# Wavefunction operators and the observable [closed]

So I got this from the exam I had yesterday. I couldn't really answer it other and it played on my mind through the night

Show that if a wave function $\psi$ , is an eigenfunction of an operator [Q], then the observable Q is sharp for that wave function.

I said that if it is an eigenfunction then performing the operator must be a whole multiple of the wavefunction and then tried to bull my way through it. What would you guys have accepted as an answer.

I mean I know that $\Delta Q=\sqrt{\overline{Q^2}- \overline{Q}^2}$

But how do I translate my knowledge across to give an answer that if the observable is sharp this reduces to 0?

-

## closed as off-topic by Manishearth♦Jul 12 '13 at 5:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Manishearth
If this question can be reworded to fit the rules in the help center, please edit the question.

Did you mean $\Delta{Q} =\sqrt{\langle Q^2\rangle-\langle Q\rangle^2}$ ? – John Rennie Jun 27 '13 at 7:21
I know it has a bar right across the first one and a bar over just the Q for the second then squared thats what it has in the textbook. I just used the form that u gave when I needed it yesterday – Jesse Ross Jun 27 '13 at 7:28

If the system is in an eigenstate of $Q$, then a measurement of $Q$ will yield the corresponding eigenvalue with probability one. An implication of this is that $Q$ has vanishing uncertainty in the state $\psi$ as you note. Here's how to show that mathematically:
If $\psi$ is a normalized eigenfunction of $Q$, then there exists some real number $q$, it's eigenvalue, for which $$Q\psi = q \psi$$ and therefore $$Q^2\psi = q^2\psi.$$ It follows that \begin{align} \langle Q\rangle^2 &= \langle\psi,Q\psi\rangle^2= q^2\langle\psi,\psi\rangle^2 =q^2\\ \langle Q^2\rangle &= \langle\psi,Q^2\psi\rangle = q^2\langle\psi,\psi\rangle=q^2 \end{align} so the uncertainty in measuring $Q$ for the state $\psi$ is \begin{align} \Delta Q = \sqrt{\langle Q^2\rangle - \langle Q\rangle^2} = \sqrt{q^2-q^2} = 0 \end{align}