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Recently I had a debate about the uncertainty principle in QFT that made me even more confused..

Because we use Fourier transforms in QFT, we should have an analogue to the usual Heisenberg uncertainty principle in QFT between the 4-vector of space-time coordinates and the conjugate momentum, but I found no reference for that, so is my guess is wrong?

We know that there is no universal Hermitian operator for time in QM, even so there is an uncertainty principle for time and energy, well, in QFT time is just a parameter, the same as the spatial coordinates, so is there an uncertainty principle for energy in QFT?

The last question made me confused regarding the energy conservation law in QFT: we use this law in QFT during calculations of propagators only (as I remember), it means we are using it with "bare" particles, while we suppose that these particles don't "interact" with vacuum fluctuations, so does that mean energy conservation law is a statistical law?

This brings to my mind the vacuum expectation value, that we suppose is zero for any observer, but it is zero statistically. At the same time we usually use Noether's theorem to deduce that energy is conserved (locally at least, and not statistically).

I believe I'm missing something here, can you please advise me?

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I similar question I found here, but unfortunately it gives no sufficient answer. – TMS Mar 5 '13 at 8:50

...uncertainty principle in QFT between the 4-vector of space-time coordinates and the conjugate momentum...

Conjugate momentum is "conjugate" to a particular generalized coordinate. Which are field values in case of QFT. Space-time coordinates (as you've noted yourself) are just parameters. So, I'm afraid, you are mixing two different things here.

... is there uncertainty principle for energy in QFT?

Yes. In QM and QFT observables are hermitian operators. If some pair of those operators do not commute -- you get an uncertainty principle. Time-energy uncertainty is a little bit tricky, and nicely explained, say, in this question.

... we use energy conservation law in QFT during calculations of propagators only

That is a strange claim. In the beginning of most QFT textbooks you can find a derivation of an energy-momentum conservation by means of Noether's theorem.

... energy conservation law is a statistical law?

Here is the list of statistical laws. As you can see there is not even a hint on energy conservation law. Therefore, either:

  • The answer is "no" and you should be satisfied with it.

  • You are for some reason (winning a debate?) inventing your own terminology. Which is a bad idea anyway.

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The uncertainty principle derivation you pointed is for QM not QFT, and it uses Schrodinger equation directly, that is not what I'm looking for, also I mentioned the "use" of conversation law in real calculations, Noether theorem has no direct use there, also I'm looking for to analyze the the question here, providing list of statistical laws adds nothing to my understandings, please check my own answer on the question, may be you will correct me there more clearly. – TMS Mar 8 '13 at 21:11
@TMS There is nothing to derive. The uncertainty principle carries over exactly from QM to QFT, you just have to be clear what the operators are. The basic operators are the fields and their conjugate momenta, which obey commutation relations as discussed in any QFT book. Write any operator you like, including the Hamiltonian, in terms of the basic operators and you can work out the commutation relations. Then apply the usual uncertainty principle to that. That's all there is to it. There are no new conceptual difficulties in QFT (at least as far as this is concerned). – Michael Brown Mar 9 '13 at 0:22
@TMS And what do you want to use for the time-energy uncertainty relation except for the Schrodinger equation? The time evolution generated by the Schrodinger equation is what the time-energy uncertainty relation is all about. – Michael Brown Mar 9 '13 at 0:24
@TMS Yes, there is a Schrodinger equation formulation of QFT. The state vectors are functionals on field configurations and the conjugate field momenta act as functional derivatives, in complete analogy to QM. This formalism is not often used because it is impractical for most things, but it would be completely natural to use it to address the time-energy uncertainty relationship. – Michael Brown Mar 9 '13 at 10:37
@TMS You can do any "picture" - Schrodinger, Heisenberg (this is the usual one in QFT), or interaction. Vaccum fluctuations don't modify energy conservation for the same reason that they don't in ordinary QM: the theory has a time translation symmetry. This means that there is a time-independent Hamiltonian which defines the energy eigenstates. In QFT these are states which contain superpositions of many field configurations (this is what we mean by "vacuum fluctuations"), just like the energy eigenstates state of a harmonic oscillator is smeared in position space. – Michael Brown Mar 9 '13 at 11:57

There is an uncertainty relation in QFT for averaged field and the corresponding momentum operators. For a detailed discussion see here. (e-Print: arXiv:1208.3647 [hep-th])

Similarly to normal QM there is no "energy-time" uncertainty relation which would have the same meaning.

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I am only familiar with QM. Pauli (or maybe Dirac) wrote that there is a symmetry: energy-time is perfectly analogous to momentum-position, and one can think of energy as the momentum a thing has as it travels thru time.

Einstein tells us that one man's space is another man's time, so one man's momentum IS another man's energy.

How can energy be conserved AND uncertain? Remember basic vector space theory: every state can be expressed as a linear combination of other states (which form a basis). So a state which is not a state of definite energy can be expressed as a linear combination of states which ARE of (different) definite energies. In each of those states, energy is conserved; the answer to your question follows.

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So if QFT is consistent with relativity (and surely it uses the idea of states as vectors), all my comments would apply to QFT as well as to QM. – Rod May 29 '15 at 13:13
Lorenz invariance is built in QFT – anna v May 29 '15 at 13:33
And in the formalism of quantum mechanics (as well as quantum field theory) energy and time are not perfectly symmetric. For instance, there is no times observable, see here… – Martin May 29 '15 at 14:42

As much as nobody answered my question, I will post what I get into, please correct/support me:

Strictly speaking, Energy Conversation in QFT is a Statistical Law, BUT (a big one!) it is statistical up to uncertainty principle (explanations follows), in other words, we never can measure this statistical behavior of the law, thus we can stick with it's precise formulation.

When we deal with bare particles, there is no field interactions with vacuum, then we always using precise energy conservation law, also when we calculating S-matrix, it is stationary treatment of the theory, averaged over infinite time, thus there we can also use energy conservation law strictly.

Vacuum interaction is basically affect on the probabilities of the output (state), but the energy is always conserved, anyway this happens only if we are dealing out of uncertainty condition, this happens because average expectation value of vacuum is always zero, again is zero up to uncertainty principle, and we never can explain what happened below uncertainty principles, and more important, what happens there will not affect the energy conservation (will only change probability of the output), and for that reason we still capable to say that all fields are Lorentz invariant (and thus energy conserved, not statistically), without specifying the scale we are talking about (within or not uncertainty principle).


The only left thing is to find/derive uncertainty principle (especially for Energy) within QFT framework,I will be glad if anybody can point for a reference.

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Actually I asked clearly to support/correct me, not to down vote, please specify your point of view. – TMS Mar 8 '13 at 21:04
+1, an upvote, i've had it with elitism. However the main point is that i do not find the answer wrong, or at least not deserving a response (if downvoted) – Nikos M. Aug 8 '14 at 1:01

protected by Qmechanic Aug 12 '15 at 15:01

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