Misconceptions about Virtual Particles - Comments

[vanhees71]

No, because it's a quantum field. Itself it is not observable in the sense that it has a classical limit, but only quantities like the four-current density, energy-momentum density, etc. define observables and have a classical limit.

That's different only for the electromagnetic field, which has a classical limit called classical electrodynamics.

 

[ftr]

No, because it's a quantum field. Itself it is not observable in the sense that it has a classical limit, but only quantities like the four-current density, energy-momentum density, etc. define observables and have a classical limit.

That's different only for the electromagnetic field, which has a classical limit called classical electrodynamics.

I am talking about that bold capital psi that we see in Lagrangian, can you measure those at specific space-time points directly.
when you solve for it for one problem and you go and measure it(assume you can) do you see that value. If another physicist solve's another problem does he measure his own value at that same point in space time. how does that work, they can't have different values if the one field is real.

 

[A. Neumaier]

he measure his own value at that same point in space time.

There cannoit be two physicists at the same point in space and time. All measuring physicists measure the fields they measure essentially at their position and time.

 

[vanhees71]

I am talking about that bold capital psi that we see in Lagrangian, can you measure those at specific space-time points directly.
when you solve for it for one problem and you go and measure it(assume you can) do you see that value. If another physicist solve's another problem does he measure his own value at that same point in space time. how does that work, they can't have different values if the one field is real.

No you can't. It's a fermionic quantum field and represented, in the path-integral approach, as a Grassmann-number valued quantity. In the operator formalism they are fermionic field operators. The measurable quantities are S-matrix elements, i.e., correlation functions, evaluated as averages over expressions built from the field operators or as functional derivatives of the generating functionals in the path-integral formalism.

The resulting S-matrix elements are transition-probability amplitudes, i.e., there modulus squared are the probabilities per unit time and volume for a specified collision to happen, given the incoming (asymptotic free) particles (usually you have two particles in the incoming state) and the outgoing (asymptotic free) particles after the collision.

 

[ftr]

There cannoit be two physicists at the same point in space and time. All measuring physicists measure the fields they measure essentially at their position and time.

No you can't. It's a fermionic quantum field and represented, in the path-integral approach, as a Grassmann-number valued quantity. In the operator formalism they are fermionic field operators. The measurable quantities are S-matrix elements, i.e., correlation functions, evaluated as averages over expressions built from the field operators or as functional derivatives of the generating functionals in the path-integral formalism.

The resulting S-matrix elements are transition-probability amplitudes, i.e., there modulus squared are the probabilities per unit time and volume for a specified collision to happen, given the incoming (asymptotic free) particles (usually you have two particles in the incoming state) and the outgoing (asymptotic free) particles after the collision.

But isn't Ψ defined over all space time. If so, the two would compute different values.

 

[A. Neumaier]

But isn't Ψ defined over all space time. If so, the two would compute different values.

What is measured is the expectation value $\langle \Psi(x)\rangle$ at the point $x$ in spacetime, for suitable operators $\Psi(x)$ (not necessarily the basic field).

 

[ftr]

What is measured is the expectation value $\langle \Psi(x)\rangle$ at the point $x$ in spacetime, for suitable operators $\Psi(x)$ (not necessarily the basic field).

Thank you Dr Neumaier, you have been most helpful. Can you please elaborate on how a conflict will not arise if Field is real in the case of the two experiments.

 

[A. Neumaier]

Thank you Dr Neumaier, you have been most helpful. Can you please elaborate on how a conflict will not arise if Field is real in the case of the two experiments.

Two different physicists will measure the expectation at different $x$, so why should they measure the same?

 

[ftr]

But doesn't the values of Ψ of A includes values at x for B experiment.

 

[A. Neumaier]

But doesn't the values of Ψ of A includes values at x for B experiment.

No, why should it? It include the value of exactly one property at $x$. Different properties correspond to different field operators constructed from the set of fundamental fields.

 

[vanhees71]

Well, the Dirac-field operator cannot be an observable, because it doesn't commute but anti-commute at space-like separated arguments, and indeed all observables are built from Dirac-field operators by local products of even numbers of dirac operators which commute at space-like distances of their arguments as it must be for observables.

 

[A. Neumaier]

Well, the Dirac-field operator cannot be an observable, because it doesn't commute but anti-commute at space-like separated arguments, and indeed all observables are built from Dirac-field operators by local products of even numbers of dirac operators which commute at space-like distances of their arguments as it must be for observables.

Yes. I had said that $\Psi(x)$ is a suitable operator made from the basic fields, not necessarily the basic field. The commutator at spacelike distances must vanish in order to be observable.It can e.g. be a component of the Dirac current.

 

[Jilang]

That is certainly wrong.That is right.Physics is not about "reality", it is about making good predictions, if different models can make the same good predictions then they are equally fine. In particular, the predictions are from quantum field theory, and QFT does not care about our words "particles" and "fields".

If physics is not concerned with reality then who is taking care of it? Please don't tell me it's the philosophers!

 

[mfb]

If physics is not concerned with reality then who is taking care of it? Please don't tell me it's the philosophers!

Let's say the philosophers think they do.
Physics cannot. Describing and predicting observations is the best we can do in hard science.

 

[jtbell]

Suppose that we were to discover the true nature of reality. How would we know that we have done it?

 

[mfb]

What does "true nature of reality" even mean?

 

[Jilang]

These questions strike me as the ones philosophers would ask.

 

[thephystudent]

 

[A. Neumaier]

If physics is not concerned with reality then who is taking care of it?

An interpretation of physics takes care of that.

There may be multiple interpretations to the same physics; in this sense, physics is agnostic to philosophy. But people are not, and have one or more philosophies abut what it all means. And they may switch from one to the other whichever is more convenient for what they do at the moment. That's why they have intuitive, figurative ways of thinking and speaking, and why they may use virtual imagery in places where it seems to help. But they all calculate by the same formulas, getting the same physics. The formulas are the essence, not the imagery.

 

[nikkkom]

What is different? After a long time, you get stable particles flying away. W* or muon are "just" our description how those particles got created.Someone in this thread argued that there was a difference between those categories I think.

Indeed. The author is in fact quite clear in defining them as different things, and he states that off-shell particles are unobservable. From "The Physics of Virtual Particles" text:

>> On-shell and off-shell particles. The mass shell of a particle of (real or complex) mass m is the 3-dimensional quadric p^2=m^2 in 4-dimensional momentum space. On-shell means that this equation holds, off-shell that this equation is violated. All observable particles are on-shell, though the mass shell is real only for stable particles. Therefore, off-shell particles (also called virtual particles; see below) are necessarily unobservable.

Either there is an actual disagreement, or you use different definitions of what phrase "off-shell" means.

 

[mfb]

I use the definition used in particle physics, which fits to the one you quoted.

If you take that literally, every particle involved in some measurement is off-shell, even if the deviation is extremely tiny. You cannot interact with on-shell particles because they have to exist forever to be guaranteed to be exactly on-shell. Typically we call particles on-shell if the deviations are so tiny that they don't matter, e. g. for muons or protons but also for other particles with a reasonable lifetime. The cutoff is completely arbitrary, however.

 

[A. Neumaier]

Typically we call particles on-shell if the deviations are so tiny that they don't matter, e. g. for muons or protons but also for other particles with a reasonable lifetime.

According to the terminology made precise in my insight article, unstable real particles are regarded as on-shell but with a complex mass (giving rise to a peak in the S-matrix elements), which is what one gets when defining them as usual as poles of the S-matrix in the second sheet. This implies an uncertainty in the measurement results. But this is of a completely different origin than what one has for virtual particles, which are off-shell with a completely unconstrained (and unmeasurable) momentum, which in reality is only an integration variable.

 

[nikkkom]

According to the terminology made precise in my insight article, unstable real particles are regarded as on-shell but with a complex mass (giving rise to a peak in the S-matrix elements), which is what one gets when defining them as usual as poles of the S-matrix in the second sheet. This implies an uncertainty in the measurement results. But this is of a completely different origin than what one has for virtual particles, which are off-shell with a completely unconstrained (and unmeasurable) momentum, which in reality is only an integration variable.

In your picture, very short-lived particles (say, top quarks), are they always virtual? Sometimes virtual, sometimes real? What is the difference, since they can never be practically directly observed in either case?

 

[A. Neumaier]

In your picture, very short-lived particles (say, top quarks), are they always virtual? Sometimes virtual, sometimes real? What is the difference, since they can never be practically directly observed in either case?

You should read the insight article itself, where I gave details.

As with any other particle, a particle is virtual if the process is represented by a Feynman diagram, and real if it is directly or indirectly observed.
Virtual particles have no life and no lifetime; they don't exist in any meaningful sense. The lifetime of a short-living particle is defined in terms of its complex mass (pole of the S-matrix) and corresponding complex momentum, while a virtual particle always has a real momentum.

 

[Jilang]

Complex mass seems..a bit complex. Isn't it just a different way of making something slightly virtual?

 

[A. Neumaier]

Complex mass seems..a bit complex. Isn't it just a different way of making something slightly virtual?

No. This has not the slightest touch of being virtual.

It is just the relativistic analogue of complex frequencies. Complex frequencies are extremely natural for describing decaying oscillations. It is used a lot in electrical engineering. http://www.cs.mun.ca/av/old/teaching/cs/notes/complexFreq_printout.pdf

 

[Jilang]

Sorry, where does relativity come into it?

 

[nikkkom]

As with any other particle, a particle is virtual if the process is represented by a Feynman diagram, and real if it is directly or indirectly observed.

What does "observed" mean, precisely?

We infer top quark existence by detecting collision products which indirectly tell us that the process they were produced in involves a top quark.

But in all cases, the diagram with that quark has it as an internal line, not an outgoing one. Thus, by your terminology (I did read your articles) that quark is always virtual: "Virtual particles. Virtual particles are defined as (intuitive imagery for) internal lines in a Feynman diagram".

So, top quarks are always virtual?

 

[A. Neumaier]

Sorry, where does relativity come into it?

Unstable particles, Feynman diagrams and hence virtual particles are all described in terms of relativistic quantum field theory.
Particles are in quantum field theory elementary oscillations of the quantum fields. Their mass is proportional to the oscillation frequency in their rest frame, according to the formula $\hbar\omega=E=mc^2$. Thus real mass is the relativistic analogue of real frequencies, corresponding to stable particles and stable oscillations, and complex mass is the relativistic analogue of complex frequencies, corresponding to decaying particles and decaying oscillations.

 

[A. Neumaier]

top quarks are always virtual?

They are virtual if you describe them in terms of internal lines of Feynman diagrams. They are real if you describe/predict them in terms of decays or resonance width (which requires a time frame that doesn't exist for virtual particles). This is technically different since a decay of a top quark is a scattering calculation with top quark in and products out, and hence has the top quark as an external line.