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Who can provide me some elegant solution for

$$\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]\qquad\text{with} \qquad\left[\hat{a},\hat{a}^{\dagger}\right]~=~1$$

other than brute force calculation?

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O lala!!! Thanks for @Prathyush and @Qmechanic !!! I got the same result with Qmechanic... I think Prathyush's suggestion should be equivalent to the my suggestion of the correspondence up to a canonical transformation. Here is my calculation (I was not confident to post it...)

$\begin{array}{c} \mbox{representation of }\left(\hat{a},\hat{a}^{\dagger}\right)\mbox{ on polynomial space }span\left\{ \frac{x^{n}}{\sqrt{n!}}\right\} _{n\ge0}\\ \hat{a}\left[f\left(x\right)\right]=\frac{d}{dx}f\left(x\right)\;;\;\hat{a}^{\dagger}\left[f\left(x\right)\right]=xf\left(x\right)\;;\;\left[\hat{a},\hat{a}^{\dagger}\right]\left[f\left(x\right)\right]=id\left[f\left(x\right)\right]\\ \left|0\right\rangle \sim 1\;;\;\left|n\right\rangle \sim x^{n}/\sqrt{n!} \end{array}$

$\begin{array}{c} \mbox{calculate the normal ordering }\left[\hat{a}^{M},\hat{a}^{\dagger}{}^{N}\right]\mbox{:}\\ \sim\left[\frac{d^{M}}{dx^{M}},x^{N}\right]=\frac{d^{M}}{dx^{M}}\left(x^{N}\star\right)-x^{N}\frac{d^{M}}{dx^{M}}\left(\star\right)\\ \sim\left\{ \overset{min\left\{ M,N\right\} }{\underset{k=0}{\sum}}\frac{N!}{\left(N-k\right)!}C_{M}^{k}\left(\hat{a}^{\dagger}\right)^{N-k}\left(\hat{a}\right)^{M-k}\right\} -\left(\hat{a}^{\dagger}\right)^{N}\left(\hat{a}\right)^{M}\\ \end{array}$

====================== One comment on 02-12-2012: The representation I was using is actually related to Bergmann representation with the inner product for Hilbert space (polynomials) being:

$\left\langle f\left(x\right),g\left(x\right)\right\rangle :=\int dxe^{-x^{2}}\overline{f\left(x\right)}g\left(x\right)\,,x\in\mathbb{R}\,,\, f,g\in\mathbb{C}\left[x\right]$

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  • $\begingroup$ I tried to map this commutator to the problem $\left[\frac{d}{dx},x\right]\circ f\left(x\right)=1\circ f\left(x\right),\,\hat{a}\sim\frac{d}{dx}\,,\,\hat{a}^{\dagger}\sim x$ but I don't know how to make this correspondence mathematically rigorous, i.e. to proof the existence of such a correspondence. $\endgroup$ – Yunlong Lian Nov 25 '12 at 9:56
  • $\begingroup$ the correspondence you write is not correct. its something like x+ip and x-ip look up wiki $\endgroup$ – Prathyush Nov 25 '12 at 10:12
  • $\begingroup$ @Prathyush If so, the correspondence should be $\hat{a}\sim x+\frac{d}{dx}\,,\,\hat{a}^{\dagger}\sim x-\frac{d}{dx}$, right? $\endgroup$ – Yunlong Lian Nov 25 '12 at 12:24
  • $\begingroup$ yes with an extra $\sqrt{\frac{1}{2}}$ factor $\endgroup$ – Prathyush Nov 25 '12 at 14:55
  • $\begingroup$ I wonder what the canonical transformation is that converts x+ip to x? $\endgroup$ – Prathyush Nov 25 '12 at 15:03
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The standard way is to use generating functions (in this case a la coherent states). Usually one would like the resulting formula to be normal-ordered.

  1. Recall the following version $$\tag{1} e^Ae^B~=~e^{[A,B]}e^Be^A$$ of the Baker-Campbell-Hausdorff formula. The formula (1) holds if the commutator $[A,B]$ commutes with both the operators $A$ and $B$.

  2. Put $A=\alpha a $ and $B=\beta a^{\dagger}$, where $\alpha,\beta\in\mathbb{C}$.

  3. Let $[a, a^{\dagger}]=\hbar {\bf 1}$, so that the commutator $[A,B]=\alpha\beta\hbar {\bf 1}$ is a $c$-number.

  4. Now Taylor-expand the exponential factors in eq. (1).

  5. For fixed orders $n,m\in \mathbb{N}_0$, consider terms in eq. (1) proportional to $\alpha^n\beta^m$.

  6. Deduce that the the antinormal-ordered operator $a^n(a^{\dagger})^m$ can be normal-ordered as $$\tag{2} a^n(a^{\dagger})^m~=~\sum_{k=0}^{\min(n,m)} \frac{n!m!\hbar^k}{(n-k)!(m-k)! k!}(a^{\dagger})^{m-k}a^{n-k}. $$

  7. Finally, deduce that the normal-ordered commutator is $$\tag{3} [a^n,(a^{\dagger})^m]~=~\sum_{k=1}^{\min(n,m)} \frac{n!m!\hbar^k}{(n-k)!(m-k)! k!}(a^{\dagger})^{m-k}a^{n-k}. $$

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