# Do the position-momentum uncertainty and time-energy uncertainty really exist in QFT?

It is well known from the Quantum Mechanics(QM) that for a particle, there is a position-momentum uncertainty relation: $$\Delta x\cdot \Delta p\geq \frac{1}{2}\hbar,$$ which basically can be derived from the commutative relation of $\hat x$ and $\hat p$ $$[\hat x,\hat p]=i\hbar.$$ There is also a time-energy uncertainty relation which is a bit subtle. But it can be well understood by imagining T as an operator, where we must introduce a parameter of the particle's worldline.

Now comes the always very confused concept of position-momentum or time-energy uncertainty relations in QFT, especially in vacuum. In QFT we only have the commutative relations: $$[\hat\phi(\vec x,t),\hat\pi(\vec x',t)]=i\hbar\delta(\vec x-\vec x').$$ Following the standard process we only can obtain the uncertainty relation about $\hat\phi(\vec x,t)$ and $\hat\pi(\vec x,t)$. There is no space-momentum uncertainty or time-energy uncertainty at all since now spacetime are simply parameters! The time and space even have no explicit physical meanings as observables in QFT, especially in vacuum. Indeed the vacuum is a stationary state just like the ground state of the harmonic oscillator in QM, which has no energy fluctuations at all.

People always take it for granted that the position-momentum or time-energy uncertainty still exists in QFT. See my another question. People always like to say things like there is a minimal length because at smaller and smaller scale there is a bigger and bigger uncertainty of energy which will eventually creates a black hole. Note in such statement, people always mean the vacuum, namely there is no particle at all. These relations are also used by many other authors without any doubt, e.g. see also A.D. Linde where he wrote:

Note, that the value of $V(\phi)$ at $t\sim t_p\sim M^{-1}_p$ can be measured with an accuracy $\sim M^{4}_p$ only due to the uncertainty principle.

Does anybody know of a derivation of the time-energy uncertainty relation (or position-momentum relation) in relativistic QFT?

• "The time and space even have no explicit physical meanings as observables in QFT". If QFT really didn't allow for position, length or time measurements then it wouldn't be so popular. It might not be immediately apparent from the formalism, but it should certainly possible to construct a toy model of a position/length measurement using QFT. – Mark Mitchison Dec 9 '15 at 18:17
• QFT is essentially a theory for many interacting particles (allowing creation/annihilation of them). So it necessarily contains single-particle quantum mechanics as a special case. What it means is that you can construct single-particle states, measure their positions and momentums, and if the energy scales are low enough so that one can forget about creation/annihilation of particles, then QFT exactly reproduces whatever you have from the single-particle QM. – Meng Cheng Dec 9 '15 at 18:27
• This being said, one should not restrict the uncertainty principles to just $x$ and $p$. It applies to all non-commuting observables, and there are many other observables than $x$ and $p$! – Meng Cheng Dec 9 '15 at 18:28
• @MengCheng Viewing QM as 0+1 QFT, then there X plays the role of field. I know that QFT should has the ability of decribing sigle low energy particle. But please remeber in that case the space or time for that single particle is very naive. Actually in that case space and time are the coordinates only for that particle. While the space and time people usually talk is general, even when there is no particle at all, for instance when talking the minimal length or spacetime foam in vacuum. – Wein Eld Dec 9 '15 at 18:35
• @WeinEld Physical rulers and clocks are "responsible" for the meaning of spacetime. If you want to see $x,p$ uncertainty from the QFT field commutators, you should try to construct a model of a physical ruler or clock whose dynamics are described by quantum field theory, and whose reading is taken by measuring some suitable field observable. Of course one way to do this (probably the simplest way) is as Meng Cheng suggests: work in the 1-particle subspace. – Mark Mitchison Dec 9 '15 at 19:37

You are correct that approached formally QFT does not have position and time observables, they are background parameters like time in QM, which seems to go against the use of localized particles. As Baker writes:"Rigorous forms of the interacting theory cannot sustain a “quanta” interpretation in which the fundamental entities are countable. And even the free theory is provably unable to describe particles localized at or around points... it so happens that in non-relativistic QM, one of the observables ($\widehat{Q}$) meets the intuitive description of a particle position operator. So for any (one-particle) state $ψ$, the inner product $(ψ,\widehat{Q}ψ)$ can be interpreted as the expected value of the particle’s position. The way is open for a particle interpretation of QM. As Malament [1996] has shown, no such position operator exists in QFT. This may be the most convincing sign that a particle interpretation of the theory cannot succeed".