Phy5645/Problem 1D sample: Difference between revisions
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:<math> | :<math> | ||
-\frac{\hbar^2}{2m} \left[ \frac{d^2\Phi(x)}{dx^2} \Delta(y) \Omega (z) + \Phi(x)\frac{d^2\Delta(y)}{dy^2} \Omega (z) + \Phi(x) \Delta (y)\frac{d^2\Omega(z)}{dz^2} \right] + \left[X(x)+Y(y)+Z(z)\right]\Phi(x) \Delta(y) \Omega (z) = E\Phi(x) \Delta(y) \Omega (z) | \begin{array} | ||
-\frac{\hbar^2}{2m} \left[ \frac{d^2\Phi(x)}{dx^2} \Delta(y) \Omega (z) + \Phi(x)\frac{d^2\Delta(y)}{dy^2} \Omega (z) + \Phi(x) \Delta (y)\frac{d^2\Omega(z)}{dz^2} \right] \\ | |||
& + \left[X(x)+Y(y)+Z(z)\right]\Phi(x) \Delta(y) \Omega (z) = E\Phi(x) \Delta(y) \Omega (z) | |||
\end{array} | |||
</math> | |||
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:<math> | :<math> | ||
\begin{array} | |||
-\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x) | -\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x) | ||
-\frac{\hbar^2}{2m} \frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2} + Y(y) | -\frac{\hbar^2}{2m} \frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2} + Y(y) \\ | ||
-\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) | & -\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) | ||
= E | = E | ||
\end{array} | |||
</math> | </math> | ||
Perfectly we can separate the right hand side in three parts, where only one depends of x, only one of y and only one of z. Then each of these parts must be equal to a constant. So: | Perfectly we can separate the right hand side in three parts, where only one depends of <math> x \!</math>, only one of <math> y \!</math> and only one of <math> z \!</math>. Then each of these parts must be equal to a constant. So: | ||
:<math> | :<math> | ||
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-\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) = E_z </math> | -\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) = E_z </math> | ||
where E_x, E_y and E_z are constant and <math> E = E_x+E_y+E_z \!</math> | where <math> E_x \!</math>, <math> E_y \!</math> and <math> E_z \!</math> are constant and <math> E = E_x+E_y+E_z \!</math> | ||
Hence the three-dimensional problem has been divided in three one-dimensional problems where the total energy E is the sum of the energies <math> E_x \!</math>, <math> E_y \!</math> and <math> E_z \!</math> in each dimension. | Hence the three-dimensional problem has been divided in three one-dimensional problems where the total energy <math> E \!</math> is the sum of the energies <math> E_x \!</math>, <math> E_y \!</math> and <math> E_z \!</math> in each dimension. |
Revision as of 00:41, 5 December 2009
(Submitted by team 1. Based on problem 3.19 in Schaum's Theory and problems of Quantum Mechanics)
Consider a particle of mass m in a three dimensional potential:
Using the Schroedinger's equation show that we can treat the problem like three independent one-dimensional problems. Relate the energy of the three-dimensional state to the effective energies of one-dimensional problem.
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The Schroedinger's equation takes the form:
Assuming that can be write like:
So,
- Failed to parse (unknown function "\begin{array}"): {\displaystyle \begin{array} -\frac{\hbar^2}{2m} \left[ \frac{d^2\Phi(x)}{dx^2} \Delta(y) \Omega (z) + \Phi(x)\frac{d^2\Delta(y)}{dy^2} \Omega (z) + \Phi(x) \Delta (y)\frac{d^2\Omega(z)}{dz^2} \right] \\ & + \left[X(x)+Y(y)+Z(z)\right]\Phi(x) \Delta(y) \Omega (z) = E\Phi(x) \Delta(y) \Omega (z) \end{array} }
Dividing by
- Failed to parse (unknown function "\begin{array}"): {\displaystyle \begin{array} -\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x) -\frac{\hbar^2}{2m} \frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2} + Y(y) \\ & -\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) = E \end{array} }
Perfectly we can separate the right hand side in three parts, where only one depends of , only one of and only one of . Then each of these parts must be equal to a constant. So:
where , and are constant and
Hence the three-dimensional problem has been divided in three one-dimensional problems where the total energy is the sum of the energies , and in each dimension.