Density operator for two-particle system Suppose that $|a\rangle$ represents the state of a single-boson, and $|a \rangle |a \rangle$ that of two identical bosons.
Is there any operator that distinguishes $|a\rangle$ and $|a\rangle|a\rangle$?
Also, what happens if $|a\rangle \langle a|$ is acted on $|a\rangle|a\rangle$? 
Is it just
$$(|a\rangle \langle a|)|a\rangle|a\rangle = |a\rangle \langle a|a\rangle |a\rangle=|a\rangle |a\rangle \, ?$$
 A: When dealing with multiple particles, you need to supply the states with an index, designating which particle is in which state, e.g.,
$$|a\rangle_1 |b\rangle_2.$$
Then, depending on whether you treat bosons or fermions, you need to symmetrize/antisymmetrize your states:
$$|a, b\rangle = (|a\rangle_1 |b\rangle_2 \pm |a\rangle_2 |b\rangle_1)/\sqrt{2}.$$
(When $a=b$ the Pauli principle for fermions immediately follows.)
One can now write a projector on the state where both particles are in state $a$ as
$$|a, a\rangle\langle a,a|.$$
When working with states that may contain different number of particles, one can opt for the second quantization notation, which counts how many particles are in each state:
$$|1_a\rangle =|a\rangle_1,\\
|1_a, 1_b\rangle =(|a\rangle_1 |b\rangle_2 + |a\rangle_2 |b\rangle_1)/\sqrt{2},\\
|2_a\rangle =|a\rangle_1|a\rangle_2,$$
to which we need for completeness to add a state with no particles:
$$|\emptyset\rangle.$$
The projector on the state where state $a$ is occupied by two particles is
$$|2_a\rangle\langle 2_a|$$
