Midterm #, Physics 5C, Spring 8. Write your responses below, on the back, or on the extra pages. Show your work, and take care to explain what you are doing; partial credit will be given for incomplete answers that demonstrate some conceptual understanding. Cross out or erase parts of the problem you wish the grader to ignore. Some potentially useful formulae are given on the back page Problem : 6 pts A particle in 3D space has a wavefunction in spherical coordinates where A and α are real constants. a Determine A ψr, θ, φ = A r e αr e iφ Solution: We must normalize the solution, which in spherical coordinates means In this case π π π π A π ψ ψr dr sin θdθdφ = A r e αr e iφ A r e αr e iφ r dr sin θdθdφ = 3 π e αr dr sin θdθdφ = 4 Since there is no dependence on θ, φ, the angular integrals just give a factor of 4π Doing the radial integral From which we find So that the normalized wavefunction is 4πA 4πA e αr dr = 5 α e αr ψ = A = α π α e αr e iφ π r = πa α = 6 b At what radius r is the probability of finding the particle at r > r equal to /? Give the result in terms of the constants given. Solution: The probability that we find the particle at r > r is P r > r = π π r 7 8 ψ r dr sin θdθdφ 9 The angular integrals just give 4π again so α P r > r = 4π r π e αr dr = α e αr = e αr α r We want probability one half so e αr = / e αr = αr = log Don t worry about the fact that the wavefunction is infinite at r = ; that won t affect the results of this problem. If you like you can imagine that ψ saturates to some finite value for very small r.
Thus r = log α c The operator associated with orbital angular momentum in the z direction is Show that the wavefunction ψ an eigenstate of ˆL z. ˆL z = i h φ Solution: We apply ˆL z to the function ˆL z ψ = i h ψ φ = i h A A φ r e αr e iφ eiφ = i h e αr = i h i Ar r φ e αr e iφ 3 4 And so ˆL z ψ = i hiψ = hψ 5 So ˆL z applied to ψ returns ψ times a constant. The eigenvalue is h. d What is the expectation value of ˆL z for this ψ? What is the uncertainty in ˆL z? Since ψ is an eigenstate, we know that a measurement will return the eigenvalue h every time. So we can say off the bat that L z = h and σ Lz =. If we want to show this explicitly not necessary to do, but a reasonable double check we can do the integrals L z = ψ ˆLz ψr dr sin θdθdφ = ψ hψr dr sin θdθdφ 6 = h ψ r dr sin θdθdφ = h 7 where in the last step we used the fact that we already normalized ψ above. We also have similarly L z = ψ ˆL z ψr dr sin θdθdφ = ψ hˆl z ψr dr sin θdθdφ 8 == ψ h ψr dr sin θdθdφ = 4 h ψ r dr sin θdθdφ = 4 h 9 and so As expected σ Lz = L z Lz = 4 h h =
Problem : pts A particle is in a semi-circular infinite well where V = inside the well grey shaded region in the figure and V = outside. In spherical coordinates, the well confines the particle at a constant radius r = R and constant polar angle θ = π/. The wavefunction thus only depends on the angle φ, i.e., ψ = ψφ a Solve for the normalized energy eigenstates of this particle. Solution: The time-independent Schrodinger equation is h m ψ + V ψ = Eψ Since the wavefunction is independent of θ, r we can drop the derivatives with respect to those functions in the Laplacian. The equation then becomes Rewriting this as h ψ mr = Eψ ψ ψ mer ψ = k ψ where k = h 3 The solution is ψ = A sinkφ + B coskφ 4 Or alternatively we could write it in terms of complex exponentials ψ = A e ikφ + B e ikφ 5 The boundary conditions give that ψ = at φ = and φ = π. These first gives ψ = B coskπ/ = B = 6 The second gives So the solutions are ψπ = A sinkπ = k = n where n =,, 3,... 7 ψ n φ = A sinnφ 8 To normalize we want π A sin nφdφ = 9 This looks a lot like a particle in a D well. If we wanted to make it look more familiar for doing the integral, we could make the substitution φ = πx A sin nπxπdx = 3 3
Apart from the extra factor of π, this is the same normalization integral we do for the D particle in a box of length L =. We know that the integration of the sin over a box gives a factor of / so A sin nπxπdx = A π = 3 And so A = /π or A = /π up to an arbitrary complex phase. The normalization is the same as a box of length π. The normalized wavefunction ψ = sinnφ 3 π In the above, we considered ψ a function of only φ, and so ψ has units of per radian. b Determine the values of energy that could be measured for the particle. Solution: From above, the energy is given by And with the quantization condition k = n the allowed energies are E = h k mr 33 E n = h n mr n =,, 3,... 34 4
Problem 3: pts Consider a particle of spin /. We will use the eigenstates of the Ŝz operator as the basis vectors, such that χ z, = χ z, = 35 where χ z, represents spin-up in the z direction, and χ z, spin down in z. The particle if is put into a uniform magnetic field pointing perpendicular to the z direction, such that the energy associated with the particle s spin is given by the Hamiltonian ɛ Ĥ = 36 ɛ where ɛ is a constant units of energy. 3a Calculate the possible values of energy that could be measured for this system. Solution: We want to find the eigenvalues of the matrix. We use the standard approach det Ĥ λî = λ ɛ ɛ λ = ɛ λ = λ = ±ɛ 37 3b Find the normalized eigenstates of the Hamiltonian Solution We find the states a Ĥ b b = ɛ a a = λ b 38 The first component of this equation gives ɛb = ±ɛa b = ±a 39 The energy eigenstates are thus v = a a v == a a 4 Normalizing these vectors such that a + a = implies a = / and so v = v = 4 3c We measure the particle s energy and find it to be in the lowest possible energy state. What is the probability that a measurement of Ŝ z returns + h/ i.e., spin up in the z direction? Solution The measurement collapses the particle to the v state. This is a superposition of Ŝz eigenstates, as we can see v = = = χ z, χ z, 4 By the quantum postulates, the probability is the coefficient in front of χ z, squared, and so it is P = /. 3d A particle is initially the χ z, state. We wait some time t and measure the z-component of spin. What is the probability that the measurement finds the particle to be in the χ z, state? Solution: The problem is essentially similar to the neutrino oscillation or rabbit-duck problem done on the homework. The eigenstates of the Hamiltonian are the stationary states. The initial state given can be written as a superposition of energy eigenstates χ z, = v + v 43 5
The time-dependence of the energy eigenstates are just given by a phase factor e iωt. Define ω = E / h and ω = E / h. The time-dependent result is χt = e ωt v + e ωt v 44 Plugging in our results for v, v χt = e ωt = e iω t + e iωt e iωt e iωt + e ωt 45 46 The probability we want is the second component squared corresponding to spin down in z. This is P z = e iω t e iωt e iω t e iωt 47 = Here ω ω = ɛ/ h ɛ/ h = ɛ/ h and so e iω t iω t e iωt iωt + 48 = cosω ω t 49 P z = cosɛt/ h 5 We check that the probability is bounded by and. If we wanted to double-check, we could also calculate that P z = + cosɛt/ h 5 So that P z + P z =. Another manipulation of the algebra would be to write the bottom componet of the array as c = e iωt e iωt = e iωt/ e iωt/ e i ωt e i ωt 5 where ω = ω ω / = ɛ/ h. This can then be written Then the probability is c = e iωt/ e iωt/ sin ωt = e iω+ωt/ sin ωt 53 P z = c = sin ɛt/ h 54 This is equivalent to the above result as can be shown from using trig identities 6
Problem 4: pts A certain thermodynamic system of fixed volume is described by the macroscopic variables U internal energy and N total number of particles. The number of microstates corresponding to a macrostate U, N turns out to be given by ΩU, N = [CU/N 3/] N 55 where C is a constant. 4a Determine an expression for the system temperature as a function of U and N. Solution: We first calculate the entropy S = Nk B log Ω = Nk B log CU/N 3/ = Nk B log C + Nk B log U 3/ Nk B log N 3/ 56 Using additional properties of the log this becomes S = Nk B log C + 3 Nk B log U 3 k B log N 57 The temperature is defined as and so T = S U = 3 N Nk B U 58 T = U 59 3 Nk B Now consider two such systems. System is initially in a macrostate U = J, N = N A while System is in a macrostate U = JN = N A. Here J denotes the energy unit Joules and N A is Avogadro s number. 4b The two systems are put into thermal contact the number of particles in each system is held fixed. Is energy most likely to flow from System to System or vice versa? Solution The temperatures are T = J = J T = J = 4 J = J 6 3 N A k B 3 N A k B 3 N A k B 3 N A k B 5 N A k B we see that T < T so energy is likely to flow from System to System. 4c After these two systems come into thermal equilibrium, what is the energy of System in Joules? Solution The temperatures will be equal in equilibrium, thus U U T = T = 6 3 N k B 3 N k B where we use U, U to represent the final energies of the systems. This gives U = U U = U N = U N N N The total energy is a constant U = U + U = 3 J and so U = U U, which gives U = U U U + = U 6 63 U + = U U = U 64 And since U = 3J we have U = 3/ J. System did indeed gain energy. 7
Problem 5: pts A tube of length L is filled with gas, which is divided into two sides by a moveable wall. Initially, side has volume V = x A and side has volume V = x A. We fix the number of particles in each side to be N = N A and N = N A. Both sides of the tube are kept at a constant temperature T. 5a If the gas is an ideal gas, find the position x in terms of L of the wall where the pressures on both sides of the wall are equal, P = P. Solution: The ideal gas law is The balance P = P then implies P = Nk BT V 65 N k B T V = N k B T V N V = N V 66 using V = xa and N = N Since x + x = L we have x = L x N x = N x x = x x = x 67 x = L x 3x = L x = 3 L 68 Now consider the problem statistically. Imagine slicing up the tube into many small elements of length x such that there are M = x / x slices on Side and M = x / x slices on Side. The number of ways of distributing N particles among M elements is Ω = N + M! M!N! 69 We assume that N and M. 5b Assuming the wall can move freely and that the ergodic hypothesis holds, calculate the location x in terms of L of the wall once equilibrium is reached. Solution The entropy of the combined system is We want to maximize this with respect to the variable x, which gives S = S + S = k B log Ω + k B log Ω 7 S x = S x + S x = 7 We are ignoring any additional degrees of freedom that may add to the number of microstates due to e.g., the energy of the particles. 8
Now x = L x so dx = dx. Using the chain rule S = S S x + x x x x So the maximize system entropy occurs when = S x S x = 7 S x = S x 73 This is analogous to the equation when we put systems in thermal equilibrium. In fact, this derivative of S is related to the statistical definition of pressure. The problem gives x = M x and x = M x. So equilibrium can be written From the given Ω, we can calculate the entropy of a system Using Sterling s approximation S M = S M 74 S = k B log Ω = k B log[n + M!] k B log[m!] k b log[n!] 75 S = N + M logn + M N + M M logm + M NlogN + N k B 76 = N + M logn + M M logm N log N 77 From this we get and so k B S M = logn + M + logm = logn + M logm 78 [ ] S N + M M = k B log M Setting this quantity equal on both sides gives [ ] [ ] N + M N + M k B log = k B log M M exponentiating 79 8 N + M = N + M M M 8 N M = N M 8 N M = N M 83 using the now that N = N N M = N M M = M Since M we can drop the s we could have done this much earlier... Multiply both sides by x From which we find 84 M = M 85 xm = xm x = x x = L x 86 As found with the ideal gas law, but from a statistical calculation. x = 3 L 87 9