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I'm stuck with the photon propagator, at chapter 5 of Mandl and Shaw QFT book. They say that since the Maxwell Lagrangian density for the free Electromagnetic field has a conjugate momenta to the field $$\frac{\partial L}{\partial(\partial_0 A^0)}=0$$ Then one cannot choose a canonical quantization.

Now in chapter 5 they say that choosing the Fermi Lagrangian density $$ L_{fermi}=-\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu - J_\mu A^\mu \tag{5.10}$$ with sign convention $(+,-,-,-)$ for the electromagnetic field the momenta is well defined $$\frac{\partial L}{\partial(\partial_0 A^\mu)}={-\dot{A}^\mu}\tag{5.11}$$ and one can quantize in a canonical way, because now it makes sense to introduce commutation relations: $$[A(x^\mu),\dot{A}^\mu (y^\mu)]=i\hbar\delta^3(x^\mu-y^\mu).\tag{5.23}$$

Now We also know that the Fermi lagrangian density is not gauge invariant, since there is the interaction term ( $A^\mu$) $$A^\mu \rightarrow A^\mu+\partial^\mu f \tag{5.7}$$ that transforms under gauge tranformation for $A^\mu$. Is it right to say that the NON Gauge invariance of the Fermi lagrangian is not a problem because the action changes but gives the same equations of motion? (this must be because of a total divergence in the action in which the Non gauge interaction term is included); I have heard this argument from my professor but i think i missed the sense. Can someone help me with this?

Sorry, i think i messed up with terms in the equations, i can try once more to ask my question: My professor sayd that the Fermi lagrangian density $$\mathcal{L_{fermi}}-\frac{1}{2}(\partial_\mu A^\nu ) (\partial_\nu A^\mu)-J_\mu A^\mu$$ is not gauge invariant, so if i perform the gauge transformation $A^\mu \rightarrow A^\mu + \partial^\mu f $ the Lagrangian density changes, in particular the term with the current $J_\mu$.

Now my question is if it is right to add a null term to the lagrangian $f\partial^\mu J_\mu$ so that i can write the transformed lagrangian : $$\mathcal{L_{fermi}}=-\frac{1}{2}(\partial_mu A^\nu ) (\partial_nu A^\mu)-J_\mu A^\mu-J_\mu \partial^\mu f - f \partial_\mu J_\mu = $$
$$=-\frac{1}{2}(\partial_\mu A^\nu ) (\partial_\nu A^\mu)-J_\mu A^\mu - \partial^\mu(f J_\mu)$$ So now the derivative does not change equations of motion derived from the action variation. Thanks to all of you for your time.

Buzz
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Federico De Matteis
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2 Answers2

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  1. Classically, Fermi's EM Lagrangian density (5.10) needs the Lorenz gauge fixing condition $$ \partial_{\mu}A^{\mu}~=~0\tag{5.13}$$ in order for its EL equation to correctly reproduce the Maxwell equations.

  2. Fermi's EM Lagrangian density (5.10) is the usual gauge-fixed EM Lagrangian density $${\cal L}_{\rm gf}~=~ -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} - \frac{(\partial_{\mu}A^{\mu})^2}{2\xi} \tag{A}$$ in the Feynman–'t Hooft gauge $\xi=1$, up to total divergence terms.

  3. For the corresponding BRST formulation, see e.g. this Phys.SE post.

  4. The gauge-fixing breaks the gauge symmetry of the action. However, one may show that the physical observables do not depend on the gauge fixing condition, cf. eg. this Phys.SE post.

Qmechanic
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It is quite possible to use the Fermi Lagrangian as a starting point of classical electromagnetic field theory. Here is my paper on this. It is correct that the resulting theory is not gauge invariant. In the paper I show that this is not an issue but rather an asset. It is straightforward to quantise this theory, see Jauch&Rohrlich, Theory of Photons and Electrons (1955), Ch2-3, p. 28: "The Lagrangian adopted here corresponds to the method of quantization developed by E. Fermi, Rendiconti d. R. Ace. dei Lined (6) 9, 881 (1929)". Apparently, Fermi never wrote down the Lagrangian himself.

Note that it is impossible to quantise the gauge invariant theory without fixing the gauge.

my2cts
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