Counting bound states in a narrow finite well: naive level counting versus the z0 criterion
A student models an electron trapped in a one-dimensional finite square well of width L = 1.0 nm and depth V_0 = 0.90 eV, with the bottom of the well at -V_0 and the top at 0. To decide how many bound states exist, she computes the infinite-square-well levels E_n = n^2 h^2/(8mL^2) and argues that only levels lying below the top of the well can be bound. Using hc = 1240 eV nm and mc^2 = 511 keV she obtains E_1 = 0.376 eV and E_2 = 1.50 eV, and therefore concludes that the well holds exactly one bound state. Her partner objects that "a finite well always has at least one bound state, and can hold more than your count", but cannot supply the criterion. The report must state the number of bound states and justify it with an explicit dimensionless parameter. Use hbar*c = 197.3 eV nm.
1. Verify the two infinite-well energies and explain why that calculation cannot by itself settle the number of bound states.
Read the worked answer
E_n = n^2 h^2/(8mL^2) = n^2 (hc)^2/(8 mc^2 L^2).
Denominator: 8 x 511000 eV x (1.0 nm)^2 = 4.088 x 10^6 eV nm^2. Numerator for n = 1: (1240)^2 = 1.538 x 10^6 eV^2 nm^2. So E_1 = 0.376 eV and E_2 = 4 x 0.376 = 1.50 eV.
The infinite-well levels are not the finite-well levels: the infinite walls push every level up, so the true finite-well energies lie below 0.376 eV and below 1.50 eV respectively. Since the second finite-well level could sit below 0.90 eV even though E_2^(infinite) = 1.50 eV lies above it, the comparison is not decisive in the direction the student assumes. The infinite-well count gives a lower bound on the number of bound states (any level with E_n^(infinite) < V_0 certainly has a finite-well counterpart below it), but it can undercount by one. A criterion built on the well itself is needed.
2. Define the parameter z_0 = (L/2)sqrt(2mV_0)/hbar, evaluate it for this well, and use the condition z_0 > n*pi/2 to determine the number of bound states.
Read the worked answer
Numerically, sqrt(2 m V_0) = sqrt(2 x 511000 eV x 0.90 eV)/c = sqrt(9.198 x 10^5) eV/c = 959 eV/c. Then (L/2) x 959 eV/c = 0.50 nm x 959 eV/c = 479.5 eV nm/c, and dividing by hbar*c = 197.3 eV nm gives z_0 = 2.43.
The successive bound states appear when z_0 crosses pi/2, pi, 3*pi/2, ... : one state exists for any z_0 > 0, a second once z_0 > 1.571, a third once z_0 > 3.142. Here z_0 = 2.43 lies between 1.571 and 3.142, so the well holds exactly two bound states, equivalently N = floor(2z_0/pi) + 1 = floor(1.547) + 1 = 2.
The second state is squeezed just below the top of the well. The student's answer of one is therefore an undercount of exactly the kind the infinite-well shortcut produces in narrow, shallow wells.
3. Assess the partner's claim that a finite well always has at least one bound state. Is it true as stated? Explain the physics and contrast with the three-dimensional case.
Read the worked answer
In one dimension the claim is true: however small V_0 and however narrow the well, the even-parity transcendental equation always has a solution, so at least one bound state survives. The physical reason is that reducing the well depth also lowers the kinetic-energy cost of spreading the wavefunction, and in 1D the spreading always wins; the parameter z_0 only needs to exceed zero.
In three dimensions the corresponding result fails: a spherical well must exceed a threshold depth (roughly set by V_0 a^2 > pi^2 hbar^2/(8m)) before it binds, because the centrifugal barrier and the r^2 weighting of the radial volume make spreading costly. So the partner is right about this 1D well but the statement must be qualified as one-dimensional; it should not be generalised.
The correct judgement for the report is: two bound states, justified by z_0 = 2.43 exceeding pi/2 but not pi, with the general 1D existence theorem quoted only as an explanation of why the count is never zero.