# $\mathfrak so(1,3)$ generators hermitian or antihermitian?

In Schwartz "Quantum field theory and the standard model" pag 160, the generators of the rotation are Hermitian, while the generators of boosts are anti-Hermitian, as an example:

$$J_1 = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ \end{matrix}\right)\,, K_1 = \left( \begin{matrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{matrix}\right)$$

So you have $$\langle \psi | \psi \rangle$$ is rotation invariant since $$e^{i\theta J_1}$$ is Unitary an then

$$\langle \psi | \psi \rangle \to \langle \psi | \left(e^{i\theta J_1}\right)^\dagger e^{i\theta J_1} |\psi \rangle = \langle \psi | e^{-i\theta J_1} e^{i\theta J_1} | \psi \rangle = \langle \psi | \psi \rangle$$

But for the boost you have that $$\left(e^{i\beta K_1}\right)^\dagger = e^{i\beta K_1}$$.

1. So why generator of boosts are chosen to be anti-Hermitian instead of Hermitian?

2. Furthermore, using Hermitian generators for rotation you get the bracket $$[J_i, J_j] = i\epsilon_{ijk}J_k$$, so in this way the $$i$$ factor makes the Lie algebra $$\mathfrak so(1,3)$$ not closed in respect to the brackets, unless you re-define the bracket with a $$-i$$ factor.

• Hi! Maybe this could be relevant Commented Oct 10, 2021 at 20:44
• Do you understand how so(1,3) differs from so(4), its compact brother? How do their respective generators correspond? For noncompact groups like SO(1,3), unitary reps are infinite dimensional, but finite dimensional reps are nonunitary. Have you gone through WuKi Tung's book? Commented Oct 10, 2021 at 21:09
• $[J_m,J_n] = i \epsilon_{mnk} J_k ~, ~ [J_m,K_n] = i \epsilon_{mnk} K_k ~, ~ [K_m,K_n] = -i \epsilon_{mnk} J_k$ is the Lie algebra of so(1,3). Check your statements... Commented Oct 10, 2021 at 21:23
• Shouldn't the $J_i$ and $K_i$ be $4\times 4$-matrices instead of $3\times 3$ ? Commented Oct 10, 2021 at 21:29
• Related/possible duplicate: physics.stackexchange.com/q/99051/50583 Commented Oct 10, 2021 at 21:42

These $$+$$ and $$-$$ signs in the generators originate from the signs in the Lorentz transformation matrices $$\Lambda_{\mu\nu}$$ (used for transforming $$4$$-vectors), which in turn originate from the signs in the Minkowski metric of spacetime.

Consider two examples:

### Rotation

A rotation around the $$x$$-axis (with angle $$\theta$$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\cos\theta&-\sin\theta\\ 0&0&\sin\theta&\cos\theta \end{pmatrix}$$ Especially notice the opposite signs of the two off-diagonal coefficients (which assures $$y'^2+z'^2=y^2+z^2$$).
This matrix can be generated by $$\Lambda=e^{-iJ_x\theta}$$ with the generator $$J_x=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&-i\\ 0&0&i&0 \end{pmatrix}$$

### Boost

A boost along the $$x$$-axis (with rapidity $$\beta$$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} \cosh\beta&\sinh\beta&0&0\\ \sinh\beta&\cosh\beta&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$ Especially notice the same signs of the two off-diagonal coefficients (which assures $$c^2t'^2-x'^2=c^2t^2-x^2$$).
This matrix can be generated by $$\Lambda=e^{iK_x\beta}$$ with the generator $$K_x=\begin{pmatrix} 0&-i&0&0\\ -i&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}$$

• Why I can't set $\Lambda=e^{K_x\beta}$, where $$K_x=\begin{pmatrix} 0&-1&0&0\\ -1&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}$$ ? In this way $K_x$ would be Hermitian and $\Lambda$ unitary Commented Oct 10, 2021 at 22:31
• @Andrea this is just a redefinition, but it will still net you a non-unitary group element, as you may check! Commented Oct 10, 2021 at 22:36
• @Andrea You could. It seems just like a convention to me. Likewise for the rotation you could set $\Lambda=e^{J_x\beta}$ with $$J_x=\begin{pmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0 \end{pmatrix}$$ Commented Oct 10, 2021 at 22:39
• @CosmasZachos you are right, may bad Commented Oct 10, 2021 at 22:51