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

  • Context: Graduate 
  • Thread starter Thread starter diraq
  • Start date Start date
  • Tags Tags
    Book Qft
Click For Summary

Discussion Overview

The discussion revolves around a potential contradiction in Weinberg's Quantum Theory of Fields regarding the simultaneous diagonalization of operators A, B, and the energy-momentum operator Pμ. Participants examine the implications of commutation relations and eigenstates in the context of quantum field theory.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions the assertion that operators A and B can be simultaneously diagonalized with the eigenstate \Psik,a,b, given that [A,Pμ]\neq 0 and [B,Pμ]\neq 0.
  • Another participant argues that while A and B do not generally commute with Pμ, they leave the state k=(0,0,1,1) invariant, suggesting that they can be simultaneously diagonalized on this state.
  • A different participant challenges the previous claims by stating that the derived commutation relations indicate non-zero results, implying a contradiction in the simultaneous diagonalization.
  • Some participants propose that there may be a sign error in the calculations, suggesting that a different momentum representation could resolve the issue.
  • One participant references external notes to suggest considering A2 + B2 instead of A and B individually, although they have not verified the commutation relations.
  • Another participant insists that there is no sign error and provides a detailed calculation of the commutation relations involving A and P1, while also inviting others to share their derivations.
  • Further discussion includes a claim that a sign error exists in a specific formula from Weinberg's text, pointing to discrepancies between different equations.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the simultaneous diagonalization of A, B, and Pμ. There is no consensus on whether a sign error exists or how to resolve the apparent contradictions in the commutation relations.

Contextual Notes

Participants reference specific equations from Weinberg's text and external notes, indicating that the discussion is highly technical and dependent on precise mathematical formulations. The validity of certain assumptions and the correctness of derived relations remain unresolved.

diraq
Messages
14
Reaction score
0
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.

Please help me on this. Thanks in advance.
 
Physics news on Phys.org
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

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

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

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.
 
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.
 
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 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)!
 
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 [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).

If possible, please post your derivation. Thanks.
 
diraq said:
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).

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!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
3K