Latest revision as of 16:47, 1 October 2024

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, e.g.:

$$(\ud{a}a)^n=\sum_{k=1}^nS(n,k)(\ud{a})^ka^k$$ where $S(n,k)$ are the Stirling numbers of the second kind (Mathematica StirlingS2[n,k]). For instance: $$(\ud{a}a)^3=\ud{a}^3a^3+3\ud{a}^2a^2+\ud{a}a\,.$$

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

Particle numbers

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

References

1. ↑ The general boson normal ordering problem. P. Blasiak, K. A. Penson and A. I. Solomon in Phys. Lett. A 309:198 (2003).
2. ↑ Combinatorics and Boson normal ordering: A gentle introduction. P. Blasiak, A. Horzela, K. A. Penson, A. I. Solomon and G. H. E. Duchamp in Am. J. Phys. 75:639 (2007).