# Infinite Cubical Well in Cartesian Coordinates

## Homework Statement

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

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

## Homework Equations

Well, I've been trying to use
$$\frac{1}{2}mv^{2} + V = \frac{1}{2m}(P^{2}_{x} + P^{2}_{y} + P^{2}_{z}) + V = E\Psi$$

## The Attempt at a Solution

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

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

$$\hat{H}_{x} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V_{x}$$

$$\hat{H}_{y} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}} + V_{y}$$

$$\hat{H}_{z} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}} + V_{z}$$

$$\Psi = X(x)Y(y)Z(z)$$

$$(\hat{H}_{x} + \hat{H}_{y} + \hat{H}_{z})(X(x)Y(y)Z(z)) = E*(X(x)Y(y)Z(z))$$

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

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Meir Achuz
Homework Helper
Gold Member
Look in a math book on "separation of variable".
You can divide each side by XYZ.
Then use the fact that YZ cancels from H_x(XYZ)/YZ.

So, if I divide by $$(X(x)Y(y)Z(z))$$ from both sides, then I end up with:

$$\hat{H} = E$$???

Would it be logical to denote energy in an dimension, like in the x-direction and y-direction and z-direction? (ie. $$E = (E_{x} + E_{y} + E_{z})$$

Meir Achuz
Homework Helper
Gold Member
Yes, you have H_x X=E_x X.
This has a simple (sin) solution since X must be zero at x=0 and a.

Well, I was a bit rushed to get my homework in on time; but for my energy equation I got:

$$E_{n} = \frac{\hbar^{2} \pi^{2}}{2m}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})$$

Does that look right? I'd still like to see if what I did was correct or not.

Last edited:
Meir Achuz
Homework Helper
Gold Member
Well, I was a bit rushed to get my homework in on time; but for my energy equation I got:

$$E_{n} = \frac{\hbar^{2} \pi^{2}}{2m}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})$$

Does that look right? I'd still like to see if what I did was correct or not.
You left out /L^2 of the side.

$$E_{n} = \frac{\hbar^{2} \pi^{2}}{2mL^{2}}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})$$???

It's important that I know this stuff because we're having a quiz on it tomorrow.