# Schrödinger equation – minimum energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.2.

We’ve seen that the time-independent Schrödinger equation can be separated into two ordinary differential equations, one in space and one in time. We’ve also seen that the solution of the spatial equation can be taken to be real. Starting with the spatial Schrödinger equation we can also derive a condition on the energy ${E}$.

$\displaystyle \frac{d^{2}\psi}{dx^{2}}=\frac{2m}{\hbar^{2}}(V-E)\psi \ \ \ \ \ (1)$

If ${E, the minimum value of the potential, then ${V-E>0}$ for all ${x}$. This means that ${\psi}$ and ${\psi"}$ have the same sign everywhere. If ${\psi}$ has a maximum, from elementary calculus we must have ${\psi"<0}$ so at the point of the maximum ${\psi}$ itself must be negative. Similarly, any minima of ${\psi}$ must occur where ${\psi}$ is positive. Therefore, ${\psi}$ cannot tend to 0 as ${x\rightarrow\infty}$ so it can’t be normalized. Thus any physically acceptable solution must have ${E>V_{min}}$. This condition is the analog of the classical condition that the energy (kinetic plus potential) of a particle can’t be less than the minimum of the potential. That is, a particle at rest at the bottom of a potential well has the lowest possible energy.

## 4 thoughts on “Schrödinger equation – minimum energy”

1. Pingback: Anonymous

2. Jack

You say “If {\psi} has a maximum, from elementary calculus we must have {\psi”<0} so at the point of the maximum {\psi} itself must be negative." The problem is that psi ~ Exp[-Abs[x]] is a counter example to the problem as stated and to your solution. You (and apparently Griffiths) assume continuity of dpsi/dx, which is not stated in the problem. My QM professor suggested R. Shankar pg 160, "Princ. of QM" as reference for a more rigorous treatment, but
I have not been able to read that yer. (Note also psi ~ Exp[-Abs[x]] is a bona fide solution, arising in the case of the negative delta function potential.)

Jack.

3. Jack

Your solution to the problem also seems to require the configuration space to be infinite — in order for the limit of psi as x-> infinity to be relevant. At a minimum a comment about compact (closed and bounded) configuration spaces (such as the infinite square well) would seem to be a necessary addition to your solution.
I’m beginning to wonder what a precise statement and solution to this problem might look like.

Jack

4. Rocky

Hello there.
Re: Jack, I think d\psi/dx should be guaranteed if not the Schrodinger equation does not make sense. (H\psi = E\psi, H=-h/2m * d^2/dx^2, how do you do second derivative when the first doesn’t exist? )
Re: OP, First i would like to thank for your solution. You said that \psi cannot go to 0 at infinity so theres not a minimum, and that does not contradict anything because a function with a point of inflection (f” = 0) can still be normalised, Just take \psi = exp[-kx], which is always positive, as an example.
I think the real reason for ( E \psi cannot be normalised ) is that it will make the coefficient K of d^2/dx(\psi) = K\psi positive. That would make the solution be \psi = Aexp[bx], b>0. Then \psi(infinite) = infinite.

Rocky