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Infinite Square Well

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Homework Statement


Consider the infinite square well described by V = 0 if 0<x<a and v = infinity otherwise. At t=0, the particle is definitely in the left half of the well, and described by the wave function,
[tex] \psi (x,0) = \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right[/tex] if 0 < x < a/2
[tex] \psi (x,0) = 0[/tex] otherwise
i. Expand [tex]\psi (x,0)[/tex] in terms of the energy eigenfunctions.
ii. Evaluate enough terms so that [tex] \psi (x,0)[/tex] is given to 5% accuracy at x = a/4.
iii. Write [tex]\psi (x,t)[/tex].


Homework Equations


Stated in part #1.


The Attempt at a Solution


I really don't know what to do, sorry. Any help will be greatly appreciated.
 

Answers and Replies

  • #2
Dick
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What ARE the energy eigenfunctions?
 
  • #3
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I don't know. That's why I'm asking.
 
  • #4
Dick
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I think the first step in 'expanding in terms of the energy eigenfunctions' might be figuring out what they are. Hint: they are solution of the time-independent Schrodinger equation with zero boundary conditions at the edges of the box. You can work them out or look them up. They should be normalized. This is what you need in the 'Relevant equations' section. Then we can start actually working on the problem.
 
  • #5
So does that mean plugging in [tex] \psi (x,0) = \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right[/tex] to the time independent Schrodnger Equation and solve?
 
  • #6
Tom Mattson
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No, it means starting from the time independent SE and solving it for the energy eigenfunctions, subject to the appropriate boundary conditions. So I guess I should ask Lisa to write down the SE and those boundary conditions, and then try to solve it. Alternatively, those eigenfuctions might be (probably are) listed in her book. If that's the case, then she can just write them down, and then find the given state as an expansion in those functions.
 
  • #7
I THINK I'm in the same class as Lisa, actually...I got the same question on my assignment! :tongue2:

Anyway, I know the eigenfunctions for an infinite square well are [tex] \psi_{n}(x)= \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right[/tex] and the corresponding eigenenergies are [tex]E_{n}=\frac{n^2\pi^2\hbar^2}{2ma^2}[/tex]. I don't know if it's any use here (It's probably useful but I just don't know what to do with them).
 
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  • #8
Gokul43201
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No, it means starting from the time independent SE and solving it for the energy eigenfunctions, subject to the appropriate boundary conditions. So I guess I should ask Lisa to write down the SE and those boundary conditions, and then try to solve it. Alternatively, those eigenfuctions might be (probably are) listed in her book. If that's the case, then she can just write them down, and then find the given state as an expansion in those functions.
Based on this response to the question "what are the energy eigenfunctions?":

I don't know. That's why I'm asking.
I would certainly recommend the former approach, else lisa may never learn what they are until it's too late.
 
  • #9
Gokul43201
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silver, I think you want to use \left( and \right) in the code for the first expression, and \hbar in the second one (and fix the parentheses for \frac{}{}).
 
  • #10
Yes, the latex didn't come out right :confused:
It should be ok now...I think.
 
  • #11
Tom Mattson
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Anyway, I know the eigenfunctions for an infinite square well are [tex] \psi_{n}(x)= \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right[/tex] and the corresponding eigenenergies are [tex]E_{n}=\frac{n^2\pi^2\hbar^2}{2ma^2}[/tex].
OK, for the sake of simplicity in typing, let's call those eigenfunctions [itex]\psi_n(x)[/itex]. Now what you're being asked to do is expand the given wavefunction in terms of those eigenfunctions. That means you're going to write down a series of the form:

[tex]\psi(x,0)=\sum_{n=1}^{\infty}c_n\psi_n(x)[/tex]

It is your job to find the [itex]c_n[/itex].
 
  • #12
Ah! I have here in my notes that..."by making use of the orthonormality of the solutions, the initial condition equation then gives us the mechanism for finding the coefficients of the series:

[tex]c_{n}=\int_{0}^{a} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \psi(x,0)dx[/tex]

Is this the right formula to use? Are the limits of integration from 0 to a? (I think it is)

So, I guess in the end I need to evaluate this integral to get [tex]c_{n}[/tex]:
[tex]c_{n}=\int_{0}^{a} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right dx[/tex]
 
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  • #13
OK, I have done the integral for [tex]c_{n}[/tex], and here's what I got:

[tex]C_{n}=0[/tex] for every n except 2. So...I guess the answer to part (a) of the question is [tex] \psi (x,0) = \sqrt{2} sin\left \frac{2 \pi x}{a} \right [/tex].

There's only one term here...? It looks a bit suspicious...because it kind of makes part (b) of the question pointless...?
 
  • #14
Tom Mattson
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[tex]C_{n}=0[/tex] for every n except 2.
I haven't worked it out, but I don't think that's right. Did you remember that [itex]\psi(x,0)=0[/itex] for [itex]a/2\leq x \leq a[/itex]?
 
  • #15
Thanks for the advice! Yes, I reworked the problem (changing the limits of integration) and I got this:

[tex]c_{n}=\int_{0}^{a/2} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right dx[/tex]

and the [tex]c_{n}s[/tex] came out to be [tex]c_{1}=\frac{2a}{3\pi}, c_{2}=\frac{a}{4}, c_{3}=\frac{2a}{5\pi}, c_{4}=0,c_{5}=\frac{-2a}{21\pi},c_{6}=0[/tex]...and so on...

So, I guess the final answer for part (a) of the question is
[tex]\sqrt{\frac{2}{a}}\frac{2a}{3\pi}sin\left \frac{\pi x}{a} \right + \sqrt{\frac{2}{a}}\frac{a}{4}sin\left \frac{2\pi x}{a} \right + \sqrt{\frac{2}{a}}\frac{2a}{5\pi}sin\left \frac{3\pi x}{a} \right + \sqrt{\frac{2}{a}}\frac{-2a}{21\pi}sin\left \frac{5\pi x}{a} \right[/tex], evaluated up to 5 terms...well, up to [tex]c_{5}[/tex] since [tex]c_{4}=0[/tex].

Now...part (b)...I guess I have to evaulate the initial wave function at x=a/4 first. Am I on the right track?
 
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  • #16
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Hi everybody :biggrin:

silverthorne, I think your approach to part (i) makes sense. However, double-check your c_n calculations - I got different values using your formula: [tex]c_{1}=\frac{4\sqrt{2}}{3\pi}, c_{2}=\frac{1}{\sqrt{2}}, c_{3}=\frac{4\sqrt{2}}{5\pi}, c_{4}=0,c_{5}=\frac{-4*\sqrt{2}}{21\pi},c_{6}=0[/tex] [EDIT: sorry about the non-LaTeX roots, my edits aren't taking effect]

Substituting these coefficients into

[tex]\psi(x,0)=\sum_{n=1}^{6}c_n\psi_n(x)[/tex]

where [tex] \psi_{n}(x)= \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right[/tex]

produces the following graph of [tex]\psi(x,0)[/tex]

http://img442.imageshack.us/img442/686/approxwavefnsh3.th.png [Broken]

which seems to converge on the given wavefunction (which is normalized).

Also, I was wondering: since the particle is definitely in the left half of the well, might we have to modify the infinite square-well energy eigenfunction by substituting L = a/2? I'm thinking we don't, since the particle's confinement is not a stationary state.
 
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  • #17
Hmm...I don't think we need to modify the energy eigenfunction because our question is still living from x=0 to x=a. OK, I am going to redo my calculations for the coefficients.

I know part c is dependent on the number of terms in part b. But the thing is I couldn't figure out what does "getting to 5% of the accuracy to a/4" means.
 
  • #18
Yep, you're right. The previous [tex]C_{n}[/tex]s I got were just the value of the integral...I forgot to multiply everything by the constant [tex]\frac{\sqrt{8}}{a}[/tex]. Thanks for the heads up. :)
 
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  • #19
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Hmm...I don't think we need to modify the energy eigenfunction because our question is still living from x=0 to x=a. OK, I am going to redo my calculations for the coefficients.

I know part c is dependent on the number of terms in part b. But the thing is I couldn't figure out what does "getting to 5% of the accuracy to a/4" means.
Evaluate enough terms of the eigenfunction expansion [tex]\Psi_e(x,t)[/tex] so that [tex]\left| 1 - \Psi \left( \frac{a}{4}, 0 \right) / \Psi_e \left( \frac{a}{4} \right) \right| < 0.05[/tex].
 
  • #20
Ah! Thank you so much!!
 

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