# Trying my luck here

1. Dec 6, 2011

### 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 [Broken]

thanks, any hints?

Last edited by a moderator: May 5, 2017
2. Dec 6, 2011

### 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.

3. Dec 6, 2011

### Fredrik

Staff Emeritus
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 $\bar Y$ 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.

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?

Last edited by a moderator: May 5, 2017
4. Dec 7, 2011

### 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.

Last edited: Dec 7, 2011
5. Dec 7, 2011

### Fredrik

Staff Emeritus
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.

Last edited: Dec 7, 2011
6. Dec 7, 2011

### 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.

P.S
$\bar P$ 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.