I've recently finished reading Mark Thomson's "Modern Particle Physics". There is a question which is not answered in his book, and to which I couldn't find on the internet in the "introductory" parts of QFT.

My problem is with the form of the Proca Lagrangian. Indeed, for a massless field (Say in QED) the lagrangian is given in the book as:

$$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}.$$

I am placing myself in the vacuum here. As we can see, there is a minus sign in front of the "kinetic" term. This does not affect dynamics in this case, but it looked already unusual to me since usually the kinetic term is taken with a positive sign. If we generalise to a massive boson, we get:

$$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}m^2A^\mu A_\mu$$

with the $(+,-,-,-)$ Minkowski sign convention. Again, the overall sign is the opposite of the usual convention, otherwise nothing really strange here.

However, my problem arises when we consider the lagrangian of spin-half particle interacting with QED (but imagining the photon has a mass, I know it's non-physical but it's an easier example for me :D). Then, the total Lagrangian becomes, if I'm not mistaken:

$$ L =i\overline{\psi}\gamma^\mu D_\mu\psi-m\psi\overline{\psi}-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A^\mu A_\mu$$

And here, the seemingly arbitrary sign isn't anymore... What is the justification for this inverted kinetic and mass sign ? I guess that for massless bosons, the sign is still arbitrary since the $F_{\mu\nu}$ term decouples from the rest of the lagrangian, but I don't think it is with the mass term (haven't done the calculations though, so shame on my lazyness if that is the answer). I know my lagrangian is not U(1) gauge invariant, so my guess is maybe that strange sign comes from the spontaneous symmetry breaking from the Higgs mechanism ?

Edit: Some precisions about what I mean by "Inverted sign" : If we take say a scalar field with a mass, it's lagrangian will be composed of a kinetic term $\frac{1}{2}\partial_\mu\phi\partial^\mu\phi$ and a mass term $\frac{1}{2}m^2\phi^2$. The lagrangian then is

$$L = T-V = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$$

with the $(+,-,-,-)$ Minkowski sign convention.

If I interpret $F_{\mu\nu}F^{\mu\nu}$ as a kinetic term, then we see that in the lagrangian I wrote up, the signs of the terms are inverted. I would like to know if this is a purely conventional choice (which I know it is, in the case of the isolated lagrangian, but not sure in the combined one), and if it's not, why it is written like that.

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    $\begingroup$ Can you clarify exactly where you think the inverted signs are? Your first equation seems like it has the usual sign to me. $\endgroup$
    – knzhou
    Mar 10, 2017 at 3:58
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    $\begingroup$ My 2 cents and guesses. The eom arising from this Proca Lagrangian are equivalently written: $(\partial_\mu \partial^\mu + m^2)A_\mu=0$ with the constraint $\partial_\mu A^\mu=0$. So maybe it's because you can reduce to a sort of Klein-Gordon form? I've seen the Proca Lagrangian also with a minus in front of the mass term, and you get the equation $(\partial_\mu \partial^\mu - m^2)A_\mu=0$ (plus the constraint), which is simply a Klein-Gordon form you usually get using the $(+,-,-,-)$ metric. I PERSONALLY think it's just a matter of convention. $\endgroup$
    – Luthien
    Mar 10, 2017 at 4:03
  • $\begingroup$ Typo: with the $(-,+,+,+)$ metric. Please, note that this is just a "hint" about something I've seen during my studies. Let's hope for someone with more knowledge to come and clarify the issue :) $\endgroup$
    – Luthien
    Mar 10, 2017 at 4:13
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    $\begingroup$ For me the scalar and Proca field have the same form of mass term. Conventionally, we can also write $L = \frac{1}{4}F^{\mu\nu}F_{\mu\nu} -\frac{1}{2}m^2A^\mu A_\mu$. In this convention your third equation may be accordingly changes to $L =-i\overline{\psi}\gamma^\mu D_\mu\psi+m\psi\overline{\psi}+\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}M^2A^\mu A_\mu$. If you are not satisfied the mass term of fermion part $m$ (at this point I'm not so convinced) you may change its sign because this sign we have chosen when we perform a square root of Klein-Gordon operator. $\endgroup$ Mar 10, 2017 at 8:12
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    $\begingroup$ May be boson and fermion parts are different. $\endgroup$ Mar 10, 2017 at 8:19

2 Answers 2


The sign is not arbitrary. It is easiest to understand when you look at the propagator, i.e. inverse equation of motion. For a scalar field in momentum space $$ \Delta_F(p) = \frac{i}{p^2-m^2} $$ For a vector field $$ \Delta^{\mu \nu}_F(p) = i\frac{-g^{\mu \nu}+\frac{p^\mu p^\nu}{m^2}}{p^2-m^2}. $$ And now look at the physical components. For a scalar field, there is only one component, and the residue of the propagator is positive. This sign corresponds to the plus sign in front of the kinetic term in the Lagrangian. For a vector field, only physical components propagate with a positive sign. The $0$'th component is unphysical, and hence it has a wrong sign. Again this is a consequence of the sign in Lagrangian and the signs in the metric tensor. In other words, for both cases - scalar and vector, only the physical degrees of freedom have positive kinetic energy.

The same argument applies to mass term. For a vector field $$ m^2 A^\mu A_\mu = m^2 A_0^2-m^2A_iA_i, $$ so the physical components have the same sign of a mass term like the scalar field. The unphysical 0'th component has opposite sign.

Causality and unitarity depend on this sign. Kinetic energy and mass have to be non-negative for physical degrees of freedom. It is easy to see that this sign choice for the Lagrangian leads to a Hamiltonian that has non-negative eigenvalues; or more precisely, that is bounded from below.

There are no physical polarization states associated with $A_0$, so this field is unphysical, it carries no energy or momentum. In the non-relativistic approximation (in certain gauge), it can correspond to instantaneous potential (the so-called potential photons in NRQED) which violate causality because they generate non-local interactions. But this is just an approximation, expansion in small velocity compared to the speed of light.
For Proca field, we can use equation of motion $$ \partial_\mu A^\mu =0 $$ to show that only three out of four components of the field are independent. This means that using 4-component vector $A$ field to describe massive vector boson, we have one unphysical field and three physical fields. Note that this equation means that the longitudinal field is noninteracting.

  • $\begingroup$ Sadly I cannot understand your propagator argument because I haven't started QFT yet, only particle physics. However, I don't understand your second point : what do you mean when you say that the 0th component is "unphysical" ? For QED for example, the 0th component is related to the electric potential, which is for me as physical as the vector potential $\endgroup$
    – Frotaur
    Mar 10, 2017 at 8:18
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    $\begingroup$ Unphysical means that it cannot propagate - there is no photon corresponding to the 0 component. Electric potential is instantaneous interaction and not a bunch of photons which carry some kinetic energy. In reality, there are no instantaneous interactions -- they are unphysical. Instantaneous interactions are just a convenient approximation in non-relativistic physics. You can use the concept of electric potential only when the source of the potential is some massive particle; you can neglect the fact that it moves (or when you consider expansion in small velocity). $\endgroup$
    – Veritas
    Mar 10, 2017 at 8:27
  • $\begingroup$ @Veritas From the Proca Lagrangian, we have the Klein-Gordon equation and a Lorentz condition. Therefore, we known there are only 3 independent fields. But how do we know the unphysical field is A_0? $\endgroup$
    – Wein Eld
    Sep 21, 2017 at 19:18

The answer sits in the fact that you want your real photons to have the correct sign. This can be realized by adding a gauge fixing term $\frac{1}{2}(\partial_\mu A^\nu)^2 = 0$.

(The additional term is due to zero because I chose a gauge where it is!)

$$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} - \frac{1}{2}(\partial_\mu A^\nu)^2 + \frac{1}{2}m^2A_\mu A^\mu$$

It is now easy to write out $F$ and perform one partial integration in order to end up with:

$$\mathcal{L} = -\frac{1}{2}\partial_\mu A_\nu \partial^\mu A^\nu + \frac{1}{2}m^2A_\nu A^\nu$$

Now we use our metric $$+ ---$$ to write out the summation over $\nu$ index:

$$\mathcal{L} = -\frac{1}{2}\partial_\mu A_0 \partial^\mu A^0 + \frac{1}{2}m^2A_0 A^0 + \frac{1}{2}\partial_\mu A_i \partial^\mu A^i - \frac{1}{2}m^2A_i A^i $$

Which is to be understood as the Lagrangian of 4 decoupled scalar fields $A_\mu$. As you already noticed one of them has "flipped" sign and is therefore unphysical. We find another unphysical field that is tangent to the momentum of the photon. It turns out that these two unphysical fields will always cancel out in calculations !

So to recapitulate, there is indeed one field with it's sign "flipped" but it always cancels out against the longitudinal photon. We end up with 2 real photon polarizations as was to be expected :)

EDIT: notice that changing to a metric with (-+++) signature which is typically done in more theoretical papers will change a whole bunch of signs !


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