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1 Problem 1. PROBLEMS from SAKURAI 4.1) Calculate the Three Lowest Energy Levels, Together with Their Degeneracies for the Follo

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1 Problem 1. PROBLEMS from SAKURAI 4.1) Calculate the Three Lowest Energy Levels, Together with Their Degeneracies for the Follo 1 SOLUTION SET 5 215B-QUANTUM MECHANICS, WINTER 2019 Problem 1. PROBLEMS FROM SAKURAI 4.1) Calculate the three lowest energy levels, together with their degeneracies for the following systems (assuming equal mass distinguishable particles) (a) Three non-interacting spin 1=2 particles in a one-dimensional box of length L The Hamiltonian for three non-interacting particles of mass m in a one-dimensional box of length L is 1 X3 H = p2 (1) 2m i i=1 Consequently the state completely factorizes and we can independently designate the spin and position eigenstates of each particle. The energy levels are independent of spin and given by π2~2 X E = 3n2 (2) ~n 2mL2 i i=1 The ground state has energy π2~2 E = 3 ; (3) (1;1;1) 2mL2 with no degeneracy in the position wave-function, but a 2-fold degeneracy in equal energy spin states for each of the three particles. Thus the ground state degeneracy is 8. The first excited states have energy π2~2 E = E = E = 3 ; (4) (2;1;1) (1;2;1) (1;1;2) mL2 with a 3-fold degeneracy in position wavefunctions and 8-fold degeneracy in spin giving a total degeneracy of 24. The second exited states have energy π2~2 E = E = E = 9 ; (5) (2;2;1) (1;2;2) (2;1;2) 2mL2 with a 3-fold degeneracy in position wavefunctions and 8-fold degeneracy in spin giving a total degeneracy of 24. (b) Four non-interacting spin 1=2 particles in a one-dimensional box of length L Four particles is exactly analogous. The ground state has energy π2~2 E = 2 ; (6) 0 mL2 with no degeneracy in the position wave-function, but a 2-fold degeneracy in equal energy spin states for each of the four particles. Thus the ground state degeneracy is 16. The first excited states have energy π2~2 E = 7 ; (7) 1 2mL2 ( ) 4 with a 1 -fold degeneracy in position wavefunctions and 16-fold degeneracy in spin giving a total degeneracy of 64. The second excited states have energy π2~2 E = 5 ; (8) 2 mL2 ( ) 4 with a 2 -fold degeneracy in position wavefunctions and 16-fold degeneracy in spin giving a total degeneracy of 96. 4.2) Which of the following commute? (a) Td and Td0 They commute because the generators of translation commute: [Pi;Pj] = 0. (b) D(^n; ϕ) and D(^n0; ϕ0) 2 They do not commute in general because the generators of rotation do not all commute: [Ji;Jk] =6 0 (c) Td and π They do not commute because parity does not commute with the generator of translations: [π; Pi] =6 0 (but fπ; Pig = 0). (d) D(^n; ϕ) and π They commute because the parity operator commutes with the generators of rotations: [Ji; π] = 0. 4.3) For Ψ an eigenstate of anticommuting operators A; B with respective eigenvalues a; b, we have that fA; Bg = 0 ) (AB + BA)jΨi = (2ab)jΨi = 0 : (9) Thus, at least one of the eigenvalues a; b must be zero. As an example, consider an eigenstate of the parity, π, an momentum operators, P, with eigenvalues π and p respectively. Then one of the two must be zero. However, in this case we can go further: because π is an invertible operators, it can not have a non-trivial nullspace. Thus, we must have that p = 0. 4.7) (a) A plane wave in three dimensions has the wavefunction 1 (x; t) = expf(ip · x − i!t)g : (10) p (2π)3=2 Thus we have that ∗ 1 (x; −t) = expf(−ip · x − i!t)g = − (x; t) : (11) p (2π)3=2 p (b) Recall (or consult Sakurai to find) that the two component eigenspinor of σ · ^n with eigenvalue +1 can be written explicitly as ( ) cos(β=2)e−iγ=2 χ (^n) = : (12) + sin(β=2)e+iγ=2 Then we have that ( )( ) +iγ=2 ∗ 0 −i cos(β=2)e −iσ χ (^n) = −i (13) 2 + i 0 sin(β=2)e−iγ=2 ( ) − sin(β=2)e−iγ=2 = (14) cos(β=2)e+iγ=2 Now we could explicitly check that this is the eigenspinor with the spin direction reversed. However, note that the two eigenspinors with reversed spin directions are orthogonal and span the Hilbert space. Thus it suffices to check that this new spinor is orthogonal to χ+: ( )( ) −iγ=2 +iγ=2 ∗ ∗ − sin(β=2)e cos(β=2)e −iσ χ (^n) · χ (^n) = 2 + + cos(β=2)e+iγ=2 sin(β=2)e−iγ=2 = − sin(β=2) cos(β=2) + sin(β=2) cos(β=2) = 0 : (15) 4.8) (a) The question as written is false. The statement is in fact only true for an energy eigenfunction. This is proved in Sakurai already as Theorem 4.2. (b) Plane wave eigenfunctions of the free particle Hamiltonian are not a violation of part (a) because the energy spectrum is degenerate (there are many states with the same energy p2=2m). 4.9) Let Z jαi = d3pϕ(p)jpi : (16) 3 Then we have that Z Θjαi = d3pϕ∗(p)Θjpi (17) and thus that Z Z Θjαi = d3p ϕ∗(p)j − pi = d3p ϕ∗(−p)jpi : (18) We conclude Θϕ(p) = ϕ∗(−p). 4.11) First recall that a non-degenerate spectrum implies that an energy eigenstate jEi is invariant under time reversal up to a phase: jE~i = ΘjEi = eiδjEi (19) and that angular momentum is odd under time-reversal: ΘLΘ−1 = −L : (20) From these two facts, we find (cf. (4.4.44) hEjLjEi = −⟨E~jLjE~i = −⟨Eje−iδLe−iδjEi = −⟨EjLjEi : (21) This can only be true if hEjLjEi = 0 : (22) iδ ∗ Recall from problem 4.8 that, because we have a nondegenerate spectrum, E(x)e = (x). If we expand the wavefunction as X X m (x) = Fl;m(r)Yl (θ; ϕ) (23) l m then this equation becomes X X X X X X iδ m ∗ m∗ ∗ ∗ − m −m e Fl;m(r)Yl (θ; ϕ) = Fl;m(r)Yl (θ; ϕ) = Fl;m(r)( 1) Yl (θ; ϕ) (24) l m l m l m We can then project onto one mode Z X X Z X X iδ m M ∗ − m −m M e Fl;m(r)Yl (θ; ϕ)YL (θ; ϕ)dΩ = Fl;m(r)( 1) Yl (θ; ϕ)YL (θ; ϕ)dΩ l m l m ) iδ − M ∗ e FL;M (r) = ( 1) FL;−M (r) (25) 4.12) The Hamiltonian 2 2 − 2 H = ASz + B(Sx Sy ) (26) can be written explicitly as 0 1 A 0 B H = ~2 @ 0 0 0 A : (27) B 0 A This matrix has eigenvectors 0 1 0 1 0 1 0 1 1 1 1 j1i = @1A ; j2i = p @0A ; j3i = p @ 0 A ; (28) 0 2 1 2 −1 4 with respective eigenvalues 0; ~2(A + B); ~2(A − B). −1 − 2 −1 2 −1 Now ΘSiΘ = Si implies ΘSi Θ = Si so ΘHΘ = H and the Hamiltonian is invariant. We can rewrite the eigenstates in terms of spin eigentstates j1i = j1; 0i 1 j2i = p (j1; 1i + j1; −1i) 2 1 j3i = p (j1; 1i − j1; −1i) : (29) 2 Then recall the convention in (4.4.77) Θjl; mi = (−1)mjl; −mi (30) so that we find Θj1i = j1; 0i = j1i 1 1 Θj2i = p (j1; 1i + j1; −1i) = p (−|1; −1i − j1; 1i) = −|2i 2 2 1 1 Θj3i = p (j1; 1i − j1; −1i) = p (−|1; −1i + j1; 1i) = j3i : (31) 2 2 5 Problem 2. SHOR'S ERROR CORRECTING CODE (a.1) Noticing that O O0 O0 O 6 O O0 i j = j i; where i = j; and ; = X; Z; (32) ZiXi = −XiZi; (33) we can check that gigj = gjgi for any pair of i; j, by verifying that the number of sites where gi has Z (X) and gj has X (Z) is even. Noticing that Z1Z2 (j000i j111i) = Z2Z3 (j000i j111i) = j000i j111i ; (34) X1X2X3 (j000i j111i) = (j000i j111i) ; (35) we can see that gi jzi = jzi for z = 0; 1. (a.2) The independence is somewhat obvious. One can explicitly show that by computing the rank of some corresponding matrix on Z2. We can use the same observation in (a.1) to show that the operators commute. Suppose j i is a codeword, that is, gi j i = j i 8i. We see that giZ j i = Zgi j i = Z j i ; (36) giX j i = Xgi j i = X j i : (37) Thus, Z j i and X j i are still codewords. (a.3) Using the observation in (a.1) (that X and Z on the same site anticommute), we can show that ZX = −XZ (notice the typo in the problem). We can also show that ⟩ ⟩ ⟩ ⟩ Z 0 = 0 ; Z 1 = − 1 ; (38) ⟩ ⟩ ⟩ ⟩ X 0 = 1 ; X 1 = 0 : (39) Thus the matrix representation. (a.4) Since the logical operators commute with the stabilizers, we can assume, without loss of generality, 0 p1 p2 p8 Z = Zg1 g2 : : : g8 ; (40) 0 q1 q2 q8 X = Xg1 g2 : : : g8 ; (41) where pi; qi 2 f0; 1g. We still have 0 ⟩ ⟩ 0 ⟩ ⟩ Z 0 = 0 ; Z 1 = − 1 ; (42) 0 ⟩ ⟩ 0 ⟩ ⟩ X 0 = 1 ; X 1 = 0 ; (43) 0 0 because gi jzi = jzi. Thus, Z and X are also good logical operators. (b.1) Noticing that g1X1 = −X1g1; gi>1X1 = X1gi>1; (44) we have g1X1 j i = −X1g1 j i = −X1 j i ; (45) gi>1X1 j i = X1gi>1 j i = X1 j i : (46) Measuring the stabilizers won't change the wavefunction since it is an eigenstate; we have the error syndrome (−; +; +; +; +; +; +; +).
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