Need a refresher on finding probabilities of a wave function.

[mateomy]

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:
<br /> \Psi = A(1 + 4cos\phi)<br />
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).

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

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.

 

[DrClaude]

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

 

[CompuChip]

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:
<br /> \Psi = A(1 + 4cos\phi)<br />
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).

<br /> \Psi = \int_{0}^{2\pi} A^2 d\phi + \int_{0}^{2\pi} A^2 16 cos^{2}\phi d\phi<br />

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
<br /> \int_0^{2\pi} |\Psi|^2 \, d\phi = \int_{0}^{2\pi} (1 + 4 \cos\phi)^2 d\phi = 1<br />
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.

 

[mateomy]

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

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...

 

[CompuChip]

Let's start with correctly determining the normalization constant, since that is giving you enough trouble right now.

You may need to take a step back in opening the brackets on the square because you are missing a factor of 2.
Also, $\cos^2 x$ does not go to zero when integrated over a $2\pi$-interval :-)