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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?

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

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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
up vote 2 down vote accepted

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}

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thank you for that. It's been playing on my mind. – Jesse Ross Jun 27 '13 at 7:30
@JesseRoss Sure thing. – joshphysics Jun 27 '13 at 7:32

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