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Wavefunction in an infinite square well

  • Thread starter jhendren
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  • #1
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Homework Statement


A wavefunction in an infinite square well in the region -L/4≤x≤3L/4 is given by ψ= Asin[(πx/L)+δ] where δ is a constant

Find a suitable value for δ (using the boundary conditions on ψ)


Homework Equations





The Attempt at a Solution


Asin[(πx/L)+δ]=?
 

Answers and Replies

  • #2
cepheid
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Welcome to PF jhendren!

Homework Statement


A wavefunction in an infinite square well in the region -L/4≤x≤3L/4 is given by ψ= Asin[(πx/L)+δ] where δ is a constant

Find a suitable value for δ (using the boundary conditions on ψ)


Homework Equations





The Attempt at a Solution


Asin[(πx/L)+δ]=?
Let's start with the bit in red above. What ARE the boundary conditions on ψ?
 
  • #3
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It does not say, that was all the info given
 
  • #4
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I beleive since there is no (n) value to just assume n=1 and that is the only thing I can grasp, and the only other way I could conceive to begin the problem is with ψ(x)=Ae^-iδt
 
  • #5
cepheid
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It does not say, that was all the info given
It doesn't say, because it is something about infinite square wells that you already know. The boundary conditions are the conditions that the wavefunction must satisfy at the ends of the well. In this case, since the potential goes to infinity beyond +3L/4 and -L/4, the particle cannot exist outside of these bounds (in order to get there, it would require infinite energy). In other words, it can't go through the walls of the well. So, there cannot be any probability of finding the particle beyond the walls. What does this mean for the value of the wavefunction at and beyond the boundaries?
 
  • #6
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Well if it is in the n=1 state shouldn't it just be a guassian and at the boundaries ψ=0?
 
  • #7
cepheid
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Well if it is in the n=1 state
Forget about n. Focus on the form of the wavefunction that you've been given in the problem.

shouldn't it just be a guassian
It's not a Gaussian, it's clearly a sine wave.

and at the boundaries ψ=0?
Sigh. Yes, ψ is 0 at the boundaries, but I'm not convinced that you understand why. I want to you re-read what I wrote (below), and make sure you understand that, "it means that ψ must be 0 at the boundaries" is the answer to the question I asked you (in red), and that the explanation for why it goes to 0 at the boundaries is in the preceding sentences.

It doesn't say, because it is something about infinite square wells that you already know. The boundary conditions are the conditions that the wavefunction must satisfy at the ends of the well. In this case, since the potential goes to infinity beyond +3L/4 and -L/4, the particle cannot exist outside of these bounds (in order to get there, it would require infinite energy). In other words, it can't go through the walls of the well. So, there cannot be any probability of finding the particle beyond the walls. What does this mean for the value of the wavefunction at and beyond the boundaries?
Once you're sure you've understood why the boundary conditions are what they are, then go ahead and apply the boundary conditions to solve the problem.
 
  • #8
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I understand why ψ behaves the way it does at the bounds, but the problem I am having is finding a value for δ, can you point me in the right direction please
 
  • #9
cepheid
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I understand why ψ behaves the way it does at the bounds, but the problem I am having is finding a value for δ, can you point me in the right direction please
Yes. What I said above was the directions: APPLY the boundary conditions. The boundary conditions on ψ(x) are that:

ψ(3L/4) = 0

and

ψ(-L/4) = 0

So plug in x = 3L/4 into the function, equate the result to 0, and see what is required for the value of δ in order to satisfy this equation. I'm pretty much spelling it out for you now..
 
  • #10
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0=Asin[(3π/4)+δ] and Asin(0)=0 or Asin(π)=0 so (3π/4)+δ=nπ so δ =(1/4)π
 
  • #11
cepheid
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0=Asin[(3π/4)+δ] and Asin(0)=0 or Asin(π)=0 so (3π/4)+δ=nπ so δ =(1/4)π
Your last step is only true for n = 1. The more general expression is:

[tex] \delta = \pi\left(n - \frac{3}{4}\right) [/tex].

EDIT: where [itex] n = 0, \pm 1, \pm 2, \pm 3, \ldots [/itex]

There is no reason to restrict n to be equal to 1, and I have no idea why you keep insisting on doing that. The whole point of applying the boundary conditions is to find out what all of the allowed values for the wavefunction are.

Now what about the other boundary condition?
 
  • #12
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δ=π(n+1/4)
 
  • #13
cepheid
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δ=π(n+1/4)
Right exactly. And you can see that, in this case, the second boundary condition doesn't really give you any new information, since if you start plugging in values for n into the two expressions, you'll see that they both generate the same list of allowed values for δ.

Consider π(n-3/4). We get

δ = ... -3π/4, π/4, 5π/4, 9π/4...

for n = 0, 1, 2, and 3, respectively.

Consider π(n+1/4). We get

δ = ... -3π/4, π/4, 5π/4, 9π/4...


for n = -1, 0, 1, and 2, respectively.

So both boundary conditions yield consistent constraints on the allowed values of δ.
 
  • #14
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Thank you
 

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