# How to calculate the probability of a particular value of an observable being measured

A particle on a ring is prepared with a wave function equal to $\frac{1}{\sqrt{\pi}}$ between $\theta=0$ and $\theta=\pi$, and $0$ between $\theta=\pi$ and $\theta=2\pi$. If a measurement of the angular momentum is made, calculate the probability of finding a particular value $l\hbar$. [The angular momentum eigenfunctions are $|l>=(2\pi)^{-1/2}e^{il\theta}]$

I've been reading the Oxford Chemistry Primer on Quantum Mechanics and they don't provide solutions. I've been struggling with this question.

I think I need to calculate the coefficient of expansion for a particular basis eigenfunction, but I'm unsure how to proceed.

• In the future, write out your question rather than including a picture of the text. That way, people can find this particular question from a search within the site or from Google. – Tyberius Sep 12 '17 at 21:58
• – Tyberius Sep 13 '17 at 0:51

In this particular case, we have a self-adjoint operator $\op{L_z}$ with eigenfunctions and eigenvalues satisfying $$\op{L_z} \psi_l(\theta) = l \hbar \psi_l(\theta) \, , \quad \text{where} \quad \psi_l(\theta) = \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{\mathrm{i} l \theta} \quad \text{and} \quad l = 0, \pm 1, \pm 2, \dotsc$$ such that any wave function $\psi$ can be expanded over orthonormal basis $\{ \psi_l(\theta) \}_{l=-\infty}^{\infty}$ of the corresponding eigenfunctions of $\op{L_z}$ as follows, $$\psi(\theta) = \sum\limits_{l=-\infty}^{\infty} c_l \psi_l(\theta) \, , \quad \text{where} \quad c_l = \braket{\psi_l}{\psi} = \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta \, .$$ And a measurement on a system in arbitrary state $\psi$ can yield any of eigenvalues $l \hbar$ with the probability given as follows $$\Pr(l \hbar) = |c_l|^{2} = c_l^* c_l \, .$$
Now, since $\psi(\theta)$ is defined to be $$\psi(\theta) = \begin{cases} \frac{1}{\sqrt{\pi}} & \text{if} \quad \theta \in [0, \pi] \\ 0 & \text{if} \quad \theta \in [\pi, 2 \pi] \end{cases} \, ,$$ the coefficient $c_l$ is equal to $$c_l = \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta = \int\limits_{0}^{\pi} \psi_l^*(\theta) 1/\sqrt{\pi} \mathrm{d} \theta + \int\limits_{\pi}^{2 \pi} \psi_l^*(\theta) 0 \mathrm{d} \theta \, ,$$ where the second term trivially vanishes leading to $$c_l = \int\limits_{0}^{\pi} \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\mathrm{i} l \theta} \frac{1}{\sqrt{\pi}} \mathrm{d} \theta = \int\limits_{0}^{\pi} \frac{1}{\sqrt{2} \pi} \mathrm{e}^{-\mathrm{i} l \theta} \mathrm{d} \theta \, .$$ And then some simple algebra leads to the final answer, which I leave to the OP, since it is far into the night for me and I'm afraid I will make a lot of stupid mistakes.