6.1 Quantum bound states

Source: Ideas from Computational Physics (Landau, Páez, Bordeianu), Ch. 6 intro–§6.1. Rewritten in original wording. Continues 5.9 Integration reference sketches. Next algorithm: 6.2 Bisection search.

Chapter 6 is about searching and fitting. Many physics problems do not hand you $x$ explicitly — they hand you a nonlinear condition $f(x)=0$ (or a table of measurements) and ask you to recover a parameter. Two themes run through the chapter:

1. Trial-and-error search — guess a value, evaluate a residual, improve the guess (bisection, Newton, …).
2. Fitting — later sections interpolate or fit tabulated data (Lagrange §6.5, splines §6.6).

The first concrete search problem is a finite square well: find the allowed energies by solving a transcendental matching condition.

The square-well eigenvalue problem

A particle in a finite square well of depth $V_0$ and half-width $a$ has even and odd bound stationary states. Continuity of $\psi$ and $\psi'$ at the wall produces a transcendental equation for the bound energy $E$.

Work in the book's convenient units

$$\hbar=1,\qquad 2m=1,\qquad a=1,\qquad V_0=10,$$

Bound energies sit below the continuum; write $E=-E_B$ with binding energy $0<E_B<V_0$. In what follows $E$ means that positive $E_B$ (book notation), so $0<E<V_0=10$. Define

$$\kappa=\sqrt{V_0-E}=\sqrt{10-E}.$$

Even parity states obey

$$\sqrt{10-E}\,\tan\sqrt{10-E}=\sqrt{E},$$

or the algebraically equivalent residual (same zeros, fewer $\tan$ poles to trip over)

$$g(E)=\sqrt{E}\,\cot\sqrt{10-E}-\sqrt{10-E}=0.$$

Odd parity uses the companion matching condition

$$\sqrt{10-E}\,\cot\sqrt{10-E}=\sqrt{E},$$

or the residual form (same zeros)

$$g_{\mathrm{odd}}(E)=\sqrt{E}\,\tan\sqrt{10-E}-\sqrt{10-E}=0.$$

(Even used $\cot$; odd flips to $\tan$ — poles sit in different places, so pick a fresh bracket.)

The numerical job is the same: find roots of a continuous residual on a safe bracket. There is no closed-form $E$ — evaluate $g$ at trial energies and hunt for sign changes (bisection in §6.2).

Drag $E$ in the explorer until the even residual crosses zero — that energy is a candidate bound state for $V_0=10$.

A safe even-root bracket

For $V_0=10$, the even residual $g(E)$ changes sign on

$$[8.0,\,8.8]:\qquad g(8.0)<0,\quad g(8.8)>0.$$

Halving that interval until the width is tiny yields the bound energy

$$E_{B}\approx 8.5927852752.$$

(Use $\cot\theta=\cos\theta/\sin\theta$ with math.sin / math.cos. Near $\sin\kappa=0$ the residual has poles — stay inside a bracket that already has opposite signs.)

Deepening the well to $V_0=20$ adds more even roots (near $\sim 6.11$ and $\sim 18.36$). The same $g(E)$ with $10\to 20$ and a fresh bracket finds them.

Takeaway

Parity	Matching idea	Residual to root-find
Even	$\kappa\tan\kappa=\sqrt{E}$	$g=\sqrt{E}\,\cot\kappa-\kappa$
Odd	$\kappa\cot\kappa=\sqrt{E}$	$g_{\mathrm{odd}}=\sqrt{E}\,\tan\kappa-\kappa$
Either	No closed $E$	Bracket → bisect (§6.2) → check $\|g\|$

Here $\kappa=\sqrt{V_0-E}$ (with $V_0=10$ in the explorer).

Quiz

Bound states and search

1 / 4

In units ħ=1, 2m=1, a=1, V₀=10, the even residual we root-find is…

Pick an answer to see if you are right.

Next: 6.2 Bisection search — turn the sign-change into a reliable algorithm. Labs follow in 6.2b.