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I don't understand why this is true

The Dirac Lagrangian is globally invariant under U(1) symmetry and the QED Lagrangian is locally invariant under U(1). Electric charge is the conservation ‘law’ associated with the U(1) symmetries so charge is conserved for both Lagrangians under U(1).

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    $\begingroup$ Can you mention where the quote is from? $\endgroup$
    – user191954
    Jul 22, 2018 at 11:03
  • $\begingroup$ Because of gauge symmetry of the latter. $\endgroup$
    – Qmechanic
    Jul 22, 2018 at 11:22

1 Answer 1

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The main difference is that Dirac Lagrangian doesn't involve photon-electron interaction which makes the local $U(1)$ symmetry possible, while the main purpose of QED is to describe the interactions of electrons and photons.

First of all, the Dirac Lagrangian reads as follows: $$ \tag{1} \mathcal{L}=i \bar\psi \gamma^\mu \partial_\mu \psi - m \bar\psi \psi $$ where global $U(1)$ symmetry is obvious. An element, $g=e^{i q \theta}$ of the symmetry group, where $\theta$ is a constant parameter, leaves the Lagrangian invariant under $$ \tag{Global U(1)} g=e^{i q \theta} \\ \psi(x) \rightarrow g \psi(x) \;\; \text{and} \;\; \bar\psi(x) \rightarrow \bar\psi (x) \, g^* $$ since it can pass the partial derivative and cancel its conjugate.

However, a local $U(1)$ transformation would not make it invariant because the element $g$ is now spacetime dependent, i.e., $$ \tag{Local U(1)} g(x) = e^{i q \theta(x)} \\ \psi(x) \rightarrow g(x) \, \psi(x) \;\; \text{and} \;\; \bar\psi(x) \rightarrow \bar\psi(x) \, g^* (x) $$ so you would have an extra term like: $$ \tag{2} \delta \mathcal{L} = - q \bar\psi \gamma^ \mu \partial_\mu \theta(x) \psi $$

Therefore, Dirac Lagrangian is not invariant under local $U(1)$ transformations. This is the starting point of the QED: how one can describe a local $U(1)$ symmetry out of the Dirac Lagrangian?

After the development of QED, the Lagrangian becomes essentially a Dirac Lagrangian plus a term consisting a photon-electron interaction, as follows: $$ \tag{3} \mathcal{L}_{QED}=i \bar\psi \gamma^\mu \partial_\mu \psi - q \bar\psi \gamma^\mu A_\mu \psi - m \bar\psi \psi - \frac14 F_{\mu \nu} F^{\mu\nu} $$ where $A_\mu (x)$ is the photon field, $F_{\mu\nu}$ is the field strength tensor of photon, and the second term is sometimes embedded in the derivative, such that, $$ \tag{4} \nabla_\mu \equiv \partial_\mu + iq A_\mu $$ becomes the new derivative, which is called the covariant derivative.

As you can see, the local $U(1)$ transformation of the second term gives the exact same extra term above, but with an opposite sign, so that they cancel each other. Because the photon field transforms as $$ \tag{5} \delta A_\mu (x) = \partial_\mu \theta(x) $$ That's why this local $U(1)$ symmetry is called a gauge symmetry, and the photon is called a gauge boson.

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  • $\begingroup$ +1 Who first wrote the QED Lagrangian and when? $\endgroup$
    – safesphere
    Jul 22, 2018 at 16:31
  • $\begingroup$ I am not sure by who and when it was written in this form. Even before Dirac it was known that the Lagrangian of classical electrodynamics should be written in such form, but until Schwinger, Tomonaga, Dyson and Feynman, it was only the Dirac part that was successfully quantized, the rest (Maxwell part) was giving infinities when applied to higher energy scales. After renormalizations are introduced, then QED was complete. $\endgroup$ Jul 22, 2018 at 17:39

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