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The uncertainty principle is a fundamental law of nature, arising from the mathematics of quantum theory. In it's most general form, it reads: $$\sigma_A^2 \sigma_B^2 ≥ (\frac{1}{2i} \langle [\hat A,\hat B] \rangle)^2$$ where $ \sigma_A $ and $ \sigma_B $ stand for the standard deviation of the observables $\hat A$ and $\hat B$, and [ ] stands for the commutator of the observables.

Now, we get the well known Heisenberg uncertainty principle by using the specific operators for position and momentum (derived from the $ Schrodinger$ $equation $) in place of $\hat A$ and $\hat B$ respectively, viz: $\hat A = x$ and $\hat B = \frac{\hbar}{i} \frac{d}{dx}$.

But since the Schrodinger equation is non-relativistic, it's fair to conclude that the Heisenberg uncertainty principle is also non-relativistic! No flaws, I guess?

Then, in light of Quantum Field Theory, what should we use in place of the observables $\hat A$ and $\hat B$ ? The use of Dirac equation is simple, but it's not a general spin equation (since it is basically a spin $\frac{1}{2}$ equation), so an answer in terms of general spin equations like the $ Joos-Weinberg$ $equation $ or the $ Bargmann-Wigner$ $equation $ would be highly helpful.

Also, if possible, the final equation of the uncertainty principle would be highly appreciated!!!

(And if it's not possible to formulate such a thing, what does the Dirac uncertainty principle look like? I would like to check my results with those of the experts out here!!)

Yuzuriha Inori
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  • https://physics.stackexchange.com/questions/191042/uncertainty-principle-in-quantum-field-theory/191064#comment716689_191064 – Constantine Black Sep 04 '17 at 09:36
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    I don't understand this question - the general Robertson-Schrödinger uncertainty relation you cite at the beginning of the question and its proof do not rely on the Schrödinger or Dirac equation, but simply follow from the general quantum mechanical rules of how expectation values are taken. So I don't understand what you mean when you ask what we "should" use in place of $A$ and $B$ - the whole point of the general relation is that you can use whatever observables you like! – ACuriousMind Sep 04 '17 at 11:06
  • @ACuriousMind Yes, you have definitely got the fact right that the general uncertainty principle follows generally from the method of taking expectation values; but my question is in another place. I wanted to ask, what should be the replacement of the operators for x and p in qft that are to used in the equation. – Yuzuriha Inori Sep 04 '17 at 11:08
  • Then your question is not about the uncertainty principle at all, and is answered e.g. in this answer by Valter Moretti – ACuriousMind Sep 04 '17 at 11:17
  • @ACuriousMind I named this because I also want the final form of the equation, not just the operators. I asked for the operators just to understand the thing. I need the equation.. Let me edit my question a bit... – Yuzuriha Inori Sep 04 '17 at 11:19

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