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.

  • 3
    $\begingroup$ $\langle \alpha |\beta \rangle$ is a scalar so it certainly commutes with a linear operator. $\endgroup$ – Ian Oct 26 '17 at 5:09

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Kosta Perhaps you should check your definition for what $\langle \alpha | \beta \rangle$ means. $\endgroup$ – angryavian Oct 26 '17 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.