- #1
student111
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Suppose one is to find the stationary states of a particle in an infinite cubic well. Inside the box the time independent SE is:
[tex] - \frac{\hbar}{2m} \big( \frac{\partial ^2 \psi}{\partial x ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } \big)= E\psi [/tex]
Using separation of variables: [tex] \psi = X(x)Y(x)Z(z) [/tex] we get:
[tex] YZ\frac{\partial ^2 X}{\partial x^2} + XZ\frac{\partial ^2 Y}{\partial y^2} + XY\frac{\partial ^2 Z}{\partial z^2} = \frac{-2mE}{\hbar ^2} XYZ [/tex]
After this one divides both sides by XYZ. My question is the following:
When dividing by XYZ one must assume that XYZ is different from zero. However the solution we obtain has lots of zeroes. Is this not a problem?
Thanks in advance
[tex] - \frac{\hbar}{2m} \big( \frac{\partial ^2 \psi}{\partial x ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } + \frac{\partial ^2 \psi}{\partial z ^2 } \big)= E\psi [/tex]
Using separation of variables: [tex] \psi = X(x)Y(x)Z(z) [/tex] we get:
[tex] YZ\frac{\partial ^2 X}{\partial x^2} + XZ\frac{\partial ^2 Y}{\partial y^2} + XY\frac{\partial ^2 Z}{\partial z^2} = \frac{-2mE}{\hbar ^2} XYZ [/tex]
After this one divides both sides by XYZ. My question is the following:
When dividing by XYZ one must assume that XYZ is different from zero. However the solution we obtain has lots of zeroes. Is this not a problem?
Thanks in advance
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