Bound-State Counting

Bound-state counting asks a simpler question than solving the spectrum:

How many normalizable energy eigenstates does this potential support?

In one dimension, the answer is often accessible before the exact energies are known. Boundary conditions, node counting, graphical eigenvalue equations, and scaling estimates can all constrain the count. This page collects the practical undergraduate tools. Detailed WKB derivations and rigorous spectral bounds belong elsewhere.

Throughout this page, assume a real one-dimensional Hamiltonian of the form

$$\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x),$$

with a potential that approaches the continuum threshold at infinity. When $V(x)\to0$ as $\lvert x\rvert\to\infty$, bound states have $E\lt0$ and square-integrable wavefunctions.

What Must Be Counted

A bound state is a normalizable eigenstate:

$$\hat H\psi_n=E_n\psi_n, \qquad \int_{-\infty}^{\infty} \lvert\psi_n(x)\rvert^2\,dx\lt\infty.$$

For potentials approaching zero at infinity, the continuum begins at $E=0$. A bound state must lie below that threshold:

$$E_n\lt0.$$

Counting bound states therefore means counting the discrete eigenvalues below the continuum threshold, not merely counting oscillatory solutions inside the attractive region. Every candidate must satisfy the matching conditions and decay at infinity.

First Qualitative Rules

Several checks come before algebra:

• A deeper or wider attractive region usually supports more bound states.
• A larger particle mass lowers kinetic-energy cost and usually supports more bound states.
• The one-dimensional ground state is nodeless under ordinary assumptions.
• The $j$th bound state has $j$ interior nodes if counting begins with $j=0$.
• A new bound state enters from threshold with a very long tail.
• For symmetric wells, parity alternates between even and odd states.

The node and parity logic is developed conceptually in Qualitative Features of One-Dimensional Bound States. Here the emphasis is counting.

Finite Square Well Counting

For the symmetric finite square well

$$V(x)= \begin{cases} -V_0, & \lvert x\rvert\lt a,\\ 0, & \lvert x\rvert\ge a, \end{cases} \qquad V_0\gt0,$$

bound states satisfy

$$-V_0\lt E\lt0.$$

Use the standard dimensionless variables

$$z=qa, \qquad z_0=a\frac{\sqrt{2mV_0}}{\hbar}.$$

The finite-well page derives

$$\kappa a=\sqrt{z_0^2-z^2}.$$

The even bound states obey

$$z\tan z = \sqrt{z_0^2-z^2},$$

and the odd bound states obey

$$-z\cot z = \sqrt{z_0^2-z^2}.$$

The right-hand side is the upper half of a circle of radius $z_0$ in the $z$-$\kappa a$ plane. Counting roots becomes a graphical problem.

Graphical Root Intervals

The even equation has one root in each interval

$$j\pi\lt z\lt j\pi+\frac{\pi}{2}, \qquad j=0,1,2,\ldots,$$

provided the interval is reached before $z=z_0$. The odd equation has one root in each interval

$$j\pi+\frac{\pi}{2}\lt z\lt(j+1)\pi, \qquad j=0,1,2,\ldots,$$

again provided the interval lies below the cutoff $z_0$.

Thus, away from exact threshold values, the finite square well has approximately

$$N = \left\lfloor \frac{2z_0}{\pi} \right\rfloor+1$$

bound states.

This formula should be read with its caveat. If $2z_0/\pi$ is exactly an integer, the highest would-be state is at threshold and is not square-normalizable. In that exact case, the count is one smaller than the expression above. Small perturbations of the well move such a threshold state either into or out of the bound spectrum.

Thresholds

The first even state exists for every attractive finite square well in one dimension. Additional states appear at threshold when $z_0$ crosses half-integer multiples of $\pi$:

$$z_0=\frac{\pi}{2}, \quad \pi, \quad \frac{3\pi}{2}, \quad 2\pi, \ldots$$

At threshold, $\kappa\to0$, so the outside decay length

$$\ell_{\text{tail}}=\frac{1}{\kappa}$$

diverges. A newly born bound state is therefore spatially large. This is why near-threshold levels are sensitive to small changes in the potential.

Scaling With Depth, Width, And Mass

The finite-well strength parameter

$$z_0=a\frac{\sqrt{2mV_0}}{\hbar}$$

contains the main scaling information:

• increasing the half-width $a$ increases $z_0$ linearly;
• increasing the depth $V_0$ increases $z_0$ like $\sqrt{V_0}$;
• increasing the mass $m$ increases $z_0$ like $\sqrt{m}$;
• increasing $\hbar$ would reduce $z_0$, reflecting stronger wave effects.

The rough count

$$N\sim\frac{2z_0}{\pi}$$

says that the number of bound states is controlled by how many half-wavelengths can fit inside the attractive region before the continuum threshold is reached.

This scaling is often more useful than the exact finite-well formula. A wide shallow well and a narrow deep well can have the same $z_0$ and therefore similar counts, even though their wavefunctions and energies differ.

Node Counting As A Check

Once a numerical or graphical calculation claims $N$ bound states, the node pattern should match:

$$\psi_0:\;0\ \text{nodes}, \quad \psi_1:\;1\ \text{node}, \quad \ldots, \quad \psi_{N-1}:\;N-1\ \text{nodes}.$$

If a purported third bound state has no node, something is wrong: either the eigenstates were not sorted by energy, the boundary conditions were applied incorrectly, or the calculation is not solving the intended one-dimensional self-adjoint problem.

For an even potential, parity gives another check. The low-lying sequence usually alternates:

$$\text{even}, \quad \text{odd}, \quad \text{even}, \quad \text{odd}, \ldots$$

Singular potentials, constrained domains, and radial reductions require more care, but the node check is extremely reliable for ordinary one-dimensional wells.

Weak Attractive Wells

One dimension is special: under broad standard conditions, an arbitrarily weak purely attractive potential can support a bound state. The Delta-Function Potential is the cleanest example. For

$$V(x)=-g\delta(x), \qquad g\gt0,$$

there is always one bound state,

$$E=-\frac{mg^2}{2\hbar^2}.$$

This should not be overgeneralized. Potentials with repulsive parts, unusual boundary conditions, higher dimensions, or singular behavior can have different threshold rules. But it is a useful warning against the intuition that a finite minimum depth is always required.

Semiclassical Counting Preview

For a smooth attractive well, a semiclassical estimate counts phase-space area. At energy $E$, the local classical momentum in the allowed region is

$$p(x;E) = \sqrt{2m\left(E-V(x)\right)}.$$

The WKB counting rule for levels below $E$ has the schematic form

$$N(E) \approx \frac{1}{\pi\hbar} \int_{x_1(E)}^{x_2(E)} p(x;E)\,dx \quad \text{endpoint correction}.$$

For the total number of bound states in a well with continuum threshold $E=0$, a rough estimate is

$$N_{\mathrm{bound}} \sim \frac{1}{\pi\hbar} \int_{V(x)\lt0} \sqrt{2m\left[-V(x)\right]}\,dx.$$

This formula captures the scaling with mass, depth, and width. It does not determine the exact integer count near threshold, and it should not be blindly applied to hard walls, delta potentials, or discontinuous wells without modified endpoint phases.

For the finite square well, the estimate gives

$$\frac{1}{\pi\hbar} \int_{-a}^{a} \sqrt{2mV_0}\,dx = \frac{2z_0}{\pi},$$

which is precisely the leading part of the finite-well count. The extra order-one information comes from matching and endpoint behavior.

The full WKB derivation is part of Bohr–Sommerfeld Quantization.

Numerical Counting Workflow

For a general one-dimensional potential, a practical workflow is:

1. Identify the continuum threshold and shift energies consistently.
2. Choose a sufficiently large numerical interval so bound-state tails are negligible at the edges.
3. Discretize the Hamiltonian or use a spectral method.
4. Count eigenvalues below the threshold.
5. Check that the wavefunctions decay before the artificial boundary.
6. Verify node ordering and, when available, parity.
7. Repeat with a larger box or finer grid to ensure the count is stable.

Numerical boxes can create fake discrete levels above threshold. Those are discretized continuum states, not bound states. The decay check is what separates genuine bound states from box artifacts.

Common Mistakes

• Counting every root of a transcendental equation without checking the energy range.
• Forgetting that a threshold state with $\kappa=0$ is not square-normalizable on the line.
• Using the infinite-well count for a finite well without accounting for tails.
• Treating the semiclassical estimate as an exact integer formula.
• Forgetting that width changes the count more strongly than depth when depth is varied only modestly.
• Missing a weakly bound state because its tail is much longer than the plotted region.
• Counting discretized continuum states from a numerical box as physical bound states.

Where This Is Used

• Finite Square Well gives the canonical graphical root equations.
• Energy Scales in One Dimension summarizes the confinement, tail, and depth-width scales behind counting estimates.
• Delta-Function Potential illustrates the one-dimensional weak-attraction lesson.
• Double Delta Potential has a second bound state only above an explicit threshold.
• Qualitative Features of One-Dimensional Bound States explains node ordering, tails, and parity checks.
• Sturm–Liouville Theory gives the mathematical framework behind real eigenvalues and ordered nodes.
• Matrix Diagonalization gives a numerical route to stable counts.
• Bohr–Sommerfeld Quantization develops the semiclassical action-counting rule.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• S. Flügge, Practical Quantum Mechanics, Springer, 1999.

Exercises

1. A finite square well has $z_0=2.3$. Estimate the number of bound states.

Solution

Use the finite-well counting guide

$$N = \left\lfloor \frac{2z_0}{\pi} \right\rfloor+1.$$

For $z_0=2.3$,

$$\frac{2z_0}{\pi} = \frac{4.6}{\pi} \approx1.46.$$

Thus

$$N=1+1=2.$$

The two states are the ground even state and the first odd state.

2. What happens at the exact threshold $z_0=\pi/2$?

Solution

The guide formula gives

$$\left\lfloor \frac{2(\pi/2)}{\pi} \right\rfloor+1 =2.$$

But the second state is exactly at threshold, where the outside decay constant is $\kappa=0$. It is not square-normalizable on the line. Therefore the actual number of bound states at the exact threshold is $1$. Just above threshold, the odd state becomes genuinely bound.

3. If the width of a finite square well is doubled while $m$ and $V_0$ are fixed, how does the rough count change?

Solution

The strength parameter is

$$z_0=a\frac{\sqrt{2mV_0}}{\hbar}.$$

Doubling the half-width $a$ doubles $z_0$. Since the leading count scales like

$$N\sim\frac{2z_0}{\pi},$$

the number of bound states roughly doubles, up to the integer rounding and threshold caveats.

4. A numerical calculation finds four bound states in a regular one-dimensional well. How many interior nodes should the highest one have?

Solution

Ordering states from the ground state as $\psi_0,\psi_1,\psi_2,\psi_3$, the node rule gives $j$ interior nodes for $\psi_j$. The fourth bound state is $\psi_3$, so it should have three interior nodes.

5. Apply the semiclassical estimate to the finite square well and compare with the finite-well count.

Solution

For the square well,

$$V(x)=-V_0 \quad \text{for} \quad -a\lt x\lt a.$$

The threshold estimate gives

$$N_{\mathrm{sc}} \sim \frac{1}{\pi\hbar} \int_{-a}^{a} \sqrt{2mV_0}\,dx = \frac{2a\sqrt{2mV_0}}{\pi\hbar}.$$

Using

$$z_0=a\frac{\sqrt{2mV_0}}{\hbar},$$

this becomes

$$N_{\mathrm{sc}} \sim \frac{2z_0}{\pi}.$$

The finite-well graphical count is approximately

$$N = \left\lfloor \frac{2z_0}{\pi} \right\rfloor+1,$$

away from exact thresholds. The semiclassical estimate captures the leading scale but not the exact integer rounding.