Spontaneous Symmetry Breaking in global $U(1)$ symmetry

I was reading SSB from Ashok Das - Lectures on Quantum Field Theory. I have a couple of questions.

First Question

$$\delta\sigma(x)=-i\epsilon[Q,\sigma(x)]=\epsilon\chi(x)$$ $$\delta\chi(x)=-i\epsilon[Q,\chi(x)]=-\epsilon\sigma(x)$$

But I get a different relation,

$$\delta\sigma(x)=i\epsilon[Q,\sigma(x)]=\epsilon\chi(x)$$ $$\delta\chi(x)=i\epsilon[Q,\chi(x)]=-\epsilon\sigma(x)$$

I don't get the "-" before the commutators. So which relation is correct?

Second Question

While deriving $H|\chi\rangle=E_0|\chi\rangle$ in equation (7.70), why does $H|0\rangle=E_0|0\rangle$? Shouldn't $H|0\rangle = 0$, like we normally expect.

Third Question

In equation (7.71) while showing that $\langle\chi|\chi\rangle\rightarrow\infty$, the author plugs in

$$\langle\chi|\chi\rangle=\int d^3x\langle0|e^{iP\cdot x}J^0(0)e^{-iP\cdot x}Q|0\rangle$$

From where does the author plug in, $e^{-iP\cdot x}$ and $e^{iP\cdot x}$?

• 1) no idea, I don't own that book (you should make the question self-contained and post the relevant equations) 2) $H|0\rangle=E_0|0\rangle$ is more general than $H|0\rangle=0$. The latter takes $E_0=0$ which defined the origin of energies, while the former does not. 3) the author is just using $\phi(x)=e^{-iPx}\phi(0)e^{+iPx}$ which is valid for any operator (because $e^{iPx}$ is the translation operator). In this case, $\phi(x)=J(x)$. – AccidentalFourierTransform Apr 17 '16 at 9:44
• @AccidentalFourierTransform: You can always search in Google Books. You will get a positive hit. – sbp Apr 17 '16 at 9:50
• Note that one should only ask one question per post, cf. this meta post. – Qmechanic Apr 17 '16 at 12:49