# Need a refresher on finding probabilities of a wave function.

After a few months off (yay summer/internships), I'm 'back in the saddle' and I'm trying to catch up with my Q-mech.

I have a wave function which is given as a particle sliding freely on a circular wire:
$$\Psi = A(1 + 4cos\phi)$$
I need to find the corresponding probabilities. So I know that I have to normalize the $\Psi$ by setting it to 1. This is split into two integrals which need to be multiplied by their complex conjugates (which are both just real values).

$$\Psi = \int_{0}^{2\pi} A^2 d\phi + \int_{0}^{2\pi} A^2 16 cos^{2}\phi d\phi$$
Eventually finding, assuming my integration wasn't messed up somewhere (spoiler, I think it was)..
$$\sqrt{\frac{1}{2\pi}}\Psi + \sqrt{\frac{1}{16\pi}}\Psi$$

So my probabilities are just the coefficients, right? I'm looking for a critique on this as well as I'm pretty sure I did something incorrectly.

Full question:

The wave function $\Psi$ where A is a normalization constant and phi is the angle the radius vector makes with the x-axis. If $L_z$ is measured, what are the possible outcomes and corresponding possibilities?

Thanks.

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DrClaude
Mentor
$$(1 + 4 \cos \phi)^2 \neq 1 + 16 \cos^2 \phi$$

CompuChip
Homework Helper
After a few months off (yay summer/internships), I'm 'back in the saddle' and I'm trying to catch up with my Q-mech.

I have a wave function which is given as a particle sliding freely on a circular wire:
$$\Psi = A(1 + 4cos\phi)$$
I need to find the corresponding probabilities. So I know that I have to normalize the $\Psi$ by setting it to 1. This is split into two integrals which need to be multiplied by their complex conjugates (which are both just real values).

$$\Psi = \int_{0}^{2\pi} A^2 d\phi + \int_{0}^{2\pi} A^2 16 cos^{2}\phi d\phi$$

The idea is good, but there is a simple mathematical error as pointed out by DrClaude.
On a more fundamental level, I think you are a bit confused what you need to set to 1. The integral gives the probability that the particle will be between ##phi = 0## and ##2\pi##, and that is what should be equal to 100%.

So what you need to solve is
$$\int_0^{2\pi} |\Psi|^2 \, d\phi = \int_{0}^{2\pi} (1 + 4 \cos\phi)^2 d\phi = 1$$
which will give you the value for A, and then you can calculate the probability of finding the particle at ##\phi \in [a, b]## by
$$\int_a^b |\Psi|^2 \, d\phi$$

So my probabilities are just the coefficients, right?
I think you are getting confused with something else. If
$$\Psi = c_1 \psi_1 + \cdots + c_n \psi_n$$
where ##\psi_i## are the eigenstates of your observable (e.g. the Hamiltonian or the position operator) then ##|c_i|^2## will be the probability of finding the particle in eigenstate i when you apply that operator - i.e. observe that property.

Ah, I see the now obvious mistake in the math...
That's embarrassing. Here's the fix of the 2-factor:
$$\int_{0}^{2\pi}A^2(1 + 8cos\phi + 16cos^2\phi)d\phi$$

In the interest of full disclosure I've updated the entire question. Because I've managed to confuse myself entirely.
(See edited post)

If I take my interval as the $2\pi$ of the wire then the last two terms go to zero. I'm kind of lost how to handle this problem...

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CompuChip