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?

  • $\begingroup$ The formulas you refer to come frome matching boundary conditions in the finite square well. $\endgroup$
    – user97139
    Commented Oct 31, 2015 at 11:50

2 Answers 2


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,

enter image description here

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

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: $\hbar=1, 2m=1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.