core37 min studyLesson 380 in Physics

Counting bound states and approaching threshold

Use parity thresholds, nodal order, and admissible variational trials to count and diagnose weakly bound states.

Before you begin

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The question for today

What you’ll understand

Count finite-well bound states while excluding zero-energy threshold solutions.
Apply the one-dimensional nodal ordering under its assumptions.
Use a negative admissible variational energy as a sufficient binding test without claiming its converse.

Knowing one bound-state energy does not reveal how many bound states a Hamiltonian has. Counting requires both spectral ordering and careful treatment of the continuum threshold. A state can spread to arbitrarily large distances as its binding tends to zero, and its limiting solution may cease to be normalizable. Variational reasoning adds another tool: an admissible negative trial energy can certify binding even before the exact eigenfunction is known.

Count matching branches rather than plot crossings

For the symmetric well of depth $V_0$ and full width $2a$ , retain $z_0=a\sqrt{2mV_0}/\hbar$ . The even ground branch has a bound root for every $z_0>0$ . Additional branches enter as $z_0$ exceeds successive values $\pi/2,\pi,3\pi/2,\ldots$ . Their parity alternates odd, even, odd. At equality the newly arriving solution has zero exterior decay constant and must be excluded from the normalizable bound set. Crossing a tangent pole is not an extra eigenvalue.

The exact count for this finite square well is $\lceil2z_0/\pi\rceil$ for positive depth, with the equality convention built into the ceiling. If $2z_0/\pi$ is an integer, the next level is precisely at threshold and remains uncounted. This formula belongs to this full-line potential and its standard matching domain. A radial half-line problem, a repulsive core, or a different boundary condition can have different thresholds and cannot inherit the formula just because its graph resembles a well.

z_0=\frac{a\sqrt{2mV_0}}{\hbar}>0,\qquad N_{\rm bound}=\left\lceil\frac{2z_0}{\pi}\right\rceil,\qquad \kappa=\frac{\sqrt{2m|E|}}{\hbar},\quad \ell_{\rm tail}=\kappa^{-1}

N_bound counts negative-energy square-integrable eigenstates of the symmetric finite well, with no spin multiplicity. A newly arriving zero-energy solution is excluded. The tail length is an amplitude decay length in metres and diverges as binding vanishes; probability density has half that decay length.↔ Scroll wide equations horizontally.

Nodes order states, not arbitrary superpositions

For a regular real one-dimensional Schrödinger problem with a connected domain and standard separated self-adjoint boundary behaviour, bound eigenvalues are simple and the nth excited eigenfunction has n interior nodes when the ground label is zero. For a reflection-symmetric full-line well, parity alternates starting with an even nodeless ground state. These statements concern ordered eigenfunctions, not the node pattern of an arbitrary superposition and not a domain disconnected by an infinite barrier.

As an excited level approaches zero from below in a finite-range full-line well, its exterior amplitude falls ever more slowly. At exactly zero, the exterior equation has constant and linear solutions. No nonzero such solution is square-integrable on an infinite half-line. A bounded threshold solution can still influence scattering strongly, but it is not an additional ordinary bound eigenstate. A resonance is likewise a scattering feature with leakage or outgoing boundary conditions, rather than a normalizable negative-energy eigenfunction simply given a finite lifetime by terminology.

Use the variational test in its correct direction

For a semibounded self-adjoint Hamiltonian, the Rayleigh quotient of a normalized admissible trial state is at least the ground spectral energy. If a short-range potential has continuum threshold zero, a negative quotient proves spectral weight below that threshold and hence a bound level under the usual decaying-potential assumptions. A positive quotient for one trial proves no absence result. Nor does unconstrained minimization over nonnormalized functions provide an energy bound: the denominator or a normalization constraint is essential.

Trials must belong to the operator's quadratic-form domain when energy is evaluated through that form. For a finite potential this ordinarily requires a square-integrable weak derivative; an arbitrary jump at a well edge is inadmissible. For the attractive delta, a continuous exponential cusp is admissible in its form domain, and the contact energy must be included. This makes the variational principle useful without pretending that every trial already satisfies the exact eigenfunction's slope jump. Boundary physics and admissibility remain part of the estimate.

E_0\le\mathcal R[\psi]=\frac{q[\psi]}{\|\psi\|^2},\qquad q[\psi]=C\int_{-\infty}^{\infty}|\psi'|^2dx-g|\psi(0)|^2\quad\text{for }V=-g\delta(x),\quad C=\frac{\hbar^2}{2m}

The first relation is the variational upper-bound principle for the ground spectral energy of a semibounded self-adjoint Hamiltonian. For the stated attractive full-line contact model, the form domain is H¹(R). C is energy·length² and g is energy·length, so the normalized quotient is an energy.↔ Scroll wide equations horizontally.

Work it through

Count states without inventing a threshold level

A symmetric finite square well has $z_0=5\pi/4$ . Count its bound states and give their parities and interior node counts in increasing energy order. Then raise its strength to exactly $z_0=3\pi/2$ . Does the state count increase at that exact value?

At $z_0=5\pi/4$ , the count argument is $2z_0/\pi=5/2$ . Its ceiling is three. The thresholds $\pi/2$ and $\pi$ have already been passed, whereas $3\pi/2$ has not. No detailed energy root calculation is needed for this count.
Ordered regular one-dimensional bound eigenfunctions have zero, one, and two interior nodes. Reflection symmetry makes their respective parities even, odd, and even. These labels count distinct spatial eigenstates; no electron-spin factor or artificial left/right degeneracy is being added.
At exactly $z_0=3\pi/2$ , the argument is three and its ceiling remains three. The next odd solution is only at E = 0, with zero exterior decay constant. It becomes a fourth normalizable bound state only when the strength exceeds that threshold.

The result

There are three bound states at both specified strengths. They are even, odd, and even with zero, one, and two interior nodes. The next odd threshold solution is not a fourth square-integrable state at exact equality.

The living notebook

Explore the model

Interactive lab

Interactive lab · root-finding

Keep numerical localization separate from physical binding

Reset and set both controls to target c = 0.25 and upper bound U = 2; predict positive root 0.5. Reset and set both controls to c = 1 and U = 2; predict root 1, then Step and inspect the bracket bound. The quadratic lab always brackets one positive algebraic root; it does not count Schrödinger levels, handle tangent poles, or decide whether a threshold wavefunction is normalizable.

Quadratic / midpointCurrent bracketPositive root

Midpoint after 0 refinements: 1.5
Bracket error bound: 1.5
Computed residual · x² − c: 0.25

Target c2Initial upper bound U3

t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA small numerical root bracket is not a physical bound-state certificate; the energy threshold and wavefunction domain need separate checks.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The variational principle places the true ground spectral energy at or below the negative trial value. With a zero essential-spectrum threshold this certifies a bound level. It does not establish exactness, uniqueness, or that every other trial has negative energy; admissibility and normalization are crucial.

Your turn

Try it yourself

For an attractive delta Hamiltonian take $C=1\,\mathrm{eV\,nm^2}$ , $g=0.20\,\mathrm{eV\,nm}$ , and normalized form-domain trial $\psi_b(x)=b^{-1/2}e^{-|x|/b}$ with $b>0$ . Derive its Rayleigh energy and minimize over b. Explain what a negative value proves and why its cusp is permitted in this calculation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The kinetic-form integral is 1/b² and the contact density is 1/b; differentiate their weighted difference with respect to b.

Compare with the solution

Normalization gives $\int|\psi_b|^2dx=1$ , and the form gives $\mathcal R(b)=C/b^2-g/b$ . Setting its derivative to zero yields $b=2C/g=10\,\mathrm{nm}$ and $\mathcal R_{\min}=-g^2/(4C)=-0.010\,\mathrm{eV}$ . This negative admissible quotient certifies binding below the zero continuum threshold. In fact this exponential family contains the exact delta bound state, so the optimized value is exact here. Its continuous cusp is allowed in H¹(R), and the delta term is included explicitly; an arbitrary discontinuous trial would not be justified by this form-domain argument.

Keep the idea

Take it with you

Finite-well state counts exclude zero-energy nonnormalizable threshold solutions.
One-dimensional nodal ordering applies to eigenfunctions under stated domain assumptions.
A negative admissible variational quotient proves binding; a failed single trial proves no absence theorem.

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