While studying canonical quantization in QFT, I observed that we quantize fields either by a commutation or an anticommutation relation \begin{equation} [\phi(x), \phi(y)]_\pm := \phi(x) \phi(y) \pm \phi(y) \phi(x) = i \delta (x - y). \end{equation}

and from this, the spin-statistics theorem tells us that we should use the commutator relation for integer spins and the anticommutator for half-integer spins.

My question is this $-$ why do we not consider more general quantization relations, for example of the following form \begin{equation} [\phi(x), \phi(y)]_q := \phi(x) \phi(y) + q \phi(y) \phi(x) = i \delta (x - y). \end{equation}

for some arbitrary constant $q$? Also, if we can consider such relations, then is it possible to argue that we should have $q = 1$ for half-integer spins and $q = -1$ for integer spin?


1 Answer 1


Your argument at the bottom checks out, in principle. There's no reason you can't consider the commutators as the $q=1$ and $q=-1$ cases of your generalised "commutator".

The answer to your question probably isn't that interesting, it's simply that the standard commutation and anti-commutation relations are what produce accurately predicting theories.

You could explore the possibilities but I suspect a lot of your theoretical predictions will be out by various powers of $q$!

Your question as it stands is a bit broad, as most "why don't we do/consider X" type questions tend to simply be answerable with "because X doesn't work". And because this is physics that is about as far as we are interested in.

Cosmos Zachos gives a bit more information on a similar question here if you're interested in reading.

  • $\begingroup$ I agree that my question's a bit broad, but my motivation was to find out if there were any mathematical reasons as to why we couldn't consider a generalized "commutator". $\endgroup$
    – Ishan Deo
    Oct 21, 2023 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.