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Trying my luck here

  1. Dec 6, 2011 #1

    MathematicalPhysicist

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    Does anybody know how prove the claim in my question here:
    http://theoreticalphysics.stackexchange.com/questions/643/a-question-from-ticcatis-red-qft-textbook [Broken]

    thanks, any hints?
     
    Last edited by a moderator: May 5, 2017
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  3. Dec 6, 2011 #2

    MathematicalPhysicist

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    I have another question from the same textbook.

    In Remark 1.6.4 he starts to show that the position operator is unique, so we take two such operators [itex]X,Y[/itex], and he assumes that [itex]Y[/itex] is given wrt the basis [itex]|k>[/itex]. The canonical commutation must be satisified for both of them, then [itex]P_r[/itex] commutes with [itex]X_s-Y_s[/itex] (which thusfar I follow). Therefore assuming that any operator can be expressed as a function of both [itex]X_r,P_r[/itex], [itex]Y_r[/itex] must be of the form [itex]X_r+f_r(P)[/itex], now the thing that I don't understand is that he argues Axiom 3 implies that [itex]f_r(P)[/itex] is of the form [itex]g(|P|^2)P_r[/itex], where axiom 3 tells us that if [itex]R[/itex] is a space rotation, then [itex]U(R)^{\dagger} X U(R) = RX[/itex] where [itex]U(R)[/itex] is the unitary representation of [itex]R[/itex] in Poincare group.

    Any help as to why Ax 3 implies it?
    I don't see it.
     
  4. Dec 6, 2011 #3

    Fredrik

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    That guy Pavel solved it for you.

    I don't understand what this means. I understand that |k> is a momentum eigenket, but that doesn't help. The exact statement in the book is "Assume that [itex]\bar Y[/itex] is the position operator with respect to the basis [itex]|\bar k\rangle[/itex]". I still don't get it. There's no basis involved in the axioms for the position operator.


    The equality in axiom 3 is [itex]U(R)^\dagger\bar X U(R)=R\bar X[/itex]. This has to mean [itex]U(R)^\dagger X^r U(R)=R^r_s X^s[/itex]. So [itex]Y^r=X^r+f^r(\bar P)[/itex] implies [itex]R^r_sY^s=R^r_sX^s+U(R)^\dagger f^r(\bar P)U(R)[/itex]. Multiply this by [itex](R^{-1})^t_r[/itex], and the result is [itex]Y^t=X^t+U(R)^\dagger(R^{-1})^t_r f^r(\bar P) U(R)[/itex]. So [tex] f^r(\bar P)=Y^r-X^r=U(R)^\dagger(R^{-1})^r_s f^s(\bar P) U(R).[/tex] Does this help, or is this the point where you are stuck?
     
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  5. Dec 7, 2011 #4

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    Do you assume that [itex]R^{-1}[/itex] commutes with [itex]U(R)^{\dagger}[/itex]?

    Assuming you do, I still don't see how from your last line you get that [itex]f_r(P)[/itex] equals [itex]g(|P|^2) P_r[/itex].

    Edit: I guess it's reasonable to argue that R and its inverse will commute with their Poincare representations, I think.
     
    Last edited: Dec 7, 2011
  6. Dec 7, 2011 #5

    Fredrik

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    There's no need to assume it. [itex]R^r_s[/itex] is the number on row r, column s, of the rotation matrix R. When it appears as a factor in a formula involving operators, it should be interpreted as the real number [itex]R^r_s[/itex] times the identity operator, and the identity operator commutes with everything.

    I assume that you can see that the converse of what you want to prove is true, i.e. that if [itex]f^r(\bar P)[/itex] is what you need it to be, then the last equality of my previous post holds. It's harder to explain why [itex]f^r(\bar P)[/itex] must be of that form. I haven't worked it out, but I think it's essential that the last equality in my previous post is supposed to hold for all R. Maybe it will help to consider a few specific choices of R.
     
    Last edited: Dec 7, 2011
  7. Dec 7, 2011 #6

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    I think I understand why if [itex]f^r(\bar{P})=g(|P|^2)P^r[/itex] then it satisfies the relation you wrote with the unitary representations, it's because the 4-momentum vector is invariant under space rotations.

    Not sure about the converse.

    P.S
    Appreciate your help. :-)
     
  8. Dec 7, 2011 #7

    Fredrik

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    [itex]\bar P[/itex] is the 3-momentum vector. Its components aren't invariant under rotation, so the components of the 4-momentum vector aren't either. But [itex]|\bar P|^2[/itex] is invariant, because a rotation by definition doesn't change the norm of any vector. Edit: To be more precise, in a notation that puts all indices downstairs (because the "one upstairs, one downstairs" notation makes it hard to handle transposes and inverses), [tex]U(R)^\dagger\bar P^2U(R)=U(R)^\dagger P_r U(R)\ U(R)^\dagger P_r U(R)=R_{rs}P_s R_{rt} P_t=(R^T)_{sr}R_{rt}P_sP_t=\delta_{st}P_sP_t=P_sP_s=\bar P^2.[/tex] I don't think there's an easy way to prove that [itex]f^r(\bar P)[/itex] must be of the form [itex]g(|\bar P|^2)P^r[/itex]. I also don't think there's an easy way to prove that [itex]Y^r-X^r[/itex] must be of the form [itex]f^r(\bar P)[/itex]. Ticciati doesn't even explain what that means. I'm going to think about this some more, but I can't guarantee that I will come up with something useful.

    The non-rigorous argument for the form of [itex]f^r(\bar P)[/itex] is that when you multiply components of [itex]\bar P[/itex] together and sum over repeated indices, the result isn't going to transform under rotations like a 3-vector unless r is the only index left. Terms like [itex]P^rP^s P_s P^t P_t[/itex] are OK, but terms like [itex]P^r P^s[/itex] aren't.
     
    Last edited: Dec 7, 2011
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