# Position of indices in QFT

I have recently started studying quantum field theory from the book Quantum Field Theory and the Standard Model by Schwartz. In chapter 2 it is said that, contrary to GR, one can ignore the index position, because we are working in Minkowski. I do understand this when the indices are contracted, nevertheless in chapter 3, we arrive to the following equality

$$\partial_\mu \left(\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\right)=0$$

and from here the energy-momentum tensor for a classical field theory is defined as

$$T_{\mu \nu}=\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\,.$$

Yet in Tong's notes and some other references (where the position of the indices is taken into account as supposed - http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf), one finds that

$$T'^\mu_ \nu=\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)}\partial_\nu \phi_n - \delta^\mu_\nu L\,.$$

Now, I can lower the index from this definition and find

$$T'_{\mu \nu}=\sum_n \frac{\partial L}{\partial(\partial^\mu \phi_n)}\partial_\nu \phi_n - g_{\mu \nu}L\,.$$

Now it seems to me that considering each convention $T'_{\mu \nu} \neq T_{\mu \nu}$. Of course, if I contract the first index the equality is restored, yet I don't see how these components are equal alone. Anyone who can help? Is this a typo or am I missing something?

• The first author is being careless with indices. Jul 8, 2016 at 20:39
• It's not a typo. In Schwartz, $F^{\mu\nu} = F_{\mu\nu} = F_{\mu}^{\ \nu}$, and so on. Jul 8, 2016 at 20:44
• I still don't understand what that means... for example, if I want to compute $T_{11}$, I would get $T_{11}=\sum_n \frac{\partial L}{\partial(\partial_x \phi_n)}\partial_x \phi_n +L$. Now I could do the same for the other definition and I get $T_{11}=\sum_n -\frac{\partial L}{\partial(\partial_x \phi_n)}\partial_x \phi_n +L$ . How are they the same? Jul 8, 2016 at 21:10
• @blackhole1511 You need to figure out where the indices go before plugging in specific components. It'll take some extra work, but you can always recover the positioning uniquely. Jul 8, 2016 at 21:23
• As a simpler case, if you wanted $A^0$ and Schwartz wrote $A^\mu = B_\mu$, you would have to raise the index to $B^\mu$ first, then extract $B^0$. Jul 8, 2016 at 21:24

In particular one can show that: $$M^{\mu}_{\alpha\beta} = (x_{\alpha}T^{\mu}_{\beta} - x_{\beta}T^{\mu}_{\alpha}) + \textrm{(Irrep)}^{\mu}_{\alpha\beta}$$ where the left hand side is the representation of a generator of Lorentz transformations and the last term on the right hand side is proportional to some sort of "internal current fields" that have to transform according to the Lorentz group as well.