# Infinite square well problem

1. May 29, 2013

### Cogswell

1. The problem statement, all variables and given/known data
Consider a particle in 1D confined in an infinite square well of width a:

$$V(x) = \begin{cases} 0, & \text{if } 0 \le x \le a \\ \infty, & \text{otherwise} \end{cases}$$

The particle has mass m and at t=0 it is prepared in the state:

$$\Psi (x,t=0) = \begin{cases} A \sin (\frac{2 \pi}{a}x) \cos (\frac{\pi}{a}x), & \text{if } 0 \le x \le a \\ 0, & \text{otherwise} \end{cases}$$

with A a real and positive number.

a. Show that $A = \frac{2}{\sqrt{a}}$

b. If the energy of the particle was measured, give the probability that it would be $E = \dfrac{9 \pi^2 \hbar ^2}{2ma}$

c. Give < $\hat{x}$ > at t=0

d. Give < $\hat{p}$ > as a function of time.

2. Relevant equations

$-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi}{dx^2} = E \psi$

3. The attempt at a solution

I got question (a) alright, but now I'm stuck on (b).

I know that you can only get certain answers for E, because the energy levels it can be in are quantised.
In the book, it says the allowed energy levels are: $E_n = \dfrac{n^2 \pi ^2 \hbar ^2}{2 m a ^2}$
The one they've given me looks similar, except it's not a $a^2$ at the bottom. Does that mean it can never be at that energy level, and that the probability is 0?

How do you figure out the probability of a particle being at a certain energy? Is there a another probability density for that as well?

2. May 29, 2013

### Dick

If there's no a^2 in the denominator of the energy they gave you that means there is a typo in the problem. Because then it doesn't even have the correct units to be an energy. Just put the a^2 in and go from there. You want the overlap integral between the prepared state and the n=3 eigenfunction.

3. May 31, 2013

### Cogswell

Sorry quantum isn't my strong point.

So let's assume that there is an a^2 at the bottom.
I can tell that it's the 3rd energy level, and so it'll be the third eigenstate.

The eigenstates are $\psi_n (x) = \sqrt{\dfrac{2}{a}} \sin \left( \dfrac{n \pi}{a} x \right)$

We want the third one, and so $\psi_3 (x) = \sqrt{\dfrac{2}{a}} \sin \left( \dfrac{3 \pi}{a} x \right)$

Now, I'm trying to visualise this.
I want the area overlap between the prepared state and the eigenfunction?

I know that Probability $= \displaystyle \int^a_0 |\Psi (x,t)|^2 dx$

Does these 2 graphs look correct?
Wolfram|Alpha - Prepared state and 3rd eigenfunction graph
(I've assumed the width of the well is 4 (a=4), just so wolfram alpha can graph it.)

And I'm looking at just the overlap?

4. May 31, 2013

### Dick

What I mean by overlap integral is that if $\phi(x)$ is your prepared state then the probability for b) is $$\vert {\int^a_0 \psi_3(x) \overline {\phi (x)} dx} \vert$$
Does that look familiar?

5. May 31, 2013

### Cogswell

Oh so the integral should be something like:

$$\displaystyle \left| \sqrt{\dfrac{2}{a}} \dfrac{2}{\sqrt{a}} \int_0^a \sin \left( \dfrac{3 \pi}{a} x\right) \sin \left( \dfrac{2 \pi}{a} x\right) \cos \left( \dfrac{\pi}{a} x\right) \right|$$

Using Wolfram Alpha I got something like $\dfrac{\sqrt{2}}{2}$

Can someone help me out with this integral? I tried using integration by parts $\displaystyle \int u v w' = u v w - \int u' v w - \int u v' w$ but I didn't get anywhere.

I know that they are orthogonal to each other so if they're all sines they'd evaluate to zero, and I can use the Kroneker delta function to evaluate them.

Also, is this how I'd evaluate all questions of these sort?
So like if I was given an energy for a harmonic oscillator or a free particle, do I just integrate the prepared state and eigenfunction multiplied together?

6. May 31, 2013

### TSny

One approach is to use the identities

sin$\alpha$cos$\beta$ = $\frac{1}{2}$[sin($\alpha + \beta$) + sin($\alpha - \beta$)]

sin$\alpha$sin$\beta$ = $\frac{1}{2}$[cos($\alpha - \beta$) - cos($\alpha + \beta$)]

7. Jun 1, 2013

### Cogswell

Thanks, that makes it so much easier. I got the same answer as Wolfram Alpha in the end (but it'll take a bit to write it here)

I'm doing part (c) of the question now, and here's where I'm at:

$$\displaystyle \dfrac{4}{a} \int^a_0 x \sin^2 \left( \dfrac{2 \pi}{a} x \right) \cos^2 \left( \dfrac{\pi}{a}x \right) dx$$

I'm using the same identities as above, and some extra identites to work it out, but I'm making a lot of careless errors. There's quite a lot of working!

The thing I'm stuck on is (d).
I know that to work out the momentum, there's 2 ways:

$< \hat{p} > = m \dfrac{d<x>}{dt}$ \displaystyle < \hat{p} > = -i \hbar \int_0^a \Psi (x,t)^* \dfrac{\partial}{\partial x} \Psi (x,t) dx $In the first case, it'll just be 0, right? In the second case, wouldn't the time factor just cancel out because it's being multiplied by its conjugate? Can anyone tell me what I should be doing for this question? 8. Jun 1, 2013 ### Dick If you follow TSny's hint you can write the prepared state as the sum of two energy eigenfunctions. Each will have its own time dependence. Products of exponentials belonging to different energy states won't cancel. 9. Jun 1, 2013 ### Cogswell I've just been going through the book and saw something that looks like that... $$\displaystyle c_n = \sqrt{\dfrac{2}{a}} \int_0^a \sin \left( \dfrac{n \pi x}{a} \right) \Psi (x,0) dx$$ It then says something about$ |c_n|^2 ## is the "probability of finding the particle in the nth stationary state"

So should I square it? And then the probability will be 0.5

10. Jun 2, 2013

### Dick

What question are you asking about? I thought we were onto the time dependent part. If so where is the time dependence?