# Example of application of creation/annihilation operators in matrix form

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: $$a^{-}\left|n\right\rangle =\sqrt{n}\left|n-1\right\rangle$$ and $$a^{+}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle$$ 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!

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I don't believe those matrices are correct. It should be $\sqrt{1}$, $\sqrt{2}$, $\sqrt{3}$ along the diagonals. – Kyle Kanos Feb 6 '14 at 14:09
you are right! I'm sorry for this – edwineveningfall Feb 6 '14 at 15:32

As mentioned in the comments. You're matrix representation of the creation 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, $$\left( \begin{array}{cccc} 0 & 0 & 0 & ...\\ 1 & 0 & 0 &...\\ 0 & \sqrt{2} & 0 &... \\ 0 &0 & \sqrt{3} & \ddots\\ \vdots &... & \ddots & \ddots \end{array} \right)$$
The kets in this space are simple: $$\left| n \right\rangle = \left( \begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right)$$ 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 $$\left| 0 \right\rangle = \left( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right)$$