. Then use λ = l, with l treated as a continuous parameter, to derive: nl () R nl () = (l + /)n 3 a. According to Feynman-Hellman now and we have with a = 4πɛ m ee so that this works out to Thus Feynman-Hellman gives Solving for R thus gives nlm l H e nlm l = E n e H e = e 4πɛ R E n = m e a n m e e 4 E n = (4πɛ ) n E n e = m ee 3 (4πɛ ) n e 4πɛ R = m ee 3 (4πɛ ) n R = m ee 4πɛ n = n a.

3 Now and H l = l(l + ) l m e R = (l + /) m e R E n l = l m e a n (l) = m e a 3 n3 where we have used ln(l) =. Putting these results into Feynman-Hellman gives (l + /) m e R = m e a n3 Solving for R gives R = (l + /)n 3 a 3

4 3. [ pts] Deriving Kramer s relation: a.) First, show via integration by parts that dr ur s u = s show that dr u r s u = s+ dr u r s+ u. dr ur s u. Then use this result to Let: Thus we have dr ur s u = f = ur dg = u dr df = (u r + r )dr g = u f dg = fg gdf = ur s u dr (u r s sur s )u Assuming the boundary term vanishes, this gives dr ur s u = dr u r s u s dr ur s u Collecting like terms then gives dr ur s u = s dr ur s u This allows us to eliminate a first-derivative at the cost of reducing the power of r by one. This result can be rewritten as dr ur s u = s dr ur s u with u u and s s + this becomes dr u r s u = s + dr u r s+ u b.) With R nl (r) = u(r)/r, the radial eigenvalue equation of the hydrogen atom becomes [ l(l + ) u = r a r + ] n a u. Use this to express dr ur s u in terms of nl () R s nl (), nl () R s nl (), and nl () R s nl (). From direct substitution we find drur u = [ l(l + ) drur s r a r + ] n a u = l(l + ) R s R s + a n a R s 4

5 c.) Set aside your result from part b.), and now integrate dr ur s u by parts, assuming the boundary terms vanish. Apply the identities derived in part a.) to each of the resulting terms in the obvious manner, and then again use the eigenvalue equation to eliminate u. With f = ur s dg = u dr df = (u r s + sur s )dr g = u we find dr ur s u = f dg = fg g df = ur s u dr u r s u s dr ur s u Assuming the boundary term vanishes and inserting the results from part a.) gives drur s u = dr u r s+ u s(s ) + R s s + inserting the eigenvalue equation on the r.h.s. gives drur s u = s + From part a.), we have dr u r s u = s drur s u = dr u r s+ [ l(l + ) r a r + n a l(l + )(s ) R s + s + ] u + dr ur s u = R s, which gives s R s (s + )a n a R s + s(s ) R s s(s ) R s d.) You now have two distinct expressions for dr ur s u. Equate them and collect like terms to derive a relation between nl () R s nl (), nl () R s nl (), and nl () R s nl (). This is called Kramer s relation. Equating the expression from b.) with the expression from c.) and collecting like terms gives Kramer s relation: a s 4 ( (l + ) s ) R s a (s + ) R s + s + n Rs =, which allows you to compute the expectation values of successive powers of R via iteration. 5

6 4. [5 pts] Spin-orbit interaction: a.) Use Kramer s relation with s =, together with your results from problem, to evaluate nl () R 3 nl (). Kramers relation with s = gives: Solving for R 3 gives a l(l + ) R 3 + a R = R 3 = With the expression from problem, this becomes a l(l + ) R R 3 = l(l + /)(l + )n 3 a 3 b.) Use this expression to show that the first-order spin-orbit interaction shift, given by the Dirac equation as nlj = e L 8πɛ Me c nlj() S R 3 nlj(), can be expressed as nlj = [ E () n ] n [j(j + ) l(l + ) 3/4] M e c. l(l + /)(l + ) We can separate the eigenstate nljm () j into the tensor product nl (R) ljm j (L S), where the state nl lives in the radial Hilbert space, and ljm j lives in the angular-momentum Hilbert space. Due to symmetry, the result cannot depend on m j, so we will drop it. The state nlj () is then an arbitrary superposition of m j states. This gives nlj () L S R 3 nlj() = nl R 3 nl lj L S lj Since J = L + L S + S we get L S = ( J L S ) which gives L S R 3 = (j(j + ) l(l + ) 3/4) a 3 n3 l(l + /)(l + ) so that nlj = [ E n () ] n (j(j + ) l(l + ) 3/4) M e c l(l + /)(l + ) 6

7 5. [5 pts] From the Dirac equation, it is found that the fine-structure of hydrogen is given by three terms: the relativistic mass correction, the spin-orbit interaction, and the Darwin term. In the atomic physics lecture notes, the Feynman-Hellman theorem is used to show that the first-order relativistic mass-correction is [ ] E () ( ) nl = n 4n M e c l + / 3. Due to symmetry, the spin-orbit interaction term is exactly zero for l =, while the combined mass-correction and Darwin terms give a net shift of = M ec α 4 ( n 4n 4 n 3 ). For l, on the other hand, the Darwin term vanishes, while the spin-orbit shift is non-zero. Consider separately the two possible cases j = l + / and j = l /, and show that for l, the full fine-structure shift is given by nj = M ec α 4 ( n 4n 4 j + / 3 ). Using the above expression for, show that this formula also works for the l = case. n Based on this formula, make a sketch of the hydrogen energy levels for n =,, 3, showing the fine structure splittings. For each n, give the degeneracy of each j sub-level, and compute the level separations in ev. For l =, we must have j = /, so the general formula agrees with the sum of the mass-correction and Darwin terms. For l, we have two terms: mass-correction, and spin-orbit. Adding them together gives, nlj = M ec α 4 [ n [j(j + ) l(l + ) 3/4] 4n 4 l(l + )(l + /) = M ec α 4 4n 4 n l + / + 3 ] ] [ n 3/4 + l + l j j + l + l 3 l(l + /)(l + ) setting j = l ± / gives, [ nlj = % n 3/4 + 3l + 3l l l /4 l / 3 ] l(l + /)(l + ) [ = % n ( )/ + l + ( )l 3 ] l(l + /)(l + ) for the case j = l + /, this reduces to [ l + l nlj = % n l(l + /))(l + ) 3 ] [ nl(l + /) = % l(l + /)(l + ) 3 ] [ n = % l + 3 ] = M ec α 4 [ n 4n 4 j + / 3 ] 7

8 while for the case j = l /, it gives [ nlj = % n l + 3l + l(l + /)(l + ) 3 ] ) ] [ ( n l + 3 = % l + l(l + /)(l + ) 3 [ n(l + /)(l + ) = % l(l + /)(l + ) 3 ] [ n = % l 3 ] = M ec α 4 [ n 4n 4 j + / 3 ] () For n =, we have l =, so that j = /. This level is shifted by and the degeneracy of the level is d = j + =. = M ec α 4 8 For n =, the l = state corresponds to j = /, while the l = state splits into j = / and j = 3/ levels. Because states with the same j, but different l are still degenerate, we see that the n = bare energy level splits into to fine-structure sublevels. The energy shifts are given by 3 = 5M ec α 4 8 = M ec α 4, 8 The fine structure splitting (the splitting is easier to measure than the absolute shifts) is therefore The degeneracies are E = 3 d = M ec α 4 3 = (j + ) = 4, where the extra factor is the number of degenerate l levels, and d 3 = j + = 4 Total number of n = states: = 8 = n, as expected. 8

9 For n = 3, the l = state again forms a j = / level, the l = state splits into j = / and j = 3/ sublevels, and the l = state splits into j = 3/ and j = 5/ sublevels. Thus there are a total of 3 sublevels, corresponding to j = /, 3/, 5/. The fine-structure energy shifts are The two fine-structure splittings are and The degeneracies of the levels are = M ec α 4 7 = M ec α = M ec α E 3,lower = = M e c α 4 3, E 3,upper = 3 5 d 3 d 3 3 d = (j + ) = 4 = (j + ) = 8 = j + = 6 Total number of n = 3 states is = 8 = n. = M e c α

10 Practice problem: This problem will not be graded and no points will be awarded. The solution will be included with the solutions to the required problems. DC Stark Effect to first-order for the n = 3 level of the hydrogen atoms: The stark shift is governed by the potential: V E = ee Z. a.) List all of the degenerate nlm () l states in the n = 3 subspace. Then use the selection rules to group the levels into closed sets of coupled states. The n = 3 manifold consists of nine states: 3,,, 3,,, 3,,, 3,,, 3,,, 3,,, 3,,, 3,,, and 3,,. From the selection rule m = m, we see that the n = 3 subspace can divided into a closed theestate manifold, and two closed two-state manifolds. They are: 3,, 3,, 3,, 3,, 3,, 3,, 3,, The remaining two states do not couple with any states in the degenerate subspace:,,,, b.) Write the matrix element 3lm () l Z 3l m l () out as an integral over r, θ, φ. Evaluate the integral for all transitions which obey the selection rules. Inserting the projector I = rdr π sin(θ)dθ π dφ rθφ rθφ gives 3lm l Z 3l m l = rdr π sin(θ)dθ π The radial wavefunction (from lecture notes) is: R nl (r) = [ ( na dφ R 3l (r) [ Y m l l (θ, φ) ] r cos θr3l (r)y m l l (θ, φ), ) ] 3 / ( (n l )! exp r ) ( ) r l ( ) r L l+ n(n + l)! na na n l na Letting Mathematica handle the integrals, we find the matrix-elements which satisfy these selection rules are (labeled as z ml,l,l ): z = 3,, Z 3,, = 3 6a z = 3,, Z 3,, = 3 3a z = 3,, Z 3,, = (9/)a z = 3,, Z 3,, = (9/)a from Z = Z we then know that: z = z = 3 6a z = z = 3 3a z = z = (9/)a z = z = (9/)a

11 c.) For each group in part a.) having more than one element, find the good eigenstates, by diagonalizing V E in the subspace of the states in the group. The good eigenstates for the m l = manifold are found by diagonalizing the 3 3 matrix (in basis { 3,,, 3,,, 3,, }): V 3 = ee z z z z The eigenvalues of this matrix are v 3 = 9ea E, v 3,, = and v 3,,3 = 9ea E. The corresponding eigenvectors are: 3,, () = 6 ( 3,, 3 3,, + 3,, ) 3,, () = ( 3,, ) 3,, 3 3,, 3 () = 6 ( 3,, + 3 3,, + 3,, ) The good eigenstates for the m l = ± manifolds are found by diagonalizing the matrices ( ) z± V 3± = ee z ± The eigenvalues of this matrix are v 3,±, = (9/)ea E and v,±, = (9/)ea E. The corresponding eigenvectors are: 3, ±, () = ( 3,, ± 3,, ± ) 3, ±, () = ( 3,, ± + 3,, ± ) In this new notation n, m l, m (), where m labels the eigenstates in a given m l manifold, the remaining two good eigenstates are: 3,, () = 3,, 3,, () = 3,, Because of the selection rules, the operator Z does not couple these states to either themselves, or any other states in the n = 3 degenerate subspace. This implies that their first-order energy shifts are zero. Thus we have defined all 9 of the good bare eigenstates { n, m l, m () }.

12 d.) Give the number of distinct sublevels that the n = 3 level splits into. State the degeneracy of each sub-level, and list the good basis states which belong to each sub-level. Include a sketch of the energy levels versus E. From first-order perturbation theory and our previous results, we find that the Stark effect splits the n = 3 level into 5 sub-levels according to: level shift degeneracy states 9ea E 3,, () (9/)ea E 3,, (), 3,, () 3 3,, (), 3,, (), 3,, () (9/)a E 3,, (), 3,, () 9ea E 3,, ()