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What is $O_l$ after all in the 10.3.1 of Weinberg's QFT book?

+ 3 like - 0 dislike
154 views

Where does it come from? What's the physical meaning of it?

And what's the meaning of $A_2(x_2),A_3(x_3),\cdots$?

What's the difference between $O_l$ and $\psi_l$ since they have the same Lorentz transformation properties? Why can $O_l$ be written the same as a free field from the start so that we do not need to renormalize it? 

It appears in section 6.4 for the first time, and I am very confused about it for a long time .

asked May 16, 2014 in Theoretical Physics by coolcty (125 points) [ revision history ]

2 Answers

+ 3 like - 0 dislike

1. $A_2(x_2),A_3(x_3),\cdots$ are just any operators depending on space-time coordinates, nothing profound.

2. $O_l$ only transforms like $\psi_l$, it can be something else, for example $O_l=\phi\psi_l$ where $\phi$ is a scalar field. It is simply not true saying "$O_l$ can be written the same as a free field from the start so that we do not need to renormalize it ". Again, it is any operator that supposedly transforms like $\psi_l$, you can take it to be free or renormalized depending on what you want. 

answered May 16, 2014 by Jia Yiyang (2,465 points) [ revision history ]
edited May 16, 2014 by Jia Yiyang

Thank you for your answer, but I am still confused about that. About 1 of your answer, \(A_2(x_2),A_3(x_3),\cdots\) appears first in 6.4. And in 10.4.19 and 10.4.10 it is used again. I wonder how they are interpreted here using the results of 6.4. .  Please forgive my ignorance for the moment.

@coolcty, In 6.4 I see the usage of notation $O_l$ but not the notation $A_2(x_2),A_3(x_3),\cdots$, but anyway in both chapters they are used in a very generic sense, very little interpretation is attached, so I don't quite understand your confusion, it'd be good if you can clarify.

.
+ 2 like - 0 dislike

$O_l(x)$ is a renormalized product of field operators at $x$. Note that $\phi(x)^2$ doesn't make sense as $\phi(x)$ is only a distribution, but after appropriate renormalization (by adding ''infinite'' linear and constant counterterms - i.e., adding finite counterterms in the cutoff theory that change with the cutoff in such a way that the limit exists when the cutoff is removed)  there is a renormalized operator for every power product (and product of derivatives). These are the $O_l(x)$. On the other hand, the $A_l(x)$ are arbitrary local fields, typically those that enter the Lagrangian.

Saying that $O_l(x)$ transforms like $\psi_l$ just expresses that $O_l(x)$ transforms according to an irreducible representation of the Lorentz group. This is done for simplicity only, and $\psi$ refers to the general irreducible fields discussed in Chapter 6.

The italic $O_l$ in (6.4.4) come from interaction terms in (6.4.1) via transition to the Heisenberg picture, and have nothing per se to do with the curly $O_l$ in Chapter 10, though they may be taken as examples of the latter. But apparently we are using different editions of Weinberg's book as the numbers you give are not consistent with mine.

answered May 23, 2014 by Arnold Neumaier (12,425 points) [ revision history ]
edited Jul 15, 2014 by Arnold Neumaier

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