Midterm II Solutions - Quantum Mechanics | PHYSICS 137A, Exams of Quantum Mechanics

Download Midterm II Solutions - Quantum Mechanics | PHYSICS 137A and more Exams Quantum Mechanics in PDF only on Docsity! Physics 137A: Quantum Mechanics I, Spring 2007 Midterm II solutions by JEM Directions: The allotted time is 80 minutes for 4 problems. Two sides of your own notes are allowed. No books or calculators are allowed, and please ask for help only if a question’s meaning is unclear. 1. (a) This operator is not Hermitian: for example, its matrix representation is 0 1 00 0 0 0 0 0  (1) which is not a Hermitian matrix since it is not equal to the complex conjugate of its transpose. Directly in Dirac notation, H† = |1〉〈2| 6= H. (b) Normalized eigenstates and eigenvalues are |3〉 with eigenvalue E2, 1√2(|1〉+ |2〉) with eigen- value E1, and 1√2(|1〉 − |2〉) with eigenvalue −E1. We did not take off for non-normalized answers or those with a different phase. (c) This initial condition projects with equal weight onto the two eigenvectors with energies E1 and −E1: |1〉 = 1√ 2 [ 1√ 2 (|1〉+ |2〉) + 1√ 2 (|1〉 − |2〉) ] . (2) The time-dependent solution, up to an overall phase that is not physically measurable, oscillates between |1〉 and |2〉 with a frequency ω = 2E1 h̄ . (3) The resulting state is equivalent to |2〉, again up to an unmeasurable overall phase, at times that satisfy ωt = 2E1 h̄ t = π, 3π, 5π, . . . , (4) or, in compact notation, t = h̄(2n+ 1)π 2E1 (5) where n is a nonnegative integer. 2. (a) For the n = 3 levels, l can take the values 0, 1, or 2. For l = 0, m is 0. For l = 1, m is -1, 0, or 1. For l = 2, m is -2, -1, 0, 1, or 2. (b) The key is to remember or show that the Hermitian conjugate of AB is B†A†. Then [A,B]† = (AB)† − (BA)† = B†A† − A†B† = BA − AB, since A and B are Hermitian. Now BA−AB = −[A,B], so [A,B] is antihermitian. (c) We want to find the ground state and its energy for H ′ = p2 2m − 2e 2 (4π0)r , (6) assuming that we know the ground state and energy of H = p2 2m − e 2 (4π0)r . (7) If we try ψ′ = ψ100(r/2) = Ce−r/(a0/2), we see that the kinetic energy is increased by a factor of 4 in going to H ′ from H: the two spatial derivatives in p2 each get multipled by 2. The potential energy has one inverse power of r, which gives an increase of 2, and also a factor of 2 in the numerator. Hence both terms increase by 4, which means that ψ′ is an eigenstate of the He+ Hamiltonian with energy 4(−13.6eV ) = −54.4eV . For an intuitive picture of this result, we can say that increasing the nuclear charge both pulls the electron closer into the nucleus and increases the potential energy at a fixed distance, thus the increase is not 2 but 4. 3. (a) [p, p2] = 0 so the kinetic energy does not contribute. Using [p, x2] = pxx− xxp = pxx− xpx+ xpx− xxp = (−ih̄)x+ x(−ih̄) = −2ih̄x, (8) we obtain [p,H] = −imω2h̄x. (9) (b) The generalized uncertainty relation gives σpσH ≥ |〈i[p,H]〉| 2 = mω2h̄|〈x〉| 2 . (10) For wavefunctions even around x = a, 〈x〉 = a, and B = mω2h̄|a|/2. 4. (a) This potential separates: V (x, y) = V1(x)V2(y). Hence the wavefunctions are products of the wavefunctions we know for a particle in a 1D box, and the energies are additive: for positive integers ψn1,n2(x, y) = 2√ ab sin( πn1(x+ a/2) a ) sin( πn2(y + b/2) b ). (11) with energies En1,n2 = h̄2π2 2m ( n1 2 a2 + n2 2 b2 ) . (12) (b) Since the potential is even, 〈x〉 = 0. We then need to find 〈x2〉, which has no y dependence so this reduces to a 1-d problem. The integral needed is∫ a/2 −a/2 x2|ψ1|2 dx = 2 a ∫ a/2 −a/2 x2 sin2(π(x+ a/2)/a) dx. (13)