Solutions to Test 2 of PHY4604-Introduction to Quantum Mechanics, Fall 2004, Exams of Physics

Download Solutions to Test 2 of PHY4604-Introduction to Quantum Mechanics, Fall 2004 and more Exams Physics in PDF only on Docsity! PHY4604–Introduction to Quantum Mechanics Fall 2004 Test 2 Solutions Nov. 5, 2004 No calculators or other materials. If you don’t recall a formula, ask and I might be able to help. If you can’t do one part of a problem, solve subsequent parts in terms of unknown answer–define clearly. Each part is worth 10 points, for max=70. Problem 1 required, attempt 1 of remaining 2 problems; circle which ones you want graded. Possibly helpful formulae and constants Pn = |〈n|ψ〉|2 Hψ = Eψ (PQ)† = Q†P † Hψ = ih̄ ∂ ∂t ψ det(H − E) = 0 p̂ = −ih̄∇ 1 = ∑ n |n〉〈n| ψ(x) = 〈x|ψ〉 T = |jtrans|/|jinc| R = |jrefl|/|jinc| T + R = 1 ψ(x, t) = ∫ dp√ 2πh̄ eipx/h̄φ(p, t) L± ≡ √ k 2 x∓ h̄√ 2m ∂ ∂x φ(p, t) = ∫ dx√ 2πh̄ e−ipx/h̄ψ(x, t) 〈r|r′〉 = δ(r− r′) H = − h̄ 2 2m ∇2 + V (r) ψ = ∑ n cnψn (O†φ, χ) = (φ,Oχ) ψ(x, t) = e−iHt/h̄ψ(x, 0) p = h̄k∫ dxeikx = 2πδ(k) [x, p̂] = ih̄ j = − ih̄ 2m (ψ∗∇ψ − ψ∇ψ∗) δ(ax) = 1 a δ(x) ∫ dxf(x)δ(x− x0) = f(x0) 1 1. Short answer. Must attempt (only) 3 of 5. (a) A Hamiltonian for a 2-level system is written as H = ((E1|1〉〈1|+ E2|2〉〈2| − t|2〉〈1| − t|1〉〈2|) where the vectors |1〉 and |2〉 represent two orthonormal states of the sys- tem which span the Hilbert space of the problem. Write down the matrix H of the problem in the |1〉, |2〉 basis, and find the eigenvalues Ea and Eb. The matrix elements 〈α|H|β〉, where α, β each can be 1 or 2, can be easily calculated to find H = [ E1 −t −t E2 ] The eigenvalue equation Hψ = Eψ can be most easily solved for the eigenvalues by looking for the characteristic polynomial of the matrix H = [ E1 − E −t −t E2 − E ] , which is found by taking the determinant. So the eigenvalues are the solutions of (E1 − E)(E2 − E)− t2 = 0, i.e. Ea,b = 1 2 ( E1 + E2 ± √ (E1 − E2)2 + 4t2 ) (b) Prove that if (ψ,Oψ) = (Oψ, ψ) for all ψ, then O is self-adjoint. The definition of the adjoint is (O†χ, φ) = (χ,Oφ), so an operator is self- adjoint if (Oχ, φ) = (χ,Oφ)∀χ, φ. This is not what we are told, but we can always expand ψ in a basis of eigenfunctions of the operator O: ψ = ∑ n cnψn. Then the equation we are given reads ∑ m,n c∗ncm(ψn,Oψm) = ∑ m,n c∗ncm(Oψn, ψm), but this is true for any ψ, so for any coefficients cn. The only way this can happen is if (ψn,Oψm) = (Oψn, ψm). Now the condition for self-adjointness can be expressed in a similar way if we expand χ = ∑ n dnψn and φ = ∑ n fnψn: ∑ m,n d∗nfm(ψn,Oψm) = ∑ m,n d∗nfm(Oψn, ψm), and now we could prove self-adjointness if (ψn,Oψm) = (Oψn, ψm), which we just happen to know from above QED. 2 (c) Suppose the system is now prepared at time t = 0 in the state ψ(x) = c(ψ0(x) + ψ ′ 1(x) + ψ ′ 2(x)). (1) What are the probabilities that a particular measurement of Q in the state ψ yields the result q0? q1? Normalization of ψ implies that c = 1/ √ 3. The probability amplitude for measuring q0 is the coefficient of ψ0 squared, i.e. 1/3. Both the other two terms are eigenfunctions corresponding to q1, so the probability is 2/3. (d) Suppose Q̂ = H; find ψ(x, t) in terms of the eigenvalues E0 and E1. ψ(x, t) = e−iHt/h̄ψ = e−iHt/h̄ 1√ 3 (ψ0 + ψ ′ 1 + ψ ′ 2) = 1√ 3 ( e−iE0t/h̄ψ0 + e−iE1t/h̄ψ′1 + e −iE1t/h̄ψ′2 ) 3. Finite step. A particle of mass m moves in the 1D potential V (x) =    0 x < 0 (I) V0 0 ≤ x ≤ L (II) 0 x > L (III) (2) Assume that an electron is approaching this potential from the left (x < 0), with 0 < E < V0. (a) Write down the solutions to the Schrödinger equation in the three regions (I, II, III) as linear combinations of (possibly complex) exponential func- tions of x. Sketch the form |ψ| must have. ψI = Ae ipx/h̄ + Be−ipx/h̄, ψII = Ce qx/h̄ + De−qx/h̄ ψIII = Fe ipx/h̄ (b) Determine the parameters which appear in the exponents in regions (I, II, III). p = √ 2mE q = √ 2m(V0 − E) 5 Probability density |ψ|2. Note the incident probability density oscillates due to interference between A and B terms. Within barrier, |ψ|2 decays monotonically. Finally, after barrier there is no reflected wave so |ψ|2 is constant. (c) State the boundary conditions on the wave function at x = 0 and x = L, and use them to determine the coefficients of the exponential terms in the wave function. Require continuity of ψ and its derivatives at the interfaces: ψI(0) = ψII(0) ψII(L) = ψIII(L) ψ′I(0) = ψ ′ II(0) ψ′II(L) = ψ ′ III(L), in other words A + B = C + D CeqL/h̄ + De−qL/h̄ = FeipL/h̄ (A−B)ip = (C −D)q q(CeqL/h̄ −De−qL/h̄) = FipeipL/h̄ (d) Write down expressions for the incident and reflected probability current density in each of the three regions, in terms of the unknown parameters of part (c). Give the definitions of the reflection and transmission coefficients for the entire potential barrier and state the condition involving them which expresses conservation of probability. j = − ih̄ 2m (ψ∗ ∂ψ ∂x − ψ∂ψ ∗ ∂x ) so jIinc = p m |A|2 jIrefl = − p m |B|2 jIItot = κ m (C∗D − CD∗) jIIItrans = p m |F |2 6 and T = |jIIItrans/|jIinc| = |F |2/|A|2 R = |jIrefl/|jIinc| = |B|2//|A|2 and we must have T+R = 1 to conserve probability. Note there is no reason to leave out the D term within the barrier for any finite L, because it poses no physical problems. In fact, you can see that without D there is no way to have a probability current density within the barrier, so you couldn’t conserve probability from one side to the other! For a more complete solution to this problem, see e.g. Gasiorowicz, Quantum Physics. 7