Negative potential infinite square well

Question

A 1D finite square well is generally defined either by \begin{equation} V(x)=\begin{cases} 0\qquad -a\leqslant x\leqslant a\\ V_0\qquad \text{otherwise} \end{cases} \tag{1} \end{equation} or \begin{equation} V(x)=\begin{cases} -V_0\qquad -a\leqslant x\leqslant a\\ 0\qquad \text{otherwise} \end{cases} \tag{2} \end{equation}

Where $a>0$ and $V_0>0$. It is just a matter of defining the zero of your potential.

Infinite square well can be regarded as the limit $V_0\to+\infty$ of $(1)$, i.e.

\begin{equation} V(x)=\begin{cases} 0\qquad -a\leqslant x\leqslant a\\ +\infty\qquad \text{otherwise} \end{cases} \tag{3} \end{equation}

In this case the wavefunction is zero outside the well and you solve Schrödinger equation inside applying boundary conditions.

What about taking the limit of $(2)$? \begin{equation} V(x)=\begin{cases} -\infty\qquad -a\leqslant x\leqslant a\\ 0\qquad \text{otherwise} \end{cases} \tag{4} \end{equation}

Since we are now dealing with infinities I'm not sure whether saying $(3)$ and $(4)$ only differ for the choice of the of the potential. Also, what can I say about the wave function inside and outside in case $(4)$?

Answer 1

A good way to approach problems with "infinities" is to study/solve them for finite values and then take an appropriate limit. So, here, you could write $$V_\infty(x) = \lim_{V_0\to\infty} V_{V_0}(x) = \lim_{V_0\to\infty} \begin{cases} 0 & x \in [-a,a] \\ V_0 &\text{otherwise} \end{cases}$$ and solve the Schrödinger equation, and then take the limit.

This implies that you can carry over the properties of the solution of the Schrödinger equation for a non-infinite $V_0$. In particular the independence of the choice of the zero of the potential (e.g. the invariance under arbitrary shifts $V(x) \to V(x) + \mathrm{const}$), implying that (3) and (4) are equivalent. If you study the wavefunctions carefully, you will see that they acquire a constant phase as you do the potential shift. But this is phase is not observable anyway (recall that states are actually elements of a projective Hilbert space). Any observable properties will remain invariant under this potential shift, in particular the (non-) vanishing of the intensity $|\psi|^2$ inside/outside the well.