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$$\left[J_i,J_j \right]=i\epsilon_{ijk}J_k$$ $$\left[J_i,M_j \right]=i\epsilon_{ijk}M_k$$ $$\left[M_i,M_j \right]=-i\epsilon_{ijk}J_k$$ where $J_i$ is the generator of rotation of Lorentz group, $M_i$ is the generator of boost of Lorentz group

In many textbook of QFT, they say that the second one implies that the generators of the boosts transform like a vector under rotations. But I can't see it explicitly. Can anyone give me the explanation.

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Vectors transform linearly, \begin{equation} x _i \rightarrow A _{ ij } x _j \end{equation} through some transformation matrix $A$.

Now consider the transformation of $ M _i $: \begin{align} e ^{ i \theta _j J _j } M _i e ^{ - i J _j \theta _j } & = \left( 1 + i \theta _j J _j - ... \right) M _i \left( 1 - i \theta _j J _j - ... \right) \\ & = M _i + i \theta _j \left[ J _j , M _i \right] + ... \end{align} Now if $ \left[ J _j , M _i \right] $ is only proportional to $ M _j $ as above then infinitesimally, \begin{align} e ^{ i \theta _j J _j } M _i e ^{ - i J _j \theta _j } & = M _i + i \theta _j \epsilon _{ jik} M _k \\ &= (\delta_{ik}+i\theta_j\epsilon_{jik})M_k\\ &= A _{i,j}M _j \end{align} for some transformation matrix $A$ as required.

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  • $\begingroup$ 4-vectors transform linearly, too. For this argument to apply shouldn't you also stress that the matrix $A$ is orthogonal, i.e. that $A \in SO(3)$? (which is easily seen to be of course, but still, just proving that you have a linear transformation isn't enough, is it?) $\endgroup$ – glS Jan 12 '15 at 9:15
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The important fact is that the change of an object $O$ under an infinitesimal transformation generated by a generator $G$ can be written in terms of their commutator (infinitesimal parameter $\alpha$):

$$O\longrightarrow O + \delta O\enspace\,\,\text{where}\enspace\,\, \delta O = i\alpha [G,\,O]$$

(to prove this, see JeffDror's answer). To interpret your first commutator $[J_i, J_j] = i\epsilon_{ijk} J_k$, put object $O=J_j$ and generator $G=J_i$, and the result on the right hand side is $\delta J = i\alpha_j\epsilon_{ijk}\,J_i$ tells you how the object $J$ transforms under the action of the generator $J$. This defines a vector. More generally, we learned $$\delta O = i\alpha_j\epsilon_{ijk}O_k\qquad\text{under $J_i$}$$ is how a vector $O$ transforms under the action of $J$. From now on, any $O$ satisfying the above is a vector.

Now move on to the next commutator $[J_i, M_j] = i\epsilon_{ijk} M_k$. To interpret this, we identify $M$ as our object and $J$ as the generator. The result on the right hand side $\delta M_i = i\alpha_j \epsilon_{ijk}M_k$ tells you that $M$ transforms precisely in the same way a vector changes.

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