Use of QED equations of motion?

[QuasiParticle]

This is perhaps a stupid question, but are the field equations, for example, for QED useful for anything? By field equations (equations of motion) I mean the equations, which are obtained by inserting the Lagrangian into the Euler-Lagrange equation. In the case of QED, one gets a Dirac-like equation with interaction terms plus a "QED version of the classical Maxwell equations". I cannot remember seeing any use for these equations in textbooks.

http://en.wikipedia.org/wiki/Quantum_electrodynamics#Equations_of_motion

 

[dauto]

Did you say you don't remember seeing Dirac's equation with an interaction or Maxwell equations in Textbooks? Are sure about that? These equations are at the very heart of electromagnetism!

 

[QuasiParticle]

Ah, right, so that is just the Dirac equation. I guess I looked at the equation in a hurry, the external potential confused me. Thanks. The others are just the Maxwell equations? Though why are they called "QED version of Maxwell equations" in the Wikipedia article? Because the four-current is due to the wave function?

What about QCD? Its equations of motion are the "Dirac equation" for QCD and "some equations for strong interaction"? Those I don't remember coming across.

 

[dauto]

Yes, those are just your garden variety Dirac and Maxwell Equations. The term on the right of the Maxwell equations are indeed the 4-current associated with the Dirac field. You're not familiar with the QCD terms because they have no classical equivalent because confinement prevents them from having any long range effects.

 

[QuasiParticle]

Thanks. You say that QCD does not have a classical equivalent, but then what equations does one obtain for QCD from the equations of motion? Are they useful for anything?

 

[tom.stoer]

The classical equations of motion define classical solutions of the theory. Of course in QFT you have quantum fluctuations, but there are well-known examples where classical solutions are used as a starting point:
- A trivial example is the solution ψ=0, A=0 which is the starting point of perturbation theory.
- Another example is the so-called instanton

http://en.wikipedia.org/wiki/Instanton

 

[edguy99]

From Penrose's Road to Reality (Chapter 25, Section 2):

It turns out that the Dirac spinor ψ, with its 4 complex components, can be represented, as a pair of 2-spinors αA and βA', one with an unprimed index and one with a primed index: ψ = (αA, βA')

* The Dirac equation can then be written as an equation coupling these two 2-spinors, each acting as a kind of ‘source’ for the other, with a ‘coupling constant’ M describing the strength of the ‘interaction’ between the two...
* More formally, break an electron field into two parts (left and right) as ψ=ψL+ψR
where ψL=(1/2)*(1−γ5)ψ and ψR=(1/2)*(1+γ5)ψ.
* These two massless fields, one left-handed and one right-handed, interact with coupling constant M equal to the mass of the electron.

These can be represented classically where 2 different things spinning are somehow held together by an interaction energy. The object you are looking at is no longer just a spinning ball, it could have multiple layers (2 in this case) that are spinning independently. That way, you can represent a larmour frequency of one axis around another and spin flips.

 

[QuasiParticle]

Thanks all! But I'm still a little confused what the Euler-Lagrange equation for QCD gives?

 

[dextercioby]

You should be able to derive them. Just write down the Lagrangian and start to differentiate it.

 

[QuasiParticle]

You should be able to derive them. Just write down the Lagrangian and start to differentiate it.

That is true. However, it would require somewhat lengthy calculations and even more work to figure out what the equations actually mean and imply. In the end I fear that some problem would arise, since I have never come across a classical version of QCD.

 

[dextercioby]

Because there is none. QCD Lagrangian is like the Dirac Lagrangian. A mathematical artifact, since there's no classical physical field available. QCD like QED are interaction theories. The system of PDE's is not integrable, thus they are not of immediate use.

 

[tom.stoer]

I don't agree. There are classical solutions of the Yang-Mills equations (no quarks, so no full QCD) for which non-trivial solutions are known.

http://en.wikipedia.org/wiki/Instanton

 

[QuasiParticle]

Because there is none. QCD Lagrangian is like the Dirac Lagrangian. A mathematical artifact, since there's no classical physical field available. QCD like QED are interaction theories. The system of PDE's is not integrable, thus they are not of immediate use.

Okay, I was slightly vague. I meant a classical theory of the strong interaction, like Maxwell in EM. I don't quite follow your last sentence.

I don't agree. There are classical solutions of the Yang-Mills equations (no quarks, so no full QCD) for which non-trivial solutions are known.

http://en.wikipedia.org/wiki/Instanton

These instanton solutions are not exactly what I was asking, right? They are interesting, though.

 

[Avodyne]

Googling "QCD equations of motion" turned up this; see sec. 2.4. Note that color-triplet indices on the quark fields are suppressed.

http://www.t39.ph.tum.de/T39_files/Lectures_files/StrongInteraction2011/QCDkap2.pdf

Eqs of motion are rarely used in QFT for the same reason that they are rarely used in QM: they're just not that useful for the quantities we're typically interested in.

 

[QuasiParticle]

Great, there are the sought-for equations. Now it would only remain to see what kinds of solutions they have and what they actually represent.