Introduction to Quantum Mechanics · Problem 2.61

Time-Independent Schrödinger Equation — David J. Griffiths

Physics — Quantum Mechanics · matrix_discretization_method

Problem

One way to obtain the allowed energies of a potential well numerically is to turn the Schrödinger equation into a matrix equation, by discretizing the variable xx. Slice the relevant interval at evenly spaced points {xj}\{x_j\}, with xj+1−xj≡Δxx_{j+1} - x_j \equiv \Delta x, and let ψj≡ψ(xj)\psi_j \equiv \psi(x_j) (likewise Vj≡V(xj)V_j \equiv V(x_j)). Then dψdx≈ψj+1−ψjΔx,d2ψdx2≈ψj+1−2ψj+ψj−1(Δx)2.(2.195)\frac{d\psi}{dx} \approx \frac{\psi_{j+1} - \psi_j}{\Delta x}, \quad \frac{d^2\psi}{dx^2} \approx \frac{\psi_{j+1} - 2\psi_j + \psi_{j-1}}{(\Delta x)^2}. \tag{2.195} (The approximation presumably improves as Δx\Delta x decreases.) The discretized Schrödinger equation reads −ℏ22m(ψj+1−2ψj+ψj−1(Δx)2)+Vjψj=Eψj,(2.196)-\frac{\hbar^2}{2m}\left(\frac{\psi_{j+1} - 2\psi_j + \psi_{j-1}}{(\Delta x)^2}\right) + V_j\psi_j = E\psi_j, \tag{2.196} or −λψj+1+(2λ+Vj)ψj−λψj−1=Eψj,where λ≡ℏ22m(Δx)2.(2.197)-\lambda\psi_{j+1} + (2\lambda + V_j)\psi_j - \lambda\psi_{j-1} = E\psi_j, \quad \text{where } \lambda \equiv \frac{\hbar^2}{2m(\Delta x)^2}. \tag{2.197} In matrix form, HΨ=EΨ(2.198)\mathsf{H}\Psi = E\Psi \tag{2.198} where (letting vj≡Vj/λv_j \equiv V_j/\lambda) H≡λ(⋱−1(2+vj−1)−1000−1(2+vj)−1000−1(2+vj+1)−1⋱)(2.199)\mathsf{H} \equiv \lambda \begin{pmatrix} \ddots & & & & \\ -1 & (2+v_{j-1}) & -1 & 0 & 0 \\ 0 & -1 & (2+v_j) & -1 & 0 \\ 0 & 0 & -1 & (2+v_{j+1}) & -1 \\ & & & & \ddots \end{pmatrix} \tag{2.199} and Ψ≡(⋮ψj−1ψjψj+1⋮)(2.200)\Psi \equiv \begin{pmatrix} \vdots \\ \psi_{j-1} \\ \psi_j \\ \psi_{j+1} \\ \vdots \end{pmatrix} \tag{2.200} (what goes in the upper left and lower right corners of H\mathsf{H} depends on the boundary conditions, as we shall see). Evidently the allowed energies are the eigenvalues of the matrix H\mathsf{H} (or would be, in the limit Δx→0\Delta x \to 0). Apply this method to the infinite square well. Chop the interval (0≤x≤a0 \le x \le a) into N+1N + 1 equal segments (so that Δx=a/(N+1)\Delta x = a/(N+1)), letting x0≡0x_0 \equiv 0 and xN+1≡ax_{N+1} \equiv a. The boundary conditions fix ψ0=ψN+1=0\psi_0 = \psi_{N+1} = 0, leaving Ψ≡(ψ1⋮ψN).(2.201)\Psi \equiv \begin{pmatrix} \psi_1 \\ \vdots \\ \psi_N \end{pmatrix}. \tag{2.201} (a) Construct the N×NN \times N matrix H\mathsf{H}, for N=1N = 1, N=2N = 2, and N=3N = 3. (Make sure you are correctly representing Equation 2.197 for the special cases j=1j = 1 and j=Nj = N.) (b) Find the eigenvalues of H\mathsf{H} for these three cases "by hand," and compare them with the exact allowed energies (Equation 2.30). (c) Using a computer (Mathematica's Eigenvalues package will do it) find the five lowest eigenvalues numerically for N=10N = 10 and N=100N = 100, and compare the exact energies. (d) Plot (by hand) the eigenvectors for N=1N = 1, 2, and 3, and (by computer, Eigenvectors) the first three eigenvectors for N=10N = 10 and N=100N = 100.

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