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In volume 1 of Weinberg QFT, equation (2.5.44), the general boost of one-particle state L(p) from standard momentum vector $K^\mu=(0,0,k,k)$ to $p^\mu$ is expressed as $$L(p)=R(\hat{p})B(\frac{|p|}{k}),$$ where $B$ means pure boost in $z$-direction and R represents pure rotation of $z$-axis into $\hat{p}$ direction.

I thought the right form should be $$L(p)= R(\hat{p})B(\frac{|p|}{k}) R^{-1}(\hat{p}).$$ Why isn’t this correct? Could some one explain the eqn(2.5.44)?

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2 Answers 2

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Originally a comment, since I haven't read the chapter in a while, but it got too long -

Conceptually it seems like the first expression makes sense as long as $B(|p|/k)$ boosts in the direction given by $K^\mu$. The $B$ operator corrects the magnitude of $k$ to that of $|p|$, and then the rotation $R$ corrects the direction. I think the expression you gave would be if one wanted to only change the magnitude of the vector without rotating it, e.g. rotate it to the orientation along which the boost acts, then boost it, and then rotate back.

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  • $\begingroup$ Thanks, your answer makes sense intuitively. Just to be sure, if the particle has mass with K^{$\mu$}=(0,0,0,k), would the second formula for L(p) also work since R^{-1}($\hat{p}$) has no effect on K? $\endgroup$
    – Wu Lee
    Commented Jul 25, 2022 at 17:45
  • $\begingroup$ If we are acting on a momentum eigenstate with reference momentum $K^{μ}=(0,0,0,k)$ then $R^{-1}(\hat{p})$ does nothing to the state, that is right. Btw the reason you have the "RBR" formula in your head is probably because boosts can be written that way. But Weinberg is not only looking at boosts, he considers lorentz transformations in general, which also include rotations (and reflections in space & time) $\endgroup$ Commented Jul 25, 2022 at 18:59
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The point of defining $L(p)= R(\hat{p})B(\frac{|p|}{k}) R^{-1}(\hat{p})$ for massive particles was only that, first, $R^{-1}(\hat{p})$ doesn't make any difference to standard momentum of the massive particle, second, using this definition only, we recover the Wigner rotation that matches just the ordinary rotation. And, you could just use the whole non-relativistic QM formalism. Weinberg makes it a point to prove this starting under equation 2.5.24.

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