How do second quantization operators act on states of "incompatible" quantum numbers?

Question

In solid state physics we might describe systems using second quantization and use the Bloch basis for the states of the quantum mechanical system. For example, to create an electron in band $n$ at $k$:
$$ a^dagger_{nk}left|0rightrangle= left|nkrightrangle $$
My question is, how would a general annihilation operator $a_{mk'}$ act on the state $left|nkrightrangle $ where $nneq m$ and more importantly $k'neq k$? Clearly, the special case where $n=m$ and $k' = k$ sets the system back to the vacuum state. But what about other states?
This can of course be generalized to other quantum mechanical systems: How do the annihilation operators act on states that the corresponding creation operators did not create?

CGS · Answer

The operator $a_{mk'}$ only works on the state k' in the m band.  If no electron is in this state within this band, then the result is 0.  That's what happens if an annihilation operator operates on an empty state (or, in your words, one that a creation operator did not create).  The operator for your n,k state would be $a_{nk}$
In second quantization all operators are written as pair of creation/annihilation operators.  For example, the occupation number operator is written as: $$hat{n_r} = c^dagger_rc_r$$ Note if the state is unoccupied the value this operator returns is 0, since the annihilation operator operates first on an empty state.  It returns 1 if the state is occupied.  Both of these results are correct.

How do second quantization operators act on states of "incompatible" quantum numbers?

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