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  Conserved charge from Ward identity

+ 2 like - 0 dislike
81 views

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:

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

where

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

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.

asked Jan 12 in Theoretical Physics by maharishi (25 points) [ no revision ]

1 Answer

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I think on the l.h.s, integral on the "pill box" can be expressed as\[\int_{pillbox} d^dx\partial_{\mu}\left \langle j_a^{\mu}\Phi(x_1)....\Phi(x_n) \right \rangle=\int_{x_1^0=t_+} d^dx\partial_{\mu}\left \langle j_a^{\mu}\Phi(x_1)....\Phi(x_n) \right \rangle\]  \[-\int_{x_1^0=t_-} d^dx\partial_{\mu}\left \langle j_a^{\mu}\Phi(x_1)....\Phi(x_n) \right \rangle\]Then, transform to surface integral, and would get l.h.s of 2.161. On the r.h.s. of 2.157, since the integral is performed on this pill box, only the first term of the summation with \(\delta(x-x_1)\) contributes. So we get the r.h.s of 2.161

answered Feb 7 by chasseurs (35 points) [ no revision ]

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