1 Commutators (10 pts)

Transcription

1 Final Exam Solutions 37A Fall 0 I. Siddiqi / E. Dodds Commutators 0 pts) ) Consider the operator  = Ĵx Ĵ y + ĴyĴx where J i represents the total angular momentum in the ith direction. a) Express both  and  in terms of the raising and lowering operators Ĵ+ and Ĵ. pts) Solution. Using Ĵx = Ĵ+ + Ĵ ) and Ĵy = i J + J ) we can directly calculate and square the result to find  = i 4 Ĵ + Ĵ +ĴiĴ+ Ĵ+Ĵ + Ĵ + Ĵ ĴiĴ+ + Ĵ+Ĵ ) = i 4 Ĵ + Ĵ ) )  = 6 Ĵ Ĵ 4 ) + 6 Ĵ +Ĵ + Ĵ Ĵ +). ) b) Calculate the expectation value of  and  with respect to the state j, m. pts) Solution. The expectation value of  is zero, since each term contains only raising or lowering operators and therefore gives a state orthogonal to the original angular momentum eigenstate. The first term in parentheses in ) gives zero in A for the same reason. So we just need to calculate j, m  j, m = 6 j, m Ĵ +Ĵ + Ĵ Ĵ + j, m which can t be simplified much. = 4 [jj + ) mm + )][jj + ) m + )m + )] + 4 [jj + ) mm )][jj + ) m )m )] The Quantum Necklace & Shell pts) a) A bead of mass m is free to slide around a circular wire of radius a. Find the eigenstates and the corresponding eigenenergies of the system. pts) Solution. Let s start by finding the right form of the Schrödinger equation. Generically we have m ψ r) + V r)ψ r) = Eψ r). 3)

2 In this case the potential is zero everywhere and we just need a good way of writing the Laplacian. Ignoring the degrees of freedom not associated with motion along the wire, we can write the Laplacian as a φ 4) which you may recognize as, e.g., the azimuthal term in the Laplacian in spherical coordinates. So the equation becomes which has solutions ma ψ φ) = Eψφ) ) ψφ) = Ae inφ + Be inφ 6) for arbitrary complex numbers A and B, where n = ma E/ plays the role of a wavevector you ll see why I named it n in a moment). The wavefunction is a function of a periodic variable, so we need to make sure the function takes the same value regardless of the label φ attached to each point. It suffices to require ψ0) = ψπ) 7) which means A + B = Ae πin + Be πin 8) for all A and B. So we need n to be an integer. The energies are therefore given by for integers n. E n = n ma 9) We still need to normalize the energy eigenstates, which I ll write as ψ n φ) = Ae inφ 0) and let the negatives ns cover the negative exponents. Normalization requires so we can take A = / π. π = A dφ = π A ) 0 Note: if you included factors of a in the normalization in a reasonable way, I won t deduct points. It s somewhat ambiguous whether the wavefunction should be per unit angle or per unit circumference or even per unit volume in which case there should also be a da factor). b) In this problem, consider a particle of mass m that is subject to the following central potential { 0 a < r < b V r) = ) elsewhere i) Write the Schrödinger equation in spherical polar coordinates for this potential. pts)

3 Solution. In the region a < r < b we have m r r ψ ) + r r and everywhere else ψ = 0. ψ r sin θ φ + r sin θ sin θ ψ ) = Eψ 3) θ θ ii) Write the reduced radial equation using the substitution Ur) = rrr). pts) Solution. We look for solutions of the form ψ r) = Rr)Y θ, φ). The equation for the radial factor Rr) is m r r R ) ll + ) + r r mr Rr) = ER 4) which with the substitution Ur) = rrr) simplifies to ll + ) Ur) + m r mr Ur) = EUr). ) iii) Consider the case of angular momentum l = 0. Find the bound state energies and eigenfunctions. 0 pts) Solution. In this case the SE is just and the solutions can be written Ur) = EUr) 6) m r Ur) = A sinkr δ) 7) where A is a normalization constant, δ is also a constant, and k = me/ is the usual wavevector. To make ψ continuous at the boundary r = a we need δ = ka or ka plus a trivial integer multiple of π). To make ψ continuous at r = b we need for some integer n, so the energies are Multiplying Ur) by /r gives kb a) = nπ 8) E n = n π mb a). 9) Rr) = A sinkr a)). 0) r iv) Normalize the radial eigenfunctions found in iii). Note: sin x) = [ cosx)]. 4 pts) 3

4 Solution. b = A a so we can choose sin kr a)dr = A [ b a) ] sinkb a)) = A b a) ) k A = b a. ) v) Write the complete, normalized wavefunctions, including radial and angular parts, for the case of l = 0. pts) ) / sinnπr a)/b a)) ψr) =. 3) 4π b a r 3 D insulators 0 pts) Consider a simple model for an insulator with D symmetry. The idea is that an electron is confined by potentials in all three dimensions, but they may have different functional forms and intensities. We consider the case of a D lattice where the electron with mass m is subject to infinite square well potentials in the x and y directions, and a harmonic potential in the z direction with force constant k. The electron can be considered to be localized close to the origin with hard wall barriers at x = ±a and y = ±a. We consider the case of very strong confinement in x and y, with a weaker confinement in z i.e., ω π / ). a) Write down a general expression for the 3D wavefunctions and eigenenergies of this system, including normalization constants and special polynomials. Note: You do not need to derive the generating function for the special polynomials, but rather simply refer to them properly. 8 pts) Solution. The solutions factor into functions that each only depend on x, y, or z: ψ r) = Xx)Y y)zz). 4) The x and y parts are the familiar infinite square well, with solutions X nx x) = { a cosπn x x/a)) n x =, 3,, 7,... a sinπn x x/a)) n x =, 4, 6, 8,... ) and similarly for Y ny y). The z part is the harmonic oscillator with ω = k/m and solutions Z nz z) = n znz! mω ) /4 mωz ) e H nz mω/ z, n z = 0,,,... 6) π as printed on the front of the exam. The H nz are the Hermite polynomials. The total energy of each eigenstate is the sum of the energies for each factor: E nx,n y,n z = π n x + n y) + ωn z + ). 7) 4

5 b) Write down the specific wavefunctions for the ground state and for all the states with the first three non-ground state energies i.e., excited states); be sure to write all possible wavefunctions for a given energy. 4 pts) Solution. In the specified limit, the smallest steps in energy come from increasing n z, so we expect the first three excited states to have n x = n y = and n z =,, 3. c) Make a table of quantum numbers and eigenenergies starting with the ground state and going to the third excited state. Indicate the number of degenerate states per energy level, if any. 4pts) n x n y n z E Solution. 0 π + ω π + 3 ω π 3 π There is no degeneracy. + ω + 7 ω d) Consider the opposite limit where ω π /. Repeat parts b) and c) in this limit. Solution. Now n z = 0 for all the lowest states, and the first few excited states have larger n x and/or n y. The ground state is the same, but the first excited energy has two states: the second excited energy has one state: and the third excited energy has two states: X x)y y)z 0 z) and X x)y y)z 0 z), 8) X x)y y)z 0 z), 9) X x)y y)z 0 z) and X x)y y)z 0 z). 30) The table looks like n x n y n z E degeneracy 0 π + ω 0 0 π + ω 0 8 π + ω π + ω 4 Angular Momentum 0 pts) Consider a particle with orbital angular momentum and in the state described by a wavefunction ψθ, φ) = 3 Y, θ, φ) + Y,0θ, φ) + Y, θ, φ).

6 a) Find ψ L + ψ 8 pts) Solution. Recall that the Y lm form an orthonormal set. This allows us to calculate, using l, m with θ, φ l, m = Y l,m θ, φ), ψ L + ψ = =, + = =, , 0 +, ) L + ) 3, 0 +, ) 3, + In the above I used L + l, m = ll + ) mm + ) l, m +. + ), ) 3, 0 +, 3) ) + ), b) If L z were measured, what possible outcomes would occur and with what probabilities? 4 pts) Solution. with probability /, 0 with probability 3/, and with probability /. c) If after measuring L z we find that the eigenvalue is, calculate the uncertainties L x and L y and their product L x L y in the post-measurement state. 8 pts) Solution. The state with L z eigenvalue has L x = L y = 0 by symmetry. Also by symmetry L x = L y, so we only need to calculate to find 3) 33) 34), L x, =, L + + L ), 3) = 4, L + + L + L + L + L L +, 36) = 4, L L +, = L x L y =. 38) Aside: we might have guessed this result at least approximately) from the uncertainty principle L x L y [L x, L y ] = i L z = which this state saturates. Just saying this isn t worth full credit, since you have to justify the assumption that this particular state minimizes the uncertainty product. There s also an even easier way of doing this problem than the above. As some of you noticed, 37) 39), L x, = L L z = ll + ) ) ) = /. 40) 6

7 Hydrogen Atom pts) An electron in a hydrogen atom is in the energy eigenstate ψ,, = Nre r a 0 Y, θ, φ). a) What is the probability per unit volume of finding the electron at r = a 0, θ = 4, φ = 60? pts) Solution. ψ a 0, 4, 60 )ψa 0, 4, 60 ) = N a 0e 3 8π sin 4 ) = N a 3 0 6eπ b) What is the probability per unit radial interval dr) of finding the electron at r = a 0? pts) Solution. To get the probability per unit radial interval we integrate the probability density over θ and φ. Since the spherical harmonic is normalized, this gives us θ φ 4) N r 4 e r a 0 Y, θ, φ) sin θdθdφ = N r 4 e r a 0 4) which we evaluate at r = a 0 to get the desired probability per unit radial interval 6 N a 4 0e. 43) c) If ˆL z and ˆL were measured, what would be the results? pts) Solution. The measurement of L z would give and the measurement of L would give. 6 Spin 0 pts) Consider a particle with spin = /. a) Find the energy levels of this particle when its Hamiltonian is given by Ĥ = E ) 0 Ŝ x + Ŝ y + E 0 Ŝz where E 0 is a constant with dimensions of energy. 8 pts) Solution. We can reduce this problem to things we already know by rewriting the first term in the Hamiltonian: Ŝ x + Ŝ y = Ŝ Ŝ z 44) so Ĥ = E 0 Ŝ ) Ŝ z + E 0 Ŝz. 4) The energy eigenstates are, m where m = /, 3/, /, /, 3/, /, with corresponding energies E m = E 0 ) ) ) 3 + m + E 0 m = E 0 4 m + m 46) 7

8 b) What is the degeneracy? pts) Solution. There are only 6 states do any of their energies repeat? m + m is quadratic, so if m could be any real number then we would certainly have degeneracy. However in this case we can just list the allowed values of m and m + m: m m + m = mm ) -/ -3/4-3/ -/4 -/ -3/4 / /4 3/ -3/4 / -/4 So there s two-fold degeneracy for two of the energies; the other two energies have degeneracy one. 8