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My problem is that I don't get, how you can calculate the Noether current under spacetime translation of the Lagrangian density of the Dirac field. I know that in the end you get the energy and the momentum as conserved quantities, but as I tried it on my own I got confused and the internet didn't provide a proper calculation, only the solution if you search long enough.

The Lagrangian density of the Dirac field is $ \mathcal{L} =\overline{\psi}(i\gamma _{\mu} \partial _{\mu} -m)\psi $

and the spacetime translation manifests through $ x'^{\mu} = x^{\mu} - \epsilon ^{\mu}$

Then we get $\Delta\psi (x) = \epsilon ^{\mu} \partial _{\mu} \psi (x)$ for the fields and $\Delta\mathcal{L} (x) = \epsilon ^{\mu} \partial _{\mu} \mathcal{L} (x)$ for the Langrangian as infinitesimal transformations. As next step you would take the formula for the noether current and then fill it in and calculated it, but that's where I get lost.

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    $\begingroup$ Specifically what have you tried? You should show the work you attempted to do. $\endgroup$
    – Triatticus
    Aug 2 at 1:30
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Lagrangian of the Dirac field is: $$ \mathcal{L} = \bar\psi [ i \gamma^\mu \partial_\mu - m ] \psi . $$

Now, let us consider an infinitesimal change $$ x^\mu \to x'^\mu = x^\mu + \epsilon^\mu . $$ The field changes as follows: $$ \psi(x) \to \psi'(x) = \psi(x) + \epsilon^\mu \partial_\mu \psi(x) . $$ Let us define $\delta\psi := \epsilon^\mu \partial_\mu \psi(x)$ . Change of Lagrangian is given by (this is true for any field !) $$ \delta\mathcal{L} = \partial_\mu (\epsilon^\mu \mathcal{L}) . $$ Since this is total derivative, you can apply Noether's theorem, and get \begin{align} j^\mu &= (\delta\psi^a) \left( \frac{\partial}{\partial (\partial \psi^a)} \mathcal{L} \right) - \epsilon^\mu \mathcal{L} \\ &= \epsilon_\nu \partial^\nu \psi^a (-(\bar\psi i \gamma^\mu)_a) - \epsilon_\nu \eta^{\mu\nu} \mathcal{L} \\ &= \epsilon_\nu [ i \bar\psi \gamma^\mu \partial^\nu \psi - \eta^{\mu\nu} \mathcal{L} ] . \end{align} I wrote indexes of Dirac spinors explicitly. Note that $\psi$ and $\bar\psi$ anti-commutes! Here we get 2 rank tensor $$ T^{\mu\nu} := i \bar\psi \gamma^\mu \partial^\nu \psi - \eta^{\mu\nu} \mathcal{L} , $$ which we call Energy-momentum tensor .

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