Is this the reason for vanishing of wavefunction beyond the infinite walls?

Merzbacher in his Quantum Mechanics says that for the "particle in a box" potential ($$V(x) = 0$$ for $$|x|\le L$$ and $$+\infty$$ otherwise),

Since the expectation value of the potential energy must be finite, the wavefunction must vanish within and on the walls of the box.

However, I don't quite get this reasoning. Why must the potential energy's expectation value be finite?

• My answer to a related question might be relevant: physics.stackexchange.com/a/587666/156895 Nov 1, 2020 at 14:36
• @J.Murray Thanks, that was enlightening!
– Atom
Nov 1, 2020 at 14:41

Let's first write the Time-independent Schrödinger equation

$$\hat{H}|\Psi\rangle=E|\Psi\rangle$$

or

$$\left(\frac{\hat{P}^2}{2m}+\hat{V}(x)\right)|\Psi\rangle=E|\Psi\rangle$$ or $$\langle\Psi|\left(\frac{\hat{P}^2}{2m}+\hat{V}(x)\right)|\Psi\rangle=E\langle\Psi|\Psi\rangle=E$$ or $$\langle\Psi|\frac{\hat{P}^2}{2m}|\Psi\rangle+\langle\Psi|\hat{V}(x)|\Psi\rangle=E$$ or $$\langle\Psi|\hat{V}(x)|\Psi\rangle=E-\frac{1}{2m}\langle\Psi|\hat{P}^2|\Psi\rangle$$

On a right hand side, You have a finite quantity and so you will expect $$\langle\Psi|\hat{V}(x)|\Psi\rangle < \infty$$ This can further be written as ( in position basis) $$\langle \hat{V}(x) \rangle=\int dx\psi^*(x)V(x)\psi(x)<\infty$$

Now for interval $$x\in [-L,L]$$, $$\langle V(x) \rangle$$ is zero as $$V(x)=0.$$ For other value of $$x$$, $$V(x)=\infty$$ and so $$\psi(x)$$ must be zero otherwise $$\langle V(x) \rangle=\infty$$.

Thus the wavefuntion (position ) must vanish outside the well. Cheers :)

• But why should $\langle\Psi | \hat P^2 | \Psi\rangle$ be finite?
– Atom
Nov 1, 2020 at 14:18
• We know that that expectation values follow the classical mechanics. So if your particle has a finite energy $E$ then it's obvious that particle must have some part of it as kinetic energy and other as potential. In either case, your kinetic energy ( and so $\langle \Psi |\hat{P}^2|\Psi\rangle$) will be finite. Nov 1, 2020 at 14:23

Young Kindaichi already provided a response to your question. What I add here is an alternative way of justifying the vanishing of the wave function in a region of infinite potential.

Let's start with some general discussion. Consider the Hamiltonian eigenvalue equation in a region of constant potential $$V$$ in the position representation:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V\psi(x)=E\psi(x).$$

For a particle with energy $$E>V$$, the general solution is:

$$\psi(x)=Ae^{ikx}+Be^{-ikx},$$ with $$k=\sqrt{2m(E-V)}/\hbar$$. For a particle with energy $$E, the solution is:

$$\psi(x)=Ae^{qx}+Be^{-qx},$$ with $$q=\sqrt{2m(V-E)}/\hbar$$. For a piece-wise potential made of constant regions, then you solve the constant potential problem in each region and match boundary conditions at the points between regions.

Going back to your question, a good way to understand the vanishing of the wave function where the potential becomes infinite is to consider this situation as a limit of the finite step potential:

$$V(x)= \begin{cases} V_0, &x>0 \\ 0, &x<0. \end{cases}$$

This is an example of a piece-wise potential discussed above, and let us call $$x<0$$ "region I" and $$x>0$$ "region II". Let the energy of our particle be $$0. In region I, the wave function reads:

$$\psi_I(x)=Ae^{ikx}+Be^{-ikx},$$

with $$k=\sqrt{2mE}/\hbar$$. In region II the solution is

$$\psi_{II}(x)=Ce^{qx}+De^{-qx},$$

with $$q=\sqrt{2m(V_0-E)}/\hbar$$.

Now let's turn to boundary conditions. We must set $$C=0$$, as otherwise $$\psi_{II}(x)$$ is not bounded as $$x\to\infty$$. This means that the solution becomes: $$\psi_{II}(x)=De^{-qx}$$. Applying boundary conditions at $$x=0$$ gives:

$$\begin{cases} \frac{B}{A}=\frac{k-iq}{k+iq},\\ \frac{D}{C}=\frac{2k}{k+iq}, \end{cases}$$

This is the general solution for a step potential.

We can now recover your situation of an infinite barrier by taking the limit $$V_0\to\infty$$. This gives $$q\to\infty$$ such that:

$$\begin{cases} \frac{B}{A}=\frac{k-iq}{k+iq}\Longrightarrow B=-A,\\ \frac{D}{A}=\frac{2k}{k+iq}\Longrightarrow D=0. \end{cases}$$

As $$D=0$$, the wave function vanishes in the region $$x>0$$ for an infinite potential.