Download Solutions to Quantum Mechanics Problem PC2130 QM1 and more Exams Quantum Mechanics in PDF only on Docsity! PC2130 QM1 0809 Solution 1. The statistical operator for an atom pointing in direction ~e is ρ = 1 + ~e 2 Prob(+n) = tr {|↑n〉〈↑n| ρ} = tr { 1 4 (1 + n · ~σ)(1 + e · ~σ) } = tr { 1 4 [1 + ~n · ~σ + ~e · ~σ + (~n · ~σ)(~e · ~σ)] } = tr { 1 4 [1 + ~n · ~σ + ~e · ~σ + ~n · ~e+ i(~n× ~e) · ~σ] } = 1 2 (1 + ~n · ~e) Similarly, Prob(−n) = tr {|↓n〉〈↓n| ρ} = tr { 1 4 (1− ~n · ~σ)(1− ~e · ~σ) } = 1 2 (1− ~n · ~e) 2. (a) If AA† = A†A = 0 and A|ai〉 = |ai〉ai, then (A− aiI)|ai〉 = 0 [〈ai|(A† − a∗i I)][(A− aiI)|ai〉] = 0 〈ai|(A†A− a∗iA− aiA† + a∗i ai)|ai〉 = 0 〈ai|(AA† − aiA† − a∗iA+ a∗i ai)|ai〉 = 0 [〈ai|(A− aiI)][(A† − a∗i I)|ai〉] = 0 which means that 〈ai|(A− aiI) = 0, or 〈ai| is an eigenbra of A with eigenvalue ai (b) Given A|a〉 = |a〉 and 〈b|A = b〈b| Taking inner product of the first equation with 〈b| and second equation with |a〉: 〈b|A|a〉 = a〈b|a〉 and 〈b|A|a〉 = b〈b|a〉 Subtracting the two, we get (a− b)〈b|a〉 = 0. Since b 6= a, 〈b|a〉 is 0. 3. (a) In term of power series, since we have (~n · ~σ)2 = 1 e(iθ~n·~σ) = 1 + (iθ~n · ~σ) + 1 2! (iθ~n · ~σ)2 + 1 3! (iθ~n · ~σ)3 + · · ·+ 1 n! (iθ~n · ~σ)n = ( 1− 1 2! θ2 + 1 4! θ4 + . . . ) + i ( θ − 1 3! θ3 + 1 5! θ5 + . . . ) (~n · ~σ) = cos θ + i sin θ(~n · ~σ) (b) With ~n = 1√ 3 (~x+ ~y + ~z) and θ = 2π3 e(i θ 2~n·~σ) = cos θ 2 + i sin θ 2 (~n · ~σ) = cos π 3 + i sin π 3 [ 1√ 3 (~σx + ~σy + ~σz) ] = 1 2 [1 + i(σx + σy + σz)] = 1 2 ( 1 + i 1 + i i− 1 1− i ) Acting on |↑z〉 1 2 ( 1 + i 1 + i i− 1 1− i )( 1 0 ) = 1 2 ( 1 + i i− 1 ) 1 4. Using 〈x|X = x〈x| and 〈x|P = −ih̄ ∂ ∂x 〈x| 〈x|XP = x〈x|P = −ih̄x ∂ ∂x 〈x| 〈x|PX = −ih̄ ∂ ∂x 〈x|X = −ih̄ ∂ ∂x 〈x|x = −ih̄x ∂ ∂x 〈x| − ih̄〈x| 〈x|[X,P ] = 〈x|XP − 〈x|PX = ih̄〈x| 5. For SHO, the Hamiltonian is H = P 2 2m + 1 2 mω2X2 E = 〈H〉 = 〈P 2〉 2m + 1 2 mω2〈X2〉 = δP 2 2m + 1 2 mω2δX2 = 1 2 ( δP 2 m + ω2δX2) ≥ √ ( δP 2 m )(ω2δX2) = √ δP 2δX2ω2 = 1 2 h̄ω 6. Using Schrödinger equation, − h̄ 2 2m d2ψ(x) dx2 − h̄ 2κ m δ(x)ψ(x) = Eψ(x) d2ψ(x) dx2 + 2κδ(x)ψ(x) = −2mE h̄2 ψ(x) Intergrate over a small region [−, ], dψ() dx − dψ(−) dx + 2κψ(0) = q2 ∫ − ψ(x)dx ≈ 2q2ψ(0) Let → 0 dψ(0+) dx − dψ(0−) dx + 2κψ(0) = 0 To the left and right of the origin, the potential term vanishes, and since the wavefunctions must vanish at infinity, ψ(x) = { Aeqx x < 0 Ae−qx x > 0 The following boundary condition must be satisfied: ψ(0+) = ψ(0−) dψ(x) dx ∣∣ 0+ − dψ(x) dx ∣∣ 0− + 2κψ(0) = 0 The first condition requires that A = B = ψ(0). The second condition gives −qψ(0)e−qx ∣∣∣ 0+ − qψ(0)eqx ∣∣∣ 0− + 2κψ(0) = 0 −qψ(0)− qψ(0) + 2κψ(0) = 0 2
