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In my QFT course, if I understood well, we define the normal ordering as the way to quantize a system where we put the creation operators at the left and the annihilation ones at the right.

For example for the following quantity:

$$ a(\vec{k})a^*(\vec{k})+b^*(\vec{k})b(\vec{k}). $$

It will give after quantization:

$$ a^\dagger(\vec{k})a(\vec{k})+b^\dagger(\vec{k})b(\vec{k}) .$$

We define the Weyl ordering as the ordering that will symmetrise the $\phi$ and $\pi$ operators.

For example if I have the following quantity before quantization :

$$ \phi(\vec{x},t) \pi(\vec{x},t).$$

I will have after quantization:

$$ \frac{1}{2}(\phi(\vec{x},t) \pi(\vec{x},t) + \pi(\vec{x},t)\phi(\vec{x},t)).$$

My question: Can we say that the weyl ordering is equivalent to symmetrise the creation and annihilation operator or it is false?

In fact in particular cases it gives the same thing but I don't know if it is true in general.

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    $\begingroup$ Yes, basically Weyl ordering is the complete symmetrization of x and ps, or s and adaggers, etc... with the same coefficient in front of every alternative ordering. Related 57299. $\endgroup$ – Cosmas Zachos Apr 19 '17 at 0:28
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    $\begingroup$ Might appreciate Terry Tao's blog. $\endgroup$ – Cosmas Zachos Apr 24 '17 at 22:52
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You may find your answer in every good book's discussion of it, but if you are not inclined to fuss, follow Fujii & Suzuki 2004:

Normal Ordering
$$:(a^\dagger )^n a^m:_N ~\equiv (a^\dagger)^n a^m,$$ Anti-Normal Ordering
$$:(a^\dagger )^n a^m:_{AN} ~\equiv a^m (a^\dagger )^n,$$ Weyl Ordering
$$:(a^\dagger )^{n} a^m:_W ~\equiv {\begin{pmatrix} n+m \\ n \end{pmatrix}}^{-1} ( \mbox{sum of all symmetric products of $n ~ a^\dagger $ and $m~ a$} ).$$

For example, when $m=n$, \begin{eqnarray} :a^\dagger a:_W ~ &=& \displaystyle \frac 12 (a^\dagger a+aa^\dagger ),\\ && \\ :(a^\dagger a)^2:_W &=& \displaystyle \frac 16 (a^\dagger a^\dagger aa+a^\dagger a a^\dagger a+a^\dagger aa a^\dagger +a a^\dagger a^\dagger a+a a^\dagger a a^\dagger +aa a^\dagger a^\dagger ),\\ && \nonumber\\ :(a^\dagger a)^3:_W &=& \displaystyle \frac{1}{20} (a^\dagger a^\dagger a^\dagger aaa+ a^\dagger a^\dagger a a^\dagger aa + a^\dagger a^\dagger aa a^\dagger a+ a^\dagger a^\dagger aaa a^\dagger + \nonumber\\ && \hspace{5mm} a^\dagger a a^\dagger a^\dagger aa+ a^\dagger a a^\dagger a a^\dagger a + a^\dagger a a^\dagger aa a^\dagger + a^\dagger aa\ a^\dagger a^\dagger a+ \\ && \hspace{5mm} a^\dagger aa a^\dagger a a^\dagger + a^\dagger aaa a^\dagger a^\dagger +a a^\dagger a^\dagger a^\dagger aa+a a^\dagger a^\dagger a a^\dagger a+ \\ && \hspace{5mm} a a^\dagger a^\dagger aa a^\dagger +a a^\dagger a a^\dagger a^\dagger a +a a^\dagger a a^\dagger a a^\dagger +a a^\dagger aa a^\dagger a^\dagger + \\ && \hspace{5mm} aa a^\dagger a^\dagger a^\dagger a+aa a^\dagger a^\dagger a a^\dagger +aa a^\dagger a a^\dagger a^\dagger +aaa a^\dagger a^\dagger a^\dagger ), \end{eqnarray} and so on. Likewise, $$ a^\dagger a^2= :a^\dagger a^2\! :_N= :a^\dagger a^2\! :_{AN}-2a=:a^\dagger a^2\! :_W -a ~. $$

  • There are, of course, lots of systematic ways to go from one ordering to another.
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