# On the derivation of Ward-Takahashi identity

I am reading Weinberg's QFT book and in 10.4 he introduced a derivation of Ward-Takahashi identity (where $$T$$ is the time ordering): \begin{align} \frac{\partial}{\partial x^\mu}T{\{J^\mu(x)\Psi_n(y)\bar\Psi_m(z)\}} &= T{ \{ \partial_\mu J^\mu(x)\Psi_n(y)\bar\Psi_m(z)\}} \\ &+ \delta(x^0 - y^0) T\{ [J^0(x),\Psi_n(y)]\bar\Psi_m(z)\} \\ &+ \delta(x^0 - z^0) T\{ \Psi_n(y),[J^0(x),\bar\Psi_m(z)]\}. \end{align} \tag{10.4.21} I understand the deltas come from time derivative of step functions in the time ordering operators,but after taking the derivatives, isn't the step function disappear? how do we still have time operators. how does $$J^0$$ appears in the second and third term on the right hand side?

I actually natively tried to express time ordering as a function of step function, $$J^\mu(x),\Psi_n(y)$$, and $$\bar\Psi_m(z)$$ and take the multivariable chain rule of it without luck.

When you have three operators, there are two step-function factors involving $$x^0$$. One with $$\theta(x^0-y^0)$$ and onew with $$\theta(x^0-z^0)$$. So, after differentiation there is still one left that determines whether the $$\bar\Psi_m(z)$$ is to the left or right of the other factors which share a common time.
It's a bit tedious to write out in detail, but the main point is that one can consistently keep the full multi-argument step function in the terms proportional to Dirac delta functions in eq. (10.4.21) if one assigns appropriate symmetry weights on diagonals & sub-diagonals. For a single-slot Heaviside step function, the symmetry weight is a $$1/2$$, cf. this Phys.SE post.