Solving for probability density (Griffith's 1.3)

[gfd43tg]

Homework Statement

Problem statement in image

Homework Equations

The Attempt at a Solution

From my reading, I know that the probability density
$$\langle x \rangle = \int x |\psi|^{2} \hspace 0.05 in dx$$

Where $\psi$ is the wave function and $\langle x \rangle$ is the probability density. So I figure that I will go ahead and do this,

$$\rho (\theta) = \int \theta |\psi|^{2} \hspace 0.05 in d \theta$$

However, how do I figure out the wave function of this? I don't have a good sense for this problem and can't connect the pieces.

 

[Orodruin]

What you have written down is the expectation value, not the probability density. It is stated in the problem what the probability density is and what properties it should have.

 

[gfd43tg]

Okay, so then I interpret the question to mean when it says that if flicked, it is equally likely to come to rest at any angle between zero and pi

$$P = \int_{0}^{\pi} \rho (\theta) \hspace 0.02 in d \theta = 1$$

So if I assume $\rho (\theta)$ is constant, then $\rho (\theta) = \frac {1}{\pi}$

 

[Orodruin]

Correct. It does not have to be harder than that :)

 

[Ray Vickson]

Okay, so then I interpret the question to mean when it says that if flicked, it is equally likely to come to rest at any angle between zero and pi

$$P = \int_{0}^{\pi} \rho (\theta) \hspace 0.02 in d \theta = 1$$

So if I assume $\rho (\theta)$ is constant, then $\rho (\theta) = \frac {1}{\pi}$

No assumption needed; that is exactly what is meant by "equally likely".

You seem to have some basic misunderstandings of probability. It usually has nothing to do with Quantum Mechanics or wave functions and the like. If we talk about probabilities in coin-tossing, there are no wave functions involved. If we talk about outcomes when spinning a roulette wheel, no wave functions are involved. When we talk about probabilities of car accidents, no wave functions are involved. When I want to know how long I have to wait for a bus, no wave functions come into the issue. Only when we talk about probabilities of events at the atomic or subatomic scale do wave functions enter into the calculation.