Seperation of variables, infinite cubic well

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SUMMARY

The discussion focuses on finding the stationary states of a particle in an infinite cubic well using the time-independent Schrödinger equation (SE). The equation is expressed as \(\frac{\hbar}{2m} \left( \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} \right) = E\psi\). By applying the separation of variables method, the wave function \(\psi\) is decomposed into \(X(x)Y(y)Z(z)\), leading to a critical examination of the assumption that \(XYZ \neq 0\) during the division process. The resolution confirms that at isolated zeros, the limit of the ratio \(\frac{X''}{X}\) remains well-behaved for eigenfunctions, thus validating the approach.

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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
 
Last edited:
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And?
 
You just have to rule out XYZ being zero everhywhere.
At an isolated zero, the limit of the ratio X"/X is well behaved for an eigenfunction.
 

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