In canonical quantization, given any two functions $f$, $g$ on phase space, one quantizes the theory by demanding that the commutator of the operators $O_f$, $O_g$ associated to $f$, $g$ is given by the Poisson bracket $$[O_f,O_g]=i\hbar O_{\{f,g\}}.\tag{1}$$ In deformation quantization, one defines the star product by $$f\star g=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{i\hbar}{2}\right)^n\Pi^n(f,g)=fg+\frac{i\hbar}{2}\{f,g\}+\mathcal{O}(\hbar^2),\tag{2}$$ where $$\Pi=\frac{\partial}{\partial q^i}\otimes\frac{\partial}{\partial p_i}-\frac{\partial}{\partial p_i}\otimes\frac{\partial}{\partial q^i}\tag{3}$$ is the Poisson bivector, satisfying $$\Pi(f,g)=\{f,g\}\tag{4}.$$ In particular, this leads to the star commutator $$[f,g]_{\star}=f\star g-g\star f=i\hbar\{f,g\}+\mathcal{O}(\hbar^2).\tag{5}$$ So, what's the difference? Usually, we take $\hbar$ to be infinitesimally small, so we would ignore the $\mathcal{O}(\hbar^2)$ terms. In this case, it seems like there's really no difference at all. So, to put it rather bluntly, why would I ever want to do deformation quantization over canonical quantization?
2 Answers
I would address what I fear is an unavoidable ambiguity of the term "deformation quantization" as it has grown to be used, implicit in your subsequent comment/question:
If I did deformation quantization of a QFT (and went to sufficiently high loop order), would I get a different answer than canonical quantization?
By this term, "deformation quantization", people usually mean two very different things:
The invertible Wigner-Weyl map, merely transcribing Hilbert space operators to (classical) phase-space functions and back. The classical SHO hamiltonian $h=(p^2 +x^2)/2$ maps to the standard QM Hamiltonian $\hat H= (\hat P^2 +\hat X^2)/2$ in this map, and so, in the sense of (2.) below, also in deformation quantization. Conversely, then, $\hat H ^n$ Wigner-maps to $h\star h \star ...\star h $ with n factors, which is not equal to $h^n$, (as you may check $h\star h = h^2 -\hbar^2/4$). But no experimental mismatches with canonical quantization have been observed, to date; so, in this sense, since a QFT is but a repackaging of an infinity of QSHOs, $\hat H ^n$ yields the above starred alternative in phase space. Using this language involves a starred expression in phase space― highly confusing if you wished to keep track of loops in the ℏ-expansion. But, make no mistake, the final amp, should you wish to reinstate ℏs, would really be the canonical quantization result, despite superficial appearances to the contrary!
A one-to-many map ("quantization") outlined in section 0.20 of this booklet, preserving Moyal Brackets but not Poisson Brackets, ("consistency" in the answer of @Qmechanic). It makes a somewhat arbitrary choice choice of Weyl-ordered operators among all other orderings, following a "bet" by Weyl. The "quantization" of $h^n$ is not $H^n$, but, instead, the Weyl transform of $h^n$, already contradistinguished from the star-power above, in 1. This would yield different amplitudes, but almost a century of success in canonical quantization QFT ensures that the simple choice of canonical quantization of it is satisfactory. Quantization is a mystery, and, in nature, the quantization choice is dictated by agreement of the choice/model with specific natural phenomena. The DQ choice here has been studied for Kerr oscillators in quantum optics and dynamical systems, as well as the quantization of Nambu Brackets, but I'd not expect you aim to go there, hence I focus on 1.
- Takeaway: 1. If you described canonical quantization in phase space, your (correlation function) answers would be identical to the ones you'd get in Hilbert space. But, if you modified your canonical commutators in an exotic DQ scheme encouraged by formal elegance, (2.), you may well, in principle, obtain slightly different amps in higher loop orders.
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