Wavefunction in an infinite square well

I do appreciate itOh, I was just starting to have fun! :-)In summary, the given wavefunction in an infinite square well can be solved for δ by applying the boundary conditions ψ(3L/4) = 0 and ψ(-L/4) = 0, which results in the expression δ = π(n - 3/4), where n is an integer. This means that δ can take on values of -3π/4, π/4, 5π/4, 9π/4, etc.
  • #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]=?
 
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  • #2
Welcome to PF jhendren!

jhendren said:

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
It does not say, that was all the info given
 
  • #4
I believe 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
jhendren said:
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
Well if it is in the n=1 state shouldn't it just be a guassian and at the boundaries ψ=0?
 
  • #7
jhendren said:
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.

jhendren said:
shouldn't it just be a guassian

It's not a Gaussian, it's clearly a sine wave.

jhendren said:
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.

cepheid said:
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
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
jhendren said:
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
0=Asin[(3π/4)+δ] and Asin(0)=0 or Asin(π)=0 so (3π/4)+δ=nπ so δ =(1/4)π
 
  • #11
jhendren said:
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
δ=π(n+1/4)
 
  • #13
jhendren said:
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
Thank you
 

1. What is a wavefunction in an infinite square well?

A wavefunction in an infinite square well is a mathematical function that describes the behavior of a particle constrained to a specific region. In this case, the particle is confined to an infinitely deep potential well with walls at either end.

2. What is the significance of the infinite square well in quantum mechanics?

The infinite square well serves as a simplified model for understanding the behavior of particles in a potential well. It allows us to study the properties of quantum particles in a controlled environment, and is often used as a starting point for more complex systems.

3. How does the wavefunction in an infinite square well relate to the energy of a particle?

The wavefunction in an infinite square well is a solution to the Schrödinger equation, and its square magnitude represents the probability of finding the particle at a certain position. The energy of the particle is related to the number of nodes in the wavefunction, with higher energy states having more nodes.

4. Can the wavefunction in an infinite square well be used to describe real particles?

The infinite square well is a simplified model and does not accurately describe real particles in the physical world. However, it can be used to make predictions about the behavior of particles in certain scenarios, and can be a useful tool for understanding quantum mechanics.

5. How does the size of the infinite square well affect the behavior of particles?

The size of the infinite square well affects the energy levels of the particle. A larger well will have more energy levels and allow for a greater range of possible energies, while a smaller well will have fewer energy levels and restrict the particle's energy states.

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