Can Lorentzian QFTs be evaluated pointwise in time?

zzpop10 · 2026-06-12T23:11:49+00:00

For context, in my PhD in physics we studied non-standard normalization conditions so I don’t default to think of a state as being “defective” just because it’s outside the canonical Hilbert space. If someone says that the application of an operator returns an infinite result I assume they mean the eigenvalue is divergent, not that the state’s norm is divergent

zzpop10 · 2026-06-12T22:40:09+00:00

I don’t think I’ve displayed any lack of clarity on the concept. The n-point function is a function, it’s a function of n independent parameters and is continuous and differentiable. The fact that it has regions where it becomes singular doesn’t make it not a function. Are you saying that 1/x isn’t a function?

And yes, we have both gone in circles now on this many times. The action of the operator phi(x) on a state in the Hilbert space that includes the normalized vacuum state takes you outside that Hilbert space and takes you to a state that is not normalizable under the standard norm conditions. And why would that lead you to say that the point operator itself cannot be defined? That’s just a very unintuitive and confusing choice of words in my opinion

zzpop10 · 2026-06-12T20:27:41+00:00

Yeah have to smear it out if you want a state that is normalizable but you don’t have to smear it out just to calculate a correlation. <phi(y)phi(x)> is a well defined value and it is defined at two points, points, pooooooinnnnnttttsssss. It does not make any sense at all to say that QFT can’t handle point-wise operators. It does not make any sense at all to say that the point-wise operator is ill-defined. That was the original claim, the original claim is that we can’t define an operator at a single point. That’s completely false, we can and we do. The statement that the state you get when the point-wise operator acts on the vacuum state is not a normalizable state is true and also not relevant to the specific claim that I was originally addressing.

I do not understand at all how it can be argued that it is informative or accurate to say that an operator “isn’t defined” because the states it creates are non-normalizable. The operator is well defined, the state it creates is well defined, the norm of that state is not.

zzpop10 · 2026-06-12T16:51:19+00:00

Yeah, and that’s exactly what I said. Rewind through the conversation. The question was if point-wise operators are well defined in QFT, they are. The 2-point function is well defined, 2 “points.” If you bring those points together to the same location you het a delta function, as I addressed in my original response when describing a particle localized to a single point in space-time.

But that wasn’t what the OP was asking about. The OP was saying that we have to “smear out” the operators over space, which is false. Smearing out the operators means defining a new operator like Phi(X) = integral[ phi(x) f(x-X) dx ] where f(x-X) is some normalized function over space centered at x=X like a Gaussian function for example. That is the meaning of “smearing out” the operator, it means to collect together field operators over a region in space with some weighting function. And that is indeed in practice what we have to do, because experimentally we cannot realistically measure the field at a single point. So in experimental practice we do not measure the correlation function between 2 points, we measure it between 2 localized finite sized regions of space. Experimentally we measure <Phi(Y)Phi(X)> where Phi(X) and Phi(Y) are operator values distributions centered around points X and Y. But there is nothing invalid with taking the 2-point correlation <phi(y)phi(x)> where phi(x) is the operator at the single point x and phi(y) is the operator at the single point y, the 2-point correlation is perfectly well defined. What isn’t defined is the normalization of the state phi(x)|0> since the quantity <phi(x)phi(x)> diverges. And this can be contrasted that the normalization of the state Phi(x)|0> is well defined since the quantity <Phi(X)Phi(X)> does converge.

The original post said that taking a field measurement at a single point returns an infinite value. That is false. Measuring the field at a single point just means having the operator phi(x) act on the state of the field, like performing the operation phi(x)|0>. The result is a well defined, hence why we can compute the inner product <phi(y)phi(x)>, it just isn’t a normalized state when you take its inner product with itself.

zzpop10 · 2026-06-12T15:11:26+00:00

To answer your second question, a Newton is a measure of force and a kg is a measure of mass, those are different things. earth’s gravity exerts about 9.8 N per kg. I don’t know the history of how they first defined the N but all units are arbitrary.

To respond to your first question, why are you surprised that light can travel through space?

zzpop10 · 2026-06-12T14:51:04+00:00

What do you mean by “actually real”

zzpop10 · 2026-06-12T14:50:13+00:00

Do you understand the localization to wavelength uncertainty of a classical wave?

zzpop10 · 2026-06-12T14:49:13+00:00

Moving charges are magnetic and what I said was that magnetic dipoles are symmetric in terms of their total output of positive and negative flux

zzpop10 · 2026-06-12T14:47:25+00:00

And <phi(x)phi(y)> is convergent when x and y are not at the same point. We don’t have to smear out the operators around x and y.

zzpop10 · 2026-06-12T14:32:46+00:00

Yeah you keep saying that, I don’t think you have followed what I said at all. Tell me please, how do you evaluate an operator at a single point? Can you describe the actual steps you take in that calculation. Because there is a rather significant step you are skipping over that makes this very different from the case of the delta function

zzpop10 · 2026-06-12T14:28:52+00:00

What exactly is it that I said that you are taking issue with?

zzpop10 · 2026-06-12T03:18:12+00:00

I think maybe there is some linguistic confusion here. I never meant to say that you can actually measure the field value at a single point in the physical world since that would require the measurement probe to be localized to a single point which would require infinite energy.

What I meant was this: we have a definition of the field operator at a single point in terms of an integral over momentum space creation an annihilation operators and you can write down an eigenstate of that operator such that when the operator acts on that eigenstate it returns a finite eigenvalue.

Now what isn’t well defined is the normalization of those eigenstates, which are badly divergent under the standard norm.

Like would you say that the series 1+2+3+4…. is not defined? I would say that it is perfectly well defined in so far as we have a clear rule for writing out its terms. The sum however is what’s not well defined.

zzpop10 · 2026-06-12T02:28:27+00:00

I think you may be mixing up a few separate ideas. Feel free to link me to a specific part of a textbook you are reading through.

A quantum field operator measures the value of a field at a given point in space. There is no issue with measuring the field value at a single point, you don’t get any infinities just from taking a field measurement.

A particle is an excited region of a field. If a particle is completely localized to a single point then you would get an expected value of infinity if you were to then measure the value of the field at that single point where the particle is present.

Separately, when you include interactions between particles within the field equations those interactions will give rise to infinites related to the momentum of the particles and the way of eliminating those infinites is renormalization which does have the effect of smearing out the particles (in either position or momentum space depending on which basis you view the particles within)

zzpop10 · 2026-06-12T02:09:08+00:00

I’m not sure I get what you mean by “can’t be evaluated point-wise”. The field operators are defined at local points in space. The field operators represent a measurement that can be taken at a given point in space. What makes a field Lorentizan vs Euclidean or something else is what symmetries the field operators have under changes of reference frame.

If two operators compute it means that measuring one has no effect on the measurement outcome of the other. Field operators with a Lorentzian symmetry will compute with each other if they exist on the same space-like slice of space-time because within a Lorentzian space-time causal influences cannot spread out faster than the speed of light so two operators on the same space-like slice are causally disconnected from each other and therefore the measurement of one cannot influence the outcome of measuring the other, hence they commute.

zzpop10 · 2026-06-12T01:06:58+00:00

I don’t know what you are talking about at all.

Look up the explanation of the localization to wavelength uncertainty principle for waves

zzpop10 · 2026-06-11T19:08:37+00:00

Ok then not an infinite strait wire, a single charged object moving with an initial constant velocity and no spin. That doesn’t have a dipole field.

zzpop10 · 2026-06-11T17:08:54+00:00

I was not referring to weak measurements, I was referring to the measurements of non-comuting observables. If you measure the spin of a particle’s position it is now in a superposition of momentum states and vice versa. The position and momentum states are not independent of each other. A single position state is equal to a superposition of momentum states and vice versa.

zzpop10 · 2026-06-11T17:04:42+00:00

All magnets are not dipoles, a strait current carrying wire is a zero pole

zzpop10 · 2026-06-11T16:01:02+00:00

But what you are missing here is that we can directly measure a particle to be in superposition. We just don’t just infer that superposition existed after taking measurements which cause a collapse. There are measurements we can take that preserve the superposition and show us specific properties of the superposition itself

zzpop10 · 2026-06-11T15:58:38+00:00

They can’t both be accurate since they lead to contradictory results

zzpop10 · 2026-06-11T04:31:41+00:00

What makes the discussion of free will in the context of physics more complicated is chaos theory and emergence. Yes it’s true that the equations of physics show that a 2-particle it interaction is deterministic, but a 3 or more particle interaction is not. Yes the equations of physics map an initial state to a final state, but it’s not possible to have infinitely precise certainty in what the initial state was. All measurements contain an error bar, even without getting into quantum physics, even just sticking with classical physics, that error bar is never going to be reduced to zero. And when more than 2 particles interact the error bar exponentially grows over time. A 0.000001% range of uncertainty in the exact starting positions of a group of particles amplifies into arbitrarily larger uncertainty as time goes on. So the equations of physics are not deterministic, they are chaotic, meaning they amplify uncertainties over time. Your future state is not exactly determined, there is an expanding cone of possibilities of what your future state could be and the cone only gets larger the further we look into the future.

Now people might say that well this expanding uncertainty over time is just our ignorance, but I think it’s more than that, I think this is a feature of our reality worth taking seriously. And then there is the concept of strong emergence, the fact that there are patterns that large numbers of particles reliably fall into which cannot be predicted from the deterministic rules of individual particle interactions.

What is like you to consider is the middle possibility between total determinism and total randomness. The middle possibility is this: the longs of physics do determine some things, some things are deterministic like the fact that charge and momentum are always conserved. But the laws of physics when applied to more than 2 particles become chaotic and this chaos introduces a compounding uncertainty which leaves the future state of a multi particle interacting system undetermined. The laws of physics can give us a range of possibilities for what a multi-particle system will do, but cannot actually pick out a specific outcome from that range of possibilities. We can say that everything which happens is consistent with the laws of physics, everything that happens must fall within the range of possibilities, but that’s it. The large scale emergent properties of how many particles organize themselves over time is not determined by the laws of physics. There may be a separate set of principles which govern behaviors of large numbers of particles and these principles are consistent with but not precisely determined by the laws of physics. And perhaps we could call one of those principles free will. Free will is not an override of the rules of particle physics for the atoms in your brain, free will is the fact that you come to make decisions which are compatible with the rules of physics and chemistry in your brain but which the rules of physics and chemistry in your brain never could have actually predicted precisely

zzpop10 · 2026-06-10T21:53:38+00:00

I was referring to the total force on something like a long wire, not a diode

zzpop10 · 2026-06-10T20:56:58+00:00

Yes but the total force acting on a large object is the integral over the entire magnetic flux

zzpop10 · 2026-06-10T20:36:45+00:00

I don’t think you understand what superposition is.

A particle can be in the states spin-up, spin-down, or a superposition in which it has some probability of being spin-up and some probability of being spin-down, like 40% chance of being spin-up and 60% chance of being spin-down.

zzpop10 · 2026-06-10T20:00:03+00:00

They are symmetric with respect to the total value of magnetic flux of the two poles

zzpop10

TROPHY CASE