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I am reading the Section 15.9 of Weinberg's book "The Quantum Theory of Fields, vol. 2". Under a shift $\delta\Psi[\chi]$ in $\Psi[\chi]$, we have

$$ \begin{split} \delta Z&=i\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big) \left(\frac{\delta_RS[\chi,\chi^{\ddagger}]}{\delta\chi_n^{\ddagger}} \right)_{\chi^{\ddagger}=\delta\Psi/\delta\chi}\left( \frac{\delta(\delta\Psi[\chi])}{\delta\chi^n}\right) \\ &=i\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big)\left\{\frac{\delta_L} {\delta\chi^n}\left(\frac{\delta_RS}{\delta\chi_n^{\ddagger}} \delta\Psi\right)-\frac{\delta_R}{\delta\chi_n^{\ddagger}} \frac{\delta_LS}{\delta\chi^n}\delta\Psi\right\}_{\chi^{\ddagger} =\delta\Psi/\delta\chi} \\ &=\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big)\left\{ \frac{\delta_RS[\chi,\chi^{\ddagger}]}{\delta\chi_n^{\ddagger}} \frac{\delta_LI_{\Psi}[\chi]}{\delta\chi^n}-i\Delta S[\chi,\chi^{\ddagger}]\right\}_{\chi^{\ddagger}=\delta\Psi/ \delta\chi}\delta\Psi[\chi] \end{split} $$

The last line is exactly the same to Eq. (15.9.33). Referring to the definition of antibracket

$$ (F,G)=\frac{\delta_RF}{\delta\chi^n}\frac{\delta_LG}{\delta\chi_n^{\ddagger}}- \frac{\delta_RF}{\delta\chi_n^{\ddagger}}\frac{\delta_LG}{\delta\chi^n} $$

we can see that the quantum master equation reads

$$ -(S,S)-2i\Delta S=0 $$

which has an extra minus sign. I are not sure whether this is a typo or not. Could someone help me to check this derivation?

Besides, I am also confused by $\delta_L$ and $\delta_R$. Any clarifications will be appreciated.

Many thanks in advance!

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Hi @soliton: Eq. (15.9.35) in Weinberg's book reads $(S,S)-2i\Delta S=0$, which is the conventional form, and without an extra minus. Is the extra minus something you derive? – Qmechanic May 25 '13 at 15:18
@Qmechanic: Yes. If Eq. (15.9.33) holds, it will have an extra minus. – soliton May 25 '13 at 15:22
up vote 5 down vote accepted

I) Let us first clarify the left and right derivatives. Left derivatives is explained between eq. (15.8.9) and (15.8.10) in Ref. 1. A left derivative means a derivative that acts from the left. E.g. if $F= \chi G$, where $G$ does not depend on $\chi$, then $\frac{\delta_LF}{\delta\chi}=G$. Similarly, a right derivative acts from the right. E.g. if $F= G\chi$, then $\frac{\delta_RF}{\delta\chi}=G$. One may then work out that left and right derivatives are equal up to a sign factor:

$$\tag{A} \frac{\delta_LF}{\delta\chi}~=~(-1)^{(|F|+1)|\chi|}\frac{\delta_RF}{\delta\chi}.$$

Here $|F|$ denotes the Grassmann parity of $F$. Note in particular that the left and right derivative of the gauge fermion $\Psi[\chi]$ are the same:

$$\tag{B} \frac{\delta_L\Psi}{\delta\chi}~=~\frac{\delta_R\Psi}{\delta\chi}, \qquad |\Psi|~=~1. $$

II) Now let us consider the Batalin-Vilkovisky formalism. We start with the full quantum master action $S[\chi,\chi^{\ddagger}]$, which depends on fields $\chi^n$ and antifields $\chi^{\ddagger}_n$.

The odd Laplacian is originally defined in eq. (16b) of Ref. 2 as

$$\tag{16b} \Delta_{BV}~:=~ \frac{\delta_R}{\delta\chi^n}\frac{\delta_L}{\delta\chi^{\ddagger}_n}. $$

Ref. 1 defines (wrongly) the odd Laplacian as

$$\tag{15.9.34} \Delta_{SW}~:=~ \frac{\delta_R}{\delta\chi^{\ddagger}_n}\frac{\delta_L}{\delta\chi^n}. $$

One may show that the two definitions (16b) and (15.9.34) are related as

$$\tag{C} \Delta_{BV}F~=~(-1)^{|F|+1}\Delta_{SW}F. $$

In particular the two definitions (16b) and (15.9.34) differ by a sign

$$\tag{D} \Delta_{BV} S~=~-\Delta_{SW} S, \qquad |S|~=~0, $$

when applied to the action $S$, which is Grassmann-even $|S|=0$.

III) The Quantum Master Equation (QME) reads in Ref. 2

$$\tag{16a} \frac{1}{2}(S,S)~=~i\hbar\Delta_{BV} S,$$

while the QME in Ref. 1 reads

$$\tag{15.9.35} \frac{1}{2}(S,S)~=~i\hbar\Delta_{SW} S.$$

So OP is right. Eqs. (15.9.34) and (15.9.35) are mutually inconsistent. There is a wrong sign in Ref. 1 in either eq. (15.9.34) or eq. (15.9.35).


  1. S. Weinberg, The Quantum Theory of Fields, Vol. 2, 1996.

  2. I.A. Batalin and G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31.

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Thanks a lot. It is very helpful. I also read the Ref. 2, but I misunderstood that the $\delta_L$ in Ref. 1 is the same to the $\delta_R$ in Ref. 2. – soliton May 26 '13 at 0:48
I have another question. Why do we define $\delta_L$ as a action from the left? This is opposite to the definition of this Wikipedia page where $f \stackrel{\leftarrow }{\partial }_x g = \frac{\partial f}{\partial x} \cdot g$. – soliton May 26 '13 at 1:02
As always different authors have different conventions. There exist different conventions for left and right derivatives. Similarly, there exist different conventions for left and right group action, and so forth. The Wikipedia page that you mention is not trustworthy in its current state. I might improve it at some point in the future. – Qmechanic May 26 '13 at 1:20

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