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.
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 $\hat{p} (4 |\alpha \rangle) = 4 (\hat{p} | \alpha \rangle)$.