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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 6: SOLUTIONS
1. The ﬁne structure consists of the relativistic mass correction, the spinorbit interaction, and the Darwin term. Based on the fact that j = ℓ + 1/2 or j = ℓ − 1/2, show that for all n, ℓ, and j , the combined ﬁnestructure eﬀect is given by Enj = −me c2
(1) α4 4n4 2n 3 − j + 1/2 2 . (1) You do not need to derive this formula, just show that it works for all cases. Answer: From the atomic physics lecture notes, eqs (57) and (58), we have found that for ℓ = 0, the spinorbit term is zero, while the combined masscorrection and Darwin terms gives a net shift of: En0 1 = −
2 (1) me c2 α4 4n4 2n − 3 2 (2) while for ℓ = 0, the Darwin term vanishes, with the combined masscorrection and spinorbit terms given a net shift of Enℓj = −
(1) me c2 α4 4n4 n (3ℓ(ℓ + 1) − j (j + 1) + 3/4) 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 . (3) For the case ℓ = 0, we have j = 1/2, so that (1) gives: En 1 = −
2 (1) me c2 α4 4n4 2n − 3 2 which agrees with (2). For the case ℓ = 0, j = ℓ + 1/2, (1) gives En(ℓ+ 1 ) = −
2 (1) me c2 α4 4n4 2n 3 − ℓ+1 2 , (4) while putting j = ℓ + 1/2 into (3) gives: Enℓj = − =− =−
(1) me c2 α4 4n4 me c2 α4 4n4 n (3ℓ(ℓ + 1) − (ℓ + 1/2)(ℓ + 3/2) + 3/4) 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 n 2ℓ2 + ℓ 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 2nℓ (ℓ + 1/2) 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 2n 3 − , ℓ+1 2 me c2 α4 4n4 me c2 α4 =− 4n4 which agrees with (4). 1 For the case ℓ = 0, = ℓ − 1/2, (1) gives En(ℓ+ 1 ) = −
2 (1) me c2 α4 4n4 2n 3 − ℓ 2 , (5) while putting j = ℓ − 1/2 into (3) gives: Enℓj = − =− =− =− which agrees with (5). Thus we have established that (1) works in all cases.
(1) me c2 α4 4n4 me c2 α4 4n4 me c2 α4 4n4 me c2 α4 4n4 n (3ℓ(ℓ + 1) − (ℓ − 1/2)(ℓ + 1/2) + 3/4) 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 n 2ℓ2 + 3ℓ + 1 3 − ℓ(ℓ + 1/2)(ℓ + 1) 2 2n ℓ2 + (3/2)ℓ + 1/2 3 − 2 + (3/2)ℓ + 1/2) ℓ(ℓ 2 2n 3 − ℓ 2 , 2 2. In lecture, we derived expressions for degenerate perturbation theory by including VD into H0 , and then applying nondegenerate perturbation theory. Now you are going to derive the same result by the more direct method. Begin from the expression (H0 + λV − Enm )nm = 0 and as usual expand Enm and nm in powers of λ. At each order you will get an equation, which you can then hit with one of three bra’s: nm, n′ m′ , where n′ = n, or nm′ , where m′ = m. Each of these cases will give a diﬀerent piece of the ﬁnal answer. Be sure to also enforce normalization at each order to obtain additional information. Following this approach at each order, compute the energy shift to thirdorder and the state to secondorder. Answer: The j th perturbation equation is
j (0) (H0 − En )nm(j ) = −V nm(j −1) + k =1 (k ) Enm nm(j −k) Assuming we are in the ‘good’ basis, we should keep in mind that Vnmnm′ = vnm δmm′ , where Vnmn′ m′ = nm(0) V n′ m′ (0) . Starting with j = 1, this gives
(0) (1) (H0 − En )nm(1) = −V nm(0) + Enm nm(0) Hitting this with nm(0)  gives
(1) Enm = Vnmnm = vnm Hitting with n′ m′ (0) , n′ = n, gives n′ m′ (0) nm(1) = − where ∆En′ n = En′ − En . Hitting with nm′ (0) , m′ = m, gives Vnmnm′ = 0, which is just the goodbasis condition. From the ﬁrstorder normalization equation, we get nm(0) nm(1) = 0. Thus we have found some, but not all ﬁrstorder results, namely:
(1) Enm = vnm (0) (0) Vn′ m′ nm , ∆En′ n nm(1) =
m′ =m nm′ (0) nm′ (0) nm(1) −
n′ =n m′ n′ m′ (0) Vn′ m′ nm , ∆En′ n where at present we have no formula for the coeﬃcients nm′ (0) nm(1) . 3 Moving to j = 2, we have
(0) (1) (2) (H0 − En )nm(2) = −V nm(1) + Enm nm(1) + Enm nm(0) Hitting this with nm(0)  gives
(2) (1) Enm = nm(0) V nm(1) − Enm nm(0) nm(1) using the ﬁrstorder results, together with Vn′ m′ nm = Vnmnm δm′ m , transforms this to
(2) Enm = − n′ =n m′ Vnmn′ m′ Vn′ m′ nm ∆En′ n and hitting with nm′ (0) , m′ = m, gives nm′ (0) nm(1) =
n′′ =n m′′ Vnm′ n′′ m′′ Vn′′ m′′ nm , (vnm′ − vnm )∆En′′ n which ﬁlls in one missing puzzle piece. Hitting with n′ m′ (0) , n′ = n, gives
(1) ∆En′ n n′ m′ (0) nm(2) = − n′ m′ (0) V nm(1) + Enm n′ m′ (0) nm(1) After putting in the previous results, this becomes n′ m′ (0) nm(2) =−
m′′ =m n′′′ =n m′′′ Vn′ m′ nm′′ Vnm′′ n′′′ m′′′ Vn′′′ m′′′ nm (vnm′′ − vnm )∆En′ n ∆En′′′ n +
n′′ =n m′′ Vn′ m′ n′′ m′′ Vn′′ m′′ nm Vn′ m′ nm vnm − ∆En′ n ∆En′′ n (∆En′ n )2 Lastly, from the normalization equation, we have nm(0) nm(2) =− 1 nm(1) nm(1) 2
2 1 =− 2 m′ =m n′′ =n m′′ Vnm′ n′′ m′′ Vn′′ m′′ nm (vnm′ − vnm − 1 2 n′ =n m′ Vn′ m′ nm 2 2 ∆En′ n To summarize our results so far, we have found the energies to second order and the states to ﬁrst order, which agree with the result from the notes:
(1) Enm = vnm , (2) Enm = − n′ =n m′ Vnmn′ m′ Vn′ m′ nm , ∆En′ n n′ m′ (0)
n′ =n m′ nm(1) =
m′ =m nm′ (0)
n′′ =n m′′ Vnm′ n′′ m′′ V + n′′ m′′ nm − (vnm′ − vnm )∆En′ n Vn′ m′ nm , ∆En′ n 4 and we have begun to construct the secondorder eigenstates: 2 Vnm′ n′′ m′′ Vn′′ m′′ nm 1 (2) (0) 1 − nm = nm − 2′ (vnm′ − vnm 2′ m =m n′′ =n m′′ n =n Vn′ m′ nm′′ Vnm′′ n′′′ m′′′ Vn′′′ m′′′ nm + n′ m′ (0) − (vnm′′ − vnm )∆En′ n ∆En′′′ n ′ ′ ′′ ′′′ ′′′
n =n m m =m n =n m m′ Vn′ m′ nm 2 ∆En′ n 2 +
n′′ =n m′′ +
m′ =m nm′ (0) nm′ (0) nm(2) Vn′ m′ n′′ m′′ Vn′′ m′′ nm Vn′ m′ nm vnm − ∆En′ n ∆En′′ n (∆En′ n )2 where we still need to compute nm′ (0) nm(2) . Lastly, for j = 3, we have
(0) (1) (2) (3) (H0 − En )nm(3) = −V nm(2) + Enm nm(2) + Enm nm(1) + Enm nm(0) Hitting this with nm(0)  gives
(3) Enm = nm(0) V nm(2) − vnm nm(0) nm(2) which simpliﬁes to
(3) Enm = n′ =n m′ Vnmn′ m′ n′ m′ (0) nm(2) which with our previous secondorder result gives Vnmn′ m′ Vn′ m′ nm′′ Vnm′′ n′′′ m′′′ Vn′′′ m′′′ nm (3) − Enm = (vnm′′ − vnm )∆En′ n ∆En′′′ n ′ ′ ′′ ′′′ ′′′
n =n m m =m n =n m +
n′′ =n m′′ which gives us the thirdorder energy shift. Vnmn′ m′ Vn′ m′ n′′ m′′ Vn′′ m′′ nm Vnmn′ m′ Vn′ m′ nm vnm − , ∆En′ n ∆En′′ n (∆En′ n )2 Hitting the j = 3 equation with nm′ (0) , m′ = m gives: nm′ (0) nm(2) = −
n′′ =n m′′ Vnm′ n′′ m′′ ′′ ′′ (0) n m nm(2) vnm Putting in our previous results gives nm′ (0) nm(2) =−
n′′ =n m′′ − + n′′′ =n m Vnm′ n′′ m′′ Vn′′ m′′ nm′′′ Vnm′′′ n′′′′ m′′′′ Vn′′′′ m′′′′ nm vnm (vnm′′′ − vnm )∆En′′ n ∆En′′′′ n m′′′ =m n′′′′ =n m′′′′ Vnm′ n′′ m′′ Vn′′ m′′ n′′′ m′′′ Vn′′′ m′′′ nm Vnm′ n′′ m′′ Vn′′ m′′ nm − vnm ∆En′′ n ∆En′′′ n (∆En′′ n )2 ′′′ 5 Results: The energies to third order are:
(0) Enm = En + λvnm − λ2 n′ =n m′ Vnmn′ m′ Vn′ m′ nm ∆En′ n Vnmn′ m′ Vn′ m′ nm′′ Vnm′′ n′′′ m′′′ Vn′′′ m′′′ nm (vnm′′ − vnm )∆En′ n ∆En′′′ n + λ3
n′ =n m′ − m′′ =m n′′′ =n m′′′ +
n′′ =n m′′ The states to secondorder are: 2 2 ′ m′ nm  ′ n′′ m′′ Vn′′ m′′ nm 1 Vn Vnm 1 + O(λ3 ) nm = nm(0) 1 + λ2 − − 2 2′ (vnm′ − vnm 2′ ∆En′ n m =m n′′ =n m′′ n =n m′ Vnm′ n′′ m′′ V + n′′ m′′ nm + nm′ (0) λ (vnm′ − vnm )∆En′ n m′ =m n′′ =n m′′ Vnm′ n′′ m′′ Vn′′ m′′ nm′′′ Vnm′′′ n′′′′ m′′′′ Vn′′′′ m′′′′ nm − − λ2 vnm (vnm′′′ − vnm )∆En′′ n ∆En′′′′ n n′′ =n m′′ m′′′ =m n′′′′ =n m′′′′ Vnm′ n′′ m′′ Vn′′ m′′ n′′′ m′′′ Vn′′′ m′′′ nm Vnm′ n′′ m′′ Vn′′ m′′ nm + − + O(λ3 ) vnm ∆En′′ n ∆En′′′ n (∆En′′ n )2 n′′′ =n m′′′ Vn′ m′ nm Vn′ m′ nm′′ Vnm′′ n′′′ m′′′ Vn′′′ m′′′ nm + λ2 − + n′ m′ (0) −λ ′n ∆En (vnm′′ − vnm )∆En′ n ∆En′′′ n n′ =n m′ m′′ =m n′′′ =n m′′′ Vn′ m′ n′′ m′′ Vn′′ m′′ nm Vn′ m′ nm vnm + − + O(λ3 ) ∆En′ n ∆En′′ n (∆En′ n )2 ′′ ′′
n =n m Vnmn′ m′ Vn′ m′ n′′ m′′ Vn′′ m′′ nm Vnmn′ m′ Vn′ m′ nm vnm − + O(λ4 ) ∆En′ n ∆En′′ n (∆En′ n )2 6 3. Show that Vhf commutes with L2 , S 2 , I 2 , but not with J 2 , Lz , Sz , Jz , or Iz . Answer: We have Vhf ∝ I · L 3(S · R)(I · R) I · S 8π + − 3+ S · I δ3 (R) R3 R5 R 3 We know that L2 commutes with Lx , Ly , and Lz , so we must have [L2 , Vhf ] = 0. Similarly, as S 2 commutes with Sx , Sy , and Sz , we must then have [S 2 , Vhf ] = 0. And lastly, as I 2 commutes with Ix , Iy , and Iz , we also have [I 2 , Vhf ] = 0. Now Lz , Sz , Jz , and Iz clearly do not commute with Vhf , as it contains also Lx , Ly , Sx , Sy ,...etc. In fact, it is not the case that [J 2 , Vhf ] = 0. This means that j is a good quantum number only with respect to the ﬁrstorder shift (which doesn’t mix states with diﬀerent j because they are nondegenerate). So this implies that at secondorder, the hyperﬁne interaction mixes states with diﬀerent j , however these shifts will be much smaller than the ﬁrstorder hyperﬁne splitting. 7 4. WignerEckert theorem: We start from the deﬁnitions U (θ ) = e−iJ ·θ/ , Ikj =
m kjm kjm, Vkj = Ikj V Ikj , where V is any operatorvalued vector. A set of three operators Vx , Vy , and Vz , form the components of a ‘vector operator’ when, for any θ, the three operators transform under rotation as an ordinary vector. This is expressed mathematically as V ′ = U (θ)V U † (θ ) = R(θ )V where R(θ ) is the usual 3 × 3 matrix describing a rotation of angle θ = θ about axis eθ = θ/θ . a.) For a rotation by φ about the zaxis, we have U Vz U † = Vz , U Vx U † = cos φVx + sin φVy , U Vy U † = − sin φVx + cos φVy . Consider an inﬁnitesimal rotation by δφ, and use these expressions to show: [Jz , Vz ] = 0 [Jz , Vx ] = i Vy [Jz , Vy ] = −i Vx Write out the six additional commutators generated by cyclic permutation of the indices. Answer: For an inﬁnitesimal rotation, can use U = (1 − i δφJz ), cos(δφ) = 1 and sin(δφ) = δφ. Dropping terms higher than ﬁrstorder in δφ gives then i i Vz + δφVz Jz − δφJz Vz = Vz , i i Vx + δφVx Jz − δφJz Vx = Vx + δφVy , i i Vy + δφVy Jz − δφJz Vy = −δφVx + Vy . which, together with cyclic permutations, leads directly to the desired commutation relations: [Jx , Vx ] = 0 [Jx , Vy ] = i Vz [Jx , Vz ] = −i Vy [Jy , Vx ] = −i Vz [Jy , Vy ] = 0 [Jy , Vz ] = i Vx [Jz , Vx ] = i Vy [Jz , Vy ] = −i Vx [Jz , Vz ] = 0 8 b.) Use the results from a.) to show: [Jz , V± ] = ± V± [J+ , V+ ] = 0 [J+ , V− ] = 2 Vz [J− , V+ ] = −2 Vz [J− , V− ] = 0 where V± = Vx ± iVy . Answer: [Jz , V± ] = [Jz , Vx ] ± i[Jz , Vy ] = i Vy ± i(−i Vx ) = (±Vx + iVy ) = ± (Vx ± iVy ) [J+ , V+ ] = [Jx , Vx ] + i[Jx , Vy ] + i[Jy , Vx ] − [Jy , Vy ] = 0 − V z + Vz − 0 =0 [J+ , V− ] = [Jx , Vx ] − i[Jx , Vy ] + i[Jy , Vx ] + [Jy , Vy ] = 0 + V z + Vz − 0 = 2 Vz [J− , V+ ] = −[J+ , V− ]† = −2 Vz [J− , V− ] = −[J+ , V+ ]† = 0 9 c.) Use [Jz , Vz ] = 0 to show that Vz kjm is an eigenstate of Jz with eigenvalue m.1 Then use [Jz , V± ] = ± V± to show that V± kjm is an eigenstate of Jz , with eigenvalue (m ± 1). From these you can establish the WignerEckert selection rules: kjmVz kjm′ = 0 kjmV± kjm′ = 0 unless m = m′ unless m = m′ ± 1 Answer: [Jz , Vz ]kjm = 0 Jz Vz kjm = Vz Jz kjm Jz Vz kjm = mVz kjm This shows that Vz kjm is an eigenstate of Jz with eigenvalue m. [Jz , V± ]kjm = ± V± kjm Jz V± kjm = V± (Jz ± )kjm Jz V± kjm = (m ± 1)V± kjm which shows that V± kjm is an eigenstate of Jz with eigenvalue (m ± 1). 1 This is not the same as saying Vz kjm ∝ kjm , because we haven’t shown that it is an eigenvalue of J s , or any of the operators represented by k. 10 d.) Since [J± , V± ] = 0 , it follows that kj (m ± 2)[J± , V± ]kjm = 0. Use this to show that kj (m±1)V± kjm kj (m ± 2)V± kj (m ± 1) = kj (m±1)J± kjm kj (m ± 2)J± kj (m ± 1) Use this to prove that kjmV± kjm′ = α± (k, j ) kjmJ± kjm′ . Answer: We start from kj (m ± 2)[J± , V± ]kjm = 0 or equivalently kj (m ± 2)J± V± kjm = kj (m ± 2)V± J± kjm now between the two operators on each side, insert the identity I = giving kj (m ± 2)J± k′ j ′ m′ k′ j ′ m′ V± kjm =
k ′ j ′ m′ k ′ j ′ m′ k ′ j ′ m′ k′ j ′ m′ k′ j ′ m′ , kj (m ± 2)V± k′ j ′ m′ k′ j ′ m′ J± kjm note that the only nonzero terms in the sum are k′ = k, j ′ = j , and m′ = m ± 1, so we can drop the summation, giving kj (m ± 2)J± kj (m ± 1) kj (m ± 1)V± kjm = kj (m ± 2)V± kj (m ± 1) kj (m ± 1)J± kjm a slight bit of algebra then gives the answer we are looking for: kj (m ± 1)V± kjm kj (m ± 2)V± kj (m ± 1) = kj (m ± 1)J± kjm kj (m ± 2)J± kj (m ± 1) Since the r.h.s corresponds to the same expression as the l.h.s, but with m → m ± 1, and the two expression being equal, implies that the l.h.s ratio is independent of m. We can then deﬁne α± (k, j ) via: kj (m ± 1)V± kjm = α± (k, j ) kj (m ± 1)J± kjm which with m → m ∓ 1 leads to kjmV± kj (m ∓ 1) = α± (k, j ) kjmJ± kj (m ∓ 1) We can replace this with kjmV± kjm′ = α± (k, j ) kjmJ± kjm′ as the matrix elements on both sides vanish unless m′ ± 1 = m. 11 e.) Hit the commutator [J− , V+ ] = −2 Vz from the left with kjm and from the right with kjm , and prove that kjmVz kjm′ = α+ (k, j ) m Use [J+ , V− ] = 2 Vz to likewise show that kjmVz kjm′ = α− (k, j ) m Use these results to prove that α+ (k, j ) = α− (k, j ) = α(k, j ). Answer: kjmJ− V+ kjm − kjmV+ J− kjm = −2 kjmVz kjm −2 kjmVz kjm = kjmVz kjm j (j + 1) − m(m + 1) kj (m+1)V+ kjm − =− + kjmVz kjm = − 1 2 1 2 j (j + 1) − m(m − 1) kjmV+ kj ( j (j + 1) − m(m + 1)α+ (k, j ) kj (m + 1)J+ kjm j (j + 1) − m(m − 1)α+ (k, j ) kjmJ+ kj (m − 1) 1 1 (j (j + 1) − m(m + 1))α+ + (j (j + 1) − m(m − 1)) 2 2 kjmVz kjm = mα+ (k, j ) [J+ , V− ] = 2 Vz from we get similarly kjmVz kjm = mα− (k, j ) which shows that α+ (k, j ) = α− (k, j ) = α(k, j ) 12 f.) Now use your previous results to prove the WignerEckert projection theorem: Ikj V Ikj = α(k, j )Ikj J Ikj State in words what this relationship means. Answer: So we have kjmVz kjm = α(k, j ) m which is equivalent to kjmVz kjm′ = α(k, j ) kjmJz kjm′ and we previously shows that kjmV± kjm′ = α(k, j ) kjmJ± kjm′ These two equations (i.e. the + and  equations) can be added or subtracted to give kjmVx kjm′ = α(k, j ) kjmJx kjm′ kjmVy kjm′ = α(k, j ) kjmJy kjm′ That the same relation holds for x,y, and z components means kjmV kjm′ = α(k, j ) kjmJ kjm′ which is essentially the WignerEckert theorem. Multiply from the left by kjm and from the right by kjm′  and then sum over m and m′ , and you recover Ikj V Ikj = α(k, j )Ikj V Ikj It means that in a subspace of welldeﬁned k and j eigenvalues, the projection of V is equal to α(k, j ) times the projection of J . 13 g.) To ﬁnd α(k, j ) ﬁrst prove that Ikj J = Ikj JIkj . Use this to prove that Ikj J · V Ikj = α(k, j ) 2 j (j + 1) From this, show that α(k, j ) = Ijk J · V Ijk 2 j (j + 1) Answer: If we factorize kjm as kj ⊗ m , then we can say Ikj = kj kj . This means Ikj JIkj = kj kj J kj kj  The components of J can be constructed from Jz , J+ and J− . Each of these pieces is diagonal with respect to k and j . A diagonal operator has the form D = n dn dn dn . Notice that for In = dn dn , we have In D = dn dn D = dn
m dm dn dm dm  = dn dn dn  from which we see that In DIn = dn dn dn dn dn  = dn dn dn  = In D So by analogy, we have Ikj J = Ikj JIkj From this we can write Ikj J · V Ikj = Ikj JIkj · V Ikj = Ikj J · Ikj V Ikj = Ikj J · α(k, j )Ikj JIkj = α(k, j )Ikj JIkj · JIkj = α(k, j )Ikj J · JIkj = α(k, j ) 2 j (j + 1) So we see that α(k, j ) = Ikj J · V Ijk 2 j (j + 1) 14 5. Let L = L1 + L2 . Use the WignerEckert projection theorem to show that ℓ1 ℓ2 ℓmℓ L1z ℓ1 ℓ2 ℓmℓ = g1 mℓ and calculate the gfactor, g1 = g1 (ℓ1 , ℓ2 , ℓ). Do the same for ℓ1 ℓ2 ℓmℓ L2z ℓ1 ℓ2 ℓmℓ , and then show that you get the necessary result for ℓ1 ℓ2 ℓmℓ (L1z + L2z )ℓ1 ℓ2 ℓmℓ Answer: We have L = L1 + L2 . According to the WignerEckert theorem, in the subspace of ﬁxed ℓ1 , ℓ2 , and ℓ, we can say: L1 = g1 L where g1 = from which it follows that ℓ1 ℓ2 ℓmℓ L1z ℓ1 ℓ2 ℓmℓ = g1 ℓ1 ℓ2 ℓmℓ Lz ℓ1 ℓ2 ℓmℓ = g1 mℓ now L · L1 = L2 + L2 · L1 1 12 = L2 + L − L2 − L2 1 1 2 2 12 = L + L2 − L2 1 2 2 which gives us g1 = By symmetry, we have g2 = 1 ℓ2 (ℓ2 + 1) − ℓ1 (ℓ1 + 1) + 2 ℓ(ℓ + 1) 1 ℓ1 (ℓ1 + 1) − ℓ2 (ℓ2 + 1) + 2 ℓ(ℓ + 1)
2 ℓ(ℓ L · L1 + 1) Since L1z + L2z = Lz , we should ﬁnd g1 mℓ + g2 mℓ = mℓ , or g1 + g2 = 1. It is clear by inspection that indeed g1 + g2 = 1. 15 ...
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This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.
 Fall '08
 MichaelMoore
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