Consider constructing the Ward identity associated with Lorentz invariance. It is possible to find a 3rd rank tensor Bρμν 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
jμνρ=TμνBxρ−TμρBxν
This ensures the symmetry of the conserved current which can be seen most easily be invoking the conservation law ∂μjμνρ=0
and
∂μTμνB=∂μ(TμνC+∂ρBρμν)=0.
Let X denote a set of n fields. The Ward identity associated with Lorentz invariance is then
∂μ⟨(Tμxρ−Tμρxν)X⟩=∑iδ(x−xi)[xνi∂ρi−xρi∂νi⟨X⟩−iSνρi⟨X⟩].
This is then equal to
⟨(Tρν−Tνρ)X⟩=−i∑iδ(x−xi)Sνρi⟨X⟩,
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:
n∑ixνin∑iδ(x−xi)∂ρi⟨X⟩−n∑ixρin∑iδ(x−xi)∂νi⟨X⟩−in∑iδ(x−xi)Sνρi⟨X⟩
and then replacing
∂μ⟨TμρX⟩=−∑iδ(x−xi)∂∂xρi⟨X⟩
for example. The result I am getting is that ⟨((∂μTμν)xρ−(∂μTμρ)xν+Tρν−Tνρ)X⟩=∑ixνi∂μ⟨TμρX⟩+∑ixρi∂μ⟨TμνX⟩−i∑iδ(x−xi)Sνρi⟨X⟩
To obtain the required result, this means that e.g
∑ixνi∂μ⟨TμρX⟩=⟨(∂μTμρ)xνX⟩,
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
Φi∈X takes on the value
xi (so
x=xi) 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