I am reading Weinberg's Quantum theory of fields Vol I and have a question about the derivation on page 71.

Right below eq. (2.5.37), it is written that $$A$$ and $$B$$ can be simultaneously diagonalized by $$\Psi_{k,a,b}$$. From the content, I inferred that $$\Psi_{k,a,b}$$ is also the eigenstate of the energy-momentum operator $$P^\mu$$ with eigenvalue $$k=(0,0,1,1)$$. But, since $$[A,P^\mu]\neq 0$$ and $$[B,P^\mu]\neq 0$$, there should not be such simultaneous eigenstate for $$A,B,P^\mu$$.

fzero
Homework Helper
Gold Member
$$A$$ and $$B$$ leave $$k=(0,0,1,1)$$ invariant, so while they don't generally commute with the momentum operator, it's true that

$$[A,P^\mu] \Psi_{k,a,b}= [B,P^\mu] \Psi_{k,a,b}= 0.$$

So $$A, B, P^\mu$$ are simultaneously diagonalizable on this state.

$$[A,P^\mu] \Psi_{k,a,b}= [B,P^\mu] \Psi_{k,a,b}= 0.$$

Thank you very much. The problem is $$[A,P^1]=-iP^3-iH=[B,P^2]=-iP^3-iH$$, so $$[A,P^1]\Psi_{k,a,b}=-2i\Psi_{k,a,b}=[B,P^2]\Psi_{k,a,b}\neq 0$$.

Avodyne
There must be a sign mistake somewhere. For example, if P=(0,0,-1,1), and everything else was unchanged, then it would work as fzero says.

I checked and there is no such sign mistake. I read from the following notes:

http://www.physics.buffalo.edu/professors/fuda/Chapter_3.pdf

Probably it is better to consider $$A^2+B^2$$ rather than $$A$$ and $$B$$ individually. But I haven't checked the commutation relations.

I checked and there is no such sign mistake. I read from the following notes:

http://www.physics.buffalo.edu/professors/fuda/Chapter_3.pdf

Probably it is better to consider $$A^2+B^2$$ rather than $$A$$ and $$B$$ individually. But I haven't checked the commutation relations.

Yes, you DID make a sign error. Verify it again (i just computed one relation and it worked out fine)!

Yes, you DID make a sign error. Verify it again (i just computed one relation and it worked out fine)!

Well, let's do $$[A,P_1]$$.

$$[A,P_1]=[J_2,P_1]+[K_1,P_1]$$. Using eq. (2.4.21) and (2.4.22), we have $$[J_2,P_1]=i\epsilon_{213}P_3$$ and $$[K_1,P_1]=-iH$$. Since $$\epsilon_{213}=-1$$, I got the result $$[A,P_1]=-i(P_3+H)=-i(P^3+P^0)$$.

Well, let's do $$[A,P_1]$$.

$$[A,P_1]=[J_2,P_1]+[K_1,P_1]$$. Using eq. (2.4.21) and (2.4.22), we have $$[J_2,P_1]=i\epsilon_{213}P_3$$ and $$[K_1,P_1]=-iH$$. Since $$\epsilon_{213}=-1$$, I got the result $$[A,P_1]=-i(P_3+H)=-i(P^3+P^0)$$.