Oscillation Theorem

Let us concentrate on the bound states of a set of wavefunctions. Let ${\frac {\hbar ^{2}}{2m}}=1$. Let $\psi _{1}\!$ be an eigenstate with energy $E_{1}\!$ and $\psi _{2}\!$ an eigenstate with energy $E_{2}\!$, and $E_{2}>E_{1}\!$. We also can set boundary conditions, where both $\psi _{1}\!$ and $\psi _{2}\!$ vanish at $x_{0}\!$.This implies that

$-\psi _{2}{\frac {\partial ^{2}\psi _{1}}{\partial x^{2}}}+V(x)\psi _{2}\psi _{1}=E_{1}\psi _{2}\psi _{1}\!$

$-\psi _{1}{\frac {\partial ^{2}\psi _{2}}{\partial x^{2}}}+V(x)\psi _{1}\psi _{2}=E_{2}\psi _{1}\psi _{2}\!$

Subtracting the second of these from the first and simplifying, we see that

${\frac {\partial }{\partial x}}\left(-\psi _{2}{\frac {\partial \psi _{1}}{\partial x}}+\psi _{1}{\frac {\partial \psi _{2}}{\partial x}}\right)=\left(E_{1}-E_{2}\right)\psi _{1}\psi _{2}\!$

If we now integrate both sides of this equation from $x_{0}\!$ to any position $x'\!$ and simplify, we see that

$-\psi _{2}(x'){\frac {\partial \psi _{1}(x')}{\partial x}}+\psi _{1}(x'){\frac {\partial \psi _{2}(x')}{\partial x}}=(E_{1}-E_{2})\int _{x_{0}}^{x'}\psi _{1}\psi _{2}dx\!$

The key is to now let $x'\!$ be the first position to the right of $x_{0}\!$ where $\psi _{1}\!$ vanishes.

$-\psi _{2}(x'){\frac {\partial \psi _{1}(x')}{\partial x}}=(E_{1}-E_{2})\int _{x_{0}}^{x'}\psi _{1}\psi _{2}dx\!\!$

Now, if we assume that $\psi _{2}\!$ does not vanish at or between $x'\!$ and $x_{0}\!$, 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 $x'\!$ and $x_{0}\!$ if $E_{2}>E_{1}\!$.