§0.3The Hamiltonian as a Matrix

The Schrödinger operator only ever touches neighbouring points. That locality is why its matrix has three diagonals, why the energies come out real, and why the whole thing runs in a browser.

Here is the single most valuable thing on this site.

Chapters 3 through 6 of Phillips work out the energy levels of a particle in a box, then a finite well, then a barrier, then a harmonic oscillator — each with its own boundary conditions, its own transcendental equation, its own special functions. It is worth following that algebra, and the site will follow it.

But all four are the same computation, and it is four lines long:

Put $x$ on a grid. Build $\hat H$ as a matrix — a finite-difference second derivative plus $V(x)$ down the diagonal. Diagonalize it. The eigenvalues are the energy levels and the eigenvectors are the stationary states.

Learn it once here and every later chapter becomes a check on something you can already compute. Every interactive plot on this site is produced by this recipe, running live in your browser.

Where the matrix comes from

A state of definite energy — a stationary state stationary state Ψ = ψ(r)e^(−iEt/ħ): definite energy, ΔE = 0, and no observable property changes with time — ever. The wave function keeps rotating in the complex plane while nothing measurable moves. defined in ch. 4 — open in glossary — satisfies the time-independent Schrödinger equation, in which the Hamiltonian hamiltonian The total-energy operator Ĥ = −(ħ²/2m)d²/dx² + V(x). Discretized on a grid it is a real symmetric tridiagonal matrix, and its eigenvalues are the allowed energies. defined in the toolkit — open in glossary $\hat H$ is the total-energy operator operator A rule that takes a function and returns a function — d/dx, or "multiply by x". Written with a hat, Â. Discretize space and it becomes literally a matrix, so every question about an operator is a linear-algebra question. defined in the toolkit — open in glossary :

That is already an eigenvalue eigenvalue A value an observable can actually be measured to have — a solution λ of Âψ = λψ. Hermitian operators have real eigenvalues, which is exactly what lets them stand for measurement outcomes. defined in the toolkit — open in glossary equation — it says “applying $\hat H$ to $\psi$ gives back $\psi$, scaled”. The only obstacle is that $\hat H$ contains a derivative, and you cannot hand a derivative to numpy.linalg.eigh. So we replace it with something we can.

Substitute the stencil and the operator becomes a matrix. Writing $t = \hbar^2/2m(\delta x)^2$, the row for grid point $i$ reads

which is one row of $\mathbf{H}\boldsymbol\psi = E\boldsymbol\psi$ with

Almost every entry is zero. Only three diagonals survive, and that is the whole shape of the problem:

The four lines

To solve a different problem, change one line. Replace V = np.zeros(N) with V = np.where(abs(x) < w/2, -V0, 0) and you have the finite well of chapter 5. Replace it with V = 0.5*m*omega**2*x**2 and you have the harmonic oscillator of chapter 6. Nothing else moves.

Try it

Every curve below is computed by exactly the recipe above, live, as you drag.

Things worth doing before you move on:

• Shrink the finite well’s depth. Watch the top level climb towards zero and then vanish. The counter tells you how many bound states are left. Chapter 5 derives the condition for this; you can find it here by dragging.
• Switch to the infinite well. Levels go as $n^2$ — the gaps get wider.
• Switch to the harmonic oscillator. Levels are evenly spaced, and the lowest one is not at the bottom of the well. That leftover $\frac{1}{2}\hbar\omega$ is the zero-point energy, and chapter 6 shows it is forced by the uncertainty principle.
• Switch to the barrier and raise it. The levels pair up into doublets. The splitting inside each pair is a direct measure of the tunnelling rate through the wall.
• Count the nodes of any selected state. State $n$ crosses zero exactly $n$ times, in every potential. That is a theorem, not a coincidence.

Natural units, and how to get back

You will have noticed every number above is of order 1, with no $10^{-34}$ in sight. That is deliberate: we set $\hbar = m = 1$ and measure lengths in a unit $a$ we choose. Doing arithmetic with $10^{-34}$ and $10^{-31}$ in floating point is a good way to lose precision for nothing.

The conversion back is one line. With $\hbar = m = 1$ and lengths in units of $a$, one unit of energy is

Two things you get for free

The widget above reports “bound states: 3” the instant you move a slider, and it does so without computing a single eigenvalue. Both of the tricks behind that are worth knowing.

Counting without solving. For a symmetric tridiagonal matrix, one $O(n)$ sweep — a Sturm sequence — counts how many eigenvalues lie below any energy you name. Since a bound state is precisely an eigenvalue below the potential at infinity, “how many bound states does this well have?” is one cheap pass, no eigensolver required. It is also what makes the sliders feel instant: the site’s solver finds each level by bisecting on that count, which is $O(n)$ per level instead of $O(n^3)$ for the whole matrix.

The node theorem. The $n$-th eigenfunction eigenfunction A function an operator returns unchanged apart from a multiplying constant: Ĥψ = Eψ. The continuous analogue of an eigenvector, and the constant E is the eigenvalue. defined in ch. 4 — open in glossary of a one-dimensional problem crosses zero exactly $n$ times, always, in any potential. Count the wiggles and you know which level you are looking at. It is also a debugging tool: if your “ground state” has a node in it, your solver has converged to the wrong thing.