# Energies and numbers of bound states in finite potential well

Hello I understand how to approach finite potential well (I learned a lot in my other topic here). However i am disturbed by equation which describes number of states $$N$$ for a finite potential well ( $$d$$ is a width of a well and $$W_p$$ is potential ):

$$N \approx \dfrac{\sqrt{2m W_p}d}{\hbar \pi}$$

I am sure it has something to do with one of the constants $$\mathcal L$$ or $$\mathcal K$$ defined this way:

\begin{align} \mathcal L &\equiv \sqrt{\tfrac{2mW}{\hbar^2}} & \mathcal{K}&\equiv \sqrt{ \tfrac{ 2m(W_p-W) }{ \hbar^2 }} \end{align}

and the transcendental equations for ODD and EVEN solutions:

\begin{align} &\frac{\mathcal K}{\mathcal L} = \tan \left(\mathcal L \tfrac{d}{2}\right) &&-\frac{\mathcal L}{\mathcal K} = \tan \left(\mathcal L \tfrac{d}{2}\right)\\ &\scriptsize{\text{transc. eq. - EVEN}} &&\scriptsize{\text{transc. eq. - ODD}} \end{align}

QUESTION: Could anyone tell me where does 1st equation come from?

• The formulas you refer to come frome matching boundary conditions in the finite square well. – user97139 Oct 31 '15 at 11:50

Inside the well, the wave functions of a bound state behave approximately like $\sin(kx)$ i.e. standing waves where $k=\pi M/d$ for an integer $M$. This contributes the kinetic energy $\hbar^2 k^2/ 2m$. The highest-lying bound states are those for which the energy left to the particle is $0-$, a small negative number, outside the well.

So the kinetic energy inside the well must be approximately equal to the height of the well $W_p$: $$W_p = \frac{\hbar^2 k^2}{2m}$$ This implies $$k = \frac{\sqrt{2mW_p}}{\hbar}$$ for the maximum allowed $k$ but I have mentioned that the spacing between the eigenstates in the $k$ space is $\pi/d$. So one has to divide the maximum $k$ above by $\pi/d$ to get the formula for $N$ you asked about.

In general, The number of bound states can be derived from Bohr-Sommerfeld quantization formula: $$\oint \sqrt(W-W_p(x))~ dx= \left(n+\gamma \right) \pi \approx N \pi$$, where N is number of bound states up to Energy $$W$$, and $$\gamma$$ is called Maslov Index. For Harmonic oscillator, It is $$1/2$$.

For a Finite square well of width $$d$$, and depth $$W_p$$ , The number of Bound states $$\color{blue}{\textrm{up to energy W }}$$ is given as,

$$N(W) \approx \pi^{-1} \left[\int _{-d/2}^{d/2} \sqrt(W+W_p) dx + \int _{d/2}^{-d/2} \sqrt(W+W_p) (-dx) \right ]\approx 2 \pi^{-1} \int _{-d/2}^{d/2} \sqrt(W+W_p) dx \color{\red}{ \approx \pi^{-1}\sqrt{W+W_p} ~d.}$$

$$\color{blue}{\textrm{ The "TOTAL" number}}$$ of Bound states can be found by setting $$W$$ equal to $$0$$ (because, the highest-lying Bound state of a potential well will be close to $$0$$, or very small negative number).

So "Total" number is given as:$$\color{\red}{ N \approx \pi^{-1}\sqrt{W_p} ~d.}$$

$$\color{green}{\textrm{Numerics}}$$: For the $$W_p=5.$$ and $$d=1.$$ Plot of transcendental equation (Last two equations of OP) for even and odd solutions as a function of $$W$$ are as,

$$\color{green}{\textrm{Which gives "ONE" bound states. }}$$

and, $$N \approx 0.712$$

Note: In general (after so- many calculations) we found a rule: the Total number of bound states for a square potential well are given as $$[N]+1$$, where [.] denotes the integer part} Reference.

$$\color{green}{\textrm{Hence, N gives "One" bound states.}}$$

I have the used: $$\hbar=1, 2m=1$$