I was reading the Wikipedia page on Proca Action. To summarize, it is almost like Maxwell action, but with a mass term because of which Proca action does NOT have gauge invariance. From the equation of motion, we see that each of the 4 modes obey Klien Gordon equation, and they also obey a "Lorenz gauge" condition $\partial_\mu B^\mu = 0$.

It is not actually a Lorenz gauge ( unlike in the case of Maxwell action ) because Proca lagrangian has no gauge invariance to begin with, and this constraint simply follows from equations of motion.

My actual question is regarding the path integral quantization of the Proca action:

Naively I would like to integrate over all possible field configurations of B to calculate the partition function: $$ Z = \int [dB] exp\{-i \int d^4 x L_{proca}(B)\} $$ But then I should integrate only over those B fields that satisfy the "Lorenz gauge" constraint. I might have to introduce a lagrange multiplier to achieve that, or perhaps some more complicated method. Wikipedia says "Quantizing the Proca action requires the use of second order constraints."

But since this constraint is coming from the equations of motions, shouldn't this constraint follow automatically from the action minimization principle? In other words, I don't think I should put any constraints while integrating over B fields. My naive guess of blindly integrating over all B fields seems more appropriate to me. Afterall, we never talk about implementing the Klien Gordon equation on the B field via constraints. Just as the Klien Gordon condition comes out naturally from sum over histories, so should the Lorenz gauge condition. Afterall, both of these conditions were derived from the equations of motion.

I want to contrast the Proca action from the Maxwell action. In Maxwell action, the gauge redundancy ( I think ) can't be inferred from the equations of motion. So it makes sense to put a guage fixing term by hand there.

Can you please give any hints or example to clear my misunderstanding here? I couldn't find any textbook that discusses path integral quantization of Proca action.

  • 1
    $\begingroup$ You do not need an extra constraint for a massive vector field. As the Wikipedia article explains correctly, the Proca equation (derived from the Lagrangian) is equivalent to the Klein-Gordon equation for all components of $B^\mu$ plus the equation $\partial_\mu B^\mu=0$. $\endgroup$
    – Hyperon
    Feb 19 at 15:39
  • $\begingroup$ The last part in the Wikipedia article might be misleading. Using path integral quantization, everything comes out automatically. $\endgroup$
    – Hyperon
    Feb 19 at 15:54
  • $\begingroup$ Is wikipedia wrong? What is the point of that comment on constrained quantization? @Hyperon $\endgroup$
    – Quillo
    Feb 19 at 16:06
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    $\begingroup$ @Quillo This remark does not refer to path integral quantization. But if you write down the field operator you have to take into account the constraint $\partial_\mu B^\mu =0$, i.e. $k_\mu \epsilon^\mu(\vec{k})$ in momentum space. $\endgroup$
    – Hyperon
    Feb 19 at 16:28

1 Answer 1


The path integral quantization of a (free) massive vector field does not pose any problems, in particular it does not require imposing any additional constraint in the functional integral. A (real) spin $1$ field $V^\mu$ is described by the Lagrangian $$\mathcal{L}= -\frac{1}{4} V_{\mu \nu}V^{\mu \nu}+\frac{M^2}{2}V_\mu V^\mu, \qquad V_{\mu \nu}=\partial_\mu V_\nu-\partial_\nu V_\mu. \tag{1} \label{1} $$ The associated generating functional is defined by $$Z[J] = \langle 0| e^{-i\int\! d^4x \, J^\mu(x)\, V_\mu(x)} |0 \rangle = \int [dV] \, e^{iS[V,J]} \tag{2} \label{2} $$ with $$ \begin{align} S[V,J]&= \int \! d^4x \,\left(-\frac{1}{4} V_{\mu \nu}V^{\mu \nu} +\frac{M^2-i\epsilon}{2} V_\mu V^\mu - J_\mu V^\mu \right) \\[5pt] &= \int \! d^4 x \left\{\frac{1}{2}V^\mu \left[\eta_{\mu \nu} \left( \square+ M^2-i \epsilon\right) -\partial_\mu \partial_\nu \right] V^\nu -J_\mu V^\mu \right\} \end{align}\tag{3} \label{3} $$ and the normalization condition $Z[0]=1$. The functional integral is evaluated by employing the usual shift of the integration variable $V_\mu = V_\mu^\prime+W_\mu$, using translation invariance of the functional measure, $[dV_\mu]=[dV_\mu^\prime]$, and choosing the field $W_\mu$ in such a way that the terms linear in the new variable of integration $V_\mu^\prime$ vanish. As a consequence, $W_\mu$ is determined by the differential equation $$\left[\eta_{\mu \nu}\left( \square+M^2-i \epsilon \right) - \partial_\mu\partial_\nu \right]W^\nu=J_\mu. \tag{4} \label{4} $$ The determination of the Green function $\Delta^{\nu \rho}(x)$ of the differential operator occuring in \eqref{4}, defined by $$ \left[\eta_{\mu \nu} \left(\square +M^2-i\epsilon \right) -\partial_\mu \partial_\nu \right]\Delta^{\nu\rho}(x)=\delta_\mu^{\;\rho} \, \delta^{(4)}(x), \tag{5} \label{5}$$ is straightforward (in contrast to the case of a massless gauge field, the inversion of the differential operator poses no problem), with its Fourier representation given by $$ \Delta^{\nu \rho}(x)=\int \!\frac{d^4k}{(2\pi)^4 }\, e^{-ik\cdot x} \,\frac{\eta^{\nu \rho}-k^\nu k^\rho/M^2}{M^2-k^2-i \epsilon}. \tag{6} \label{6} $$ The (unique) solution of \eqref{4} is thus given by $$W^\mu(x)=\int \! d^4 y \, \Delta^{\mu \nu}(x-y)\, J_\nu(y), \tag{7} \label{7}$$ finding $$Z[J]=e^{-\frac{i}{2}\int\! d^4x \, d^4 y \, J_\mu(x) \, \Delta^{\mu \nu}(x-y) \, J^\nu(y)} \tag{8} \label{8}$$ for the generating functional.

  • $\begingroup$ Thanks. Is there a way to check ( for a more complicated theory in my case ) that the path integral takes care of all the constraints coming from equations of motion? Perhaps by checking that the operator between the V has no zero mode??? $\endgroup$
    – baba26
    Feb 20 at 1:53
  • 1
    $\begingroup$ It is often easier to study the differential operator in Fourier space, where it becomes a multiplication operator. See e.g. physics.stackexchange.com/q/798235 for the general method to find the inverse of such an operator acting on a vector field (checking at the same time if it exists). Similarly, this can be down for the more complicated case of a Rarita-Schwinger field. $\endgroup$
    – Hyperon
    Feb 20 at 6:23

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