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  Constructing Ward identity associated with conserved currents

+ 2 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user CAF
asked Jul 16, 2014 in Theoretical Physics by CAF (100 points) [ no revision ]
Most voted comments show all comments
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.

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user ACuriousMind
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!

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user CAF
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.

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user ACuriousMind
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.

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user ACuriousMind
I see how the l.h.s is obtained. For the r.h.s I have$$ \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho} X\rangle - \sum_i x^{\rho}_i \partial_{\mu}\langle T^{\mu \nu} X \rangle - i\sum_i \delta(x-x_i)S^{\nu \rho}\langle X \rangle$$

This post imported from StackExchange Physics at 2014-07-20 09:29 (UCT), posted by SE-user CAF
Most recent comments show all comments
Sorry! I didn't mean to enforce it upon you, I just got the impression you knew. I think the fields $\phi_i$ can be treated as stochastic or random variables and so maybe $x^{\rho}$ can be taken outside the $\langle .. \rangle$. But then that means I would need $\sum_i x^{\rho}_i = x^{\rho}$, which I don't think is true. (The sum of all positions at which the fields are evaluated does not constitute the whole space).

This post imported from StackExchange Physics at 2014-07-20 09:30 (UCT), posted by SE-user CAF

Bump !

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