# How are the creation and annihilation operators constructed for a single mode?

I'm reading Weinberg's QFT, and he defines the creation and annihilation operators as \begin{align} (a_k)_{n_1',n_2',\dots,n_1,n_2,\dots}&=\sqrt{n_k}\delta_{n_k',n_k-1}\prod_{j\ne k}\delta_{n_j',n_j}\\ (a_k^\dagger)_{n_1',n_2',\dots,n_1,n_2,\dots}&=\sqrt{n_k+1}\delta_{n_k',n_k+1}\prod_{j\ne k}\delta_{n_j',n_j}\\ \end{align} where I think he might have made a typo in the second equation (the original was $$\delta_{n_{k'}'n_k+1}$$ for the first delta). He then gives the matrix representations for a single normal mode as

$$\begin{equation*} a=\begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \dots\\ 0 & 0 & \sqrt{2} & 0 & \dots\\ 0 & 0 & 0 & \sqrt{3} & \dots\\ 0 & 0 & 0 & 0 & \dots\\ \dots & \dots & \dots & \dots & \dots \end{bmatrix} \qquad a^\dagger=\begin{bmatrix} 0 & 0 & 0 & 0 & \dots\\ \sqrt{1} & 0 & 0 & 0 & \dots\\ 0 & \sqrt{2} & 0 & 0 & \dots\\ 0 & 0 & \sqrt{3} & 0 & \dots\\ \dots & \dots & \dots & \dots & \dots \end{bmatrix} \end{equation*}$$

My question is, how are these constructed? It seems to force $$n_k=k$$, but as far as I know for bosons $$n_k$$ can be any number. Furthermore, doesn't each $$n_k$$ correspond to the number of quanta per mode? He then says that these operating on a column vector with integer components $$n_1,n_2,\dots$$ represent a state with $$n_k$$ in each normal mode $$k$$. Then $$a_k$$ will just lower/raise $$n_k$$ by one unit. However, I do not see this from the matrix that he has given. It seems that if I have a column vector arbitrary integer entries they all get mixed up when either $$a$$ or $$a^\dagger$$ acts on them, instead of just $$n_k$$.

The product operator also seems quite contradictory to me, since it seems to say it's non-zero only if $$n_j'=n_j$$, but for a single value $$n_k'=n_k-1$$ must be true in order for it to be non-zero. In that case how is the matrix not a single entry, since for the next entry the product is immediately no longer fulfilled.

Could someone explain this to me please?

Just ignore all the $$k$$ subscripts and drop the product over $$j\ne k$$, as it is empty $$(a)_{n',n} = \sqrt{n}\delta_{n',n-1}$$ which is the matrix Weinberg gives. If you apply this matrix to a vector with one non-zero component in the $$n$$th place, you will see it results in vector with a single non-zero component in the $$n-1$$th place, that is it reduces $$n$$ by 1.
As for the $$j\ne k$$ terms, remember that $$\delta_{i,j}$$ is just a representation of the identity matrix, so this term is saying it does nothing to the $$n_j$$ with $$j\ne k$$, as you would expect.
Similar considerations apply to $$a^\dagger$$.
• What I mean with dropping the product over all the $j\ne k$, when you only have the one component there are no $j \ne k$ and so you should treat it as 1. When you only have one mode which energy level you are in is determined by the number of quanta, so your second question does not really make sense unless you have more modes (in which case $a_k$ only acts on one mode and ignores the others). If a given mode is in a superposition of different vales of $n$, then you can pull it apart by linearity and let $a$ act on each component separately Feb 28, 2022 at 13:30
• I was just thinking, the operator doesn't seem to change the actual n number? For example, if I have 1 in the second row (1st excited state), then when I raise it it has $\sqrt{2}[0,0,1,\dots]$, but shouldn't $|2\rangle=[0,0,2,\dots]$? Feb 28, 2022 at 15:37