Quantum Mechanics in Three Dimensions


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1 Physics 342 Lecture 21 Quantum Mechanics in Three Dimensions Lecture 21 Physics 342 Quantum Mechanics I Monday, March 22nd, 21 We are used to the temporal separation that gives, for example, the timeindependent Schrödinger equation. In three dimensions, even this timeindependent form leads to a PDE, and so we consider spatial separation, familiar from E&M Three Copies Our onedimensional replacement: p x i x can be generalized to three dimensions in the obvious way. Cartesian coordinates have no preferential directions, so we expect the threedimensional replacement: p i 21.1 applied to the Hamiltonian. In addition, our wavefunctions become functions of all three coordinates: Ψx, y, z, t or any other equivalent set spherical, cylindrical, prolate spheroidal, what have you. We can work out the commutation relations for the three obvious copies of our onedimensional: [x, p x ] = i, but what about the new players: [x, y] and [x, p y ] and obvious extensions involving z and p z? The coordinate operators clearly commute, [x, y] =, and the momentum operators will as well, by crossderivative equality: [p x, p y ] = 2 2 x y 2 = y x Finally, mixtures of coordinates and momenta that are not related, like [x, p y ] will commute since x y = throwing in a test function to make the 1 of 11
2 21.1. THREE COPIES Lecture 21 situation clear, we have [x, p y ]fx, y = x f i y x fx, y = y i x f y x f =. y 21.3 Tabulating our results, the threedimensional commutation relations read letting r i be r 1 = x, r 2 = y, r 3 = z for i = 1, 2, 3 [r i, p j ] = i δ ij [r i, r j ] = [p i, p j ] = The wavefunction itself must now be interpreted as a full threedimensional density, with Ψr, t 2 dτ the probability per unit volume of finding a particle in the vicinity of r at time t. The normalization condition becomes a volume integral, as do all expectation values: 1 = Ψ 2 dτ r = Ψ r Ψ dτ 21.5 p = Ψ i Ψ dτ. where the integration is over all space, and dτ is the volume element. In Cartesian coordinates, dτ = dx dy dz, but it takes different forms depending on how we ve parametrized in cylindrical coordinates, for example, dτ = s ds dφ dz. For a finite volume, we have the obvious interpretation Ψ 2 dτ 21.6 Ω is the probability of finding the particle in the volume defined by Ω. To find the probability flowing into and out of this volume, we can use the probability conservation statement in three dimensions. Let ρ = Ψr, t 2, then we had J = i 2 m [Ψ Ψ Ψ Ψ ], and ρ t = J d dt Ω ρ dτ = dω J da 21.7 where the lefthand side is the rate of change of probability inside the volume Ω, and the righthand side is the amount of probability flowing out through the boundary of the volume dω, reminiscent of the electromagnetic charge conservation equation which has identical form. 2 of 11
3 21.1. THREE COPIES Lecture Two Dimensional Example Suppose we retreat to two dimensions and take the simplest possible potential other than the free particle case an infinite square well. Our twodimensional potential can be specified via: { < x < a and < y < a V x, y = otherwise 21.8 We must solve Schrödinger s equation, with boundary conditions. In this dimensionally expanded setting, the boundary conditions for ψx, y are: ψ, y = ψa, y = ψx, = ψx, a =, 21.9 and the timeindependent Schrödinger s equation on the interior of the boundary reads m x y 2 ψx, y = E ψx, y The temporal separation has already occurred here, we had the usual: Ĥ Ψx, y, t = i Ψx,y,t t, and took Ψx, y, t = ψx, y φt. Then we can separate with constant E, leading to 21.1 and φt = e i E t. If we use multiplicative separation in 21.1: ψx, y = Xx Y y, then we can write the above as X x Xx + Y y = 2 m E Y y The usual separation argument holds, on the left, we have two functions, one only depending on x, one only on y, and together, these must equal the constant on the right. Then each term must individually be equal to a constant, and a solution to follows from a solution to the set of three equation: X x = k 2 Xx Y y = l 2 Y y k 2 l 2 = 2 m E of 11
4 21.1. THREE COPIES Lecture 21 The ODEs in X and Y imply sinusoidal dependence on x and y so we have Xx = A cosk x + B sink x Y y = F cosl y + G sinl y, together with our side constraint on the sum of k 2 and l 2. It s time to impose the boundary conditions: The wave function reads: ψx, y =A cosk x + B sink x F cosl y + G sinl y Take x =, our boundary condition requires that ψ, y =, and this immediately tells us that A =. Similarly, for y =, we learn that F =. After imposing these, our wave function looks like ψx, y = B G sink x sinl y We absorb the product B G into a single constant: P = B G, and we are ready to set the boundary at a. In order to satisfy ψa, y = P sink a sinl y = for all y, we must have k a = m π for integer m. On the yside, we want ψx, a = P sink x sinl a =, and this gives l a = n π for integer n. The wave function, satisfying Schrödinger s equation and all boundary conditions, is: m π x n π y ψ mn x, y = P sin sin for m, n Z + a a m E mn = 2 π 2 n π M a a I have denoted mass with M in the above with P left for overall normalization. As always, we can normalize the individual wavefunctions. In this case, we have, with probability one, a particle of mass M localized to the region x, y [, a] [, a]. Then 1 = a a P 2 sin 2 m π x a sin 2 n π y a 2 dx dy = P 2 a of 11
5 21.2. ORBITAL MOTION AND CLASSICAL MECHANICS Lecture 21 so that P = 2 a. Check the units, we now have ψ mnx, y ψ mn x, y a probability density in two dimensions. As always, the general solution is a linear combination of the particular solutions: Ψx, y, t = m=1 n=1 Emn t i c mn ψ mn x, y e 21.2 and the set {c mn } are just waiting for an initial ψx, y to be provided, at which point they can be set. There are a couple of important differences between the one dimensional infinite square well and this twodimensional form. The most noticeable is the degeneracy associated with energy. In one dimension, the energies were related to an integer n via: E n = n2 π m a 2, so for each n, there was an energy. In the twodimensional case, the independent and orthonormal wave functions ψ 21 x, y and ψ 12 x, y both have the same energy. It is interesting to think about what a measurement of energy does in this context. Suppose we measure the energy E = 5 π2 2 2 m a 2, then we know that the wave function collapses to the associated eigenstate but which associated eigenstate ψ 12 x, y or ψ 21 x, y? In addition, there is the question do we have one particle in two dimensions, or two particles in one dimension? How could we tell? And what, really, is the distinction? 21.2 Orbital Motion and Classical Mechanics Consider the classical mechanics form of the Lagrangian governing, for example, orbital motion in a spherically symmetric gravitational field: L = 1 2 m ṙ2 + r 2 θ2 + r 2 sin 2 θ φ 2 Ur Ur = G M m r We have written the Lagrangian in terms of spherical coordinates, related to the Cartesian ones in the usual way: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ of 11
6 21.2. ORBITAL MOTION AND CLASSICAL MECHANICS Lecture 21 We can form the Hamiltonian as usual, the canonical momenta are: p r = L ṙ = m ṙ p θ = L θ = m r2 θ p φ = L φ = m r2 sin 2 θ φ, and then the Hamiltonian is just the Legendre transform of L H = ṙ p r + θ p θ + φ p φ L, p r,p θ,p φ where the evaluated at sign reminds us to evaluate the Hamiltonian in terms of the momenta which we can do by inverting rather than the time derivatives of the coordinates. Written out, we have H = 1 2 m p 2 r + p2 θ r 2 + p 2 φ r 2 sin 2 θ + Ur The first term, is of course, p p in spherical coordinates. Regardless, our usual observation is that we can set motion into the x y plane by requiring θ = and θ = π 2, then the Hamiltonian simplifies: H = 1 p 2 r + p2 φ 2 m r 2 + Ur, and finally, we know from the equations of motion 1 that H φ = ṗ φ = so that p φ is a constant. The problem of orbital motion reduces to a onedimensional Hamiltonian the radial coordinate is the only interesting equation of motion, and it is governed by an effective potential. If we write: H = p2 r 2 m + Ur + p2 φ 2 m r then we may as well be considering the onedimensional problem: Find the motion rt given a Hamiltonian that is a function only of p r and r p φ, 1 This is also an example of Noether s theorem the lack of dependence on φ means that φ itself is ignorable, and we can add a constant to φ without changing the Lagrangian at all that symmetry implies conservation, in this case, of the zcomponent of angular momentum. 6 of 11
7 21.3. QM IN SPHERICAL COORDINATES Lecture 21 as we have just established is a constant. We mention all of this because the same basic breakup occurs when we consider the quantum mechanical analogue of the above, wavefunctions governed by a spherically symmetric potential. That we should be interested in such a restrictive form for the potential is dictated by the ultimate problem we wish to solve over the next few days: the Hydrogen atom, with its Coulombic potential. Note that the above use of p r, p θ and p φ induces the question: What are the quantum mechanical operators associated with these classical momenta? We will sidestep the issue for now, but be on the lookout for a return later on QM in Spherical Coordinates Without going through the explicit transformations to spherical coordinates, we note that Schrödinger s equation still reads: and now the Hamiltonian is H = p p 2 m H Ψr, t = i Ψr, t, t + Ur Ĥ = 2 2 m 2 + Ur using the usual replacement, and assuming that the potential Ur depends only on the distance to the origin i.e. Ur = Ur is spherically symmetric. Now, we already know the Laplacian in spherical coordinates: 2 Ψr, θ, φ, t = 1 r 2 r 2 Ψ + 1 r r r 2 sin θ Ψ 1 2 Ψ + sin θ θ θ r 2 sin 2 θ φ Assuming that the potential Ur is timeindependent, we can once again separate the temporal and spatial portions of the wavefunction, Ψr, t = ψr φt and recover the timeindependent Schrödinger equation Ĥ ψ = E ψ φt = e i E t, allowing us to focus on the spatial portion ψr. 7 of 11
8 21.3. QM IN SPHERICAL COORDINATES Lecture 21 We can separate further: Let ψr, θ, φ = Rr Θθ Φφ, then the Hamiltonian operator can be written as 1 R d r 2 dr dr dr + 1 sin θ sin 2 θ Θ }{{} θ,φ d sin θ dθ + 1 d 2 Φ dθ dθ Φ dφ 2 2 m r2 2 Ur = 2 m r2 2 E The angular and radial portions are now separated we will call the separation constant l l + 1 for reasons which should be familiar from, for example, electrodynamics The Angular Equations Set the angular terms equal to the negative of the separation constant: 1 sin θ d sin 2 sin θ dθ + 1 d 2 Φ θ Θ dθ dθ Φ dφ 2 = l l Now, one of the terms depends only on Φ, so we will set it equal to a constant, call it m 2 : 1 d 2 Φ Φ dφ 2 = m2 Φφ = A e i m φ + B e i m φ, and by allowing m to be positive or negative, we can recover either solution, so we will just set: Φφ = e i m φ where we float the normalization since this is multiplicative separation, we know there will be an overall constant that we can set at the end. At this point, we see that the particulars of this form for Φφ will not play a role in the probability density, since Ψ 2 will have Φ Φ = 1 built in. Still it is usual in terms of E&M in particular, to require that Φφ = Φφ + 2 π, which implies that m is an integer, although this will also be forced on us from the rest of the angular separation. For the Θθ portion, we have: sin θ d sin θ dθ + l l + 1 sin 2 θ m 2 Θ = dθ dθ 8 of 11
9 21.3. QM IN SPHERICAL COORDINATES Lecture 21 To bring this into more familiar form, let z = cos θ, then d dθ = dz dθ dz = 1 z 2 d dz, and the above becomes thinking of Θ as a function of z, now 1 z 2 d 1 z 2 dθz +l l+1 1 z 2 m 2 Θz =, dz dz or d 1 z 2 dθz + l l + 1 m2 dz dz 1 z 2 Θz = which is the standard form for the associated Legendre equation 2. Its solutions are known as associated Legendre Polynomials, and are defined in terms of the Legendre polynomials via Pl m z = 1 z2 m /2 d m dz m P lz Since this is a second order differential equation, there is, of course, another solution associated Legendre Q functions, but these blow up at z = ±1. As a further restriction, these functions are only defined for integer m [ l, l] for a given l which must then also be an integer. We have, finally, the form for the angular solution at least, separable solution for Schrödinger s equation with spherically symmetric potential: d Θθ Φφ e i m φ P m l cos θ Looking ahead to the full normalization for the probability density, we have introducing a factor A in front of the angular terms: 1 = = π 2π Ψ 2 dτ = π R R r 2 dr A 2 R r R Θ Θ Φ Φ r 2 sin θ dφ dθ dr 2 π A 2 Pl m cos θ Pl m cos θ sin θ We can normalize the angular part by itself 3, by setting A such that π 2 π A 2 Pl m cos θ Pl m cos θ sin θ = See, for example, Riley, Hobson & Bence p Notice that the m = case gives back Legendre s equation. R 3 Keep in mind that this will mean we need to separately normalize the radial part, R R r 2 dr = 1. 9 of 11
10 21.3. QM IN SPHERICAL COORDINATES Lecture 21 It turns out that the value of A is, 2 l m! A = ɛ 4 π 1 + m! ɛ { 1 m m 1 m > With this normalization, the angular solutions are called spherical harmonics, denoted Yl m θ, φ spherical since they are in spherical coordinates, harmonic because they are associated with solutions of Laplace s equation: 2 F =. Written out: Yl m 2 l m! θ, φ = ɛ 4 π 1 + m! ei m φ Pl m cos θ These are orthonormal in the sense that 2π Yl m θ, φ Y m θ, φ dω = l π Y m l θ, φ Y m θ, φ sin θ dθ dφ l = δ ll δ mm of 11
11 21.3. QM IN SPHERICAL COORDINATES Lecture 21 Homework Reading: Griffiths, pp Problem 21.1 Griffiths 4.1. Here, you will work out the relevant, Cartesian, commutators for position and momentum. Feel free to use indexed notation, and in particular, the relation: = δ ij think of x x x = 1, and y =. Problem 21.2 x i x j Griffiths 4.2. Finding the stationary states of a threedimensional infinite well. Problem 21.3 The ordinary differential equation: żt + α zt = is solved by zt = z e α t for initial value z = z. In this problem, we will develop the series solution Frobenius method in preparation for our work to come. a. Start by setting zt = t p j= c j t j for unknown p. Take the derivative of this function as a series with respect to t. b. Insert zt and your expression for żt in 21.46, collecting all [ like powers of t into a single expression, i.e. fill in the...s in: t p... t 1 + ] j=... tj =. c. You should have a lone t 1 term, use this to set p =, and, from the independence of powers of t j, develop the recursion relation between the coefficients c j+1 and c j. Solve the recursion, starting with c = z, and compare with the Taylor series expansion of z e α t. 11 of 11
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