Problem 2.2 of Griffiths' Intro to Quantum Mechanics states that

"$E$ must exceed the minimum value of $V(x)$ for every normalizable solution to the time-independent Schrödinger equation."

The problem has rearranged the TISE slightly, to show the form of the ODE:

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

I saw a solution online that made the claim that if $V_{min} > E$, or $E < V_{min}$ then our general solution is (with $k>0$):

$$\frac{d^2 \psi}{dx^2}=k\psi$$


I find this to be an odd argument since $V$ is a function of $x$. If we make this restriction I'm not sure how we can generally prove the problem statement.

Anyways - assuming a constant $V$, then integrating $|\psi|^2$ from $-\infty$ to $\infty$ diverges, regardless of the coefficients. So presumably this is the reason $E$ must be greater than $V_{min}$.

I saw another video which states that we only need to know that the curvature is positive (i.e. $V(x) > E$) in order to know that it is not normalizable.

These arguments seems weak to me since we could always make an infinite potential well, where for some defined region such as $0\leq x \leq a$ the potential is zero. Then the integral is finite within that region and zero outside the well, and therefore the wavefunction is normalizable.

Take another case where $E > V_{min}$ and $V_{min} = 0$. Without defining the well, we know the general form of $\psi$:

$$\psi(x) = c_1 sin(\sqrt{k}x) + c_2 cos(\sqrt{k} x)$$

Integrating $|\psi|^2$ from $-\infty$ to $\infty$ won't work either. It's only when we introduce the infinite potential well that it is normalizable.

Are the above answers sufficient? What am I missing to be able to generally prove the statement of the problem, that $E > V_{min}$ for all normalizable solutions?


2 Answers 2


It is just matter of math. Even if one may also try to find physical interpretations of this result a posteriori.

Let us focus on the identity $$\frac{d^2 \psi}{dx^2}=\frac{2m}{\hbar^2}[V(x)-E]\psi\:.$$

Suppose that $E< V_{min}$ and there is a normalized eigenfunction $\psi$ with eigenvalue $E$. In particular, $$\frac{\hbar^2}{2m} \psi(x)^*\frac{d^2 \psi}{dx^2} = (V(x)-E) \psi(x)^*\psi(x) \geq 0.$$ Therefore the integral of it is positive. The integral is finite by hypothesis because the vector is assumed to stay in the domain if kinetic energy operator $-\frac{\hbar^2}{2m}\frac{d^2 }{dx^2}$ and thus the integral of the left-hand side is finite. Notice also that the integal of the righ-hand side cannot vanish, otherwise $\psi$ would be zero almost everywhere otherwise the vector would not be normalized to one.

On the other hand, since $\psi$ solves the initial eigenvalue Schroedinger equation, the found integral, up to a positive factor, should be equal to the integral $$\int \psi(x)^* \frac{d^2 \psi(x)}{dx^2}dx\:.$$ This integral cannot be positive, just by integrating by parts(*). We conclude that the said $\psi$ cannot exist.

(*) More rigorously, we are assuming that $\psi\in D(T)$ where $T=PP/2m$ is the kinetic energy observable. Therefore $$\langle \psi|T\psi\rangle = -\frac{\hbar^2}{2m}\int \psi(x)^* \frac{d^2 \psi(x)}{dx^2}dx\:.$$ The derivatives are in the weak sense here. Since $D(T)\subset D(P)$ and the momentum $P$ is selfadjoint, $$\langle \psi|T\psi\rangle = \frac{\hbar}{2m}\langle \psi|PP\psi\rangle = \frac{\hbar}{2m}\langle P\psi|P\psi\rangle \geq 0.$$

  • $\begingroup$ Why is it not possible for $(-\int \frac{d\psi^{*}}{dx} \frac{d\psi}{dx}dx)$ to be positive? $\endgroup$
    – michael b
    Sep 22, 2022 at 22:39
  • $\begingroup$ The only way I know to reason this is on the physical grounds that $<T>$ (expectation of kinetic energy) is positive. Regardless, this is exactly the type of answer I was looking for. Thanks. $\endgroup$
    – michael b
    Sep 23, 2022 at 3:09
  • 2
    $\begingroup$ I see it now. Since $\frac{d\psi^{*}}{dx}=(\frac{d\psi}{dx})^{*}$ it's mathematically impossible for the expression I wrote in my first comment to be positive due to the fact that $zz^{*}=|z|^2$ $\endgroup$
    – michael b
    Sep 23, 2022 at 3:30
  • $\begingroup$ Yes, that is the reason. $\endgroup$ Sep 23, 2022 at 5:10

Think about a similar classical problem to understand this. Suppose you have a mass m that can be sitting on the ground anywhere along the $x$ axis. Suppose each point along the axis is at a different altitude.

Then you have a function, $V(x)$, that gives the potential energy of the mass if it is at $x$.

Suppose you know the total kinetic + potential energy, $E$, of the mass. You are asked to find where the mass can be. The answer is anywhere that $V(x) < E$.

Suppose that the ground is at high altitude everywhere. $V_{min} > E$. Then you can't solve the problem. The mass can't be anywhere with that $E$.

Things are a little different with quantum mechanics. There are solutions that are classically impossible. E.G. if $V(x) < E$ some places and $V(x) > E$ others, you can find the probability that the mass is in a region where $V(x) > E$.

But if $V_{min} > E$, you can't solve the problem. You can find solutions, but these solutions are not normalizable. That is, they do not describe a physically real situation.


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