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| (Submitted by team 1. Based on problem 3.19 in Schaum's Theory and problems of Quantum Mechanics)
| | The Schrödinger equation takes the form, |
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| Consider a particle of mass m in a three dimensional potential:
| | :<math>-\frac{\hbar^2}{2m}\frac{d^2\psi(x,y,z)}{dx^2}+\left(X(x)+Y(y)+Z(z)\right)\psi(x,y,z)=E\psi(x,y,z).</math> |
| <math>V(x,y,z) = X(x)+Y(y)+Z(z)\!</math> | |
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| 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.
| | Let us assume that <math>\psi</math> has the form, <math>\psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z).\!</math> Then |
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| The Schroedinger's equation takes the form:
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| :<math>-\frac{\hbar^2}{2m}\frac{d^2\psi(x,y,z)}{dx^2}+\left(X(x)+Y(y)+Z(z)\right)\psi(x,y,z)=E\psi(x,y,z)</math>
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| Assuming that <math>\psi\!</math> can be write like:
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| :<math>\psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z) \!</math>
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| So,
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| :<math> | | :<math> |
| \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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| Dividing by <math>\psi(x,y,z) \!</math> | | Dividing by <math>\psi(x,y,z),\!</math> we obtain |
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| :<math> | | :<math> |
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| </math> | | </math> |
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| We can perfectly separate the right hand side into three parts, where it will only depend on <math> x \!</math>, or on <math> y \!</math> or only on <math> z \!</math>. Then each of these parts must be equal to a constant. So: | | We may now separate the left-hand side into three parts, each depending on only one of the three coordinates <math>x,\,y,</math> and <math>z.\!</math> Each of these parts must be equal to a constant. Therefore, |
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| :<math> | | :<math> |
| -\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x) = E_x </math> | | -\frac{\hbar^2}{2m}\frac{d^2\Phi(x)}{dx^2} + X(x)\Phi(x) = E_x\Phi(x) </math> |
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| :<math> | | :<math> |
| -\frac{\hbar^2}{2m} \frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2} + Y(y) = E_y </math> | | -\frac{\hbar^2}{2m}\frac{d^2\Delta(y)}{dy^2} + Y(y)\Delta(y) = E_y\Delta(y) </math> |
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| :<math> | | :<math> |
| -\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{d^2\Omega(z)}{dz^2} + Z(z)\Omega(z) = E_z\Omega(z), </math> |
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| where <math> E_x \!</math>, <math> E_y \!</math> and <math> E_z \!</math> are constants 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 constants and <math> E = E_x+E_y+E_z.\!</math> |
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| | Hence, the three-dimensional problem has been divided into 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. |
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| Hence, the three-dimensional problem has been divided into 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.
| | Back to [[Motion in One Dimension#Problem|Motion in One Dimension]] |
The Schrödinger equation takes the form,
Let us assume that 
has the form, 
Then
![{\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/a6ebb1eebcc6ff52d7a5c0c7d9ce717c22cecdaa)
Dividing by 
we obtain
We may now separate the left-hand side into three parts, each depending on only one of the three coordinates 
and 
Each of these parts must be equal to a constant. Therefore,
where 
, 
and 
are constants and 
Hence, the three-dimensional problem has been divided into three one-dimensional problems where the total energy 
is the sum of the energies and in each dimension.
Back to Motion in One Dimension