# Bose algebra

Here are collected some results related with the operator $a$ which commutation with its adjoint $\ud{a}$ reads:

$$\tag{1}[a,\ud{a}]=1$$

(cf. Fermi algebra)

I have written a Mathematica piece of code to compute such correlators automatically. I found out in this way formulas already known in the literature, in particular from the work of Blasiak[1][2]. You can download it to play with it (check or compute results useful in your daily quantum algebra). This is detailed in this blog post.

## Creation & annihilation

The basic rules are:

\begin{align*} a\ket{n}&=\sqrt{n}\ket{n-1}\,,&\bra{n}\,&a=\bra{n+1}\sqrt{n+1}\,,\\ \ud{a}\ket{n}&=\sqrt{n+1}\ket{n+1}\,,&\bra{n}\,&\ud{a}=\bra{n-1}\sqrt{n}\,, \end{align*}

of which a general expression can be drawn:

$$a^{\dagger i}a^j a^{\dagger k}\ket{n}={(n+k)!\over(n+k-j)!}\sqrt{(n+i+k-j)!\over n!}\ket{n+i+k-j}\,.$$

Some particular cases:

\begin{align*} \kern-1cm{(\mathrm{for}~i\le n+j)}\kern1cm a^i{\ud{a}}^j\ket{n}&={(n+j)!\over\sqrt{n!}\sqrt{(n+j-i)!}}\ket{n+j-i}\,,\\ \kern-1cm{(\mathrm{for}~i\le n)}\kern1cm a^{\dagger j}a^i\ket{n}&={\sqrt{n!}\sqrt{(n+j-i)!}\over(n-i)!}\ket{n+j-i}\,. \end{align*}

## Normal ordering

A general result, based on Wick theorem:

$$a^n\ud{a}^m=\sum_{k=0}^{\min(n,m)}k!\binom{m}{k}\binom{n}{k}\ud{a}^{m-k}a^{n-k}$$

Another one, inferred from my Mathematica notebook above, that could certainly be proved by recurrence or directly from combinatorics:

$$(a\ud{a})^n=\sum_{k=0}^n \left\{\begin{matrix} n\\k \end{matrix} \right\} \ud{a}^ka^k$$

where $\left\{\begin{matrix} n\\k \end{matrix} \right\}$ are the Stirling partition numbers (the number of ways to partition a set of $n$ objects into $k$ non-empty subsets). It is easy to obtain similar results in terms of generalized Stirling and Bell numbers for $(a^k\ud{l})^n$, which are results already provided by Blasiak.

### Some useful particular cases

$$a\ud{a}^n=\ud{a}^na+n\ud{a}^{n-1}$$ $$a^n\ud{a}=\ud{a}a^n+na^{n-1}$$

### Some recurrent expressions

A fairly general result (forall $k, l\in\mathbb{N}$, including $0$):

$$aa^{\dagger k}a^la^\dagger=a^{\dagger(k+1)}a^{l+1}+(k+l+1)a^{\dagger k}a^l+kla^{\dagger(k-1)}a^{l-1}$$

I maintain a long list of Bose algebra expressions in canonic notations to assist me in my computations. You can glance at it to find particular and redundant cases of the above.

## Commutation

A compilation of useful results derived from Eq.~(1):

We define $\hat n\equiv\ud{a}a$.

 $[a^n,\ud{a}]=na^{n-1}$ $[a,\ud{a}^n]=n\ud{a}^{n-1}$ $[\ud{a},a^n]=-na^{n-1}$ $[\ud{a}^n,a]=-n\ud{a}^{n-1}$ $[a^2,\ud{a}^n]=2n\ud{a}^{n-1}a+n(n-1)\ud{a}^{n-2}$ $[\ud{a}^2,a^n]=-2n\ud{a}a^{n-1}-n(n-1)a^{n-2}$ $[\hat n,a^m]=-ma^m$ $[a,(\hat n)^k]=\left(\sum_{i=0}^{k-1}\binom{k}{i}\hat n^i\right)a$ $[\ud{a}^ma^n,\hat n]=(n-m)\ud{a}^ma^n$ Special cases below $[a,\hat n]=a$ $[a,\hat n^2]=(2\hat n+1)a$ $[a,\hat n^3]=(3\hat n^2+3\hat n+1)a$ $[a,\hat n^4]=(4\hat n^3+6\hat n^2+4\hat n+1)a$

## Particle numbers

• $a\ud{a}=\hat n+1$.
• $\ud{a}^2a^2=\hat n^2-\hat n$.