Constructing Ward identity associated with conserved currents

Consider constructing the Ward identity associated with Lorentz invariance. It is possible to find a 3rd rank tensor $B^{\rho \mu \nu}$ antisymmetric in the first two indices, then the stress-energy tensor can be made symmetric. Once done, the conserved current coming from the classical analysis is of the form

$$j^{\mu \nu \rho} = T_B^{\mu \nu}x^{\rho} - T_B^{\mu \rho}x^{\nu}$$

This ensures the symmetry of the conserved current which can be seen most easily be invoking the conservation law $$\partial_{\mu}j^{\mu \nu \rho} = 0$$ and $$\partial_{\mu}T_B^{\mu \nu} =\partial_{\mu} (T^{\mu \nu}_C + \partial_{\rho}B^{\rho \mu \nu}) = 0.$$

Let $X$ denote a set of $n$ fields. The Ward identity associated with Lorentz invariance is then

$$\partial_{\mu} \langle (T^{\mu}x^{\rho} - T^{\mu \rho}x^{\nu})X\rangle = \sum_i \delta(x-x_i)\left[ x^{\nu}_i \partial^{\rho}_i - x^{\rho}_i\partial^{\nu}_i\langle X \rangle - iS^{\nu \rho}_i \langle X \rangle\right].\tag{1}$$

This is then equal to

$$\langle (T^{\rho \nu} - T^{\nu \rho})X \rangle = -i\sum_i \delta (x-x_i)S^{\nu \rho}_i\langle X \rangle,$$

which states that the stress tensor is symmetric within correlation functions, except at the position of the other fields of the correlator.

My question is: how is this last equation and statement derived?

I think the Ward identity associated with translation invariance is used after perhaps splitting (1) up like so:

$$\sum_i^n x^{\nu}_i \sum_i^n \delta(x-x_i)\partial^{\rho}_i \langle X \rangle - \sum_i^n x^{\rho}_i \sum_i^n \delta(x-x_i)\partial^{\nu}_i \langle X \rangle - i\sum_i^n\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$ and then replacing $$\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X \rangle = -\sum_i \delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$

for example. The result I am getting is that $$\langle ((\partial_{\mu}T^{\mu \nu})x^{\rho} - (\partial_{\mu}T^{\mu \rho})x^{\nu} + T^{\rho \nu} - T^{\nu \rho})X \rangle = \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle + \sum_i x^{\rho}_i \partial_{\mu} \langle T^{\mu \nu} X \rangle - i\sum_i\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$ To obtain the required result, this means that e.g$$\sum_i x^{\nu}_i \partial_{\mu} \langle T^{\mu \rho}X \rangle = \langle(\partial_{\mu}T^{\mu \rho})x^{\nu} X \rangle,$$ but why is this the case? Regarding the statement at the end, do they mean that when the position in space $x$ happens to coincide with one of the points where the field $\Phi_i \in X$ takes on the value $x_i$ (so $x = x_i$) then the r.h.s tends to infinity and the equation is then nonsensical?

• A few notational quibbles: $\partial_i^\mu$ shall denote $\partial^\mu$ w.r.t. $x_i$, i.e. the argument of the i-th field in $X$, right? And what is $S^{\mu\rho}_i$? Also, just because $\partial_\mu T^\mu_\rho$ vanishes classically, this does not mean that $\partial_\mu\langle T^\mu_\rho X\rangle$ vanishes quantumly - this is precisely what the Ward identities tell you. – ACuriousMind Jul 17 '14 at 12:04
• Yes. $S^{\nu \rho}_i$ is the spin operator for the i-th field. Thanks for clearing that up, but I am now unsure of what permits the cancellation of the first two terms. Thanks ACuriousMind! – CAF Jul 17 '14 at 12:21
• You've got everything you need, just apply the product rule to the l.h.s. of $(1)$ and use your replacement from the last Eq. – ACuriousMind Jul 17 '14 at 12:30
• Well, you can always delete them later. I'm not sure where your problem lies now: Do product rule on the l.h.s. of $(1)$ and on the r.h.s. do the "splitting up" you did in your OP. Then use the replacement from the last Eq. in your OP and cancel the terms on both sides. What remains is the Eq. you wanted to show. – ACuriousMind Jul 17 '14 at 12:54
• Your last equation in your question is not true. However, when $\delta(x-x_i)$ is added into the summation on the l.h.s., it will be correct since $\delta(x-x_i)x_i^{\mu} =\delta(x-x_i)x^{\mu}$ – mastrok Feb 3 '15 at 7:24

This step is wrong: $$\langle ((\partial_{\mu}T^{\mu \nu})x^{\rho} - (\partial_{\mu}T^{\mu \rho})x^{\nu} + T^{\rho \nu} - T^{\nu \rho})X \rangle = \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle + \sum_i x^{\rho}_i \partial_{\mu} \langle T^{\mu \nu} X \rangle - i\sum_i\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$ because $$\sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle\neq \sum_i \delta(x-x_i) x^{\nu}_i \partial_i^{\rho}\langle X \rangle ,$$ and the spliting are wrong too $$\sum_i \delta(x-x_i) x^{\nu}_i \partial_i^{\rho}\langle X \rangle \neq\sum_i x^{\nu}_i\sum_i \delta(x-x_i) \partial_i^{\rho}\langle X \rangle .$$ The right way to proceed here is to note that the distribution $$\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X \rangle = -\sum_i \delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$ acting on a test function $g(x)$, when composed with another function $f (x)$, gives: $$\int g (x)f (x)\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X\rangle= - \int g (x)\sum_i f (x_i)\delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$ in the sense of distribution, we have the equality: $$f (x)\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X\rangle=-\sum_i f (x_i)\delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$ In your case, the function $f(x)$ that composes with you distribution is $x^{\rho}$ and $x^{\nu}$. Note that $x^{\mu}$ is a c-number, and can jump out of the expectation value. Now, in eq. 1, the terms on the l.h.s. originated by the derivative acting only on $\langle T^{\mu\nu}X\rangle$ cancels the first two terms on the r.h.s. because they are the same distribution.
Now you have that $\langle T^{\mu\nu}(x) X\rangle$ is symmetric if $X$ don't have fields at $x$. This implies that as operator, $T^{\mu\nu}$ is symmetric.