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My professor gave me the following derivation for the full generator of the Lorentz transformations. The starting point is to consider a subgroup of the conformal group that leaves the origin fixed (as done so in Di Francesco, for example). Denote by $J_{\mu \nu}$ the full generator and so $J_{\mu \nu}\Phi(0) = S_{\mu \nu}\Phi(0)$, where $S_{\mu \nu}$ is the component of the full generator transforming only the field spin components.

Infinitesimally, $\Phi'(0) = (1 + \epsilon S_{\mu \nu})\Phi(0)$ but also $\Phi'(0) = e^{-ix' \cdot P}\Phi'(x')$. This I understand. The next line is the first one I am not so sure of. He then writes $$\Phi'(0) = e^{-ix' \cdot P}(1+\epsilon J_{\mu \nu})\Phi(x)$$ How is this true? He seems to be using the transformation $\Phi'(x') = D \Phi(x)\,\,\,(1)$, but using $D$ as the full generator. If I had read that statement a few months back, I would have accepted it fine, but I posted a question here earlier that discussed the fact that in $(1)$, $D$ is the spin component of the generator because $x'$ and $x$ represent the same physical point in the space, they are just two different representations of the same point relative to two different coordinate systems. So the statement seems to be contradicting what I thought I understood. That was my first concern.

My second concern is less fundamental (more of a computational problem) and I can edit it into my answer later. Many thanks!

Edit: So assuming the last line to be correct, we proceed: $$\Phi'(0) = e^{-ix' \cdot P}(1 + \epsilon J_{\mu \nu})e^{i x \cdot P}\Phi(0) \approx (1 + \epsilon \underbrace{e^{-ix \cdot P}J_{\mu \nu}e^{i x \cdot P}})\Phi(0)$$ using the fact to zeroth order $x' \approx x$. The brace can be evaluated using the Hausdorff formula: \begin{align}e^{-ix \cdot P}J_{\mu \nu}e^{i x \cdot P} &\approx J_{\mu \nu} + [J_{\mu \nu}, ix^{\sigma}P_{\sigma}] \\&= J_{\mu \nu} + i\left[x^{\sigma}[J_{\mu \nu},P_{\sigma}] + [J_{\mu \nu}, x^{\sigma}]P_{\sigma}\right] = J_{\mu \nu} + x_{\mu}P_{\nu} - x_{\nu}P_{\mu},\end{align} where the second commutator on the RHS there vanishes. Now compare with $\Phi'(0) = (1+\epsilon S_{\mu \nu})\Phi(0)$ as written at the top and we have the result $J = L + S$. My question is: why does this second commutation relation vanish?

I should say I am going through 'Conformal field theory' Di Francesco section 4.2.1. The above argument was an alternate argument my professor posed when we sat down and agreed that equation (4.26) in Di Francesco is simply wrong.

CAF
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  • Maybe a typo, the formula should be something like $\Phi'(0) = e^{-ix \cdot P}(1+\epsilon J_{\mu \nu})\Phi(x)$ – Trimok Aug 21 '14 at 11:54
  • Hi Trimok. Later on (in fact I have now edited the question) we approximated $x' \approx x$ in the infinitesimal transformation. – CAF Aug 21 '14 at 12:02
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    The commutators like $[J_{\mu\nu},x^\alpha]$ vanish because $x$ is a vector of $c$-numbers, so in each representation, it is simply proportional to the unit matrix that commutes with all the actual generators. Here $x$ is not meant to be something like the $x$ operator acting on a wave function in QM. – Luboš Motl Aug 21 '14 at 14:04
  • @LubosMotl: Thanks very much! So $x$ here is a parameter, not an operator, so commutes with $J$. That answers my second question. – CAF Aug 21 '14 at 14:19
  • Hi, thanks. I wasn't quite able to figure out what the first question is. The full transformation is composed of two things: finding the right $x'$ for a given $x$, and evaluating the components at the new point (that's the orbital part); and mixing the components with each other (that's the spin part). The full generator may be divided in this way. If we specialize to transformations that do not change $x$, the orbital part is trivial so it may be subtracted or added without spoiling the formula. – Luboš Motl Aug 21 '14 at 15:06
  • Also, the most general transformation always fixes some point, so it may be obtained from a transformation fixing the origin, by conjugating it with a translation by $Y$ which is the point that is fixed. – Luboš Motl Aug 21 '14 at 15:06
  • My understanding is that in the formula, $\Phi'(x') = D\Phi(x)$, where $D$ is some representation of the lie group (Lorentz in this case), $x'$ and $x$ correspond to the same physical point in space (see, for example, Greiner 'Field Quantisation, P.40 footnote) So we may write $\Phi'(P) = D\Phi(P)$ in which case, only the spin components are being changed so that is why I thought we should not have the full generator present in $\Phi'(x') = (1 + \epsilon J_{\mu \nu})\Phi(x)$ used in the proof. – CAF Aug 21 '14 at 15:18
  • Take the case of a spinless field. Then $\Phi'(x') = \Phi(x)$ which is the case when $D = 1 + \epsilon \cdot S$ is unity. If we replace $S$ here by $J$, the result $\Phi'(x') = \Phi(x)$ is not so clear. – CAF Aug 21 '14 at 15:20
  • Or indeed, perhaps the usage of $J$ here is one in which it is meant that the orbital part is realized trivially. – CAF Aug 21 '14 at 15:26
  • @LubosMotl: Do you have any further comments with regard to what I wrote above? – CAF Aug 23 '14 at 11:17
  • There are several issues, I've mentioned some of them, and one must be careful about the conventions and strategy used by someone else. If you have a problem with someone's derivation, try to check the "big claims" only and derive it yourself - all the hard ideas of the derivation may clearly be seen even in the derivation following slightly different strategy, conventions, and active vs passive distinction. I feel that if you can't derive it yourself even while looking at the "template derivation", it's useless to follow someone else's derivation, anyway. – Luboš Motl Aug 23 '14 at 13:56
  • @LubošMotl: Sorry for late reply, but I got it, thanks. You are right, I derived it myself and realized that the two derivations are equivalent for an infinitesimal transformation $x' \approx x$ which I didn't invoke at another point in the argument presented above. – CAF Aug 31 '14 at 16:06
  • I was wondering, since you seem to be familiar with Di Francesco's book, if you could explain how equation (2.174) on P.46 comes about. The authors say it is obtained by inspection, but I just don't see it. Thanks. – CAF Aug 31 '14 at 16:07
  • If it is preferred for this to answered elsewhere then here would be good, since I asked the same question earlier on:http://physics.stackexchange.com/questions/129173/constructing-conserved-current-given-the-lagrangian – CAF Aug 31 '14 at 16:57

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