m (→Commutation) |
(Update a formula, publish a notebook and comment on results already existing) |
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$$\label{eq:TueSep4123411CEST2012}[a,\ud{a}]=1$$ | $$\label{eq:TueSep4123411CEST2012}[a,\ud{a}]=1$$ | ||
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| + | {{I}} have hacked 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{{cite|blasiak03a}}{{cite|blasiak07a}}. You can '''[[:File:boson-normal-ordering.nb|download it]]''' to play with it (check or compute results useful in your daily quantum algebra). | ||
== Creation & annihilation == | == Creation & annihilation == | ||
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$$a^n\ud{a}^m=\sum_{k=0}^{\min(n,m)}k!\binom{m}{k}\binom{n}{k}\ud{a}^{m-k}a^{n-k}$$ | $$a^n\ud{a}^m=\sum_{k=0}^{\min(n,m)}k!\binom{m}{k}\binom{n}{k}\ud{a}^{m-k}a^{n-k}$$ | ||
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| + | 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: | Some useful particular cases: | ||
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*$a\ud{a}=\hat n+1$. | *$a\ud{a}=\hat n+1$. | ||
*$\ud{a}^2a^2=\hat n^2-\hat n$. | *$\ud{a}^2a^2=\hat n^2-\hat n$. | ||
| + | |||
| + | == References == | ||
| + | |||
| + | <references/> | ||
Contents |
Here are collected some results related with the operator $a$ which commutation with its adjoint $\ud{a}$ reads:
$$\tag{1}[a,\ud{a}]=1$$
I have hacked 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).
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:
$$ \begin{align*} \kern-1cm{(\mathrm{for}~i\le n+j)}\kern1cm a^i{\ud{a}}^j\ket{n}&={(n+j)!\over\sqrt{n!}\sqrt{(n+j-i)!}}\ket{n+j-i}\,,\\ \kern-1cm{(\mathrm{for}~i\le n)}\kern1cm a^{\dagger j}a^i\ket{n}&={\sqrt{n!}\sqrt{(n+j-i)!}\over(n-i)!}\ket{n+j-i}\,. \end{align*} $$
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:
$$a\ud{a}^n=\ud{a}^na+n\ud{a}^{n-1}$$ $$a^n\ud{a}=\ud{a}a^n+na^{n-1}$$
A compilation of useful results derived from Eq.~(1):
We define $\hat n\equiv\ud{a}a$.
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Special cases below
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