Way to check probability of being in eigenstate?

1. Mar 11, 2009

ballzac

Hi,
I've got a problem that is to calculate the normalisation constant and then the probability of obtaining an energy measurement of $$E_n$$ for an infinite square well.

I often find out that I have made a mistake along the way which has made the problem ten times more complicated than it should be. In this case I have about a page of calculations, and even if I was doing it the right way, there could easily be an error in there somewhere that would be problematic for me. I was wondering if there was a way to check the answer.

The initial wave function is
$$\Psi(x,t=0)=A(a^2-x^2)$$

for a square well from -a to a, where A is the normalisation constant.

and my Probability is

$$P_n=\frac{15}{n^2\pi^2a^4}(\frac{4a^2}{n^2\pi^2}+(-1)^{(n-1)}\frac{4a}{n\pi}+1)$$

when n is odd, and $$P_n=0$$ when n is even

Apart from anything else, this seems extremely complicated for such a simple wave function, so I am inclined to think I've made a mistake. Can anyone help me figure out if I need to go back and start from scratch? If need be I can post all my workings out, but being so long it will take a little while to type up, so I will only post it if requested. Thank you.

2. Mar 12, 2009

kreil

the coefficients $|c_n|^2$ tells you the probability that a measurement of the energy will yield the value $E_n$

since the expectation value of the energy is

$$<H> = \Sigma_{n=1}^{\infty} |c_n|^2 E_n$$

$$c_n= \int \psi^*(x) \psi(x) dx$$

which is why you first need to normalize for A using:

$$\int_{-a}^a | \psi(x) |^2 dx = |A|^2 \int_{-a}^{a} (a^2-x^2)^2 dx = 1$$

Last edited: Mar 12, 2009
3. Mar 12, 2009

LongLiveYorke

Well, just looking at your expression for probability, we know you made a mistake somewhere because it depends on a, which is a unit-full quantity. Since probability is unitless, it shouldn't depend on the value of a. And in your quantity in the parentheses, the units don't match up, since one is proportional to a^2 and the other a.

So, I would go back and double check your math. In the end, the a's should cancel. If you have access to mathematica, you can use it to check your integral, or you can find some sort of online integrator to check your math.

4. Mar 13, 2009

ballzac

Ah, yes. That all makes sense to me. Apart from the fact that when I get the right answer it will probably 'look' right, I was thinking that maybe I could use my value for $$c_n$$ and the eigenfunctions to approximate the infinite series and see if it is converges towards the original wave function.

5. Mar 13, 2009

ballzac

Help! I'm still doing something wrong. It looks like such a simple problem, but I must be making a basic mistake somewhere.
I put this into mathematica:

Integrate[A (a^2 - x^2) cos (n pi x/(2 a)), {x, 0, a}]

Is it the correct integral for the problem at hand?
I am assuming here that n is odd.

The output is
1/8 a^3 A cos n pi

Okay, I put

Integrate[A^2 (a^2 - x^2)^2, {x, -a, a}]

into mathematica. Is this the correct integral for normalising the wave function?

The output is

(16 a^5 A^2)/15

This must equal unity, so I rearrange this to get.

$$A=\frac{\sqrt{15}}{4}a^{-5/2}$$

Okay, substituting this into the result from the previous integral, gives

$$c_n=\frac{\sqrt{15a}}{32}cos(n\pi)$$

so

$$P_n=c_n^2=\frac{15a}{32^2}cos^2(n\pi)$$

This still depends on a, and as LLY stated, this cannot be the case, and as the integrals were evaluated by computer, either I have made a simple error in the integrals that I cannot spot, or I completely misunderstand how to solve this type of problem. Thanks in advance for the help.

6. Mar 13, 2009

Hurkyl

Staff Emeritus
If you're going to pay attention to units, you have to pay attention to all of them. In particular, your expression for the eigenfunction uses the convention that wavefunctions are unitless... but your expression for the inner product requires that wavefunctions have units (length)-1/2!

7. Mar 13, 2009

ballzac

That was really dumb of me, though I'm glad it was something stupid like that and not that I was misunderstanding the problem. Thank you so much for the help. I now have

$$P_n=\frac{15}{32^2}cos^2(n\pi)$$

which looks much, much better.

8. Mar 13, 2009

ballzac

oops. I am only just getting familiar with mathematica and had entered the equation incorrectly. Now it's looking a bit more complicated, but correct I suspect.

EDIT: Actually the result of the integral looked complex, but due to the fact that n is odd, a lot of factors disappeared, then a trigonometric factor disappeared because P_n is the SQUARE of c_n.

My final answer is $$P_n=\frac{960}{n^6\pi^6}$$
Hopefully it is correct now. The maths seems to work out, e.g. it is always less than one, and it looks like it would add up to one over all n.

Last edited: Mar 13, 2009