# Lorentz algebra and its generators

I'm reading Maggiore's book A Modern Introduction to Quantum Field Theory and I'm getting a bit confused when he writes about Lorentz algebra:

$$K^i = J^{i0},$$

$$J^{i}=\frac{1}{2}\epsilon^{ijk}J^{jk},$$

$$[J^{i}, J^{j}] = i\epsilon^{ijk} J^{k},$$

$$[J^{i}, K^j] =i\epsilon^{ijk} K^k.$$

Then he states that $K^i$ is a spatial vector due to the last commutation relation. Is that the way a spatial vector transform under the $SO(3)$ algebra? If yes why?

From Claude Cohen-Tannoudji, Volume 2, X.D.1:

(...) an observable $\textbf{V}$ is a vector if its three components $V_x, V_y$ and $V_z$ in an orthonormal frame $Oxyz$ satisfy the following commutation relations: $$\tag{4-a} [J_x,V_x] = 0$$ $$\tag{4-b} [J_x,V_y] = i \hbar V_z$$ $$\tag{4-c} [J_x, V_z] = -i \hbar V_y$$ as well as those obtained by cyclic permutation of the indices $x,y$ and $z$.

In your notation, these relations can be more compactly written as $$\tag{1} [J_i,V_j] = i \epsilon_{ijk} V_k$$ or (in a more formal, less rigorous way) $$\textbf{J} \times \textbf{V} = i \hbar \textbf{V}.$$

In other words, (1) are the defining relations of a vector operator $\textbf{V}$.

Other information about vector operators can be found on this wikipedia article and this physics.se answer.

• Now it's clearer... I was not thinking of $K^i$ as being a vector OPERATOR. – Worldsheep Dec 30 '14 at 9:32
• @Worldsheep well yes. In your notation, both $K^i$ (thought of as $\{ K^i \}_{i=1,2,3}$, not as the i-th component) and $J^i$ are vector operators, as they both satisfy the defining commutation relations. That $J^i$ satisfies them is of course no surprise, as they coincide in this case with the commutation relations defining the $\mathfrak{su}(2) \approx \mathfrak{so}(3)$ Lie algebra generated by the angular momentum operators. – glS Dec 30 '14 at 9:37

You can think of the $J^i$, $i=1,2,3$ as rotation by $\pi/2$, or to be more precise $[J^i, K^j]$ as a rotation of $K^j$ around the $i$th axis by $\pi/2$. Then, e.g., $[J^1,K^2] = iK^3$, which corresponds to the fact that, if you rotate around the $x$ axis, you are rotating vectors on the $y-z$ plane. Hence $K^2$ is rotated onto $K^3$ (similarly, $K^3$ is rotated onto $-K^2$. This is the behaviour of spatial vectors under rotation.

• I understand the formal analogy you are drawing and the compatibility of this "$\pi/2$ rotation" view with the commutation relations, but I'm curious: is this more than a coincidence? If so (how it probably is) why do the generators corrispond to a rotation of $\pi/2$ (instead of any other angle)? – glS Dec 30 '14 at 13:56
• I'll think about it – Phoenix87 Dec 30 '14 at 14:25