Section 6: Measurements, Uncertainty and Spherical Symmetry Solutions

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1 Physics 143a: Quantum Mechanics I Spring 015, Harvard Section 6: Measurements, Uncertainty and Spherical Symmetry Solutions Here is a summary of the most important points from the recent lectures, relevant for either solving homework problems, or for your general education. This material is covered in Chapter 3 of [?]. The nature of quantum measurement as best we know it is as follows: consider a Hermitian operator Q, and a wave function in the state Ψ = q n Ψ q n, 1 where q n are the orthonormal eigenvectors of Q. After a measurement, the wave function will collapse onto the state q n with probability q n Ψ. The uncertainty principle states that if, for a Hermitian operator A: with averages taken as A = Ψ A Ψ, then σ A A A σ A σ B 1 [A, B]. 3 An important application is Heisenberg s uncertainty principle, which says σ x σ px. 4 The Schrödinger equation for a particle of mass m in 3 spatial dimensions is i Ψ t = m Ψ + V rψ. 5 We solve this just as is 1 dimension, by solving the time-independent Schrödinger equation Ĥ Ψ = E Ψ. This time Ψ corresponds to a function of all spatial coordinates, Ψx, y, z. The normalization condition is 1 = d 3 r Ψ. 6 In the case where V is only a function of the distance to some origin: V = V x + y + z = V r, then we know that stationary states ψ are given by in spherical coordinates 1 ψ nlq = ur r Yq l θ, φ. 7 1 It is conventional to call the parameter q here m. But for now this can be confusing with the mass m, and so I will avoid mixing the two ms when possible. 1

2 Y q l are the spherical harmonics. The function ur obeys the equation d u m dr + V r + ll + 1 mr u = Eu. 8 u is defined for 0 r <. The boundary conditions on u are typically that ur = 0 = 0, so that ψ is continuous. Problem 1 BB84 Quantum Cryptography: Quantum mechanics gives us a way to send a string of bits 01001, etc., to a friend, with the absolute confidence that nobody else has read the message. The clever trick that we use is based on the disruptive nature of quantum measurement. Suppose that Alice wants to send her string of bits to her friend Bob, across a possibly insecure line of communication. However, suppose that she can send quantum bits. The basic idea is that Alice will send 4 bits, each with equal 1/4 probability, at times t = 0, 1,,...: ψ 1 = 0, ψ = 1, ψ 3 = , ψ 4 =. Now, Bob recieves this string of quantum bits. He has no idea what to expect, and so with equal probability he makes a measurement with one of the two Hermitian operators: σ z = , σ x = a What are the eigenvalues and eigenvectors of σ z and σ x? Try the quantum states above! Solution: Let s try σ z ψ 1 = = = 0 = ψ 1. So ψ 1 is an eigenvector of σ z with eigenvalue 1. Analogously, we can find ψ is an eigenvector of σ z with eigenvalue 1. Since σ z is a matrix, these are all eigenvalues. Now let s try, following the hint, ψ 3,4 with σ x : σ x ψ 3 = σ x ψ 4 = = = ψ = = ψ 4. So ψ 3 is the eigenvector of σ x with eigenvalue 1, and ψ 4 the eigenvector with eigenvalue 1. b There are 8 possibilities: Alice has sent ψ 1,,3,4 and Bob measures with σ z,x. Each is equally likely. Describe what quantum states Bob will have after the measurement, with what probability. Show that Bob does not alter the quantum state if Alice has sent ψ 1, and Bob measures with σ z, or if Alice has sent ψ 3,4 and Bob measures with σ x. Solution: Let s get four possibilities out of the way. If we make a measurement of a Hermitian operator on a quantum state which is an eigenstate, the quantum state is unchanged and we measure that eigenvalue with probability 1. So if we measure ψ 1 or ψ with σ z, we get +1 or 1 respectively; similarly, if we measure ψ 3 or ψ 4 with σ x, we get +1 or 1 respectively. But what if we measure ψ 1 with σ x? Well, we need to compute the overlap of ψ 1 with the eigenvectors of σ x, ψ 3,4 : ψ 3 ψ 1 = 0 0 = ψ 4 ψ 1 = 1.

3 After the measurement, we get with equal probability the state ψ 3,4, and a measurement of the appropriate eigenvalue. An analogous argument works for ψ. And since the measurement of ψ 3,4 with σ z must return either ψ 1,, we can take the above results and immediately conclude that we re equally likely to get either ψ 1,, each with probability 1/, in this case. The next step in the protocol is that Alice publicly sends Bob a string of zs and xs. The n th letter in the string is a z if she sent ψ 1, at time n, and an x otherwise. Bob compares Alice s string with what he measured, and he returns to her a list of all times n at which he measured with the appropriate σ z,x operator. As we showed in part b, if Bob has directly received a quantum state from Alice, then it is these states which are unaltered. Bob now sends a list of the quantum states he has measured to Alice. Now, however, suppose that at each time step, Bob does not receive a bit directly from Alice. Instead, there s a quantum eavesdropper Eve who has been listening in. For our purposes, that means that at each time step, Eve makes a measurement on Alice s quantum state before Bob does. She uses the same two operators. As Eve also won t know which random states Alice is sending, she must, like Bob, choose her operators randomly. c What is the probability that Alice has sent the state ψ i, and Bob receives the state before his measurement ψ j? Make a table of these probabilities, Pi j. You should be able to exploit results from part b to do this quickly! Solution: Eve behaves just like Bob, and is equally likely to pick to measure with σ x or σ z. From part b it is straightforward to conclude: P1 1 = P = P3 3 = P4 4 = 1 P1 = P 1 = P3 4 = P4 3 = 0 P1 3, 4 = P 3, 4 = P3 1, = P4 1, = 1 4. d What is the probability ρ that Alice has sent ψ 1, and Bob measured with σ z, or that Alice has sent ψ 3,4 and Bob measured with σ x, and Bob s measurement of the quantum state disagrees with what Alice prepared? Solution: The probability that the quantum state passes through Eve unaltered is 1/, and altered is 1/. If the state is altered, it is altered because the measurement has been taken in a different basis. Bob is thus equally likely to measure the correct or incorrect result. So ρ = 1 1 = 1 4. If Alice and Bob disagree on at least a fraction ρ of the quantum states when they compare, then Alice knows that there is an eavesdropper in the channel. Otherwise she knows that nobody has been listening, and the channel is secure. She may now securely send her message to Bob across this channel. This scheme was developed in [?]. Problem The Energy-Time Uncertainty Principle, Redux: Let Ψ0 be the initial state of some quantum system with Hamiltonian H. Suppose that at some time τ, Ψτ Ψ0 = 0. 3

4 a Show that τ obeys the following inequality, where C is an O1 number: σ H τ C, where σ H is the uncertainty in the Hamiltonian of the state Ψ0. The simplest way to proceed is as follows: consider Xt e i H t/ Ψt Ψ0, and find an upper bound for ReXt. Solution: Let us explicitly write down the abstract evolution for the wave function. For simplicity we assume that there is a discrete basis of energy eigenstates, and we get Ψt e ient/ c n n. So we find, following the suggestion above: [ [ ] Xt = e i H t c n e n ] ient/ cm m = e ien H t/ c n. Using that cos y 1 y /, we find ReXt = c n cos E n H t c n 1 1 En H t = 1 σ H t. Now if Ψτ Ψ0 = 0, then ReXτ = 0. From our above inequality, this can t happen until which gives us C =. 0 1 σ H τ = σ H τ, b A more serious calculation [?] gives C = π/. Construct a quantum state Ψ0 and a Hamiltonian H such that the inequality above is realized with this value of C. Solution: Consider a harmonic oscillator of angular frequency ω in the following state: Ψ0 = The argument generalizes to many other choices this is just one case. We know that Ψt = 1 0 e iωt/ + 1 e 3iωt/ = e iωt e iωt/. At time τ = π/ω, we have Ψτ = i 0 1, Now ω σh = 1 3 ω + Ψτ Ψ0 = i = i = 0. And we find that the energy-time uncertainty principle is satisfied: [ 1 ω + 3 ω ] = 1 ω + 9 ω ω = ω σ H τ = ω π ω = π. 4

5 Problem 3 Quantum Wires: Consider a very thin strip of a semiconductor, with a square cross section of side length a. Electrons of mass m kg are approximately constrained to move in the three dimensional potential { 0 z, y a/ V x, y, z. otherwise There will be electrons in energy eigenstates for every E < µ in the system; µ > 0. semiconductor we can estimate µ 1 ev J. a What are the eigenvectors and eigenvalues of the Hamiltonian? Solution: The eigenvalue equation is If we make the ansatz then we find the equation m x + y + z ψ = Eψ. ψx, y, z = XxY yzz, [ X m X + Y ] Y + Z XY Z = EXY Z. Z For a typical X /X is a function only of X; likewise for Y and Z. It is only possible for this equation to be satisfied if X X = E Y x, Y = E Z y, Z = E z, with E x, E y and E z constants, and E = E x + E y + E z. Now we have three 1d problems to analyze. In the x direction, there are no constraints at all, and so we know from 1d that Xx = e ikxx and E x = kx/m. In the y direction, we are constrained to the region y < a/, and this is just the particle in a box. The wave functions are Y y = a sin ny π y a, E y = n yπ a ma. An identical result holds for Zz: Zz = a sin nz π z a, E z = n zπ a ma. Thus the eigenvectors/eigenvalues are ψ kx,ny,n z = ny π a eikxx sin y a nz π sin z a, E = a a m k x + ny π a nz π +. a b For what value of a will the dynamics of all electrons be describable by the Hamiltonian of a one dimensional free particle in the potential V x = constant? Compare your answer to the typical interatomic spacing, 0. nm. Solution: We need for all E < µ eigenstates to have n z = n y = 1. We want for the lowest energy in the say n y =, n z = 1 state to be positive. This requires that π µ < π + 4 = 5 π m a a ma. 5

6 a < 5 π mµ. Plugging in for the numbers above, we estimate a > 1 nm. direction! This is about 5 atoms thick in each Wires of width smaller than this, made of typical materials, will behave as effectively one dimensional quantum systems. This width is too small to be fabricated cleanly in most labs. But recent materials such as carbon nanotubes have allowed for realistic, effectively one dimensional quantum systems to be created and used for engineering purposes. Problem 4 Binding to an Impurity in a Metal: Consider a particle of mass m in three dimensions in the potential V r = αδr R, where α, R > 0. This is a crude model for attractive interactions between a free electron and an impurity in a metal. a Find all bound states of this potential with l = 0 angular momentum, including both the stationary state and the energy. Does a bound state always exist? Solution: We write our bound state as ψr = ur/r, where d u αδr Ru = Eu. m dr Look for states with E < 0. Then defining me κ, we find that demanding normalizability: { A sinhκr r < R ur Be κr r > R. The boundary condition that u0 = 0 follows from demanding that the wave function ψ is smooth as r 0. Continuity at r = R gives us that We also require that A sinhκr = Be κr. αur = αbe κr = u R + u R = κbe κr Aκ coshκr m m mα = κ 1 + cothκr. Does this always have a solution? Well, if κr is small, we have mα = κ 1 κr + Oκ0 = κr +. R This only has a solution if α > mr. You can check by numerically plotting the right hand side that at larger κ our conclusions are unaltered. 6

7 b Compare to what happens for the δ well in one dimension. Solution: Unlike in one dimension, here there s not always a bound state. If it does exist, then the energy is given by solving the above equation for κ which can t be done exactly, and the wave function is as found above. But if the impurity is weak enough, it can t trap any electrons in three dimensions. This is a crude model that suggests the following result, which turns out to be rigorous. In one dimension, any amount of impurities at all can trap electrons in a metal and localize them, which means that it is very hard to maintain an electrical current namely, a one dimensional metal becomes an insulator for any impurity density. However, we must have a finite strength of impurities modeled by α in a three dimensional metal before we get an insulator. [1] D. J. Grififths. Introduction to Quantum Mechanics Prentice Hall, nd ed., 004 [] C. H. Bennett and G. Brassard. Quantum cryptography: Public key distribution and coin tossing, Proceedings of IEEE International Conference on Computers, Systems and Signal Processing [3] L. Vaidman. Minimum time for the evolution to an orthogonal quantum state, American Journal of Physics