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

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In summary: 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!
  • #1
diraq
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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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  • #2
[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.
 
  • #3
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].
 
  • #4
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.
 
  • #5
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.
 
  • #6
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)!
 
  • #7
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.
 
  • #8
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.
 
  • #9
Thanks!
 

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

What is Weinberg's QFT book?

Weinberg's QFT book, also known as "The Quantum Theory of Fields", is a three-volume series written by theoretical physicist Steven Weinberg. It provides a comprehensive and in-depth study of quantum field theory, a fundamental framework in theoretical physics.

Who is the target audience for Weinberg's QFT book?

The book is primarily aimed at advanced graduate students and researchers in theoretical physics. It assumes a strong background in mathematics and quantum mechanics.

What sets Weinberg's QFT book apart from other textbooks on the subject?

Weinberg's QFT book is known for its rigorous mathematical treatment of quantum field theory, as well as its focus on conceptual understanding rather than just mathematical formalism. It also covers a wide range of topics, including modern developments such as string theory and supersymmetry.

Is Weinberg's QFT book suitable for self-study?

While the book can be used for self-study, it is highly recommended to have a strong background in mathematics and physics before attempting to read it. It can be challenging for beginners, but it is a valuable resource for those looking to deepen their understanding of quantum field theory.

Are there any prerequisites for reading Weinberg's QFT book?

As mentioned before, a strong background in mathematics and physics is necessary for understanding this book. In particular, knowledge of calculus, linear algebra, and quantum mechanics is essential. It is also helpful to have some familiarity with special relativity and classical field theory.

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