|
|
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| \begin{align} | | \begin{align} |
| -\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] \\ | | -\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)
| | + \left[X(x)+Y(y)+Z(z)\right]\Phi(x) \Delta(y) \Omega (z) = E\Phi(x) \Delta(y) \Omega (z) |
| \end{align} | | \end{align} |
| </math> | | </math> |
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| -\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{align} | | \end{align} |
Revision as of 00:43, 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.
______________________________________________________________________________________________________________________________________
The Schroedinger's equation takes the form:

Assuming that
can be write like:

So,
![{\displaystyle {\begin{aligned}-{\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d872507c0bc8e381acd1c5438825d97a0f199143)
Dividing by

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.