Oscillation Theorem

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Quantum Mechanics A
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The most fundamental equation of quantum mechanics; given a Hamiltonian \mathcal{H}, it describes how a state |\Psi\rangle evolves in time.
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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.\!

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