Let $O$ be a single-particle observable for a system, and $|L\rangle$ and $|R\rangle$ two orthonormal eigenstates of $O$. You may imagine that the system consists in two photons, and $|L\rangle$ and $|R\rangle$ represent some orthogonal polarization states of each photon.
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Then what is the physical meaning of the operator $|L\rangle\langle L|$ (or $|R\rangle\langle R|$ for that matter)? Is it an observable on $H$? If so, what kind of observable is it? More precisely, what is its relation with respect to $O$, $|L\rangle$, or $|R\rangle$ from a physical point of view?
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Relatedly, let $\hat{L} \equiv |L\rangle\langle L|$ and $\hat{L}^{2} \equiv \hat{L} \otimes I + I \otimes \hat{L} $.
Then is it physically meaningful to think about the expected value of $L^{2}$ with respect to, say, $|L\rangle|L\rangle$? In other words, does $\langle L|\langle L|\hat{L}^{2}|L\rangle|L\rangle$ have any non-trivial physical meaning?
Thanks for reading.
Best Answer
A projector is an observable - you can directly check that it is Hermitian $|L\rangle\langle L|^\dagger = |L\rangle \langle L|$. As to interpretation - a projector onto a single state will measure the value $1$ for definite if the system is in that state. If the system is in an orthogonal state it will measure $0$. Therefore you can think of projectors as operators whose measurement corresponds to asking a binary question. Any measurement you can think of can be approximated by a series of binary questions and so its not surprising that any observable can be decomposed into such projectors.
As for your second question: I don't see why not. The notation $L^2$ is confusing though - I'd stick to calling this $L_1+L_2$ or similar. Note that this operator is not a projector. It's still Hermitian, and it's a reasonable thing to consider if you have two subsystems on which $L$ is itself sensible to consider.