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  Comparison of massive and massless QED: what is the main difference?

+ 5 like - 0 dislike

Let's have massive spinor QED:
\tag 1 L = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + \mu^{2}A^{2} + \bar{\Psi}(i\gamma^{\mu}D_{\mu} - m)\Psi , \quad D_{\mu} \equiv \partial_{\mu} - ieA_{\mu}

This theory doesn't have local gauge symmetry because of existence of mass term for $A$ field, but has global invariance under transformations $\Psi \to e^{i \alpha}\Psi$, thus in two cases (massive and massless QED) current $J_{\mu} = \bar{\Psi}\gamma_{\mu}\Psi$ is conserved. I want to establish the differences between massive and massless spinor QED.

1. It seems to me that the derivation generalized Ward identity,
(l - k)_{\mu}\Gamma^{\mu}(l, k) = iS^{-1}(k) - iS^{-1}(l),

isn't depend on the value of photon mass, because the derivation is started from calculation of divergence of correlator $\partial^{\mu}_{x}\langle \hat{T} \left( J_{\mu}(x)\Psi (y)\bar{\Psi}(z)\right)$, and $J_{\mu}$ is conserved and has equal commutation relations $[J_{\mu}(\mathbf x), \Psi (\mathbf y)]$ in both cases. Thus all charges in both theories are renormalized equivalently, $e \to \sqrt{Z_{3}}^{-1}e$ ($Z_{3}$ denotes $A$ field renormalization constant).

2. Identity $k^{i}_{\mu}M^{\mu}(p_{1},...,p_{n}, k_{1},..., k_{i},...,k_{m}) = 0$, where $k^{i}$ corresponds to external photon momentum and $M_{\mu}$ corresponds to the amplitude with $m$ external photonic lines without polarization vector $\epsilon^{\mu}(k_{i})$, $M =\epsilon^{\mu}(k_{i})M_{\mu} $, is also hold for both cases. Thus longitudinal polarizations don't take role as in- and out-states.

Thus, if I haven't missed something, the main difference between massive and massless QED is in the pole structure of propagators:
D_{\mu \nu}^{\text{massive}}(p) \sim \frac{1}{p^{2} - \mu^{2}}, \quad D_{\mu \nu}^{\text{massless}}(p) \sim \frac{1}{p^{2}},

and in corresponding nonrelativistic limits (Coulomb law and Yukawa law correspondingly).


1. Does massive QED have problem with renormalizability due to the mass term? It seems to me that $(2)$ makes it renormalizable, but I'm not sure.
2. If the first question has negative answer, than suppose following idea: mass term of gauge field $A$ in $(1)$ could be obtained from theory with scalar field through Stueckelberg mechanism. This theory is renormalizable, but is completely equivalent to $(1)$ when we neglect the scalar field part. Thus the situation may be following: the theory $(1)$ is explicitly renormalizable if we able to use $R_{\epsilon}$ gauge.

3. Finally, if the first question has positive answer, are two theories completely equivalent up to the pole structure of the propagator.

asked Sep 2, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Sep 2, 2015 by NAME_XXX

I will only add that a non-vanishing but arbitrarily small mass for the photon is indistinguishable from a zero mass. In other words, for any experiment that potentially distinguish  a massless from a photon with  mass, there exists a non-zero value for the mass of the photon that prevents the experiment from ruling out a massive photon. That is, the best we can experimentally do is to put upper bounds to the mass of the photon.

2 Answers

+ 4 like - 0 dislike

Massive QED is renormalizable.

The main difference to standard (massless) QED is that it has longitudinal degrees of freedom (suppressed by powers of the photon mass) because there is no gauge invariance.

Another difference is that it has no infrared divergences. Therefore a small hoton mass is often used to regularize the latter in standard QED. See, e.g., Weinberg's book on QFT, Vol. 1.

answered Sep 2, 2015 by Arnold Neumaier (15,787 points) [ no revision ]
Most voted comments show all comments

@NAME_XXX: Of course they participate in interactions. These interactions disappear only in the limit where the mass tends to zero.

In massless QED, the remnant of the longitudinal states is the nonpropagating Coulomb-like part of the field. In massive QED, the longitudinal part also propagates, though with speed $<c$.

Then this longitudinal part may be observed as non-longitudinal from another reference frame. The only way I can imagine that is propagating with electron, i.e., being attached, not radiated. "Longitudinal" refers to the specific equations, not to polarizations of propagating heavy photons.

Then this longitudinal part may be observed as non-longitudinal from another reference frame.

Yes. There is nothing special about the longitudinal modes, since there is no gauge invariance. The whole field propagates, and the distinction between transversal (polarization) modes and longitudinal modes is useful only in the massless limit.

The longitudinal part of a vector is Lorentz invariant, hence observer-independent. It is the part that is parallel to the direction of motion of the particle. 

Most recent comments show all comments

@ArnoldNeumaier : but they interact only as intermediate states (according to the statement 2 in my question), don't they?

@NAME_XXX: There are Fock states corresponding to the longitudinal states of massive QED. In conventional perturbative renormalization theory they appear in the asymptotic LSZ formula. In this sense they are not only intermediate states.

+ 3 like - 0 dislike

I guess my brain has some quirks today, I keep reverting my statements. I hope this is the last edit.

The "third" and "longitudinal" polarization indeed enters the reactions, because it is longitudinal only in the spatial sense! I.e., we have the polarization vectors $\epsilon^\mu_{(\lambda)},\,\lambda=-1,0,1$ and for all of these we have space-time transversality $\epsilon^\mu_{(\lambda)} k_\mu =0$. So the $k_\mu M^\mu$ does not change anything.

There is a slight inconsistency in the nomenclature which causes the whole confusion here, "longitudinal" is sometimes used in the sense of "spacetime longitudinal", i.e. proportional to the four-vector $k^\mu$. However, in the vector field sense, we only have a "spatial longitudinal" mode which means that in the helicity basis $\epsilon^i_{(0)} \sim k^i$ with the full four-vectors remaining "space-time transversal" $\epsilon^\mu_{(0)} k_\mu =0$.

Hence, my previous discussion of "longitudinal" modes was faulty because of this misidentification of the mode. The "space-time longitudinal" mode of a vector field indeed does show up sometimes in the quantization process and indeed can be neglected as a "ghost". But the "spatial longitudinal" mode is physical, emerges from "QED" reactions (interacts with the spinor current), and thus cannot be completely neglected.

After getting rid of terms which do not interact with the spinor current $\bar{\psi} \gamma^\mu \psi$, the propagator of the massive field turns out to look similar as the massless-photon propagator. But the massive-photon propagation does in fact include the third mode! Also, as remarked by Arnold Neumaier, the amplitudes for the production of longitudinal photons gets suppressed by powers of $m$ so we can use $m \to 0^+$ as an infrared regularization of QED. However, the longitudinal mode would show in long-term equilibria for finite $m$. (Obviously, one can put the mass so low that it would take the age of the universe to reach an equilibrium photon gas with longitudinal photons taking up any significant amount.)

One could maybe try to modify the Hilbert space of in and out states of the massive vector field so that "spatial longitudinal" polarizations have zero norm and they can be a quotient of the physical states. For this, the commutation relations for the "creation" and "annihilation" operators would have to be modified similarly to the Gupta-Bleuer quantization.

However, when you would take a look at the in and out states you would find out that the norm of states is not Lorentz-covariant. The only covariant completion is including the third polarization with the usual norm and "correct" commutation relations. This is probably not the only place where Lorentz covariance would be violated and the resulting theory might turn out quite exotic.

answered Sep 2, 2015 by Void (1,645 points) [ revision history ]
edited Sep 3, 2015 by Void

Massive photons move with smaller velocity?

@No_real_name Yes, but the mass could be so miniscule that there would be basically no difference between the speed of a massive photon and a massless photon. Even with the probably massive neutrinos we see them arriving earlier from astrophysical events because unlike photons they are less reactive and the reactivity slows down the massless photons more than the mass "slows down" the neutrinos.

But the statement 2 from my question (about $k_{\mu}M^{\mu} = 0$) solve the problem of experimental evidence of longitudinal polarizations, doesn't it? The only problem with longitudinal polarizations, maybe, is that they make some contribution in matrix elements as intermediate states.

Could you please prepend your answer with a statement of where you start? (strict approach to what?)

@NAME_XXX @ArnoldNeumaier Sorry for muddling around, please see the newest edit.

It's not my business anymore, but EMW in different media may propagate slower than in vacuum, and they may have three "independent" polarizations.

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