# Matrix Representation of Lorentz Group Generators

Let $$\Lambda^{\alpha}{}_{\beta}$$ denote a generic Lorentz transformation.

Then, an infinitesimal transformation can be written like

$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + \omega^{\mu}{}_{\nu}$$

where

$$\omega^{ij} = \epsilon^{ijk}\theta_k$$

$$\omega^{i0} = - \omega^{0i} = \delta^i$$

where $$i,j,k$$ run from 1 to 3 and $$\delta^i$$ is a parametre related with boosts. Then, an infinitesimal transformation has a matrix representation

$$\begin{pmatrix} 1 & -\delta_1 & -\delta_2 & -\delta_3\\ -\delta_1 & 1 & \theta_3 & -\theta_2\\ -\delta_2 & -\theta_3 & 1 & \theta_1\\ -\delta_3 & \theta_2 & -\theta_1 & 1 \end{pmatrix}$$

However, we can also write

$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu}$$

where $$J_{\alpha \beta}$$ are the generators of the group. I want to prove that $$J_{01}$$ is of the form

$$\begin{pmatrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

My problem is in understanding the notation in

$$\Lambda^{\mu}{}_{\nu} = \delta^{\mu}{}_{\nu} + i\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu}$$

For example, I tried to compute $$\left(J_{01} \right)^{0}{}_{1}$$ by doing

$$\Lambda^{0}{}_{1} = \delta^{0}{}_{1} + i\frac{\omega^{01}}{2}\left(J_{01} \right)^{0}{}_{1}$$

$$\Leftrightarrow - \delta_1 = 0 -i \frac{\delta_1}{2}\left(J_{01} \right)^{0}{}_{1}$$

which yields $$\left(J_{01} \right)^{0}{}_{1} = -2i$$, which is not correct. What am I doing wrong?

• Note that the factor of two in the expression for a Lorentz transformation is there to prevent overcounting since $J_{01}=-J_{10}$. – Charlie Feb 18 at 21:41
• What you are doing wrong is not summing over repeat indices α and β . – Cosmas Zachos Feb 18 at 22:06

In the following text, we use $$\eta_{\mu\nu} = \text{diag}(-1,+1,+1,+1)$$.

For an infinitesimal homogeneous Lorentz transformation, we have

$${\omega^\mu}_\nu = \begin{pmatrix} 0 & \zeta_1 & \zeta_2 & \zeta_3 \\ \zeta_1 & 0 & -\theta_3 & \theta_2 \\ \zeta_2 & \theta_3 & 0 & -\theta_1 \\ \zeta_3 & -\theta_2 & \theta_1 & 0 \end{pmatrix},$$

and

\begin{aligned} {\Lambda^\mu}_\nu &= {\delta^\mu}_\nu + {\omega^\mu}_\nu \\ &= {\delta^\mu}_\nu + \eta^{\rho\mu}\omega_{\rho\nu} \\ &= {\delta^\mu}_\nu + \eta^{\rho\mu}\omega_{\rho\sigma}{\delta^\sigma}_\nu \\ &= {\delta^\mu}_\nu + \frac{1}{2}\omega_{\rho\sigma}(\eta^{\rho\mu}{\delta^\sigma}_\nu - \eta^{\sigma\mu}{\delta^\rho}_\nu) \\ &= {\delta^\mu}_\nu + \frac{i}{2}\omega_{\rho\sigma}{(S_V^{\rho\sigma})^\mu}_\nu \\ \end{aligned}

where the vector representation of the generators are defined as

$${(S_V^{\rho\sigma})^\mu}_\nu \equiv -i({\eta}^{\rho\mu}{\delta^\sigma}_\nu - {\eta}^{\sigma\mu}{\delta^\rho}_\nu).$$

Note that $$\omega_{\rho\sigma}$$ and $$S_V^{\rho\sigma}$$ are antisymmetric in the indices $$(\rho\sigma)$$, and

$${(S_V^{0i})}^\dagger = - S_V^{0i}, \quad {(S_V^{ij})}^\dagger = S_V^{ij}.$$

If we define

$$\boldsymbol\theta \equiv (\theta_1, \theta_2, \theta_3), \quad \boldsymbol\zeta \equiv (\zeta_1, \zeta_2, \zeta_3),$$ $$\mathbf{J} \equiv (S_V^{23},S_V^{31},S_V^{12}),\quad \mathbf{K} \equiv (S_V^{01},S_V^{02},S_V^{03}),$$

then

$${\Lambda^\mu}_\nu = {\delta^\mu}_\nu - i{{(\boldsymbol\theta \cdot \mathbf{J} + \boldsymbol\zeta \cdot \mathbf{K})}^\mu}_\nu.$$

The explicit expressions of the matrices $$J_i$$ and $$K_i$$ are

$$J_1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \end{pmatrix}, \quad J_2 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \end{pmatrix}, \quad J_3 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ $$K_1 = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad K_2 = \begin{pmatrix} 0 & 0 & i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \quad K_3 = \begin{pmatrix} 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \end{pmatrix}$$

Thanks to @Charlie and @Cosmas Zachos I was able to find the correct answer.

It simply suffices to develop the sum

$$\frac{\omega^{\alpha \beta}}{2}\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = -\delta_1 \left(J_{01} \right)^{\mu}{}_{\nu} - \delta_2 \left(J_{02} \right)^{\mu}{}_{\nu} - \delta_3 \left(J_{03} \right)^{\mu}{}_{\nu} + \theta_3\left(J_{12} \right)^{\mu}{}_{\nu} - \theta_2 \left(J_{13} \right)^{\mu}{}_{\nu} + \theta_1\left(J_{23} \right)^{\mu}{}_{\nu}$$

where I used

$$\left(J_{\alpha \beta} \right)^{\mu}{}_{\nu} = - \left(J_{\beta \alpha } \right)^{\mu}{}_{\nu}$$

and the other properties mentioned above.