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So I have, what may very well be, a trivial question. I'm working through Griffiths Quantum Mechanics text, problem 3.21.

Show that the projection operators are idempotent: $\hat p ^2 = \hat p$. Determine the eigenvalues of $\hat p$, and characterize its eigenfunctions.

My linear algebra is very weak. I wasn't required to take an advanced course in it. Rather, I was given a (very) brief introduction to linear algebra during my differential equations courses. My lack of experience in linear algebra is beginning to haunt me.

So here's my question.

I begin the problem by noting: $$ \begin{align} \hat p ^2 |\beta\rangle & = \hat p \hat p |\beta\rangle \\ & = \hat p \langle\alpha|\beta\rangle |\alpha\rangle \end{align} $$

I got this far from the projection operator the book gives. My confusion lies in the following step. I can see easily enough that if I'm able to to do the following operations, I can easily show the idempotence. I'm just not sure why you're able to do this.

Beginning again, $$ \begin{align} \hat p ^2 |\beta\rangle & = \hat p \hat p |\beta\rangle \\ & = \hat p \langle\alpha|\beta\rangle |\alpha\rangle \\ & = \langle\alpha|\beta\rangle \hat p |\alpha\rangle \ \text{(This step)} \\ & = \langle\alpha|\beta\rangle \langle\alpha|\alpha\rangle |\alpha\rangle \\ & = \langle\alpha|\beta\rangle |\alpha\rangle \\ & = \hat p |\beta\rangle \end{align} $$

The step marked "this step" confuses me. Up until here, I haven't seen any formal description regarding the distribution of operators, at least not in this fashion. Does it have something to do with the $\langle\alpha|\beta\rangle$ inner product? If so, why?

On another note, I'm unsure how to determine the eigenvalues of $\hat p$ here, and what it means to 'characterize' eigenfunctions.

Please be gentle on terminology, I'm still building my linear algebra vocabulary.

Thank you.

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    $\begingroup$ $\langle \alpha |\beta \rangle$ is a scalar so it certainly commutes with a linear operator. $\endgroup$ – Ian Oct 26 '17 at 5:09
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It is not as scary as you think. $\langle \alpha \mid \beta\rangle$ is a number (scalar), so you can move it around. It's like saying $\hat{p} (4 |\alpha \rangle) = 4 (\hat{p} | \alpha \rangle)$.

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  • $\begingroup$ Oh. Well, that is easy enough. This may be another trivial question, but why is the notation, $\langle\alpha|\beta\rangle$, a scalar? $\endgroup$ – Kosta Oct 26 '17 at 5:11
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    $\begingroup$ @Kosta Perhaps you should check your definition for what $\langle \alpha | \beta \rangle$ means. $\endgroup$ – angryavian Oct 26 '17 at 5:13

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