quantum mechanics - Finite Square Well Inside an Infinite Square Well

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

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

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.: $$\frac{-k}{\sqrt{z_0 - k^2}} \coth(k(b-a)) ~.$$

Answer

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

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 $b0$(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.