# Help with infinite square well

1. Oct 21, 2006

### kcirick

Question:
A particle of mass m moves in 1-D infinite square well. at t=0, its wave function is $\Psi\left(x,t=0\right)=A\left(a^{2}-x^{2}\right)$. Find the probability that the particle is in the energy eigenstate $E_{n}$. Does the probability change with time?

What I have so far:
So far I just found the normalization constant for the wave function at t=0:

$$\int|\Psi\left(x,t=0\right)|^{2}dx=1$$
$$...A=\frac{1}{4}\sqrt\frac{15}{a^{5}}$$

So we have $$\Psi\left(x, t=0\right)=\frac{1}{4}\sqrt\frac{15}{a^{5}}\left(a^{2}-x^{2}\right)$$

Now, because this is infinite square well, $\Psi=0$ at the boundary. From that, we can find out the energy eigenstate:
$$E_{n}=\frac{n^{2}\hbar^{2}\pi^{2}}{8ma^{2}}$$ (Derived in class)

But what to do from there? I don't know...

2. Oct 21, 2006

### Staff: Mentor

Hint #1: Any wave function can be expressed as a linear combination of energy eigenstates:

$$\Psi = c_1 \Psi_1 + c_2 \Psi_2 + ...$$

where (if the functions are all normalized) $c_1^* c_1$ is the probability of being in eigenstate $\Psi_1$, etc.

Hint #2: If the eigenstates are normalized, what is the value of

$$\int {\Psi_m^* \Psi_n} dx$$

(a) when m = n, and (b) when m ≠ n?

3. Oct 21, 2006

### kcirick

$$\Psi = c_1 \psi_1 + c_2 \psi_2 + ...$$
where:
$$\psi_1=\frac{1}{\sqrt{a}}cos\left(\frac{n\pi}{2a}\right)x, n=1,3,5,...$$
$$\psi_2=\frac{1}{\sqrt{a}}sin\left(\frac{n\pi}{2a}\right)x,n=2,4,6,...$$
Because it is an infinite square well.

Then according to a note that I have,
$$c_1=\int^{a}_{-a}\Psi(x)\frac{1}{\sqrt{a}}cos\left(\frac{n\pi}{2a}\right)xdx$$
$$c_2=\int^{a}_{-a}\Psi(x)\frac{1}{\sqrt{a}}sin\left(\frac{n\pi}{2a}\right)xdx$$
where:
$$\Psi\left(x, t=0\right)=\frac{1}{4}\sqrt\frac{15}{a^{5}}\left(a^{2}-x^{2}\right)$$

But where did that relation come from? Can someone help me out?

Also, to jtbell: How is the probability of being in eigenstate $\psi_1 = c_1*c_1$? Can you please explain? Thank you!

Last edited: Oct 21, 2006
4. Oct 21, 2006

### Staff: Mentor

I'd prefer to write that as $cos\left(\frac{n\pi}{2a}x\right)$, and similarly for the other one with the sine. Most people would read your version as "calculate $\frac{n\pi}{2a}$, find the cosine of the result, and finally multiply by x", which is not the same as "calculate $\frac{n\pi}{2a} x$ and find the cosine of the result."

I think those should be:

$$c_n=\int^{a}_{-a}\Psi(x)\frac{1}{\sqrt{a}}cos\left(\frac{n\pi}{2a}x\right)dx, n = 1,3,5...$$

$$c_n=\int^{a}_{-a}\Psi(x)\frac{1}{\sqrt{a}}sin\left(\frac{n\pi}{2a}x\right)dx, n = 2,4,6...$$

Both of the above relations can be written as

$$c_n=\int\Psi^*_n(x)\Psi(x)dx$$

To derive this, start with your first equation at the top, multiply both sides by $\Psi^*_n(x)$, and integrate:

$$\int \Psi^*_n \Psi dx = c_1 \int \Psi^*_n \Psi_1 dx + c_2 \int \Psi^*_n \Psi_2 dx + ...$$

Now take note of hint #2 in my earlier posting. Your textbook almost certainly gives the answer to this hint, and possibly so do your notes. Look for the words "orthogonal" or "orthonormal".

I think the usual way to show this (or at least make it plausible) is to calculate the expectation value of the energy for $\Psi$, using the expansion given by your first equation at the top. Have you studied expectation values yet?

5. Oct 21, 2006

### kcirick

Thanks jtbell. Things are starting to become clear, but still the entire QM concept isn't very clear to me, and may never will be...

Anyways, I calculated $c_n$ for both even n and odd n. For the odd n, I get a very nice answer:
$$c_n = \left(-1\right)^{\frac{n-1}{2}}\frac{4\sqrt{15}}{n\pi}, n=1,3,5,...$$

But for the even n, I get $c_n=0$ (because sin is an anti-symmetric function). Is this normal?

We did do Expectation values in class, but all I really know is that:
$$\left<E\right>=\int\Psi^*E\Psi dx$$
That is the next part of my question, I will get to that eventually.

I really appreciate this help, because to me QM stuff doesn't really make sense to me. Anyway, thank you!

6. Oct 21, 2006

### Staff: Mentor

That's exactly what I would expect, because your given $\Psi$ is symmetric about the origin (or "even"). It can't contain any contributions from anti-symmetric ("odd") eigenstates.

7. Oct 21, 2006

### kcirick

Well, when m ≠ n, the integral is equal to 0, but neither my textbook nor my lecture notes tells me the answer when m = n. From intuition, my guess is 1, but is it correct? If it is, then I can do the following:

$$P = \int \Psi^* \Psi dx$$

Since $\Psi = c_{n}\psi_{n} [/tex] for energy eigenstate [itex]E_{n}$,

$$P = \int c_{n}^*\psi_{n}^* c_{n}\psi_{n}dx$$
$$P = c_{n}^* c_{n} \int\psi_{n}^* \psi_{n}$$
$$P = c_{n}^* c_{n}$$

Is this allowed?

8. Oct 23, 2006

### Staff: Mentor

Yes, but you shouldn't need pure intuition to realize this. If m = n, then the integral is just the normalization integral for the function in question. If the function is normalized already, this integral is...?

No, in general, you have to assume that $\Psi$ is a linear combination of all the energy eigenstates:

$$\Psi = c_1 \psi_1 + c_2 \psi_2 + ... + c_n \psi_n + ...$$

$$P = \int \Psi^*\Psi dx = \int (c_1^* \psi_1^* + c_2^* \psi_2^* + ... + c_n^* \psi_n^* + ...)(c_1 \psi_1 + c_2 \psi_2 + ... + c_n \psi_n + ...) dx$$

Now multiply this out. You can't write the whole thing out, of course, because there are an infinite number of terms (n goes to infinity), but if you go out to, say, n = 3, that should be enough to see the pattern.

Last edited: Oct 23, 2006