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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1\!} be an eigenstate with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_1\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2\!} an eigenstate with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_2\!} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_2>E_1\!} . We also can set boundary conditions, where both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2\!} vanish at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\!} .This implies that
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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\!}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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\!}
Subtracting the second of these from the first and simplifying, we see that
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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\!}
If we now integrate both sides of this equation from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\!} to any position Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x'\!} and simplify, we see that
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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\!}
The key is to now let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x'\!} be the first position to the right of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\!} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1\!} vanishes.
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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\!\!}
Now, if we assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2\!} does not vanish at or between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x'\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2\!} must vanish at least once between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x'\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0\!} if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_2>E_1\!} .