I was wondering how it would sound like the creation/annihilation of particles that we usually do in the context of Dirac formalism, with matrices and vectors. As a reminder we know that:
\begin{equation}
a^{-}\left|n\right\rangle =\sqrt{n}\left|n-1\right\rangle
\end{equation}
and
\begin{equation}
a^{+}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle
\end{equation}
The operators are defined as:
\begin{equation*}a^{-}=\left(\begin{array}{ccccc}
0 & 1 & 0 & 0 & …\\
0 & 0 & 1 & 0 & …\\
0 & 0 & 0 & 1 & …\\
0 & 0 & 0 & 0 & …\\
… & … & … & … & …
\end{array}\right)
\end{equation*}
and
\begin{equation*}a^{+}=\left(\begin{array}{ccccc}
0 & 0 & 0 & 0 & …\\
1 & 0 & 0 & 0 & …\\
0 & 1 & 0 & 0 & …\\
0 & 0 & 1 & 0 & …\\
… & … & … & … & …
\end{array}\right)
\end{equation*}
So how is the form of the ket $\left|n\right\rangle$ in terms of a vector? It must a tensor product of vectors for sure. Moreover the matrix $a^{-}$ (or $a^{+}$) times $\left|0\right\rangle$ (probably a vector with all its entries equal to $0$?) must be something that gives a vector with an entry increased of $1$. But, to be rigorous, a prefix should be added to the matrix in order to indicate that it is acting on a precise factor of the tensor product which made up $\left|n\right\rangle$.
An example of a multiplication between a creation/annihilation matrix for a vector would be highly appreciated.
If the question is not clear let me know!
Best Answer
As mentioned in the comments. Your matrix representation of the creation and annihilation operators is incorrect. This is easy to see since \begin{align} a ^\dagger _{ nm} & = \left\langle n \right| a ^\dagger \left| m \right\rangle \\ & = \sqrt{ m + 1 } \delta _{ n , m + 1 }. \end{align} Thus we have, \begin{equation} \left( \begin{array}{cccc} 0 & 0 & 0 & ...\\ 1 & 0 & 0 &...\\ 0 & \sqrt{2} & 0 &... \\ 0 &0 & \sqrt{3} & \ddots\\ \vdots &... & \ddots & \ddots \end{array} \right) \end{equation}
The kets in this space are simple: \begin{equation} \left| n \right\rangle = \left( \begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right) \end{equation} where $1$ is at the $ n +1 $'th spot down the vector. Its easy to see that acting on this vector we get a result consistent with the relation for the raising operator above.
The vacuum state is then just given by \begin{equation} \left| 0 \right\rangle = \left( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right) \end{equation}