I could never understand the choice of gauge in QED. Let's say I know that $A_{\mu}$ has 4 components, hence 4 degrees of freedom. For, say, a photon I need only two. Let's say I pick Lorentz gauge and set

$\partial_{\mu} A^{\mu} = 0$

What does it change? I know, that it makes the equations of motion symmetric, but how can I see explicitly that I have 3 degrees of freedom now?

For a photon, one usually goes further and choses $A^0 = 0$ and $\nabla \cdot \vec{A} = 0$. Somehow it reduces the number of dof to 2... I can't see all of that. I mean I do understand that the constraints should reduce the number of dof in the system, but there has to be some systematic approach, like, say, Lagrangian multipliers in Class. Mech., not just "I want to do this cuz it looks cool and makes my life easier"=(

  • $\begingroup$ The systematic approach was first described by Dirac in "Lectures on Quantum Mechanics". For a more recent and comprehensive treatment of the subject have a look at "Quantization of Gauge Systems" by Henneaux and Teitelboim. $\endgroup$
    – Phoenix87
    Sep 4, 2016 at 20:12
  • $\begingroup$ This is covered in detail in Chapters 2 and 5 in Weinberg I. $\endgroup$ Sep 4, 2016 at 21:18
  • $\begingroup$ Generally an equation would remove a degree of freedom, because either it fixes it as in $\mathbf{A}^0=0$ or it expressed one degree of freedom in terms of the others. $\endgroup$ Sep 6, 2016 at 4:13

1 Answer 1


In usual theories the number of freedom of a system can be obtained by looking at the number of variables in relation to the number of equations describing the system. In the case of classical electrodynamics one would be tempted to derive the equation of motion for the photon from the Lagrangian ${\scr{L}}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ and try to constrain the 4 components of the $A_\mu$ field to two.

However, the $A_\mu$ field experiences gauge invariance $A_\mu \rightarrow A_\mu - \partial_\mu \alpha$. The values $\alpha$ can be choosen freely in the Lagrangian and this gauge invariance is responsible for a redundancy in the description of the system, the true number of degrees of freedom remains hidden. To find out the true physical degrees of freedom the system needs to be quantized and the gauge redundancy must be isolated. This is done through the Gupta-Bleuler formalism in the QED. The more general procedure is called Fadeev-Popov quantization and is also applicable to non-Abelian theories.

The main point in the quantization procedure is to write the photon field as a Fourier decomposition with annihilation and creation operators $a$ and $a^\dagger$: $$A_\mu =\int \frac{d^4k}{(2\pi)^4}\sum_{\lambda=0}^3(e^{-ikx}a^\dagger(k)\epsilon_\mu(k, \lambda) + e^{ikx}a(k)\epsilon_\mu^*(k, \lambda)) .$$

The former four degrees of freedom of the system are now in the 4 linearly independent polarization vectors $\epsilon(k)$. The Lorentz gauge $\partial_\mu A^\mu =0$ must now be imposed on the quantum level, hence on the Hilbert space giving $k_\mu \epsilon^\mu= 0$. This is restricting the possible polarizations of the photon by eliminating the longitudinal polarizaiton. Hence, one degree of freedom is getting lost.

By continuing the procedure and using the massless condition $k^2=0$ one can make another possible polarization decouple from the physical degrees of freedom and leaving the system with only 2 pyhsical transverse polarizaions. The process of the quantization is higly non-trivial and so is the counting of degrees of freedom.

  • $\begingroup$ Why do we think that photons have 2 dof? Experiment? What if there is more, but we just don't have tools to register them? And it somehow magically linked to zero mass... Usually the reason for imposing constrains in QM and QFT is symmetries. Gauge invariance is not a symmetry in real (3+1)D space, but some kind of symmetry in parameter space, which I don't understand, cannot imagine. $\endgroup$
    – MsTais
    Sep 7, 2016 at 13:25
  • $\begingroup$ Gauge invariance hides the true number of degrees of freedom and leads to a redundancy in the physical description. Therefore it is more of a symmetry of the mathematical expressions, not that much of a physical symmetry... QFT is able to describe the nature with 2-dof-photons quite well. What would be the reason to assume there are other dof's? $\endgroup$
    – Statics
    Sep 7, 2016 at 13:56
  • $\begingroup$ Symmetry of mathematica expressions comes from physical symmetries, symmetries in parameter space, etc. They don't just show up typically. The reason of me doubting that is the same as always: the fact that we don't see something happening is not the proof that it doesn't happen. I find the fact that $A^{\mu}$ coming out from electrodynamics and naturally living in 4D, and yet not describing the photon correctly without some tricks, a great irritation=( If something fundamental is not general, it very much might be false. $\endgroup$
    – MsTais
    Sep 7, 2016 at 20:20
  • 1
    $\begingroup$ The gauge symmetry does not describe anything physical, since it does not change the physics for different gauge group parameters $\alpha$. The electrodynamical $A_\mu$ does not describe the full picture correctly since the description is on the classical level. In order to account for the quantum nature of the particles the description needs to be quantized. One can view the imposing of gauge invariance as a trick to obtain the desired conserved quantities. Nevertheless it remains a very efficient way to describe the nature of QED. $\endgroup$
    – Statics
    Sep 8, 2016 at 8:37

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