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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?

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2 Answers 2

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  1. Note that the 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. The Lagrangian density (5.10) is the usual gauge-fixed Lagrangian density $${\cal L}_{\rm gf}~=~ -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} - \frac{(\partial_{\mu}A^{\mu})^2}{2\xi} $$ 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.

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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 derovative does not change equations of motion derived from the action variation. Thanks to all of you for your time.

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