# Oscillation Theorem

Let $\psi_1\!$ be a bound state of a given potential with energy $E_1\!$ and $\psi_2\!$ be another bound state with energy $E_2\!$ such that $E_2>E_1\!$. Let $x_0\!$ be a point at which both $\psi_1\!$ and $\psi_2\!$ vanish; this is guaranteed to happen at least for $x_0\to\pm\infty.\!$ We will now prove that, between any two points $x_0\!$ and $x_1\!$ at which $\psi_1\!$ vanishes, there must be at least one point at which $\psi_2\!$ vanishes. Let us begin by writing down the Schrödinger equation for each wave function:

$-\frac{\hbar^2}{2m}\frac{d^2 \psi_1}{dx^2}+V(x)\psi_1=E_1\psi_1\!$
$-\frac{\hbar^2}{2m}\frac{d^2 \psi_2}{dx^2}+V(x)\psi_2=E_2\psi_2\!$

Multiplying the first equation by $\psi_2\!$ and the second by $\psi_1,\!$ subtracting the second equation from the first, and simplifying, we see that

$\frac{d}{dx}\left(-\psi_2\frac{d\psi_1}{dx}+\psi_1\frac{d\psi_2}{dx}\right)=\frac{2m}{\hbar^2}\left(E_1-E_2\right)\psi_1\psi_2.\!$

If we now integrate both sides of this equation from $x_0\!$ to any position $x'\!$ and simplify, we see that

$-\psi_2(x')\frac{d\psi_1(x')}{dx}+\psi_1(x')\frac{d\psi_2(x')}{dx}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2\,dx\!$

The key is to now let $x'=x_1\!$ be the first position to the right of $x_0\!$ where $\psi_1\!$ vanishes.

$\psi_2(x_1)\frac{d\psi_1(x_1)}{dx}=\frac{2m}{\hbar^2}(E_2-E_1)\int_{x_0}^{x_1}\psi_1\psi_2\,dx\!$

Now, if we assume that $\psi_2\!$ does not vanish at or between $x_0\!$ and $x_1\!$, then it is easy to see that the left hand side of the previous equation has a different sign from that of the right hand side, and thus it must be true that $\psi_2\!$ must vanish at least once between $x_0\!$ and $x_1\!$ if $E_2>E_1.\!$