852HW6_Solutions - PHYS852 Quantum Mechanics II, Spring...

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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 spin-orbit 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 fine-structure 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 spin-orbit term is zero, while the combined mass-correction 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 mass-correction and spin-orbit 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 non-degenerate 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 third-order and the state to second-order. 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 good-basis condition. From the first-order normalization equation, we get nm(0) |nm(1) = 0. Thus we have found some, but not all first-order 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 first-order 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 second-order 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 second-order 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 third-order 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 second-order 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 first-order shift (which doesn’t mix states with different j because they are nondegenerate). So this implies that at second-order, the hyperfine interaction mixes states with different j , however these shifts will be much smaller than the first-order hyperfine splitting. 7 4. Wigner-Eckert theorem: We start from the definitions U (θ ) = e−iJ ·θ/ , Ikj = m |kjm kjm|, Vkj = Ikj V Ikj , where V is any operator-valued 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 z-axis, 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 first-order 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 Wigner-Eckert selection rules: kjm|Vz |kjm′ = 0 kjm|V± |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 kjm|V± |kjm′ = α± (k, j ) kjm|J± |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 non-zero 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 kjm|V± |kj (m ∓ 1) = α± (k, j ) kjm|J± |kj (m ∓ 1) We can replace this with kjm|V± |kjm′ = α± (k, j ) kjm|J± |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 kjm|Vz |kjm′ = α+ (k, j ) m Use [J+ , V− ] = 2 Vz to likewise show that kjm|Vz |kjm′ = α− (k, j ) m Use these results to prove that α+ (k, j ) = α− (k, j ) = α(k, j ). Answer: kjm|J− V+ |kjm − kjm|V+ J− |kjm = −2 kjm|Vz |kjm −2 kjm|Vz |kjm = kjm|Vz |kjm j (j + 1) − m(m + 1) kj (m+1)|V+ |kjm − =− + kjm|Vz |kjm = − 1 2 1 2 j (j + 1) − m(m − 1) kjm|V+ |kj ( j (j + 1) − m(m + 1)α+ (k, j ) kj (m + 1)|J+ |kjm j (j + 1) − m(m − 1)α+ (k, j ) kjm|J+ |kj (m − 1) 1 1 (j (j + 1) − m(m + 1))α+ + (j (j + 1) − m(m − 1)) 2 2 kjm|Vz |kjm = mα+ (k, j ) [J+ , V− ] = 2 Vz from we get similarly kjm|Vz |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 Wigner-Eckert projection theorem: Ikj V Ikj = α(k, j )Ikj J Ikj State in words what this relationship means. Answer: So we have kjm|Vz |kjm = α(k, j ) m which is equivalent to kjm|Vz |kjm′ = α(k, j ) kjm|Jz |kjm′ and we previously shows that kjm|V± |kjm′ = α(k, j ) kjm|J± |kjm′ These two equations (i.e. the + and - equations) can be added or subtracted to give kjm|Vx |kjm′ = α(k, j ) kjm|Jx |kjm′ kjm|Vy |kjm′ = α(k, j ) kjm|Jy |kjm′ That the same relation holds for x,y, and z components means kjm|V |kjm′ = α(k, j ) kjm|J |kjm′ which is essentially the Wigner-Eckert 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 well-defined 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 Wigner-Eckert projection theorem to show that ℓ1 ℓ2 ℓmℓ |L1z |ℓ1 ℓ2 ℓmℓ = g1 mℓ and calculate the g-factor, 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 Wigner-Eckert 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.

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