# Is the purpose of the equations of motion in Quantum Field Theory simply to implement Lorentz invariant constraints?

+ 3 like - 0 dislike
341 views

A massless spin $1$ particle has 2 degrees of freedom. However, we usually describe it using four-vectors, which have four components. Hence, somehow we must get rid of the superfluous degrees of freedom. This job is done by the Maxwell equations. To quote from Gilmore's "Lie Groups, Physics, and Geometry"

“In some sense, Maxwell's equations were a historical accident. Had the discovery of quantum
mechanics preceded the unification of electricity and magnetism, Maxwell's equations might not have
loomed so large in the history physics. … Since the quantum description has only two independent
components associated with each four momentum, there are four dimensions worth of linear
combinations of the classical field components that do not describe physically allowed states, for each
four momentum. Some mechanism must be derived for annihilating these superpositions. This
mechanism is the set of equations discovered by Maxwell. In this sense, Maxwell's equations are an
expression of our ignorance.”

An analogous argument can be made for the Dirac equation. For example at page 444 in Spin in Particle Physics by Elliot Leader he writes At least for me this seems like a quite unconventional way to interpret the famous EOM's of QFT and I'm having problems to understand it fully. (I've read the accompanying sections in boths books, but wasn't fully able to grasp what they mean.) Therefore my questions:

• How does this interpretation of the equations of motion come about? (If it is correct, after all)
• Is there a textbook or paper that discusses this pedagogically?

The EMF contains not only propagating photons, but also a time dependent (Coulomb-like) near field, so the number of degrees of freedom is not 2 in general case.

@VladimirKalitvianski: The electromagnetic field has 3 degrees of freedom at each point, but the associated particles (photons) are asymptotic concepts and have only 2 degrees of freedom.

@ArnoldNeumaier: You write this as if I said something wrong.

+ 0 like - 0 dislike

Weinberg's quantum field theory book (Volume 1) treats this in detail.The point is that under weak assumptions, interacting unitary representations of the Poincare group together with locality and the cluster decomposition property (all three very fundamental properties) force everything to be as it is.

answered Nov 18, 2017 by (14,059 points)

@ArnoldNeumaier: I recognize importance of "fundamental properties", but I disagree with the uniqueness of the theory formulation. Let us remember the Classical Electrodynamics of a point particle. First we think that the radiation reaction force "must follow from", say, $L_{int}\propto A\cdot j$. Then we subtract a huge self-induction term from it ($\delta m \ddot{x}$) and obtain a term $\propto \dddot{x}$ in the equations of motion. This is no good either, so we replace it with a term $\propto \dot{F}(x,\dot{x},t)$, all of our radiation reaction terms being of good dimension and transformation properties. Our final choice is dictated with "physical acceptability" of our last equation, although it does not conserve energy momentum, strictly speaking.

@VladimirKalitvianski: In classical electrodynamics there are no massless particles, hence CED has nothing to say in the present context.

I am talking about QED, not about classical electrodynamics. QED is fully determined by the required representation of the Poincare group and renormalizability, leaving only mass and charge as free parameters.

@ArnoldNeumaier: It is good that you recognize the variety of radiation reaction terms in the Classical Electrodynamics. It is good that you recognize the necessity of the mass and charge renormalizations in QED. It is still bad that you forget to mention the field strength renormalizations and the necessity of the soft diagram summation. The latter necessity is due to impossibility of considering the coupling to the soft modes perturbatively. It also means the necessity to take the soft modes into account exactly. After that the equations of motion and the interaction remainder change which means at least non uniqueness of the theory formulation.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.