Source of Dirac Field: Classical & Quantum Explanation

[jjustinn]

Classically as well as quantum-mechanically, the source of the Maxwell field is the electron/four-current (Dirac field), so the use of the Green Function propagator for the Maxwell field makes perfect sense: the Maxwell field is inhomogenous in the presence of matter.

But what about the source of the Dirac field? Classically, I suppose it's empirically considered as a given (I guess since it's exactly locally conserved, ∂J=0, any initial configuration would be sufficient), but the Dirac equation is still homogenous (if it weren't, ∂J≠0, Right?)...so then why does the Dirac Green Function propagator play such a fundamental part in QED? What is the "source" here -- the Dirac field's version of the Maxwell field's four-current?

A naive guess is that the source is the quantum vacuum, or generically "energy", but that seems not only imprecise, but very specific to QED (and therefore somewhat circular; when the inhomogenous form of the Dirac equation was introduced into QED, they surely didn't use QED as a justification) -- and anyway, it seems the same formulation should apply equally well classically.

What am I missing -- any thoughts?

 

[dauto]

There isn't a source. Electron number is conserved. photon number isn't

 

[jjustinn]

There isn't a source. Electron number is conserved. photon number isn't

Eh, ignoring the fact that electron / photon number is a quantum phenomena, wouldn't that just mean the source (e.g. The delta function inhomogeneity you get when applying the propagator to the Dirac operator, which violates the [homogenous] Dirac equation) would have to produce electron/positron pairs?

 

[dauto]

may be you should provide a specific equation for context. I'm not sure I understand your question.

 

[jjustinn]

may be you should provide a specific equation for context. I'm not sure I understand your question.

So call the Dirac operator, (λ∂ - m), D. The Dirac equation is then Dψ = 0. The propagator S would be the "inverse" of D, that is, DS(x - y) = δ(x - y), so the solution of the inhomogenous equation Dψ(x) = J(x) would be ψ(x) = ∫dyS(x - y)J(x)... But that obviously doesn't make sense when J(x) = 0, as required by the Dirac equation.

 

[dauto]

It seems that you're talking about the free Dirac equation (no interactions) before quantization so ψ is not an operator, it's a wavefunction. So you're right ψ(x) = ∫dyS(x - y)J(x) is a solution of the inhomogenous equation Dψ(x) = J(x). Note that I said "a solution", not "the solution" because you can always add a solution to the homogeneous equation to a solution to the inhomogeneous equation to find a different solution to the inhomogeneous equation. ψ'(x) = ψ(x) + ψ₀(x) where Dψ₀(x) = 0. Now if you set J(x) = 0 , you get ψ'(x) = ψ₀(x) as expected.

 

[jjustinn]

It seems that you're talking about the free Dirac equation (no interactions) before quantization so ψ is not an operator, it's a wavefunction. So you're right ψ(x) = ∫dyS(x - y)J(x) is a solution of the inhomogenous equation Dψ(x) = J(x). Note that I said "a solution", not "the solution" because you can always add a solution to the homogeneous equation to a solution to the inhomogeneous equation to find a different solution to the inhomogeneous equation. ψ'(x) = ψ(x) + ψ₀(x) where Dψ₀(x) = 0. Now if you set J(x) = 0 , you get ψ'(x) = ψ₀(x) as expected.

So this is getting deep into QED, which I was hoping to avoid, but in any case let's see where it takes us:

First, what interacting Dirac equation isn't homogenous, second-quantized or otherwise? If we're sticking to the electron-photon system, the interacting equation is just ∂→∂ - A, right? So the A term will still multiply the wavefunction (operator or otherwise), just like the mass term in the free equation...so you still end up with a homogenous equation (explicitly, γ∂ψ - γAψ - mψ = 0).

So, secondly, while it's true that you can add homogenous solutions to inhomogenous solutions of inhomogenous equations, you cannot do so for homogenous equations -- which, as far as I can tell, the Dirac equation is, whether second-quantized or not (though hopefully we can steer clear of arguments that depend on that).

 

[dauto]

So this is getting deep into QED, which I was hoping to avoid, but in any case let's see where it takes us:

How can you avoid QED when your question is about QED? :-)

First, what interacting Dirac equation isn't homogenous, second-quantized or otherwise? If we're sticking to the electron-photon system, the interacting equation is just ∂→∂ - A, right? So the A term will still multiply the wavefunction (operator or otherwise), just like the mass term in the free equation...so you still end up with a homogenous equation (explicitly, γ∂ψ - γAψ - mψ = 0).

This equation isn't linear despite the appearances because A is also a dynamic field itself dependent on the currents produced by the electron field. That's why the interacting equation cannot be solved by elementary methods and is usually solved by some perturbation technique

So, secondly, while it's true that you can add homogenous solutions to inhomogenous solutions of inhomogenous equations, you cannot do so for homogenous equations

Sure can. That's the whole point of expanding the solution into normal modes - Fourier transform

-- which, as far as I can tell, the Dirac equation is, whether second-quantized or not (though hopefully we can steer clear of arguments that depend on that).

No such luck. Note that despite the similarity of the equations, appearances can be deceiving, Second quantization is doing something quite different than the failed first quantization Dirac theory. In first quantization ψ is just a series of complex functions equivalent to the wavefunction in the Schrodinger equation. In second quantization ψ are the dynamic variables equivalent to the coordinates x, y, and z in the Schrodinger equation. That means that ψ are operators and must satisfy some commutation relations with their associated momenta (obtained from the Lagrangian). When one expands ψ in normal modes, the coefficients of each mode behave as creation and destruction operators. It turns out that if you destroy a particle in one location and create another elsewhere, the amplitude for that process called a propagation can be calculated and turn out to be closely related to the Green's function!

 

[jjustinn]

How can you avoid QED when your question is about QED? :-)

This equation isn't linear despite the appearances because A is also a dynamic field itself dependent on the currents produced by the electron field. That's why the interacting equation cannot be solved by elementary methods and is usually solved by some perturbation technique

I must be missing something: where did linearity come in?

Sure can. That's the whole point of expanding the solution into normal modes - Fourier transform

Huh? The point of Fourier expansion is so that you can add inhomogenous solutions to homogenous ones?
If y=A and y=B solve Dy = 0, so does A+B. If y=C solves Dy = 1, then y = A + B + C is a solution of the inhomogenous equation (distributive law: D(A + B + C) = DA + DB + DC, DA = DB = 0, DC = 1), but linear combinations containing C will never solve the homogenous equation, no matter how you transform it.

No such luck. Note that despite the similarity of the equations, appearances can be deceiving, Second quantization is doing something quite different than the failed first quantization Dirac theory. In first quantization ψ is just a series of complex functions equivalent to the wavefunction in the Schrodinger equation. In second quantization ψ are the dynamic variables equivalent to the coordinates x, y, and z in the Schrodinger equation. That means that ψ are operators and must satisfy some commutation relations with their associated momenta (obtained from the Lagrangian). When one expands ψ in normal modes, the coefficients of each mode behave as creation and destruction operators. It turns out that if you destroy a particle in one location and create another elsewhere, the amplitude for that process called a propagation can be calculated and turn out to be closely related to the Green's function!

I recall from Greiner and other sources that the second-quantized operators still satisfied the (homogenous) Dirac equation (same for the KG/Maxwell fields), but I could be wrong -- do you have a link stating otherwise? That would be great.

The closest I could find to what you are saying was this (http://en.wikipedia.org/wiki/Static...gral_formulation_of_virtual-particle_exchange) / this (http://en.wikipedia.org/wiki/Common...s_with_differential_operators_in_the_argument), where they show how the path integral of the exponential of the action leads to a solution involving the propagator (eg the unit solution of the inhomogenous field equation)...but to get there, they have to add a "disturbance" (e.g. An explicit source term) Jψ to the action...now, at first glance, I thought that was the answer, because there the source multiples the field, just like in the minimally-coupled Maxwell/Dirac equation...but then I remembered that was the *action*, not the field equation...so that term is the equivalent of replacing the 0 on the RHS of the Dirac equation with J...which gives me an equivalent way of asking the same question: where does that J come from?

 

[dauto]

I must be missing something: where did linearity come in?

The point was just to show that the solution for the interacting system cannot be obtained by elementary methods alone.

Huh? The point of Fourier expansion is so that you can add inhomogenous solutions to homogenous ones?
If y=A and y=B solve Dy = 0, so does A+B. If y=C solves Dy = 1, then y = A + B + C is a solution of the inhomogenous equation (distributive law: D(A + B + C) = DA + DB + DC, DA = DB = 0, DC = 1), but linear combinations containing C will never solve the homogenous equation, no matter how you transform it.

We've been talking past each other. I agree with what you said. I was talking about adding to solutions of the homogeneous equation. (A + B in your example), and you've been talking about adding a inhomogeneous solution to a homogeneous one (A + C in your example).

I recall from Greiner and other sources that the second-quantized operators still satisfied the (homogenous) Dirac equation (same for the KG/Maxwell fields), but I could be wrong -- do you have a link stating otherwise? That would be great.

The closest I could find to what you are saying was this (http://en.wikipedia.org/wiki/Static...gral_formulation_of_virtual-particle_exchange) / this (http://en.wikipedia.org/wiki/Common...s_with_differential_operators_in_the_argument), where they show how the path integral of the exponential of the action leads to a solution involving the propagator (eg the unit solution of the inhomogenous field equation)...but to get there, they have to add a "disturbance" (e.g. An explicit source term) Jψ to the action...now, at first glance, I thought that was the answer, because there the source multiples the field, just like in the minimally-coupled Maxwell/Dirac equation...but then I remembered that was the *action*, not the field equation...so that term is the equivalent of replacing the 0 on the RHS of the Dirac equation with J...which gives me an equivalent way of asking the same question: where does that J come from?

Yes, it satisfies the same equation, formally, but the nature of the field themselves is different. In first quantization the fields are complex functions, while in second quantization they are operators. The operators may satisfy some (anti)commutation relations that complex functions simply cannot do. So even though they are formally satisfying the "same equation", they are actually different solutions in a fundamental level. Right now I'm at work, so I cannot give you a better answer than that. I'll post a better answer later this weekend