Superposition of Stationary States in Quantum Mechanics, Assignments of Chemistry

Download Superposition of Stationary States in Quantum Mechanics and more Assignments Chemistry in PDF only on Docsity! OPTI 511, Spring 2009 Problem Set 4 Prof. R.J. Jones 1. Superpositions of two stationary states A stationary state is a quantum state (or wavefunction) with a probability density distribution that is independent of time. The modes (or eigenfunctions, or eigenstates) of time-independent potential wells are stationary states. Since the probability density for a stationary state is independent of time, the expectation values 〈x〉 and 〈p〉 for a stationary state must also be independent of time. (Convince yourself that this must be true!) In general however, particles confined to a potential well are represented by superpositions of eigenstates of the potential. While the eigenstates of a potential are stationary states, superpositions of them are usually not stationary states. In this problem, you will examine some consequences of a simple superposition of just two eigenstates. Suppose that a particle of mass m in the 1-D square well of length L (with infinite-potential walls at x = 0 and x = L) has an initial wave function (at time t = 0) that is an equal superposition of the two lowest-energy normalized eigenstates ψ1(x) and ψ2(x). On page 18 of the notes, we discussed ψ3(x), which was a sinusoidal function with 3 half-waves spanning the distance from x = 0 to x = L. Here, you’ll work with the two longer-wavelength (and hence lower-energy) modes. We’ll write the superposition at time t = 0 in the following way: Ψ(x, 0) = A[ψ1(x) + ψ2(x)]. (a) Normalize Ψ(x, 0). That is, find A. This is very easy if you exploit the orthonormality of ψ1(x) and ψ2(x). Once Ψ(x, 0) is normalized at t = 0, it stays normalized for later times — if you doubt this, check it explicitly after doing part (c). (b) Find the time-dependent superposition Ψ(x, t) by writing it as an equal superposition of the two time-dependent wave functions Ψ1(x, t) and Ψ2(x, t) for the two eigenstates. Write this expression in terms of ψ1(x), ψ2(x) and the energy E1 of the ground state. (You’ll need to write the energy eigen- value E2 of the state ψ2(x) in terms of E1.) Don’t yet write out the functional forms of ψ1(x) and ψ2(x). (c) Find |Ψ(x, t)|2. Express your answer in terms of L, sinusoidal functions of position, and a co- sinusoidal function of time instead of the time-dependent complete this part of the problem, you will now need to substitute the correct normalized eigenfunctions for ψ1(x) and ψ2(x). (d) Roughly sketch |Ψ(x, t)|2 for t = 0, or better yet, plot it with a computer program. Your sketch should not be symmetric with respect to the center of the well! (e) When you compute 〈x〉 as a function of time, you will find that it is time dependent, and in fact it oscillates back and forth in the potential well. Before doing any computations, make a guess at the amplitude of oscillations (the maximum extent of 〈x〉 away from the center of the well). If you guess an amplitude greater than L/2, if your guess does not have units of length, STOP and try to figure out what you don’t understand. Now compute 〈x〉(t), the position expectation value as a function of time. What is the angular frequency of oscillations back and forth in the well? What is the computed amplitude of the oscillation? (f) Ehrenfest’s Theorem states that expectation values obey Newton’s Second Law, F = ma. Or in other words, expectation values can be used in classical physics expressions. As a consequence, we can write 〈p〉 = m · d〈x〉/dt, making 〈p〉 often quite easy to compute. (Note that this technique does not apply to finding 〈p2〉.) Use Ehrenfest’s Theorem to compute 〈px〉(t), the time-dependent expectation value of momentum along the x direction. Do not solve another integral to find 〈px〉(t). 1 2. Superpositions of multiple stationary states (a) For any arbitrary time-independent confining potential, not just a particle in a 1-D infinite square well, any wave function associated with a bound particle can be written as a linear superposition of energy eigenstates ψn(~x): Ψ(~x, t) = ∑ n cnψn(~x)e−iEnt/h̄, where the summation extends over all quantum numbers n, and each En is the energy eigenvalue associated with the eigenstate ψn. Using this general expression for Ψ(~x, t) (rather than the specific two-state superposition used in problem 1), find the expectation value 〈E〉 for the total energy of the particle represented by Ψ. Express your answer in terms of cn and En only. To do this problem, you should use ih̄ ddt as the operator for total energy. This is justified because we know the specific time dependence of our states (the exponential terms), whereas we know nothing specific about the spatial parts of the wavefunctions. Since the time dependence is explicitly given, we can work with the time-representation of the energy operator. Keep proper track of eigenstate labels within the double summation (If you don’t see why you need a double summation over two dummy indices, you need to try to figure out why one is necessary.) This problem should be quick and only take a few steps. Hint: the eigenfunctions ψn(~x) are all orthonormal (orthogonal and normalized) in three dimensions. Integration is not necessary in this problem! (b) What is 〈E〉 for the specific two-state square-well superposition state used in problem 1? Express your answer in terms of E1. Your answer should be independent of time. (c) What are the possible results of a measurement of energy (in terms of E1) of the particle in the square well of problem 1, again assuming the same two-state superposition? Is the answer from part (b) a possible measurement result? 2 5. Wavepacket Spreading Consider an arbitrary 1D wavefunction Ψ(x, t) that represents a particle of mass m trapped in a time-independent 1D potential well V (x). If we know the exact form of the wavefunction at a specific time, say t = 0, then we can determine the wavefunction for all times (as long as the potential well remains independent of time - that is, as long as no forces act on the particle). Specifically, we’d write Ψ(x, t) = ∑ n cnψn(x) exp (−iEnt/h̄), (1) where ψn and En are the energy eigenfunctions and eigenvalues associated with V (x). The values cn are the expansion coefficients that must be properly chosen such that Ψ(x, t) matches the known Ψ(x, t = 0) at time t = 0. However, in free space (that is, V (x) = 0 everywhere), there is not a discrete set of energy eigenvalues and eigenfunctions. The discrete sum above must be replaced by an integral: Ψ(x, t) = ∫ ∞ −∞ dk φ(k)eikx exp (−iEkt/h̄). (2) In this integral over all wavenumbers k (where k = p/h̄), we interpret the following: First, the eikx takes the place of ψn(x) in the discrete sum, since in free space, plane waves are the solutions to our free-particle time-independent Schrödinger Equation. The total energy E is now just kinetic energy, so Ek = h̄2k2/(2m). And finally, φ(k), which is the Fourier transform of Ψ(x, t = 0), takes the place of the expansion coefficients cn. φ(k) describes the amplitude of the continuously varying momentum components much like the discrete expansion coefficients cn specified the components of each energy eigenstate. (And φ(k) is independent of time in free space: as long as forces don’t act on a particle, its momentum components remain unchanged.) Now suppose that you are given an initial wavepacket Ψ(x, 0), and you want to find out how it evolves in free space - in other words, you want to determine Ψ(x, t) for all times t > 0. The first thing you have to do is to determine φ(k), which tells you about the momentum components of Ψ(x, 0): φ(k) = ∫ ∞ −∞ dxΨ(x, 0)e−ikx. (3) Once φ(k) is known, you can find Ψ(x, t) by evaluating the integral Ψ(x, t) = ∫ ∞ −∞ dk φ(k) eikx exp (−ih̄t/(2m) · k2), (4) which is the same as equation 2 once Ek is replaced with h̄2k2/(2m) (this is important to see, since k is the integration variable). Although most of the problems in OPTI511 deal with discrete energy eigenstates (and hence involve the discrete superpositions first mentioned above), wavepacket spreading in free space is nevertheless an important problem and is the subject of the following questions. The math involved is not very tricky. However, there are lots of signs, exponents, and variables to keep track of, so be careful! Make your mathematical expressions neat and easy to follow, explicitly keep track of your variable substi- tutions, and check your work often and carefully. We now consider a harmonic oscillator potential well V = (1/2)mω2x2. In a previous homework prob- lem, you showed that one of the energy eigenstates of this potential well is given by ψ0(x) = Ae−x 2/(2a2), where a = √ h̄ mω and A is a normalization coefficient. Suppose that a particle of mass m is confined in this potential well, and is in the ground state ψ0. At time t = 0, the potential V is suddenly 5 removed, and the initial spatial wavepacket Ψ(x, 0) = Ae−x 2/(2a2) is left to evolve in free space for times t > 0. (In case this problem seems silly or unrealistic to you, it’s not: millions of rubidium atoms are subject to this exact process nearly every day in a lab on the 5th floor of OSC. They are so cold, and their energies are so low, that they are described as starting out in the lowest energy eigenstate of the confining harmonic potential before the potential is suddenly removed. And yes, there is a good reason for doing this, but we won’t get into that here...) (a.) Determine A, or just write down the answer if you know it. You may assume it is real and positive. (b.) Determine φ(k), using equation 3 above. It’s OK to use a shortcut if you see one - just explain what you’re doing and what your reasoning is. If you do the integral, it becomes fairly straightforward when you complete the square of the terms in the exponent. (c.) To check your work from (b), make sure that φ(k) is normalized: ∫∞ −∞ dk|φ(k)|2 had better be equal to 1! (d.) Now find Ψ(x, t) by applying the integral of equation 4. As opposed to the initial state Ψ(x, 0), Ψ(x, t) will be complex and may look a little bit funny to you. However, you should check that Ψ(x, t) reduces to Ψ(x, 0) when you insert 0 for t. If it doesn’t, check your math. Hint: your answer should have the term (1+ ih̄t ma2 ) appearing twice - once under a square root, and once in the exponential term. Simplify your answer by substituting √ h̄ mω for a in the two spots where you have h̄t ma2 (but not in the spots where a doesn’t appear along with a t). (e.) Evaluate |Ψ(x, t)|2. This quantity should be real! (f.) It should be immediately clear that |Ψ(x, t)|2 is a Gaussian distribution. It should also be clear that the Gaussian “radius” of |Ψ(x, t)|2 is time-dependent. Define the radius such that at t = 0 it has the value a. For t > 0, you should note that the radius is a(1 + ω2t2) 1 2 . Letting α(t) = a(1 + ω2t2) 1 2 be an expression for the time-dependent Gaussian radius, what other similar expression does this resemble? (g.) Define P|x|<a(t) as the probability that a position measurement of the particle will give a result between x = −a and x = a at time t, if the measurement is to be made at time t. Evaluate P|x|<a(0) and P|x|<a(t = 10/ω). To do this, you should make use of the error function definition in the notes on page 2 of appendix G (erf(x) about a third of the way down the page, but make sure you interpret the labels x and t in this expression properly!). Significance of this problem: even if you know pretty well the region in which you’d expect to find the particle at some initial time, if you leave it alone in free space (so that it doesn’t interact with anything, and no forces are present), the wavepacket will spread. Eventually, you no longer have a good idea of where the particle is. This is called spatial wavepacket spreading, and partially addresses some questions that have come up in class, such as the meaning of short in the question “what happens if you measure a particle’s position (localizing it such that the wavepacket is narrow), then wait for a short time before measuring it again?” Now you know the procedure for figuring out the answer! 6