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. 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

$$a\ud{a}a\ud{a}^k=\ud{a}^{k+1}a^2+(1+2k)\ud{a}^ka+k^2\ud{a}^{k-1}$$

$$a^l\ud{a}a\ud{a}=\ud{a}^2a^{l+1}+(1+2l)\ud{a}a^l+l^2a^{l-1}$$

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

$$a^\dagger aa^{\dagger k}a^{l}=a^{\dagger(k+1)}a^{l+1}+ka^{\dagger k}a^l$$

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

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

$$(a^\dagger a)^2a^{\dagger k}a^l=a^{\dagger(k+2)}a^{l+2}+(1+2k)a^{\dagger(k+1)}a^{l+1}+k^2a^{\dagger k}a^l$$

$$a^{\dagger k}a^l(a^{\dagger}a)^2=a^{\dagger(k+2)}a^{l+2}+(1+2l)a^{\dagger(k+1)}a^{l+1}+l^2a^{\dagger k}a^l$$

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

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