Infinite Square Well Width

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I am new to quantum mechanics so I am just trying to get an understanding of the infinite square well. I have been reading a lot of material and I see a lot of times that the barriers of the well say -L/2 and L/2. I know that outside the well to the left is -infinity and to the right is infinity. Does the left barrier always have to be zero? The examples I am finding all have it that when. Can you just arbitrarily pick the width of the well? Also, the result you get from the well is the probably the particle is going to be in that spot inside the well, correct? Does that mean if I have my barriers at 1 and 2 the result will be the probably that it will be between 1 and 2 or will it actually give me where between one and two it is going to be? Thanks for all your help in advance!
 

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  • #2
G01
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I'm having a hard time following what your actual questions are but I'll give it a go. I'll try to paraphrase your questions above my answers (in bold). If I do not have your question correct, please tell me.

1. Some books have the well spanning from -L/2 to L/2, others from 0 to L. Does it matter?


In short no. In both cases, the width of the well is the same. It is L. The wave functions you get out of the Schrodinger equation will look slightly different (i.e. the phase will be shifted) because the boundary conditions are different, but physically, it's the same problem modeled in two different coordinate systems, so all observables you measure will be the same.

2. What does the result of solving the Shrodinger Equation tell you?

By solving the Sch.E.Q., you get the wavefunction, which contains all the information available about the system.

If you multiple the wave function by its complex conjugate, you will get a probability density.

If you integrate this probability density between two points, you will get the probability of finding the particle between those two points.
 
  • #3
Thanks for your response. The part about the well spanning is they have the barrier's points are actually -L/2 and L/2 with the width of the well as L. Why do they divide it by two? I hope this makes sense.

I guess when I integrate by the two points are those two points the barriers on the well or can they be two points within the well?
 
  • #4
G01
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Thanks for your response. The part about the well spanning is they have the barrier's points are actually -L/2 and L/2 with the width of the well as L. Why do they divide it by two? I hope this makes sense.
If the well's boundaries were at -L and L, the total with of the well would be 2L. Thus, if we want a well, centered at 0, to have a total width of L, it should extend 1/2L to the right, and 1/2L to the left. So, it's boundaries are at L/2 and -L/2.

I guess when I integrate by the two points are those two points the barriers on the well or can they be two points within the well?
They can be any two points, either on the boundaries or not.

By definition, the particle has to be inside the infinite well. Thus, if you pick the boundaries as your end points, you will get a result of 1 or 100% probability for finding the particle in that range. (Assuming your wave function is normalized, of course.)
 
  • #5
Thank you so much for you help!
 

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