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Need a refresher on finding probabilities of a wave function.

  • Thread starter mateomy
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  • #1
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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:
[tex]
\Psi = A(1 + 4cos\phi)
[/tex]
I need to find the corresponding probabilities. So I know that I have to normalize the [itex]\Psi[/itex] 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).

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

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 [itex]\Psi[/itex] where A is a normalization constant and phi is the angle the radius vector makes with the x-axis. If [itex]L_z[/itex] is measured, what are the possible outcomes and corresponding possibilities?




Thanks.
 
Last edited:

Answers and Replies

  • #2
DrClaude
Mentor
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$$
(1 + 4 \cos \phi)^2 \neq 1 + 16 \cos^2 \phi
$$
 
  • #3
CompuChip
Science Advisor
Homework Helper
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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:
[tex]
\Psi = A(1 + 4cos\phi)
[/tex]
I need to find the corresponding probabilities. So I know that I have to normalize the [itex]\Psi[/itex] 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).

[tex]
\Psi = \int_{0}^{2\pi} A^2 d\phi + \int_{0}^{2\pi} A^2 16 cos^{2}\phi d\phi
[/tex]
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
[tex]
\int_0^{2\pi} |\Psi|^2 \, d\phi = \int_{0}^{2\pi} (1 + 4 \cos\phi)^2 d\phi = 1
[/tex]
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.
 
  • #4
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Ah, I see the now obvious mistake in the math...
That's embarrassing. Here's the fix of the 2-factor:
[tex]
\int_{0}^{2\pi}A^2(1 + 8cos\phi + 16cos^2\phi)d\phi
[/tex]

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 [itex]2\pi[/itex] of the wire then the last two terms go to zero. I'm kind of lost how to handle this problem...
 
Last edited:
  • #5
CompuChip
Science Advisor
Homework Helper
4,302
47
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 :-)
 
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