# Width of a finite potential well that has exactly 7 bound states

Currently we are working on finding solutions for the one-dimensional time-independent Schrodinger equation where we have a particle, in our case a nucleon, in a finite potential well. We arrived at the following relationship to determine the potential that would give rise to exactly $N$ bound states in a well of width $L$ although we were more given this rather then shown where this comes from.

$$\frac{\hbar^2(N-1)^2\pi^2}{2mL^2}<V_0<\frac{\hbar^2N^2\pi^2}{2mL^2}$$

I have done some reading on this relationships comes from but I am still slightly fuzzy on it so any clarification would be helpful. However, my main question is how to find $L$ given some potential and 7 bound states. If I isolate $L^2$ in the above equation I arrive at the following.

$$\frac{\hbar^2(N-1)^2\pi^2}{2mV_0}<L^2<\frac{\hbar^2N^2\pi^2}{2mV_0}$$

When I plug in our values ($V_0 = 40\space MeV$, $N = 7$, and $m = 938.9265\space MeV$) my solution for $L$ comes out in units of seconds. I was informed that I could arrive at an expression with $mc^2$ in it which would return a solution in units of meters however I am not sure how to go about this. I think my issue arises from the fact that I'm not fully aware of where this relationship comes from but it may just be that I am missing something obvious so any input would be appreciated.

• Your mass is given in units of energy. This is common in high-energy physics, and in order to get the mass in the units of mass you have to divide by $c^{2}$. – eranreches Nov 12 '17 at 19:11
• You seem to be using the eigenvalues of an infinite square well in a situation with a finite square well; this is OK as an initial approximation but it will only be good as a guideline and it does not accurately represent the finite square well. (That said, the finite square well is an approximation to begin with, so there's only so much accuracy to be had in any case.) – Emilio Pisanty Nov 12 '17 at 19:29

The biggest issue is that your eigenvalue equation is that of an infinite well and so you have no chance of supporting finitely many bound states.

To proceed you need to use a symmetric finite well $$V(x)=\left\{ \begin{array}{cc} -V_0&-a\le x\le a\\ 0&\vert x\vert>a\, . \end{array}\right. \tag{1}$$ The solutions break nicely into even and odd parity solutions, which are found by numerically solving a transcendental equation. The lowest energy (ground) state will be even parity, and so will the third, fifth and 7th. The odd parity states will be the second, 4th, 6th, 8th etc. in energy.

If $V_0$ is fixed, you will need to adjust the width $2a$ so the eight bound state, i.e. the fourth solution with odd parity, does not appear while the 7th bound state (i.e. the fourth even solution) does. This will actually give you a range of possible width $2a$.

Note: a possibly more realistic scenario would be to use a 3D rectangular well. This will cut down on the number of distinct solutions you need to search as the solutions will be identical along all 3 directions. However, this comes with counting problems.

If you want to get fancy, you might want to use a spherical 3D well with $r$ replacing $x$ in (1), $r\ge 0$ and an infinite wall at $r=0$. The solutions for a finite spherical well use spherical Bessel and Haeckel functions and are not so easy to find, although the procedure is basically the same as the Cartesian well.

Each of these (more realistic) 3D extensions comes with energy degeneracy. If $$\Psi_{n_xn_yn_z}(x,y,z)=\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)$$ is a Cartesian solution, then the ground state will have $n_x=n_y=n_z=1$, the next excited states have $(n_x,n_y,n_z)=(2,1,1),(1,2,1)$ or $(1,1,2)$ so that’s $4$ states up to there, and the next set will have $(n_x,n_y,n_z)=(2,2,1),(2,1,2)$ or $(1,2,2)$ to give you exactly $7$ states. Now, if you well is wide enough to accommodate a solution along $x$ for $n_x=2$, then the state $(n_x,n_y,n_z)=(2,2,2)$ can also occur and your width would accommodate $8$, not $7$ states. Including spin does not help: if you assume each $(n_x,n_y,n_z)$ level can accommodate two spin states, you get into the same kind of trouble as soon as you allow any $n=2$ solution.

This problem does not disappear in the spherical case. The ground state will have $n=1,\ell=0$ and will accommodate $2$ states (accounting for spin). The next level is an $n=1,\ell=1$ level that will accommodate $2\times 3=6$ levels (again accounting for spin). I think the next after is $n=2, \ell=0$ again which would accommodate $2$ states: here again you cannot get only $7$ states.