I am going through the derivation of the Ward identities in chapter 2 of Di Francesco, Conformal Field Theory and I am not sure how they go from equation 2.157:

$$\frac{\partial}{\partial x^{\mu}}\langle j^{\mu}_a(x)\Phi(x_1)\ldots\Phi(x_n)\rangle = -i\sum_{k = 1}^n\delta(x - x_i)\langle\Phi(x_1)\ldots G_a\Phi(x_i)\ldots\Phi(x_n)\rangle$$

to equation 2.161:

$$\langle Q_a(t_+)\Phi(x_1)Y\rangle - \langle Q_a(t_-)\Phi(x_1)Y\rangle = -i\langle G_a\Phi(x_1)Y\rangle$$


$$Q_a = \int\!\mathrm{d}^{d-1}{\bf x}\,j^0_{\mu}(x)$$ $$Y = \Phi(x_2)\ldots\Phi(x_n)$$

and $t_{\pm} = x_1^0 \pm \varepsilon$. The authors say that 2.161 follows from integrating 2.157 in a "pill box", with time running from $t_-$ to $t_+$ and ${\bf x}$ encompassing all space, expect small volumes centered about ${\bf x}_2, \ldots, {\bf x}_n$.

It doesn't seem obvious to me that the left-hand side of 2.161 (involving the $Q$) follows from this procedure. In particular, I am not sure why we should ignore the surface term. I also wonder how explicitely the left-hand side of 2.161 would be changed if we included some other ${\bf x}_i$'s in the integration volume.


1 Answer 1


They volume integrate the Ward identity between two hypersurfaces of constant $t$ ($t_+$ and $t_-$). It is important that this volume only contains the $\pmb x_1$ insertion, because it is the only one that is going to contribute on the right (because the delta functions vanish except the one that falls in the integration volume)

So the right-hand side is clear, the only contribution in the sum comes when $k=1$.

The left-hand side, being a volume integral of a divergence, gives you the surface flux integral, through Stokes' theorem. It is actually what it is written there, the $t_+$ face minus the $t_-$ face.

If there where more $x_i$ insertions inside the volume, the LHS would be the same, but the RHS would contain a sum of each term inside.

This is the way of translating into path integral language the action of a conserved charge (through the commutator) to a given operator for any quantisation of the theory. Moreover, you can deform the contour so that it is a sphere surrounding the $x_1$ insertion, so that action of charges are just given by spheres in the path integral formulation.

  • $\begingroup$ Do you mean the LHS would be strictly the same? But if at the same time the RHS is different, isn't there a contradiction? As for applying Stokes' theorem, a priori you also get contributions from the surfaces encompassing the small volumes you remove. Sure, this doesn't count if $j(x)$ is a regular function (as opposed to a distribution) but generally speaking, it may be problematic. $\endgroup$
    – John Smith
    Jan 11, 2018 at 23:22
  • $\begingroup$ I meant that the LHS would still have the same form. This is because either you change the volume you are integrating in to include another insertion, so that the surfaces are now different, or either you move the insertion inside the oroginal volume, in which case the function you are integrating is different. $\endgroup$
    – amrgoloh
    Jan 12, 2018 at 8:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.