# Quantum Mechanics Infinite Well Help

1. Nov 9, 2005

### NIQ

Hello all,
I just ran across this forum thanks to google and decided to register and ask for some help on my homework!
I'm in 2nd year university and we have a problem set due on Friday and this is the question on it:
I started to do part a) but it seems very wrong. What I did was try to normalize the function to unity so that
A^2*∫(from -a to a) (a^2 - x^2)^2dx = 1
and solving for A i found
A = sqrt((15*a^5)/(8*a^10 - 15*a^8 - 10*a^4 + 3))
This value seemed way too complicated but I kept going anyways hoping it would simplify, what I did next was try to calculate A_n such that
A_n = ∫(from 0 to a) dx*Ψ(x,0)*Ψ_n(star)(x,0) = ∫(from 0 to a) dx*Ψ*sin((n*pi*x)/a)
but then once again I got some fairly complicated function so I stopped.
Can anyone help me out here? I am very lost!
Thanks!

2. Nov 10, 2005

### Jelfish

I'm a bit confused as to how you got your value for A.
So you have:
$$A^2 \int^a_{-a} (a^2-x^2)^2 dx=1$$
$$A^2 \int^a_{-a} (a^4+x^4-2a^2 x^2) dx=1$$
$$A^2 [a^4 x + \frac{x^5}{5}-\frac{2a^2x^3}{3}]_{-a}^a =1$$
$$A^2 [2a^5 + \frac{2a^5}{5}-\frac{4a^5}{3}]=1$$
$$A^2 [\frac{16a^5}{15}]=1$$
$$A^2=\frac{15}{16a^5}$$
$$A=\sqrt{\frac{15}{16a^5}}$$
I'm not particularly sure what you're doing for the second part. My guess (and I'm actually learning this stuff right now too) is that you should find A(k) (the amplitude function, i.e. the transform from x to k) and then relate the probability of finding E_n with finding the corresponding k_n where

$$k_n = \sqrt{\frac{2m}{\hbar ^2}E_n}$$
and
$$E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}$$

Of course, I could be completely wrong. I would extend that definition onto the third part by finding the expectation value of k.

Maybe someone else can be more helpful.

Last edited: Nov 10, 2005
3. Nov 10, 2005

### NIQ

Ok the first part helped, I see what I did wrong, as for the rest of it I did something similar to what my professor showed us.
We learned that: $$P_n = |A_n|^2$$
And to find $$A_n$$ you have to do the following:
$$A_n = \int^a_{-a} \psi(x) \psi^*_n dx$$
where $$\psi^*_n = A (a^2 - x^2) sin(\frac{n \pi x}{a}) dx$$
after some rough integration and simplifying I got:
$$A_n = \frac{60 n \pi + 120 cos(n \pi) + 15 n^2 \pi^2}{8 n^3 \pi^3}$$
And so if you square this you get $$P_n$$

To find $$<E>$$ this is what I did:
$$<E> = \frac{<p^2>}{2 m}$$
$$<E> = \frac{-\hbar}{2 m} \int^\infty_{-\infty} \psi^* (x) \frac{d^2}{dx} \psi (x) dx$$

$$\cdots$$

$$<E> = \frac{-\hbar}{2 m} \sqrt{\frac{15}{16 a^2}} (-x^2) \vert^\infty_{-\infty}$$
Which doesn't seem like a good final answer... because it gives me $$0$$ right?

Gonna try to figure out what I did wrong.

Thanks for the help though!

Last edited: Nov 10, 2005
4. Nov 11, 2005

### lightgrav

<E> depends on the Energy of each of the "single" square-well wave functions,
that is, on "n" - it's an average of how much of each wave you actually have.