Weinberg's path integral for fermions in Volume 1

Question

In sec. 9.5 of Weinberg's QFT 1, he introduces operators $Q$ and $P$ satisfying $$ \{Q,P\}=i \tag{9.5.1} $$ $$ \{Q,Q\}=\{P,P\}=0 \tag{9.5.2} $$ and eigenstates $|q\rangle$: $$ Q|q\rangle =q|q\rangle \tag{9.5.14} $$ where $q$ is a Grassmann number. The Grassmann $q$ is taken to satisfy $$ \{Q,q\}=\{P,q\}=\{q',q\}=0. \tag{9.5.16} $$ The two basis states in the Hilbert space are $|0\rangle$ and $|1\rangle$, satisfying $$ Q|0\rangle=0\tag{9.5.8} $$ $$ P|0\rangle=|1\rangle.\tag{9.5.9} $$ However, there is a problem when I try to evaluate $$\langle 0|QPq|0\rangle.$$

1. on the one hand, using $\{Q,P\}=i$, I get $iq$;

2. on the other hand, using $$\langle 0|Q=i\langle 1|\tag{9.5.12}$$ and $\{P,q\}=0$, I get $-iq$.

Answer 1

Hints:

1. OP's first method is correct. For clarity, it should probably be noted that it implicitly uses that the fermionic eigenvalue $q$ commutes with the bosonic state $|0\rangle$.

2. For OP's second method remember that the fermionic eigenvalue $q$ anticommutes with the fermionic state $|1\rangle$. This gives an additional minus sign.