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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.

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    $\begingroup$ 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. $\endgroup$ – Tyberius Sep 12 '17 at 21:58
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    $\begingroup$ Related: en.m.wikipedia.org/wiki/Semicircular_potential_well $\endgroup$ – Tyberius Sep 13 '17 at 0:51
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$$ \newcommand{\ket}[1]{\,\lvert{#1}\rangle} \newcommand{\bra}[1]{\langle{#1}\rvert\,} \newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle} \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle} \newcommand{\op}[1]{\hat{#1}} $$

Solution should be a pretty straightforward application of the Born rule.


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.

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    $\begingroup$ Mister Wildcat, welcome back! We missed you. ;D $\endgroup$ – Mithoron Sep 13 '17 at 0:57

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