Oscillation Theorem: Difference between revisions
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{{Quantum Mechanics A}} | {{Quantum Mechanics A}} | ||
Let <math>\psi_1</math> be a bound state of a given potential with energy <math>E_1</math> and <math>\psi_2</math> be another bound state with energy <math>E_2</math> such that <math>E_2>E_1</math>. Let <math>x_0</math> be a point at which both <math>\psi_1</math> and <math>\psi_2</math> vanish; this is guaranteed to happen at least for <math>x_0\to\pm\infty.</math> We will now prove that, between any two points <math>x_0</math> and <math>x_1</math> at which <math>\psi_1</math> vanishes, there must be at least one point at which <math>\psi_2</math> vanishes. Let us begin by writing down the Schrödinger equation for each wave function: | |||
:<math>-\frac{\hbar^2}{2m} | :<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi_1}{dx^2}+V(x)\psi_1=E_1\psi_1\!</math> | ||
:<math>-\frac{\hbar^2}{2m} | :<math>-\frac{\hbar^2}{2m}\frac{d^2 \psi_2}{dx^2}+V(x)\psi_2=E_2\psi_2\!</math> | ||
Multiplying the first equation by <math>\psi_2</math> and the second by <math>\psi_1,</math> subtracting the second equation from the first, and simplifying, we see that | |||
:<math>\frac{ | :<math>\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.\!</math> | ||
If we now integrate both sides of this equation from <math>x_0 | If we now integrate both sides of this equation from <math>x_0</math> to any position <math>x'</math> and simplify, we see that | ||
The key is to now let <math>x'\!</math> be the first position to the right of <math>x_0\!</math> where <math>\psi_1\!</math> vanishes. | :<math>-\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\!</math> | ||
:<math>-\psi_2( | |||
The key is to now let <math>x'=x_1\!</math> be the first position to the right of <math>x_0\!</math> where <math>\psi_1\!</math> vanishes. | |||
:<math>-\psi_2(x_1)\frac{\partial \psi_1(x_1)}{\partial x}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x_1}\psi_1\psi_2\,dx\!</math> | |||
Now, if we assume that <math>\psi_2 | Now, if we assume that <math>\psi_2</math> does not vanish at or between <math>x_0</math> and <math>x_1</math>, 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 <math>\psi_2</math> must vanish at least once between <math>x_0</math> and <math>x_1</math> if <math>E_2>E_1.</math> |
Revision as of 15:51, 25 April 2013
Let be a bound state of a given potential with energy and be another bound state with energy such that . Let be a point at which both and vanish; this is guaranteed to happen at least for We will now prove that, between any two points and at which vanishes, there must be at least one point at which vanishes. Let us begin by writing down the Schrödinger equation for each wave function:
Multiplying the first equation by and the second by subtracting the second equation from the first, and simplifying, we see that
If we now integrate both sides of this equation from to any position and simplify, we see that
The key is to now let be the first position to the right of where vanishes.
Now, if we assume that does not vanish at or between and , 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 must vanish at least once between and if