# Why must E exceed Vmin(x) for normalizable solutions?

1. Nov 2, 2015

### RicardoMP

Hi! I've been studying the time-independent Schrödinger equation and the infinite square well and was faced with this problem from Griffith's "Introduction to Quantum Mechanics". Rewriting the equation this way $$\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V(x)-E]\psi$$, I have to show that E must exceed the minimum value of V(x), otherwise $$\psi$$ and its second derivative will always have the same sign, which would compromise its normalization. Why is that? I've been thinking for awhile and can't come up with a good proof that such function cannot be normalized.

2. Nov 2, 2015

### jfizzix

As part of an answer, we could imagine the following.

If the function $\psi(x)$ and its second derivative $\psi''(x)$ had the same sign, then what would happen at a critical point $x_0$ where the first derivative $\psi'(x_0)=0$?

If $\psi(x)$ were positive at this critical point, then $\psi''(x)$ would have to be as well, so the wavefunction would be concave up.
Since $\psi''(x_0)>0$, the values of $\psi(x)$ near $x_0$ would be larger than $\psi(x_0)$, and $\psi''(x)$ would never change sign.

As a result, the wavefunction would just get larger and larger farther away from the critical point. Since the wavefunction couldn't reach zero for extreme values of $x$, the wavefunction would not be normalizeable.

For a more complete proof, you could also consider whether a wavefunction could exist that had no critical points, and also what happens if the critical point(s) has a negative or zero sign.

3. Nov 6, 2015

### RicardoMP

I see! Makes perfect sense! Thank you very much!

4. Nov 7, 2015

### vanhees71

It's much easier to understand in the operator formalism, i.e., not expressing it in terms of the position representation (wave functions).

$$\hat{H}=\hat{\vec{p}^2}{2m} + V(\hat{\vec{x}}).$$
If now $V$ is bounded from below, then so is the Hamiltonian, because the kinetic energy is positive semidefinite, and there's a ground state. Without loss of generality we can assume that the minimum value of $V$ is $0$. Assume there's an energy eigenstate with $E=0$, it must then be an eigenvector of $\hat{\vec{p}}$ with eigenvalues $\vec{p}=0$, but a momentum eigenstate cannot be a normalizable state, because the momentum operator has a purely continuous spectrum. Thus if there is a bound ground state, its energy eigenvalue cannot be 0 but must be >0.

5. Nov 7, 2015

### PeroK

Just to add to @jfizzix's answer. The following properties of a real-valued $\Psi$ are sufficient:

1) $\Psi$ is continuous on $\mathbb{R}$.

2) $\lim_{x \to \pm \infty} \Psi(x) = 0$

One way to start a formal proof is to note that, unless $\Psi = 0$, $\exists x_0$ such that $\Psi(x_0) \ne 0$. And, without loss of generality, you can assume $\Psi(x_0) > 0$. From there, you can use the intermediate and mean-value theorems to show the existence of a local maximum with a positive value - and that contradicts the "same-sign" property.

The case for a complex-valued $\Psi$ follows from the real-valued case.