Probabilities Inside Cubic 3D Infinite Well

AI Thread Summary
The discussion revolves around calculating the probability of finding an electron in a specific region of a cubic 3D infinite well, given certain quantum states. The normalized wave function is provided, and participants discuss the necessity of integrating to find the probability, particularly focusing on the y-dimension range of 1/3L to 2/3L. There is clarification that for a cubic well, the dimensions are equal (Lx = Ly = Lz), and the conditions for x and z are inherently satisfied. Participants also confirm that if the ranges were different, a triple integral would be required to compute the probability accurately. The conversation emphasizes the importance of integration in probability calculations for quantum systems.
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Homework Statement



An electron is trapped in a cubic 3D infinite well. In the states (nx,ny,nz) = (a)(2,1,1), (b)(1,2,1) (c)(1,1,2), what is the probability of finding the electron in the region (0 ≤ x ≤ L, 1/3L ≤ y ≤ 2/3L, 0 ≤ z ≤ L)?

Homework Equations



My normalized wave function in the box is:

\psi _{(x,y,z)} = \left( \frac{2}{L}\right)^{\frac{3}{2}}sin\left(\frac{n_{x} \pi x}{L_{x}}\right)sin\left(\frac{n_{y} \pi y}{L_{y}}\right)sin\left(\frac{n_{z} \pi z}{L_{z}}\right)

And my probability is found by |ψ2|

The Attempt at a Solution



Without integrating I am not sure how to proceed on this problem. If I had exact values for L, obviously it would be fairly straight forward.

My problem is I don't know what to put in for the normalized L (since it isn't axis specific) and my axis specific L's give ranges.
 
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Since this well is cubic, isn't Lx=Ly=Lz?

The first and last conditions are always going to be satisfied, so all you have to worry about is 1/3L ≤ y ≤ 2/3L. You find that by integrating the one-dimensional solution from 1/3L to 2/3L. It shouldn't be a hard integral, especially if you use Wolfram Alpha. For part b, you shouldn't need to integrate; think about the symmetry of the wavefunction and you'll get the answer.
 
Ugh...how soon we forget.

I don't have the probability right. You nailed it. I was looking at an example in the book and they skipped over the probability calculation so I wrong assumed there wasn't any integration. No wonder it didn't make sense.

What do you mean by "The first and last conditions are always going to be satisfied..."

And say I had the same problem my ranges were all like the center one where it isn't just 0 to L. Would I set up a triple integral over the different variable ranges?
 
erok81 said:
What do you mean by "The first and last conditions are always going to be satisfied..."

I mean that 0 ≤ x ≤ L and 0 ≤ z ≤ L are always true, so you don't need to worry about them.

And say I had the same problem my ranges were all like the center one where it isn't just 0 to L. Would I set up a triple integral over the different variable ranges?

Yes.
 
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