Is There a Contradiction in Weinberg's QFT Book on Simultaneous Eigenstates?

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

Please help me on this. Thanks in advance.
 
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[tex]A[/tex] and [tex]B[/tex] leave [tex]k=(0,0,1,1)[/tex] invariant, so while they don't generally commute with the momentum operator, it's true that

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

So [tex]A, B, P^\mu[/tex] are simultaneously diagonalizable on this state.
 
fzero said:
[tex] [A,P^\mu] \Psi_{k,a,b}= [B,P^\mu] \Psi_{k,a,b}= 0.[/tex]

Thank you very much. The problem is [tex][A,P^1]=-iP^3-iH=[B,P^2]=-iP^3-iH[/tex], so [tex][A,P^1]\Psi_{k,a,b}=-2i\Psi_{k,a,b}=[B,P^2]\Psi_{k,a,b}\neq 0[/tex].
 
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 [tex]A^2+B^2[/tex] rather than [tex]A[/tex] and [tex]B[/tex] individually. But I haven't checked the commutation relations.
 
diraq said:
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 [tex]A^2+B^2[/tex] rather than [tex]A[/tex] and [tex]B[/tex] 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)!
 
Careful said:
Yes, you DID make a sign error. Verify it again (i just computed one relation and it worked out fine)!

Well, let's do [tex][A,P_1][/tex].

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

If possible, please post your derivation. Thanks.
 
diraq said:
Well, let's do [tex][A,P_1][/tex].

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

If possible, please post your derivation. Thanks.


The mistake is in your first line, A = J_2 - K_1. If you would have cared, you would notice that Steven made a sign error in formula 2.5.33 if you compare with 2.4.17.
 
Thanks!
 

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