QMI PRELIM Problem 1. All problems have the same point value. If a problem is divided in parts, each part has equal value. Show all your work.

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1 QMI PRELIM 013 All problems have the same point value. If a problem is divided in parts, each part has equal value. Show all your work. Problem 1 L = r p, p = i h ( ) (a) Show that L z = i h y x ; (cyclic for other components) x y L i ɛ ijk r j p k = i hɛ ijk r j k. So L z = i h(x y y x ) = i h(y x x y ). (b) Show that the commutator [L x, L y ] = i hl z (and cyclic), and consequently that we can write L L = i h L. Also, show [L z, r ± ] = ± r ± h, where r ± = x±iy We know [r i, p j ] = i h δ ij. So: [L i, L j ] = ɛ ikl ɛ jmn [r k p l, r m p n ] = ɛ ikl ɛ jmn (r m [r k, p n ]p l + r k [p l, r m ]p n ) = i h ɛ ikl ɛ jmn (r m p l δ kn r k p n δ lm ) = i h (( δ ij δ lm + δ im δ lj )r m p l ( δ ij δ kn + δ in δ kj )r k p n ) = i h (r i p j r j p i ) = i hɛ ijk L k. 1

2 Hence [L x, L y ] = i hl z (et cyc). Then: ( L L) k = ɛ ijk L i L j = 1 ɛ ijk(l i L j L j L i ) = 1 ɛ ijk[l i, L j ] = 1 i h ɛ ijkɛ ijl L l = 1 i h (δ kl)l l = i h L k So L L = i h L. (c) Show that [L, L z ] = 0, where L = L x + L y + L z. [L, L z ] = [L x, L z ] + [L y, L z ] = i h ( L x L y L y L x + L y L x + L x L y ) = 0. (d) Show that L ± L = L L z ± hl z, where L ± = L x ± il y. L ± L = L x + L y ± il y L x il x L y = L L z ± i[l y, L x ] = L L z ± hl z. Problem Let ψ be an eigenfunction of L : L ψ = l(l+1) h ψ. Since [L, L z ] = 0, we can make ψ an eigenfunction of L z as well: L z ψ m = m h ψ m. L and L z are defined as in Problem 1. (a) With raising operator L + = L x + il y, and results from Problem 1, show that L + ψ m generates an eigenfunction of same L but with L z eigenvalue of (m+1) h.

3 L z (L + ψ m ) = (L z L x + il z L y )ψ m = (L x L z + i hl y + i(l y L z i hl x ))ψ m = (L + L z + hl + )ψ m = (m+1) h (L + ψ m ). (b) Compute the non-zero matrix elements of L +. From part (a) we know L + l m = c + l m+1, so l m+1 L + l m = c +, all other matrix elements 0. So c + = h l(l + 1) m(m + 1). c + = l m L L + l m = l m (L L z hl z ) l m = h (l(l + 1) m h m) 3

4 Problem 3 Consider a spin vector σ n = σ x cos α sin β + σ y sin α sin β + σ z cos β, and: ( ) ( ) ( ) i 1 0 σ x = σ 1 0 y = σ i 0 z =. 0 1 (a) Determine the eigenvalues of σ n. (Hint: compute σ n). ( ) cos β e σ n = iα sin β e iα sin β cos β = σ n = ( ) σ n has an eigenvalue +1, so σ n could have eigenvalues ±1. Need to check each. The eigenvalues are ±1. det(σ n 1) = (cos β 1)( cos β 1) sin β = 1 cos β sin β = 0 det(σ n + 1) = (cos β + 1)( cos β + 1) sin β = 1 cos β sin β = 0 (b) Pure rotation by an angle γ of the z-axis in the x z plane leads to the rotation matrix ( ) cos γ sin γ sin γ cos γ Interpret this result by comparing to rotation of ordinary vectors, e.g. l, r. Pure spin rotation by an angle γ of the z-axis in x z plane (β=0), leads to a rotation matrix as given. Comparing to rotation of ordinary vectors, this is seen to correspond to spin rotations by half angles. Problem 4 4

5 At fixed principal quantum number n and orbital quantum number l, the Hamiltonian of an alkali atom can be modelled as H = A + B L S, A, B = constants. (a) Diagonalize this Hamiltonian for l = 1 and l =. Use: L S = j(j +1) l(l+1) 3/4. Hence for j = l+1/ we find L S = l/, while for j = l 1/ we find L S = l/ 1 (b) Which other operator(s) built in terms of L and S commutes with the Hamiltonian? J = L + S (and functions thereof). (c) The Hilbert space of states for l = 1 is the product of two spaces, V = V L V S. V L is the orbital angular momentum space (dimension 3) and V S is the spin space (dimension ). By definition, given a pure state ψ V, the reduced density matrix in V L is ρ L = tr S ψ ψ. The entanglement entropy is defined as S = tr ρ L log ρ L. Compute the entanglement entropy of the j = 1/ state ψ = S L 3 1 ρ L = diag (/3, 1/3, 0); S = (1/3) log 3 + (/3) log(3/) = log 3 (/3) log. The maximal entanglement entropy is log 3. Problem 5 Consider the scattering of a particle of energy E in a square well potential V = 0 for x < 0 and x > b, V = V 0 > 0 for 0 < x < b. Assuming that the transmission coefficient T 1 (thick barrier approximation), obtain its value in terms of the above parameters. You may use the 1 S L. 5

6 definitions me k = h k m(v0 E) = h With this obtain the value of the transmission probability T = transmitted amplitude incident amplitude. Hψ = Eψ = ( p + V )ψ m p = m(e V ), ψ = ψ(x)e iet/ h ψ + m (E V )ψ = 0 h In barrier region, from r = 0 to r = b, V = V 0 k 1 h m(e V ) Outside of barrier region, for r 0, r b, k = 1 h me. Say alpha partile incident from Region I: Region I: ψ I = De ikx + Ee ikx Region II: ψ II = Be k x + Ce k x Region III: ψ III = Ae ikx At x = 0, ψ I = ψ II or D + E = B + C and: ψ I x = ψ II x = Dikeikx Eike ikx = Bk e k x Ck e k x so: ik(d E) = k (B C), D E = k (B C) = ik (B C)) ik k At x = b, ψ II = ψ III, Ae ikb = Be k b + Ce k b, and ikae ikb = k Be k b k Ce k b, which we solve: Be k b = Ae ikb Ce k b ikae ikb = k Ae ikb k Ce k b k Ce k b Ae ikb (ik k ) = k Ce k b = C = 1 ik k Aeikb ( )e k b k = A eikb+k b ( ik + 1) k 6

7 Solve for B: Be k b = Ae ikb [ A eikb+k (1 i k k )]e k b = Ae ikb A eikb (1 ik k ) = Ae ikb [1 1 ik (1 k )] = B = A )b e(ik k (1 + ik k ) From D+E = B+C and D E = ik ik ik (B C) or D = B(1 )+C(1+ ) k k k we obtain for a thick barrier, B 0: and finally: D = 1 ik (1 + A D = k )[A 4e ikb k b (1 + ik ik )(1 ) k k A D = (1 ik k )eikb+k b ] 16e k b [1 + ( k k ) ][1 + ( k k ) ]. Problem 6 Consider the Hamiltonian of a one-dimensional particle with mass m and potential V = A/x. A can be either positive or negative. (a) Prove that the spectrum is unbounded below for ma < 1/4 (Hint: study how the potential energy and the kinetic energy scale when the parameter b 0 in the test wave packet ψ = b 1/ a x a exp( x /b ), with a > 1/ but arbitrarily close to 1/: a = 1/ + ɛ +.) dx dψ a b, dx ma ψ mab. For any ma < 1/4 one 0 dx 0 x can choose an a such that a < ma, so the potential drives the energy to. (b) When ma > 1/4, it can be proven that the spectrum is bounded below (do not try to prove it!). In this case, is the spectrum continuous or discrete? What is the lowest value of the energy? (Hint: under the scaling 7

8 x y = λx, the Hamiltonian transforms as H λ H.) Because of the scaling, if ψ(x) is an eigenstate of H with eigenvalue E, ψ(λx) has eigenvalue E/λ. So the spectrum is continuous and the lowest eigenvalue is E = 0. 8