The Schrödinger Equation in Three Dimensions Particle in a Rigid Three-Dimensional Box (Cartesian Coordinates) To illustrate the solution of the time-independent Schrödinger equation (TISE) in three dimensions, we start with the simple problem of a particle in a rigid box. This is the three- dimensional version of the problem of the particle in a one-dimensional, rigid box. In one dimension, the TISE is written as 2 d 2ψ (x) − +U (x)ψ (x) = Eψ (x). (1) 2m dx2 In three dimensions, the wave function will in general be a z function of the three spatial coordinates. In the Cartesian y coordinate system, these coordinates are x, y, and z. Also, the potential energy U will in general be a function of all 3 x 2 d 2ψ coordinates. Now, in the 1-D TISE, the term − Lz 2m dx2 Lx 2 2 2 px kx can be identified with the kinetic energy = of the Ly 2m 2m 2 d 2ψ particle because − = [E −U ]ψ. [Try, for example, the free-particle wave function 2m dx2 ψ = Aei(kx−ωt).] 2 2 2 In three dimensions, the KE is(px + py + pz ) / 2m , so we suspect that additional second derivative terms will be needed to represent the additional kinetic energy terms. In fact, the TISE in three dimensions is written as 2 ⎛ ∂2ψ (x, y, z) ∂2ψ (x, y, z) ∂2ψ (x, y, z) ⎞ − ⎜ 2 + 2 + 2 ⎟ +U (x, y, z)ψ (x, y, z) = Eψ (x, y, z). (2) 2m ⎝ ∂x ∂y ∂z ⎠ ∂ψ (x, y,z) The symbol represents the partial derivative of ψ (x, y,z) with respect to x, which is ∂x simply the derivative of ψ (x, y,z) with respect to x with y and z held constant. Thus, ∂ψ ifψ = x2 yz, then = 2xyz . ∂x In the particle in a box problem, U(x, y, z) = 0 inside the box and U = ∞ at the walls.1 So, inside the box, the TISE becomes 2 ⎛ ∂2ψ ∂2ψ ∂2ψ ⎞ − ⎜ 2 + 2 + 2 ⎟ = Eψ. (3) 2m ⎝ ∂x ∂y ∂z ⎠ 1 Inside the box, 0<x<Lx, 0<y<Ly, and 0<z<Lz. (See figure of box above.) 1 To solve partial differential equations (the TISE in 3D is an example of these equations), one can employ the method of separation of variables. We write ψ (x, y,z) = X(x)Y (y)Z(z), (4) where X is a function of x only, Y is a function of y only, and Z is a function of z only. Substituting for ψ in Eq. (3) yields 2 ⎡ d 2 X d 2Y d 2Z ⎤ Y (y)Z(z) X (x)Y (z) X (x)Y (y) EX (x)Y (y)Z(z). (5) − ⎢ 2 + 2 + 2 ⎥ = 2m ⎣ dx dy dz ⎦ Note that the partial derivatives have disappeared, since X, Y, and Z are functions of one variable 2 only. Dividing both sides by − X (x)Y (y)Z(z) yields 2m ⎡ 1 d 2 X 1 d 2Y 1 d 2Z ⎤ 2mE . (6) ⎢ 2 + 2 + 2 ⎥ = − 2 ⎣ X dx Y dy Z dz ⎦ For a given solution, E is a constant. Further, note that the first term in the square brackets is a function of x only, the second term is a function of y only, and the third term is a function of z only. For Eq. (6) to be valid for all values of x, y, and z, each of the three terms must be constant. If this were not true, and, for example, all the terms were variable, then if one held y and z constant and changed x, the sum on the left hand side of the equation would change, violating the equation. Eq. (6) therefore becomes three separated ordinary differential equations: 1 d 2 X = −k 2 , (7) X dx2 x 1 d 2Y = −k 2 , (8) Y dy2 y and 1 d 2Z = −k 2 , (9) Z dz2 z with 2 2 2 2mE kx + ky + kz = . (10) 2 2 2 2 The separation constants are written as −k x , −k y , and −k z in analogy with the 1-D particle in a box problem. Eqs. (7), (8), and (9) are identical to the equation obtained in the 1-D problem and the boundary conditions are the same. For example, X(x) = 0 at x = 0 and x = Lx since the wave functions cannot penetrate the wall. The boundary condition at x = 0 leads to X(x) = A1sin kxx. nxπ The boundary condition at x = Lx leads to kx = , where nx = 1, 2,… Similar solutions are Lx obtained for Y(y) and Z(z). Hence, we find that 2 ⎛ nxπ x⎞ ⎛ nyπ y⎞ ⎛ nzπz⎞ ψ (x, y,z) = Asin⎜ ⎟ sin⎜ ⎟ sin⎜ ⎟ , (11) ⎝ Lx ⎠ ⎝ Ly ⎠ ⎝ Lz ⎠ and from Eq. (10), we find the energy 2 2 2 2 ⎡ 2 ⎤ π ⎛ nx ⎞ ⎛ ny ⎞ ⎛ nz ⎞ E = ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥. (12) 2m ⎢ L ⎜ L ⎟ L ⎥ ⎣⎝ x ⎠ ⎝ y ⎠ ⎝ z ⎠ ⎦ The integers nx, ny, and nz are called quantum numbers. These quantum numbers specify the quantum states of the system. Thus, we can label the wave functions and energies according to the values of the quantum numbers: ; E . ψ nx ny nz nx ny nz The ground state energy is 2 2 2 2π 2 ⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ E111 = ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥. (13) 2m ⎢ L ⎜ L ⎟ L ⎥ ⎣⎝ x ⎠ ⎝ y ⎠ ⎝ z ⎠ ⎦ If the box is a cube, then Lx = Ly = Lz = L, and the energy becomes 2 2 π 2 2 2 E = n +n +n . (14) nxnynz 2mL2 ( x y z ) 32π 2 The ground state energy is then E = . 111 2mL2 Degeneracy For the next higher energy level up from the ground state, there are 3 distinct wave functions or quantum states of the cubical box that have this energy:ψ 211, ψ 121, ψ 112 . These quantum states are distinct because their probability densities are different. (E.g., 2 2 2 ⎛ 2π x⎞ 2 ⎛ π y⎞ 2 ⎛ πz⎞ ψ = A sin sin sin ≠ 211 ⎝⎜ L ⎠⎟ ⎝⎜ L ⎠⎟ ⎝⎜ L ⎠⎟ 2 2 2 ⎛ π x⎞ 2 ⎛ 2π y⎞ 2 ⎛ πz⎞ 2 A sin sin sin = ψ .) The occurrence of more than one distinct ⎝⎜ L ⎠⎟ ⎝⎜ L ⎠⎟ ⎝⎜ L ⎠⎟ 121 quantum states (wave functions) with the same energy is called degeneracy. In the case of the particle in a rigid, cubical box, the next-lowest energy level is three-fold degenerate. 32π 2 E = E = E = . (15) 211 121 112 mL2 . Note that the ground state is non-degenerate. Further, the three-fold degeneracy of the first- excited state is removed if Lx ≠ Ly ≠ Lz . Degeneracy is intimately connected to symmetry: the 3 greater the symmetry of the system, the higher the degeneracy of the quantum states. [See www.falstad.com for simulations.] Contour Maps Contours represent lines of equal probability density. See the contour maps for the 2-D rigid box in the Taylor et al. textbook. The Three Dimensional Central-Force Problem Ultimately, our aim is to solve the TISE for an atomic system. The simplest atomic system is a one-electron atom in which a single electron is bound to the nucleus. In this case, the potential energy kZe2 U(x, y,z) = U(r) = − . (16) r The potential energy is a function of only the radial distance –e r. The corresponding force is directed along – r . Such a force (in this case, the electrostatic force) is called a central force. +Ze Note that U is spherically symmetric since it depends only on r. Thus, classically, in a one-electron atomic system in which no external forces act, one does not expect the energy E to depend on the angular position of the electron, only on the radial distance from the nucleus. Because of the spherical symmetry, the solution to the TISE is tractable if we use spherical polar coordinates rather than Cartesian coordinates. In the spherical coordinate system, the coordinates are r, θ, andφ , where r is the radial distance, θ is the polar angle, and φ is the azimuthal angle. For a spherically symmetric potential energy z U(r), the TISE cannot be solved using P separation of variables in Cartesian coordinates. In spherical coordinates, however, the TISE is r separable, and this is the solution we shall z = r cosθ θ study. It turns out that even though the TISE is y separable in spherical coordinates, the solution is complicated and we shall not go into all the φ mathematical details of the solution. Rather, we x = r cosθ cosφ shall concentrate on the physics. x y = r cosθ sin φ In spherical coordinates, the TISE for a spherically symmetric potential is given by 2 ⎛ 1 ∂2 1 ∂ ∂ψ 1 ∂2ψ ⎞ (r ) ⎛sin ⎞ U (r) E , (17) − ⎜ 2 ψ + 2 ⎜ θ ⎟ + 2 2 2 ⎟ + ψ = ψ 2µ ⎝ r ∂r r sinθ ∂θ ⎝ ∂θ ⎠ r sin θ ∂φ ⎠ 4 where ψ ≡ ψ (r,θ,φ)and the mass is taken to be the reduced mass µ . [Recall that this accounts for the slight motion of the nucleus.] Using separation of variables, we seek a solution of the form ψ (r,θ,φ) = R(r)Θ(θ)Φ(φ). (18) Substituting in the TISE and separating yields 1 d 2Φ = −m2 , (19) Φ dφ 2 l 2 1 d ⎛ dΘ⎞ ⎡ ml ⎤ (20) ⎜ sinθ ⎟ + ⎢l(l + 1) − 2 ⎥Θ = 0, sinθ dθ ⎝ dθ ⎠ ⎣ sin θ ⎦ and d 2 2µ ⎡ l(l +1)2 ⎤ (rR) U (r) E rR 0, (21) 2 − 2 ⎢ + 2 − ⎥ = dr ⎣ 2µr ⎦ 2 where in Eq. (19), the separation constant is −m and in Eq. (20), the separation constant is l(l + 1) . Note that neither the θ nor the φ equation contains U(r).