Proving a Claim in Ticcati's Red QFT Textbook: Seeking Advice and Hints

[MathematicalPhysicist]

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

thanks, any hints?

 

[MathematicalPhysicist]

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

Any help as to why Ax 3 implies it?
I don't see it.

 

[Fredrik]

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

thanks, any hints?

That guy Pavel solved it for you.

he assumes that Y is given wrt the basis |k>.

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 \bar Y is the position operator with respect to the basis |\bar k\rangle". I still don't get it. There's no basis involved in the axioms for the position operator.

...now the thing that I don't understand is that he argues Axiom 3 implies that f_r(P) is of the form g(|P|^2)P_r, where axiom 3 tells us that if R is a space rotation, then U(R)^{\dagger} X U(R) = RX where U(R) is the unitary representation of R in Poincare group.

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

 

[MathematicalPhysicist]

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

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

Edit: I guess it's reasonable to argue that R and its inverse will commute with their Poincare representations, I think.

 

[Fredrik]

There's no need to assume it. R^r_s 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 R^r_s 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 f^r(\bar P) is what you need it to be, then the last equality of my previous post holds. It's harder to explain why f^r(\bar P) 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.

 

[MathematicalPhysicist]

I think I understand why if f^r(\bar{P})=g(|P|^2)P^r 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. :-)

 

[Fredrik]

\bar P is the 3-momentum vector. Its components aren't invariant under rotation, so the components of the 4-momentum vector aren't either. But |\bar P|^2 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), 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. I don't think there's an easy way to prove that f^r(\bar P) must be of the form g(|\bar P|^2)P^r. I also don't think there's an easy way to prove that Y^r-X^r must be of the form f^r(\bar P). 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 f^r(\bar P) is that when you multiply components of \bar P 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 P^rP^s P_s P^t P_t are OK, but terms like P^r P^s aren't.