Help with D. Tong example on Noether in QFT In this lectures, example 1.3.2 on page 14 concludes that the Noether current is 
 
But how can the current be a two index object when it is defined in eq. (1.38), which is 

as a one index object? If I apply the formula I obtain something of the form $j^\mu$. Can someone make the calculations explicitly?
 A: One Noether current $j^\mu$ corresponds to one one-parameter symmetry.
In case of translations in spacetime, there are 4 independent translations. One for time translation, three for spatial.
One can choose an arbitrary constant vector $a^\mu$ to represent the direction of translation. Then associated to the transformation $$ x^\mu\mapsto x^\mu +\epsilon a^\mu $$ is the Noether current $j^\mu_{a}$ (the $a$ index is for the four vector $a$, it is not a component index! ) which is given by $$ j^\mu_a=\frac{\partial\mathcal L}{\partial\partial_\mu\phi}a^\nu\partial_\nu\phi-a^\nu\delta^\mu_\nu\mathcal L. $$
As you can see, this expression is linear in $a^\nu$, so there is a tensor field of type (1,1), $T^\mu_{\ \nu}$ such that $$ j^\mu_a=T^\mu_{\ \nu}a^\nu. $$
The conservation equation is only for $j^\mu_a$, but it is valid for all $a^\nu$ constant vectors. In particular, it is valid for $$ a^\nu\equiv e^\nu_{(\rho)}\equiv \delta^\nu_{(\rho)}, $$ the basis vectors, all of them.
So we have $$ \partial_\mu j^\mu_{e_{(\rho)}}=\partial_\mu (T^\mu_{\ \nu}e^\nu_{(\rho)})=\partial_\mu(T^\mu_{\ \nu}\delta^\nu_\rho)=\partial_\mu T^\mu_{\ \rho}=0, $$ so we have the conservation equation valid also for the entire $T^\mu_{\ \nu}$.
A: In this case you can think of left hand side as representing a vector of currents, each component of which has an index ν. So then your conservation equation applies to each component of j (that is for μ = 0,1,2...D):
$$\partial_\nu (j^\mu)^\nu = 0 ~ \forall ~ \mu$$
or $\partial_\nu T^{\mu\nu} = 0 ~ \forall ~ \mu$
A: It's OK to have multiple indices on the Noether current; you just need to know how to lose all but one of them to form the most general conserved current that results.
The Killing vectors satisfy $\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0$. (Replace all $\nabla$s in what follows with $\partial$s if you only care about flat spacetime.) Since $T^{\mu\nu}=T^{\nu\mu}=0$, $\nabla_\mu\xi_\nu T^{\mu\nu}=0$. The result $\nabla_\mu T^{\mu\nu}=0$ then implies $\nabla_\mu j^\mu=0,\,j^\mu:=\xi_\nu T^{\mu\nu}=\xi^\nu T^\mu_\nu$. Thus each $2$-index current gives a vector space of $1$-index conserved currents, whose dimension is equal to the number of linearly independent Killing vectors. This is why an $n$-dimensional is called maximally symmetric if its number of linearly independent Killing vectors is maximal (i.e. $\frac{1}{2}n(n+1)$).
A: I think it is explained somewhat in David Tong's notes.
The symmetry under consideration is spacetime translation in a direction $\varepsilon^\mu$. For example, $\varepsilon^\mu=(1,0,0,0)^\mu$ would be the time translation symmetry. To draw an analogy with David Tong's notes, $\epsilon^\mu=\lambda \varepsilon^\mu$ would be the infinitesimal quantity, where $\lambda$ is the infinitesimal amount you want to translate in the $\varepsilon^\mu$ direction.
If you compare to the previous expressions, you get the conserved (finite) four-vector:
$$j^\mu=\frac{\partial L}{\partial \partial_\mu \phi} \varepsilon^\nu \partial_\nu \phi-\varepsilon^\mu L$$
But a trick very commonly used: if $A^\mu v_\mu=B^\mu v_\mu$ for every covector $v$, then $A^\mu=B^\mu$ outright. In our case, Noether directly gives that for every epsilon:
$$\varepsilon^\nu j^\mu_{\ \ \nu}=\varepsilon^\nu\left(\frac{\partial L}{\partial \partial_\mu \phi} \partial_\nu \phi-\delta^\mu_\nu L\right)$$
hence, we actually have a conserved $(1,1)$ tensor:
$$j^\mu_{\ \ \nu}=\frac{\partial L}{\partial \partial_\mu \phi} \partial_\nu \phi-\delta^\mu_\nu L$$
A: $T^{\mu \nu}$ is conserved because the Lagrangian is invariant under space-time translations, meaning that it is invariant under space translations and time translations. Naturally, all the quantities conserved should be inside $T^{\mu \nu}$.
For example, $Q^0 = \int d^3x T^{00}$ is the first one, and $Q^0$ is the Energy of the system (conserved because of time translation symmetry).
Another quantity is $P^i = \int d^3x T^{i0}$ which is the conserved momentum due to space translation symmetry.
The current is expressed in the form of a tensor to put the conservation laws (plural) in the covariant way $\partial_{\mu}T^{\mu \nu} = 0$.
Writing it explicitly, we have $\partial_{0}T^{0 \nu} + \partial_{i}T^{i \nu} = 0$ which is a vector equation. The 0th component would be the conservation of energy, and the jth component, the conservation of each component of the momentum.
Note that if the conserved current had only one index, you could only express one equation (which would mean one conserved quantity). The conserved current is a tensor because it describes multiple conserved quantities (or conservation equations) in a single object.
