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## Homework Statement

Use separation of variables in cartesian coordinates to solve the infinite cubical well (or "particle in a box"):

[tex]V(x,y,z) = \{^{0, if x, y, z are all between 0 and a;}_{\infty , otherwise.}[/tex]

## Homework Equations

Well, I've been trying to use

[tex]\frac{1}{2}mv^{2} + V = \frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = E\Psi[/tex]

## The Attempt at a Solution

[tex]\frac{1}{2}mv^{2} + V = \frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = E\Psi[/tex]

[tex]\frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{x} + \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}} + V_{y} + \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}} + V_{z}[/tex]

[tex]\hat{H}_{x} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{x}[/tex]

[tex]\hat{H}_{y} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}} + V_{y}[/tex]

[tex]\hat{H}_{z} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}} + V_{z}[/tex]

[tex]\Psi = X(x)Y(y)Z(z)[/tex]

[tex](\hat{H}_{x} + \hat{H}_{y} + \hat{H}_{z})(X(x)Y(y)Z(z)) = E*(X(x)Y(y)Z(z))[/tex]

So getting it here doesn't really tell me what the energies are. Can I cancel out [tex]\Psi[/tex]?