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

[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.

 

[jfizzix]

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.

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.

 

[RicardoMP]

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.

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

 

[vanhees71]

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

The Hamiltonian reads
$$\hat{H}=\hat{\vec{p}^2}{2m} + V(\hat{\vec{x}}).$$
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.

 

[PeroK]

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.

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.