quantum field theory - Energy momentum tensor from Noether's theorem

in the book Quantum Field Theory by Itzykson and Zuber the following derivation for the stress-energy tensor is proposed (p.22):

Assume a Lagrangian density depending on the spacetime coordinates $x$ only through fields and their gradients. Under a translation we have

$$\mathcal L (x+a)=\mathcal L[\phi_i(x+a),\partial_\mu\phi_i(x+a)].$$

Consider an infinitesimal $x$-dependent transformation

$$\delta\phi_i=\delta a^\mu(x)\partial_\mu\phi_i(x),$$ $$\delta\partial_u\phi_i(x)=\delta a^\nu \partial_\nu\partial_\mu\phi_i(x)+\partial_\mu[\delta a^\nu (x)]\partial_\nu \phi_i(x).$$

The proof then proceeds with variation of the action and integration by parts. But why do we consider an $x$-dependent local transformation instead of a global transformation?

Answer

This is a clever method used to derive Noether's current for any global symmetry; for the translational symmetry, it produces the stress-energy tensor.

We have to consider a local transformation because the variation of the action, $\delta S$, vanishes for the global transformation because the global transformation is by definition a symmetry:

$$\delta S = 0$$ This value of $\delta S$ would follow "tautologically" and we couldn't deduce anything new out of it.

It follows that if we "generalize" the symmetry transformation rules and make them $x$-dependent, i.e. the transformation will be specified by $\delta a(x)$ (and $a$ is a vector for translations, but may be scalar for other symmetries), then $\delta S$ will be nonzero but it will inevitably depend on derivatives of $a$ only; for constant choices of $a$, we must get zero (because it's the global symmetry). For actions that only depend on first derivatives of the fields, the variation of the action will inevitably have the form

$$S = \int (\partial_\mu a) j^\mu d^d x$$ where $j^\mu$ is some particular function of the fields or other degrees of freedom (and their derivatives). Note that this form is inevitable: $\delta S$ has to be linear in $a$ and/or its derivatives, but it must vanish for $a={\rm const}$ so there can't be any terms of the form $\int ab\,\,d^dx$, i.e. terms proportional to undifferentiated $a$. There are also no higher-derivative terms if the action didn't have higher derivatives of fields to start with.

Now, the argument that $j^\mu$ is a current is simple. When equations of motion are satisfied, $\delta S =0$ for any variation of fields, whether it's a symmetry or not. In particular, $\delta S = 0$ holds for the "generalized" or "localized" global symmetry given by $a(x)$ which is no longer an exact symmetry, so $\delta S$ is the nonzero expression above. But by integration by parts, $\delta S = 0$ means

$$0 = \int a(x)\cdot \partial_\mu j^\mu(x)\,\, d^d x$$ which vanishes iff $\partial_\mu j^\mu$ vanishes at each point. This proves that $j^\mu$ obtained in this way is a conserved current; its integral has to be a conserved quantity. You could ask why someone invented this method. He or she invented it because he or she was creative and smart. What's important for everyone else is to check the arguments above and see that one may derive a conserved current in this way. The original author of the method was able to "see" the whole argument in his or her head.

(I say "her" as well to pay some tribute to Noether who didn't quite invent this elegant method – her papers were messy – but she invented the whole relationship between symmetries and conservation laws.)