Oscillation Theorem: Difference between revisions
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{{Quantum Mechanics A}} | {{Quantum Mechanics A}} | ||
Let us concentrate on the bound states of a set of wavefunctions | Let us concentrate on the bound states of a set of wavefunctions. Let <math>\psi_1\!</math> be an eigenstate with energy <math>E_1\!</math> and <math>\psi_2\!</math> an eigenstate with energy <math>E_2\!</math>, and <math>E_2>E_1\!</math>. We also can set boundary conditions, where both <math>\psi_1\!</math> and <math>\psi_2\!</math> vanish at <math>x_0\!</math>.This implies that | ||
:<math>-\psi_2\frac{\partial^2 \psi_1}{\partial x^2}+V(x)\psi_2\psi_1=E_1\psi_2\psi_1\!</math> | :<math>-\frac{\hbar^2}{2m}\psi_2\frac{\partial^2 \psi_1}{\partial x^2}+V(x)\psi_2\psi_1=E_1\psi_2\psi_1\!</math> | ||
:<math>-\psi_1\frac{\partial^2 \psi_2}{\partial x^2}+V(x)\psi_1\psi_2=E_2\psi_1\psi_2\!</math> | :<math>-\frac{\hbar^2}{2m}\psi_1\frac{\partial^2 \psi_2}{\partial x^2}+V(x)\psi_1\psi_2=E_2\psi_1\psi_2\!</math> | ||
Subtracting the second of these from the first and simplifying, we see that | Subtracting the second of these from the first and simplifying, we see that | ||
:<math>\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\!</math> | :<math>\frac{\partial}{\partial x}\left(-\psi_2\frac{\partial \psi_1}{\partial x}+\psi_1\frac{\partial \psi_2}{\partial x}\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\!</math> to any position <math>x'\!</math> and simplify, we see that | 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 | ||
:<math>-\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_2dx\!</math> | :<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}+\psi_1(x')\frac{\partial \psi_2(x')}{\partial x}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!</math> | ||
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. | 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{\partial \psi_1(x')}{\partial x}=(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!\!</math> | :<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}=\frac{2m}{\hbar^2}(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!\!</math> | ||
Now, if we assume that <math>\psi_2\!</math> does not vanish at or between <math>x'\!</math> and <math>x_0\!</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'\!</math> and <math>x_0\!</math> if <math>E_2>E_1\!</math>. | Now, if we assume that <math>\psi_2\!</math> does not vanish at or between <math>x'\!</math> and <math>x_0\!</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'\!</math> and <math>x_0\!</math> if <math>E_2>E_1\!</math>. |
Revision as of 13:46, 25 April 2013
Let us concentrate on the bound states of a set of wavefunctions. Let be an eigenstate with energy and an eigenstate with energy , and . We also can set boundary conditions, where both and vanish at .This implies that
Subtracting the second of these 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 .