# Why is the probability of measuring an eigenvalue its coefficient squared?

1. Jun 22, 2011

### timn

1. The problem statement, all variables and given/known data

This is an example from Gasiorowicz's Quantum Physics. "Example 3-1" is a particle in an infinite potential-well, but that should not matter.

2. Relevant equations

3. The attempt at a solution

Why is P(-2) (which I suppose is the probability that the eigenvalue -2 is measured) the coefficient squared?

The sum of the squares of the coefficient should be normalised, so it makes sense, but I don't understand why.

To figure out how much an eigenfunction contributes to the probability function -- psi^2 -- I'd square psi as follows:

$$\left( \frac{N}{4}\sqrt{2\pi}(u_ {-2}+2u_0+u_2) \right)^2 = \frac{N^2\pi^2}{8}(u_ {-2}^2+2u_0^2+u_2^2+2u_{-2}u_2+4u_{-2}u_0+4u_0u_2)$$

followed by being completely lost.

Could anyone explain this or make it seem plausible for me?

Edit: For reference, the answer is P(-2)=1/6.

2. Jun 22, 2011

### G01

You can start this way if you want. Once you take the (absolute) square of psi, then remember that psi is normalized so let:

$$\int\left( \frac{N}{4}\sqrt{2\pi}(u_ {-2}+2u_0+u_2) \right)^2 d\phi = \frac{N^2\pi^2}{8}\int(u_ {-2}^2+2u_0^2+u_2^2+2u_{-2}u_2+4u_{-2}u_0+4u_0u_2)d\phi=1$$

Use the othogonality of the eigenfunctions to simplify the integral. Notice that terms like $u_n^*u_m$ with $m\not= n$ will integrate to zero. Thus after all the integrations are done, you should be left with:

$$\frac{2\pi N^2}{16}\int(|u_2|^2+4|u_0|^2+|u_{-2}|^2)d\phi=1$$

Simplify this by integrating. You will see that the sum of the absolute squares of the coefficients will be equal to one, and thus it also follows from your line of reasoning that the squares of the coefficients are the correct measure of probabilities for different eigenstates.

Last edited: Jun 22, 2011
3. Jun 22, 2011

### timn

Aha! I had forgotten about the orthogonality. Also, I had an epiphany when I realised that $$A_n = \int_0^a u_n^*(x) \psi(x) dx$$ is just a projection.

Thank you!