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Consider the standard Dirac Lagrangian, $\mathcal{L}=\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$, and a transformed one differing by a total derivative

$$ \mathcal{L}'=\mathcal{L}-\frac{i}{2}\partial_{\mu}\left(\overline{\psi}\gamma^{\mu}\psi\right). $$

The energy-momentum tensor computed from the Dirac Lagrangian can be shown to be $T^{\mu\nu}=i\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi$. Given that, it should be possible to prove that the energy-momentum tensor computed from $\mathcal{L}'$ is given by $T'^{\mu\nu}=T^{\mu\nu}-\frac{i}{2}\partial^{\nu}\left(\overline{\psi}\gamma^{\mu}\psi\right)$.

I was trying to do it but I can't finish it. I'm using the standard formula for the energy-momentum tensor,

$$ T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\psi\right)}\partial^{\nu}\psi-\eta^{\mu\nu}\mathcal{L} $$

and doing,

$$ \begin{array}{ll} T'^{\mu\nu} & =\frac{i}{2}\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi-\eta^{\mu\nu}\left(\mathcal{L}-\frac{i}{2}\partial_{\mu}\left(\overline{\psi}\gamma^{\mu}\psi\right)\right)\\ & =T^{\mu\nu}-\frac{i}{2}\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi+\frac{i}{2}\partial^{\nu}\left(\overline{\psi}\gamma^{\mu}\psi\right)\\ & =T^{\mu\nu}+\frac{i}{2}\left(\partial^{\nu}\overline{\psi}\right)\gamma^{\mu}\psi\\ & =? \end{array} $$

Edit:

Following @Quantum spaghettification's suggestion I get

$$ \begin{array}{ll} T'^{\mu\nu} & =\frac{\partial\mathcal{L}'}{\partial\left(\partial_{\mu}\psi\right)}\partial^{\nu}\psi+\frac{\partial\mathcal{L}'}{\partial\left(\partial_{\mu}\overline{\psi}\right)}\partial^{\nu}\overline{\psi}-\eta^{\mu\nu}\mathcal{L}'\\ & =\frac{i}{2}\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi-\frac{i}{2}\gamma^{\mu}\psi\partial^{\nu}\overline{\psi}-\eta^{\mu\nu}\left(\mathcal{L}-\frac{i}{2}\partial_{\mu}\left(\overline{\psi}\gamma^{\mu}\psi\right)\right)\\ & =T^{\mu\nu}-\frac{i}{2}\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi-\frac{i}{2}\gamma^{\mu}\psi\partial^{\nu}\overline{\psi}+\frac{i}{2}\partial^{\nu}\left(\overline{\psi}\gamma^{\mu}\psi\right)\\ & =T^{\mu\nu}-\frac{i}{2}\gamma^{\mu}\psi\partial^{\nu}\overline{\psi}+\frac{i}{2}\left(\partial^{\nu}\overline{\psi}\right)\gamma^{\mu}\psi\\ & =? \end{array} $$

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  • $\begingroup$ For starters, think about: 1. How $\psi$ and $\bar{\psi}$ are related. 2. Left & right derivatives and ordering of Grassmann-odd fields. $\endgroup$ – Qmechanic Jul 1 '18 at 10:38
  • $\begingroup$ @Qmechanic I haven't yet learn what you mention in (2). Can't I do it just with algebra? $\endgroup$ – johani Jul 1 '18 at 12:04
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Never mind the anticommuting nature of the Dirac fields: if you're stuck to this problem, you probably didn't get there anyways. Just define $T^{\mu\nu}$ as

$$ T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\ \partial^{\nu}\psi+\partial^{\nu}\bar{\psi}\ \frac{\partial \mathcal{L}}{\partial_{\mu}\bar{\psi}}-\eta^{\mu\nu}\, \mathcal{L} $$

This is the specific form of the tensor written in Someone's answer for the case of the Dirac field. Notice the order of the factors in the second term: $\bar{\psi}$ is a row vector, hence it must be written to the left. The definition in Q. spaghettification's answer overlooked this, otherwise it is the very same as that.

Now, your first definition was plain incorrect. In the edit, the definition was correct apart from the order in the second term. As for the third line in your edit, there is a mistake: it should read

$$ T'^{\mu\nu}=T^{\mu\nu}-\frac{i}{2}\bar{\psi}\gamma^{\mu}\partial^{\nu}\psi-\frac{i}{2}(\partial_{\nu}\bar{\psi})\gamma^{\mu}\psi+\frac{i}{2} \eta^{\mu\nu} \partial_{\sigma}(\bar{\psi}\gamma^{\sigma}\psi)\quad (\star) $$

because the indices of the gamma matrix and derivative in the last term are saturated: you cannot contract them with those of the metric. Now, the second and third terms are the derivative you are looking for:

$$ -\frac{i}{2}\bar{\psi}\gamma^{\mu}\partial^{\nu}\psi-\frac{i}{2}(\partial_{\nu}\bar{\psi})\gamma^{\mu}\psi=-\frac{i}{2}\partial_{\nu}(\bar{\psi}\gamma^{\mu}\psi) $$

As for the last term, use the Dirac equations to write

$$ i\partial_{\sigma}(\bar{\psi}\gamma^{\sigma}\psi)=\bar{\psi}(i\gamma^{\sigma}\partial_{\sigma}\psi)+(i\partial_{\sigma}\bar{\psi}\gamma^{\sigma})\psi=m\bar{\psi}\psi-m\bar{\psi}\psi=0 $$

In case you didn't know, the Dirac equation for $\bar{\psi}$ is

$$ i\partial_{\sigma}\bar{\psi}\gamma^{\sigma}=-m\bar{\psi} $$

As the last term in $(\star)$ is zero and the second and third term make up the divergence, you got your result.

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  • $\begingroup$ Thank you for helping me understand. A couple of comments: I don't quite understand the reason for changing the order in that second term for the case of the Dirac field (I'll take that to move on and try to learn why later); I guess this never happens for scalar fields, but could it happen when we work with a Lagrangian for the $\overrightarrow{A}$ field? As for the last term, I was guessing I couldn't use the same indices, but since I was out of ideas I let it pass... Thanks. $\endgroup$ – johani Jul 1 '18 at 16:31
  • $\begingroup$ For no deep reason. Just look at Someone's definition (which is valid for any kind of field, be it scalar, spinor, vector, and so on) and keep in mind that we are working with numbers. $\partial^{\nu}\phi_{i}$, being a (position-dependent) number, commutes with the derivative of the Lagrangian, thus you can freely reverse the order of the factors. The Dirac field is a complex vector with 4 components, thus you should really write it as $\psi_{i}$, with $i=1,\dots,4$. The same is true of $\bar{\psi}_{i}$. However, ... $\endgroup$ – Giorgio Comitini Jul 1 '18 at 16:42
  • $\begingroup$ ... when you write the expression in implicit form (e.g. $\sum_{i}\bar{\psi}_{i}\psi_{i}=\bar{\psi}\psi$) the row vector ($\bar{\psi}$) must be written on the left, while the column vector ($\psi$) must be written on the right. This is the same as simple matrix multiplication. You wouldn't write $\psi\bar{\psi}$, because this wouldn't give you a number, it would give a 4x4 matrix. $\endgroup$ – Giorgio Comitini Jul 1 '18 at 16:45
  • $\begingroup$ It does happen for $n$-tuples of scalar fields too, if you leave the indices implicit: you would write $\Phi^{\dagger}\Phi$, not $\Phi\Phi^{\dagger}$. $\endgroup$ – Giorgio Comitini Jul 1 '18 at 16:47
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The general canonical energy-momentum tensor is defined by these components : $$\tag{1} T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi_i )}\partial^{\nu}\phi_i-\eta^{\mu\nu}\mathcal{L}, $$ where $\phi_i$ are all the independent components of the fields in the game. Since the Dirac $\psi$ has 8 real numbers (4 complex components), you also need to add the contributions from $\overline{\psi}$.

Also, it is very important to remember that the canonical energy-momentum tensor is defined up to a divergence $\partial_{\alpha} \, \Theta^{\alpha \mu \nu}$, where $\Theta^{\alpha \mu \nu} = -\, \Theta^{\mu \alpha \nu}$. This gives you the posibility to find a symetrical version of $T^{\mu \nu}$ without changing the physics (total energy and momentum in the fields).

Take note that your lagrangian : $$\mathcal{L}=\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi,$$ isn't a real number. While it is not really a problem, it is usually preferable that the field action be defined as a real number :

$$\tag{2} \mathcal{L_{\mathrm{real}}}= \frac{i}{2} \Big( \, \overline{\psi} \, \gamma^{\mu} \,(\partial_{\mu} \, \psi \,) - (\partial_{\mu} \, \overline{\psi} \,) \, \gamma^{\mu} \, \psi \, \Big) - m \, \overline{\psi} \, \psi.$$

The full symetric energy-momentum is then

$$\tag{3} T^{\mu \nu} = \frac{i}{4} \Big( \, \overline{\psi} \, \gamma^{\mu} \, (\partial^{\nu} \, \psi \,) + \overline{\psi} \, \gamma^{\nu} \, (\partial^{\mu} \, \psi \,) - (\partial^{\mu} \, \overline{\psi} \,) \, \gamma^{\nu} \, \psi - (\partial^{\nu} \, \overline{\psi} \,) \, \gamma^{\mu} \, \psi \, \Big)$$

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  • $\begingroup$ Please explain how you get to this energy-momentum tensor, as I tried the same and got what's up in my edit. $\endgroup$ – johani Jul 1 '18 at 14:12
  • $\begingroup$ @johani, unfurtunately, the symetrization procedure is pretty long and involves a lot of algebric manipulations with the gamma matrices. It's a good exercice, but it's a difficult one! $\endgroup$ – Cham Jul 1 '18 at 15:13

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