Download Introduction to Quantum Mechanics Practice Test 2 Solutions for PHY4604 Fall 2004 and more Exams Physics in PDF only on Docsity! PHY4604–Introduction to Quantum Mechanics Fall 2004 Practice Test 2 solutions October 4, 2004 These problems are similar but not identical to the actual test. One or two parts will actually show up. 1. Short answer. • You are told that a potential can be written V (x) = 10∑ n=1 Vnx 2n. All Vn ≥ 0. What can you say about the solutions ψ to Hψ = Eψ? Since the potential is an even function of x, the parity operator commutes with the Hamiltonian. This means the eigenstates can be classified accord- ing to parity. • What is tunnelling? Draw a picture describing the process, and describe explicitly a situation which distinguishes classical physics from quantum mechanics. When a particle approaches a potential barrier higher than its total energy, in classical physics it must be reflected with 100% probability. In quan- tum, mechanics, however, it is transmitted to the other side with some probability, provided the barrier is not infinite. • Write the Dirac expression 〈α|H|α〉 as an explicit integral in three dimen- sions, assuming that 〈r|α〉 represents a wave function ψα(r). If |α〉 is an eigenvector of H with eigenvalue Eα, evaluate the integral. 〈α|H|α〉 = ∫ d3rψ∗α(r)Hψα(r) • Identify: Ehrenfest’s theorem. Expectation values obey classical laws, e.g. 〈p/m〉 = d〈x〉/dt. • If A and B are self-adjoint, is the combination A + iB self-adjoint, anti- self-adjoint, unitary, antiunitary, none of the above, or more than one of the above? Defend your answer. (A + iB)† = A† − iB† = A− iB, 1 which is neither self-adjoint nor anti-self-adjoint. For it to be unitary we would need (A + iB)(A − iB) = A2 + B2 + i[A,B] = 1, which we don’t know anything about, so we can’t say. None of the above is therefore the best answer. • Prove that if (ψ,Oψ) = (Oψ, ψ) for all ψ, then O is self-adjoint. We did this in class. Expand ψ in an orthonormal basis: ψ = ∑ n cnψn. Then the relation we are given means ∑ n,m c∗mcn(ψm,Oψn) = ∑ n,m c∗mcn(O†ψm, ψn), which since the relation is true for all ψ, means the cn’s are independent, i.e. the only way you can satisfy the equation is to have Omn = O∗nm, where Omn ≡ (ψm,Oψn). To prove self-adjointness, we need (χ,Oφ) = (Oχ, φ)) ∀χ, φ. If we put χ = ∑n fnψn and φ = ∑ m gmψm, we can expand (χ,Oφ) = ∑ m,n f ∗ngmOnm = = ∑ m,n f ∗ngmO∗mn = (Oχ, φ)) QED • Prove that eigenvectors of a Hermitian operator corresponding to distinct eigenvalues are orthogonal. Let |1〉, |2〉 be eigenvectors of Hermitian O corresponding to eigenvalues o1 and o2. Self-adjoint implies 〈1|O|2〉 = 〈O1|2〉, ,i.e. o2〈1|2〉 = o1〈1|2〉, where I wrote o1 instead of o ∗ 1 since we know eigenvalues of Hermitian operators are real. Now if o1 6= o2 the two vectors must be orthogonal, 〈1|2〉 = 0. • Find the expansion coefficient of the first excited SHO state in the function (x2 + x20) −2 It’s zero, since the first excited SHO state is odd, but the function you are trying to expand is even. 2
