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Infinite Cubical Well in Cartesian Coordinates

  • Thread starter Rahmuss
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  • #1
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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]?
 

Answers and Replies

  • #2
Meir Achuz
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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.
 
  • #3
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So, if I divide by [tex](X(x)Y(y)Z(z))[/tex] from both sides, then I end up with:

[tex]\hat{H} = E[/tex]???
 
  • #4
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Would it be logical to denote energy in an dimension, like in the x-direction and y-direction and z-direction? (ie. [tex]E = (E_{x} + E_{y} + E_{z})[/tex]
 
  • #5
Meir Achuz
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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.
 
  • #6
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Well, I was a bit rushed to get my homework in on time; but for my energy equation I got:

[tex]E_{n} = \frac{\hbar^{2} \pi^{2}}{2m}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})[/tex]

Does that look right? I'd still like to see if what I did was correct or not.
 
Last edited:
  • #7
Meir Achuz
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Well, I was a bit rushed to get my homework in on time; but for my energy equation I got:

[tex]E_{n} = \frac{\hbar^{2} \pi^{2}}{2m}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})[/tex]

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.
 
  • #8
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[tex]E_{n} = \frac{\hbar^{2} \pi^{2}}{2mL^{2}}(n^{2}_{x} + n^{2}_{y} + n^{2}_{z})[/tex]???

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

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