# Infinite Square Well strategy

1. Feb 2, 2006

### cepheid

Staff Emeritus
I need a little help with the strategy on this question. My work is below the problem description.

A particle of mass m is in an infinite square well of width a (it goes from x = 0 to x = a). The eigenfunctions of the Hamiltonian are known to be:

$$\psi_{n}(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n \pi}{a}x}\right)$$

The corresponding eigenvalues are:

$$E_n = \frac{\hbar^2 n^2 \pi^2}{2ma^2}$$

At time t = 0, the particle has wave function $\Psi(x,0) = Ax(a-x)$ where $A = (30/a^5)^{1/2}$.

a) $\Psi(x,0)$ can be written as a linear combination of the eigenfunctions of the Hamiltonian. Let us use the symbol cn to denote the expansion coefficients. Obtain a general expression for the cn in terms of $\pi$.

Here is my work so far. Based on what we know of Fourier series,

$$c_n = \int{\psi_n^*(x) \Psi(x,0)\, dx} = \sqrt\frac{2}{a}}\sqrt\frac{30}{a^5}} \int_0^a{\sin\left(\frac{n \pi}{a}x}\right)(ax - x^2)\, dx}$$

$$= \sqrt\frac{60}{a^4}} \int_0^a{x\sin\left(\frac{n \pi}{a}x}\right)\, dx}\, - \, \sqrt\frac{60}{a^6}} \int_0^a{x^2\sin\left(\frac{n \pi}{a}x}\right)\, dx}$$

...and now I have to integrate by parts. Is this the correct method? The reason I hesitate is that it looks like the final answer will not be only in terms of pi. There will still be some a's left, surely! Yet, the next part asks me to calculate c1, c2, and c3 to seven significant figures! This isn't looking good...I must be missing something.

Last edited: Feb 2, 2006
2. Feb 2, 2006

### Meir Achuz

Change the integration variable to z=ax. Then the constant a will drop out.

3. Feb 3, 2006

### nrqed

You are doing it perfectly correctly. Your answer will not contain "a" (it should not since the c_n are dimensionless constants!). You can easily see this by simply looking at the dimension of your integrands. The first has dimension of length squared and the second of length cube, so that wll cancel the factors of a in the normalization constants. As the other poster said, to see this, just change your variable of integration to z=x/a then you will see the a disappearing completely (don`t forget to change your limits of integration too ;-) )

Hope this helps

Pat

4. Feb 4, 2006

### Meir Achuz

You're right. I should have said z=x/a.

5. Feb 5, 2006

### cepheid

Staff Emeritus
I integrated without changing any variables and found that the a's did indeed drop out. Thanks for your help. I just needed to keep going on this question.