Ok here's a potential I invented and am trying to solve:

$$ V(x) = \begin{cases} -V_0&0<x<b \\ 0&b<x<a \\ \infty&x>a \\ \end{cases}$$

and $V(-x) = V(x)$ (Even potential)

I solved it twice and I got the same nonsensical transcendental equation for the allowed energies:
$$ \frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = \tan(b \sqrt{z_0-k^2}). $$

, where $k=\sqrt{-2mE}/\hbar$ and $z_0 = 2mV_0/\hbar^2$

The problem is that when I take the limit as $b→a$ (the ordinary infinite square well) I get a division by 0.

So is there something fundamentally wrong with trying to solve this potential? Is it wrong to have an Infinite potential and bury some of it under the 0 (negative potential) ? Note: I am solving it for negative energies (bound bound states?).

EDIT by CZ, for readability and analytic continuation of l.h.side, cf. Gilbert et al.: $$\frac{-k}{\sqrt{z_0 - k^2}} \coth(k(b-a)) ~. $$

  • $\begingroup$ When you take the limit of $b\rightarrow a$, it is not quite an ordinary infinite square well because the energy is below the bottom of the well. As you take that limit, the wave function approaches something like a delta function. $\endgroup$
    – George G
    Commented Jul 4, 2014 at 17:15
  • $\begingroup$ centered where? $\endgroup$
    – user120404
    Commented Jul 4, 2014 at 17:33
  • 1
    $\begingroup$ You must take the limit correctly, i.e. only at the end of all computations. It is, in general, wrong to assume that in intermediary steps you will have well defined quantities. This can happen when in the limit you are taking has some degeneracy issues with it. In this case you are collapsing different boundary conditions onto each other. A simple way to avoid problems is to put $b = a - \epsilon$ and expand in powers of $\epsilon$. to leading order the formula for k will contain an $\frac{1}{\epsilon}$ term, and then allows you to solve for $k$ $\endgroup$ Commented Jul 4, 2014 at 17:42
  • $\begingroup$ Would you please demonstrate what you mean because I don't understand how to apply what you are saying. $\endgroup$
    – user120404
    Commented Jul 4, 2014 at 18:51
  • 2
    $\begingroup$ @user120404 : You must have done an error, because supposing $V_0=0$, we are in the standard infinite square well, and we know that, in this case, $E$ has to be positive, so your definition of $k$ ($k=\sqrt{-2mE}/\hbar$) is not correct, and $\sqrt{z_0-k^2}= \sqrt{-k^2}$ makes no sense too. $\endgroup$
    – Trimok
    Commented Jul 5, 2014 at 11:42

2 Answers 2


The potential you have written is perfectly legitimate, let's do the derivation from scratch: $$ V(x) = \begin{cases} -V_0&0<x<b \\ 0&b<x<a \\ \infty&x>a \\ \end{cases}$$

and $V(-x) = V(x)$ the Hamiltonian will commute with the parity operator $[\hat{H},\hat{\mathscr{P}}] = 0$, therefore the eigenstates of the energy will be eigenstates of the parity operator too. Also, we can solve the problem in the interval $[0,a]$ with $b<a$ since the eigenfunctions of $\hat{H}$ are parity eigenstates too. Now and this is the most important thing: an infinite well can ONLY have bound states, so we have to look for bound states of the infinite well and adapt this fact to work with the finite well inside (which adimts "scattering" (positive energy) states). This basically means that $E>0$ (you can easily check that if you started from the condition to get the bound states inside the finite well you would get an impossible condition for the infinite one at the point $x=a$)

Region $[0,b]$:

The Shrodinger equation is $$-\frac{\hbar^2}{2m}\nabla^2\psi-V_0\psi= E\psi$$ with $V_0>0$ in one dimension the solution will be: $$\psi_+(x)=C_1 cos(\omega x)\\ \psi_-(x)=C_1 sin(\omega x)$$ we will work with parity $+$. for comodity. With $\omega=\sqrt{\frac{2m}{\hbar^2}(V_0+E)}$.

Region $[b,a]$:

Here the Shrodinger equation is: $$\frac{\hbar^2}{2m}\nabla^2\phi= E\phi \\ \phi(a)=0$$ notice that $\phi(a)=0$ is the foundamental condition that "tells" the particle that there is an infinite well, and this condition is also the reason that $E>0$ infact if $E<0$ you would have $\phi_+= A_1 \cosh(kx)$ which only has immaginary solutions for $\phi(a)=0$ or the useless $k=0$. The solution for $E>0$ here will be: $$\phi_+(x)=A_1 cos(k x)\\ \phi_-(x)=A_1 sin(k x)$$ with $k=\sqrt{\frac{2m}{\hbar^2}E}$, now since $\phi_{\pm}(a) = 0$ we have that $k$ must be: $$k_+=\frac{(2n+1)\pi}{2 a} \\ k_-=\frac{n\pi}{a} $$.

Now we have to match the solutions in $x=b$. (we have to match the derivatives too since the general solution must be $C^2$)

$$\psi_+(b)=\phi_+(b) \\ \psi'_+(b)=\phi'_+(b)$$ this leads to (for the parity even eigenstates):

$$C_1 cos(\omega b) = A_1 cos(k b)\\ C_1 \omega sin(\omega b) = A_1 k sin(k b)$$ The solution reads: $$\tan(\omega b) = \frac{k}{\omega} \tan(k b) $$ which can be written as : $$\tag 1 \omega \tan(\omega b) = \frac{(2n+1)\pi}{2a} \tan\left( \frac{2n+1}{2a}\pi b \right)$$

Now, if you do $b \rightarrow a$ you get that $\tan\left(\frac{2n+1}{2}\pi\right) \rightarrow \infty$ which means that $cos(\omega b)=0$ which means that $\omega = \frac{2n+1}{2 b}\pi $ (also $k$ vanishes) which is the usual energy for the infinite well.

The limit $b \rightarrow 0$ is consistent too. Consider equation $(1)$ written as $$\omega=k \frac{\tan\left(kb\right)}{\tan\left(\omega b\right)}$$ now take the limit $b \rightarrow 0$ on both sides $$\ \omega = k\cdot \frac{k}{\omega}\longrightarrow \omega = k$$ and you are left with $k = \frac{(2n+1)\pi}{2 a}$ which is the usual infinite well energy.


I do not see the problems you appear to be worried about--perhaps others! To start with, with the infinite walls, all states are bound, with positive or negative energies. You might as well take your 0 of energy at the very bottom of the potential, and a step up at $V_0$, instead, the physics being the same. It would also spare you needless grief to work in units s.t. ħ=1 and m=1/2. In that case, there are 3 types of solutions, for $E<V_0$, $E=V_0$, and $E>V_0$. You might consult a very similar problem in Gilbert et al, EurJPhys 26 (2005) 815–825.

Your limit b → a appears fine: The left-hand side blows up, and so therefore the tangent on the r.h.side must too, so the cosine in the denominator must vanish, and bK = n π/2 for some odd n, while K is the unique "inner" wavenumber, K=nπ/2b, familiar from the infinite square well. Might choose to flip the sign of your energy as Trimok points out.

Other limits you may have to check involve b → 0 when the r.h.side vanishes, and the zero of the energy loses its significance, so the "outer" wavenumber can take an i in front of it, so the left hand side vanishes by Euler's formula, too.

The limit $V_0$ → 0 is more interesting, and with care should yield a constraint equation only depending on a, but you may have to tweak it and use more convenient variables to see that. Remember, the momentum at the outer, infinite, wall is discontinuous. Try to find just the ground state.

There is an even more interesting limit you should consider, with the energy flush with the top of the step, so k, E → 0 in your language, and $E=V_0$ in mine. The outer wavefunction is a straight line, e.g., $A(x-a)$ on the right, and you get the condition collapse to a condition between $a,b$ and $z_0$.


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