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In Quantum field theory by M. Srednicki, we find on p. 337 the statement regarding the Maxwell Lagrangian

The action we seek should be Lorentz invariant, gauge invariant, parity and time-reversal invariant, and no more than second order in derivatives. The only candidate is $S=\int d^4x\mathcal{L}$, where $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+J^\mu A_\mu.$$

Let us ignore the source term for now and expanding the field-strength tensor $F^{\mu\nu}$, we find $$ \mathcal{L} = - \frac{1}{2} \left(\partial^\mu A^\nu\right) \left(\partial_\mu A_\nu\right) + \frac{1}{2} \left(\partial^\mu A^\nu\right) \left(\partial_\nu A_\mu\right) .\tag{1} $$

How do I arrive at eq. (1) from first principles?

  1. We know that the action of a relativistic field theory must be Lorentz-invariant scalar. We can construct Lorentz-invariant scalars by contracting Lorentz-invariant tensors.

  2. From polarization experiments, we should be able to conclude that photons are spin-1 particles. Hence, we are looking at vector fields $A^\mu$.

  3. Photons should be massless otherwise the QED Lagrangian is not invariant under local gauge transformations which would then spoil electric charge conservation. Hence, our Lagrangian cannot contain a mass term and the Lagrangian should be invariant under local gauge transformations.

  4. Photons don't self-interact directly (though they can interact through vacuum fluctuations). Hence, there are no power terms, e.g., $(A_\mu A^\mu)^2$, in the Lagrangian.

  5. With the assumptions so far, the action contains only derivatives of $A^\mu$. From dimensional analysis, see Constraints on scalar field theories in $n$ dimensional spacetime, we reduce the Lagrangian to $$ \mathcal{L} = c_1 \left( \partial_\mu A^\nu \right) \left( \partial^\mu A_\nu \right) + c_2 \left( \partial_\mu A^\mu \right) \left( \partial_\nu A^\nu \right) + c_3 \left( \partial_\mu A^\nu \right) \left( \partial_\nu A^\mu \right) \tag{2}. $$

  6. For the action to exist, we need the vector field $A^\mu$ to be integrable, i.e., to falloff sufficiently rapid at infinity. Hence, boundary terms are zero and we can perform partial integration to move around the derivatives. We thereby see that we can rewrite the third term as the first term. The argument was used in C. de Rham. Massive Gravity. 2014. and we reduced the Lagrangian density to $$ \mathcal{L} = c_1 \left( \partial_\mu A^\nu \right) \left( \partial^\mu A_\nu \right) + c_2 \left( \partial_\mu A^\mu \right) \left( \partial_\nu A^\nu \right) \tag{3}. $$

Comparing eq. (3) with eq. (1), we are left to show that $c_1=-1/2$ and $c_2=-c_1$.

We still have not used the time-reversal and parity invariance.

How do we see that time-reversal and parity invariance require $c_1=-c_2=-1/2$?

According to this post on physicsforums.com, parity transformations only change the field argument, i.e.,

$$ \hat{A}(x,t) \xrightarrow{x\to -x} \hat{A}(-x,t) \tag{4} $$

but inserting eq. (4) into the action and substituting the integration variable to "undo" the parity transformation does not change anything (or I missed something).

The same would be true for the time-reversal.

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    $\begingroup$ parity and time-reversal preclude $\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$, which you missed. The overall coefficient is arbitrary, you can relabel $A\to c_1^{-1/2}A$ to remove $c_1$. So you only have one more coefficient to fix; gauge invariance? $\endgroup$ Oct 11, 2021 at 16:31
  • $\begingroup$ Possible duplicate: Deriving Lagrangian density for electromagnetic field and links therein. $\endgroup$
    – Qmechanic
    Oct 11, 2021 at 16:43
  • $\begingroup$ Too bad that this Lagrangian, though itself gauge invariant, does not give gauge invariant Noether currents. In fact, such a lagrangian does not exist. $\endgroup$
    – my2cts
    Oct 11, 2021 at 17:09
  • $\begingroup$ @my2cts can you elaborate? Doesn't the QED Lagrangian (Dirac + Maxwell) give rise to a Noether current representing particle (and thereby charge) conservation? $\endgroup$
    – bodokaiser
    Oct 11, 2021 at 18:56
  • $\begingroup$ @bodokaiser They do give Noether currents for charge-current, energy-momentum and angular momentum, but these are not gauge invariant. $\endgroup$
    – my2cts
    Oct 11, 2021 at 20:03

1 Answer 1

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Applying gauge invariance is the last step. Consider an infinitesimal gauge transformation $A_\mu \to A_\mu + \partial_\mu \alpha$. Then the change in the Lagrangian, to first order in $\alpha$, is $$\Delta S = \int d^4x \, 2 c_1 (\partial_\mu \partial^\nu \alpha)(\partial^\mu A_\nu) + 2 c_2 (\partial_\mu \partial^\mu \alpha) (\partial_\nu A^\nu).$$ We may integrate the first term by parts twice, to give $$\Delta S = \int d^4x \, 2 c_1 (\partial_\mu \partial^\mu \alpha)(\partial^\nu A_\nu) + 2 c_2 (\partial_\mu \partial^\mu \alpha) (\partial_\nu A^\nu)$$ from which it immediately follows that $c_1 + c_2 = 0$. Finally, to set the specific values of $c_1$ and $c_2$, we conventionally scale $A_\mu$ by a constant so that the kinetic term is canonically normalized.

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  • $\begingroup$ Thinking a bit more about this, I wonder if there is a way to show $c_1+c_2=0$ without invoking local gauge symmetry. Coming from the "theory direction" and assuming the Dirac Lagrangian to be established, we know that local gauge symmetry is required by charge conservation. However, what if argue from the "experimental view"? We can perform experiments to infer that photons have spin-1 and no mass but this is not sufficient to find $c_1+c_2=0$? $\endgroup$
    – bodokaiser
    Oct 12, 2021 at 18:16

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