I agree that $$\hat a|0\rangle=0$$ But then, based on the above, the following should hold $$\hat a_k |N_1,...,N_{k-1},0,N_{k+1},...\rangle=|N_1\rangle\oplus\cdots\oplus |N_{k-1}\rangle\oplus \hat a_k|0\rangle \oplus|N_{k+1}\rangle\cdots$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=|N_1\rangle\oplus\cdots\oplus |N_{k-1}\rangle\oplus 0 \oplus|N_{k+1}\rangle\cdots$$

However, I've seen QFT textbooks claim that the calculation should result in $0$. $$\hat a_k |N_1,...N_{k-1},0,N_{k+1},...\rangle=0$$ Then does this mean that $$|N_1\rangle\oplus\cdots\oplus |N_{k-1}\rangle\oplus 0 \oplus|N_{k+1}\rangle\cdots=0\tag{*}$$ I don't see how what we've derived makes sense. Perhaps someone could explain why (*) makes sense mathematically?

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    $\begingroup$ It's a tensor product, not a direct sum. That's your error. $\endgroup$ – DanielSank Oct 1 '14 at 4:48
  • $\begingroup$ It is customary to mark an answer as accepted once one of them is acceptable to you. If none are acceptable please indicate where you need further information. $\endgroup$ – DanielSank Oct 4 '14 at 16:34

Suppose I have two quantum modes $|\cdot \rangle_A$ and $|\cdot \rangle_B$. Suppose mode $A$ is in state

$$\sum_n a_n|n\rangle_A$$

and mode $B$ is in state

$$\sum_m b_m|m\rangle_B . $$

Then the state of the combined system is

$$\sum_{n,m} a_n b_m \left( |n\rangle_A \otimes |m\rangle_B \right). $$

Note that the coefficients are multiplied. Now consider the special case where mode $A$ is in state

$$|0\rangle_A . $$

Then the system state is

$$ \sum_m b_m \left( |0\rangle_A \otimes |m\rangle_B \right) = |0\rangle_A \otimes \left( \sum_m b_m |m\rangle_B \right) . $$

If we act $a_A$ on this thing we get

$$ a |0\rangle_A \otimes \left( \sum_m b_m |m\rangle_B \right) = 0 \otimes \left( \sum_m b_m |m\rangle_B \right) = 0. $$

If your question really boils down to what those little $\otimes$ symbols really mean then that might make another good question. You can roughly (and quite successfully) think of them as a thing which multiplies numbers and concatenates vector spaces (i.e. concatenates the spaces of the various quantum modes). It's the multiplication of the numbers that leads to the zero in this problem.

This discussion of second quantisation may be useful.

P.S. Before the theory train demolishes my house, I know that this summary of how tensor products work is rough and incomplete. This is intentional.


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