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where \left\{\begin{matrix} n\\k \end{matrix} \right\} are the
 
where \left\{\begin{matrix} n\\k \end{matrix} \right\} are the
 
Stirling partition numbers (the number of ways to partition a set of
 
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.
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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.:
 +
 
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(\ud{a}a)^n=\sum_{k=1}^nS(n,k)(\ud{a})^ka^k
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where S(n,k) are the Stirling numbers of the second kind ([[Mathematica]] <tt>StirlingS2[n,k]</tt>). For instance:
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(\ud{a}a)^3=\ud{a}^3a^3+3\ud{a}^2a^2+\ud{a}a\,.
  
 
=== Some useful particular cases ===
 
=== Some useful particular cases ===

Latest revision as of 16:47, 1 October 2024

Contents

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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.

  • [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.

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).