What corresponds to a static field in QFT

[kof9595995]

For example for quantized EM field the Hilbert space is made up of photon states which corrrespond to EM waves classically. However what corresponds to static EM field in QFT? It can't live in the Hilbert space of photons because any superposition of photon states will still correspond to a traveling EM wave (wave packet).

 

[Bill_K]

In the Hilbert space of free-particle photon states, all the photons are transverse. There are also photons with longitudinal and timelike polarizations. Even after imposing a gauge condition, these photons can exist as virtual particles, and do appear in the interaction j^μA_μ, and therefore can't be completely dispensed with.

 

[kof9595995]

Ahh, Ok, I thought some gauge condition would rule out longitudinal photons completely. Thanks.

 

[DrDu]

I think Bill_K's statement holds e.g. in the Lorentz gauge. In Coulomb gauge, the static longitudinal field isn't quantized at all (and also not expressed in terms of photons), but identical to the classical EM value.

 

[Meir Achuz]

For example for quantized EM field the Hilbert space is made up of photon states which corrrespond to EM waves classically. However what corresponds to static EM field in QFT? It can't live in the Hilbert space of photons because any superposition of photon states will still correspond to a traveling EM wave (wave packet).

A static EM field has no real photons. The energy between two charges can be calculated by perturbation theory involving virtual photons.
To order alpha, this gives the classical EM interaction potential.
To higher order, it gives the Lamb shift, etc.

 

[Lapidus]

A static EM field has no real photons. The energy between two charges can be calculated by perturbation theory involving virtual photons.
To order alpha, this gives the classical EM interaction potential.
To higher order, it gives the Lamb shift, etc.

That's what I learned, too.

But then I read sometimes, especially here at this site, virtual photons do not exist. What's up with this?

 

[A. Neumaier]

That's what I learned, too.

But then I read sometimes, especially here at this site, virtual photons do not exist. What's up with this?

In quantum field theory, a static field is a stationary e/m field whose expectation is independent of time. Photon states arise from the quantization of the fluctuations of the e/m field and are always transverse and real.

Longitudinal or virtual photons never enter the picture. Both are fictions stemming from an inadequate interpretation of the QFT formalism that takes perturbative expansions literally.

See also the entry ''Virtual particles and Coulomb interaction'' in Chapter A7 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#virtcoul

 

[kof9595995]

I think Bill_K's statement holds e.g. in the Lorentz gauge. In Coulomb gauge, the static longitudinal field isn't quantized at all (and also not expressed in terms of photons), but identical to the classical EM value.

I think you're right, but we'll still need to quantize the longitudinal EM field if we want to calculate everything quantum mechanically, right? I suppose we'll again get something like longitudinal photons if we do this?

 

[DrDu]

Arnold Neumaier or tom stoer are the specialist on that topic. I think in qft the Coulomb law is implemented as a restriction in the process of quantization. Nevertheless, in the Coulomb gauge, there is no conjugate momentum corresponding to the longitudinal part of the vector potential A. Hence you also don't have to quantize it as there is no field-momentum commutator.

 

[A. Neumaier]

I think you're right, but we'll still need to quantize the longitudinal EM field if we want to calculate everything quantum mechanically, right? I suppose we'll again get something like longitudinal photons if we do this?

In covariant quantization (which is presented in most modern textbooks) one gets ''longitudinal photons'' in a Krein space with an indefinite inner product. These lack any sensible interpretation in terms of probabilities (which needs a Hilbert space, i.e., a definite inner product).

As DrDu mentioned, there are no longitudinal photons in the Coulomb gauge. See the older textbook by Bjorcken and Drell for a treatment of QED in this gauge.

 

[kof9595995]

In covariant quantization (which is presented in most modern textbooks) one gets ''longitudinal photons'' in a Krein space with an indefinite inner product. These lack any sensible interpretation in terms of probabilities (which needs a Hilbert space, i.e., a definite inner product).

As DrDu mentioned, there are no longitudinal photons in the Coulomb gauge. See the older textbook by Bjorcken and Drell for a treatment of QED in this gauge.

Thanks, but I don't have time to go through it now, so just want to confirm some points from you
1."Covariant quantization" refers to Lorentz guage?
2. Even if we want to calculate static field interaction quantum mechanically, we can still leave the unquantized longitudinal field alone in Coulomb gauge?

 

[A. Neumaier]

Thanks, but I don't have time to go through it now, so just want to confirm some points from you
1."Covariant quantization" refers to Lorentz guage?

Yes. One gets different kinds of virtual stuff in different gauges - showing that there is no reality behind it.

2. Even if we want to calculate static field interaction quantum mechanically, we can still leave the unquantized longitudinal field alone in Coulomb gauge?

I don't understand the question.

 

[kof9595995]

I don't understand the question.

I mean, as DrDu said, longitudinal field(or static field) component is not quantized, so when we want to calculate, let say coulomb repulsion between two positive charges, we have to use the longitudinal field, right? But longitudinal field is still classical, so how does it make sense in QFT description?

 

[A. Neumaier]

I mean, as DrDu said, longitudinal field(or static field) component is not quantized, so when we want to calculate, let say coulomb repulsion between two positive charges, we have to use the longitudinal field, right? But longitudinal field is still classical, so how does it make sense in QFT description?

The Coulomb repulsion is a term in the Hamiltonian, not a quantum field.

 

[kof9595995]

I haven't learned QED yet, but do you mean we first write H in terms of field, and it's a sum of several terms, and one of the terms gives rise to Coulomb interaction? If so it seems counter-intuitive because in classical EM Coulomb interaction is a longitudinal field, in QFT we quantized transverse field only(coulomb gauge), yet it gives a correspondence in the Hamiltonian?

 

[A. Neumaier]

I haven't learned QED yet, but do you mean we first write H in terms of field, and it's a sum of several terms, and one of the terms gives rise to Coulomb interaction? If so it seems counter-intuitive because in classical EM Coulomb interaction is a longitudinal field, in QFT we quantized transverse field only(coulomb gauge), yet it gives a correspondence in the Hamiltonian?

The transverse part is in the field, the longitudinal part in the Hamiltonian.
http://en.wikipedia.org/wiki/Coulomb_gauge#Coulomb_gauge

 

[kof9595995]

The transverse part is in the field, the longitudinal part in the Hamiltonian.
http://en.wikipedia.org/wiki/Coulomb_gauge#Coulomb_gauge

Thanks for the reference, this seems to answer my question:"It is particularly useful for "semi-classical" calculations in quantum mechanics, in which the vector potential is quantized but the Coulomb interaction is not."

 

[kof9595995]

So I found this in sakurai's book: "In 1930 E. Fermi was able to show that {A_\parallel }(longitudinal field) and {A_0} together give rise to the instantaneous static Coulomb interactions... "
Here what does "instantaneous" mean?

 

[A. Neumaier]

So I found this in sakurai's book: "In 1930 E. Fermi was able to show that {A_\parallel }(longitudinal field) and {A_0} together give rise to the instantaneous static Coulomb interactions... "
Here what does "instantaneous" mean?

It means ''at any fixed time''.

 

[kof9595995]

Why need he phrase it this way?Is there anything subtle about this?

 

[A. Neumaier]

Why need he phrase it this way?Is there anything subtle about this?

I don't see any. But I don't have the book, so can't check.

 

[kof9595995]

Thanks. From what I've read so far I don't see any either, I'll update it in this post if I find any.

 

[kof9595995]

The Coulomb repulsion is a term in the Hamiltonian, not a quantum field.

Now I think I know where my confusion was. If QED is somehow exactly solvable, then jμAμ( which gives static field interaction) should indeed change the form of the quantized field. Now that we can't solve QED exactly, and we're mostly interested in asymptotic states, so we keep the free field prescription in quantization and reconstruct the quantized interaction by a term in Hamiltionian.
What I had in mind was actually an exactly solved quantum field, but didn't realize it.

 

[cbetanco]

Yes. One gets different kinds of virtual stuff in different gauges - showing that there is no reality behind it.

And what of jets that have a unique flavor signature in collider experiments? I just don't see how you can say there is no reality behind the "virtual stuff" when we can scatter off of, say, virtual b quarks in the proton and produce b jets. Surely the flavor tagging does not depend on the gauge...

 

[A. Neumaier]

And what of jets that have a unique flavor signature in collider experiments? I just don't see how you can say there is no reality behind the "virtual stuff" when we can scatter off of, say, virtual b quarks in the proton and produce b jets. Surely the flavor tagging does not depend on the gauge...

jets are real particles produced by real quark interactions. Nothing virtual at all.