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In Ashok Das Lectures on QFT book, pg. 135-136, he stated the following hermicity properties for the Lorentz group generators:

$$ {J_i}^\dagger=-J_i\,,\quad{K_i}^\dagger=K_i \tag{4.45}\label{4.45} $$

where $J_i$ are the infinitesimal rotation generators and $K_i$ are the infinitesimal boost generators.

$J_i$ and $K_i$ are defined in the coordinate representation as $$J_i=-\frac{1}{2}{\epsilon_{i}}^{jk}M_{jk}\,,$$ $$K_i=M_{0i}\tag{4.41}\label{4.41}$$ where $M_{\mu\nu}=x_\mu\partial_{\nu}-x_\nu \partial_\mu$ and ${\epsilon_i}^{jk}$ is the raised levi civita tensor.

I know how to take the complex transpose of a matrix. But how does one take the complex transpose of something like $J_i=-\frac{1}{2}{\epsilon_{i}}^{jk}M_{jk}$, which is not a matrix, to show that the first two equations are correct?

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Reference : My answer here Vector product in a 4-dimensional Minkowski spacetime.

The only thing you need is to prove that $\:M_{jk}\:$ is a $4\times4$ antisymmetric matrix. My suggestion is to see this matrix as the outer (also known as vector or cross) product of real 4-vectors by the analogy of the outer (vector) product of 3-vectors in the Euclidean real space $\mathbb{R}^3$. In the latter this product is essentially a $3\times3$ antisymmetric matrix and as such a kind it has 3 independent elements, that's why we consider it also as a vector in $\mathbb{R}^3$. In the 4-dimensional case the outer product is a $4\times4$ antisymmetric matrix and as such a kind it has 6 independent elements, it's a 6-vector.

(Note : I suggest first to take a look in $\boldsymbol{\S 2,3,4}$ of my answer in the above link)

So consider (the covariant version of) the position 4-vector \begin{align} & \mathbf{X} \boldsymbol{=} X_\mu\boldsymbol{=} \begin{bmatrix} x_0\vphantom{\dfrac{a}{b}}\\ x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \boldsymbol{+}x^0\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}x^1\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}x^2\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}x^3\vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}c\,t \:\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\mathbf{x}\:\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-01a}\label{A-01a}\\ & \text{with transpose} \nonumber\\ & \mathbf{X}^{\boldsymbol{\top}}\boldsymbol{=} \begin{bmatrix} x_0 & x_1 & x_2 & x_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} x_0 & \hphantom{x_1} & \boldsymbol{-}\mathbf{x}^{\boldsymbol{\top}} & \hphantom{x_3} & \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-01b}\label{A-01b} \end{align} and also (the covariant version of) the 4-gradient vector operator

\begin{align} & \boldsymbol{\Box\!\!\!\!\Box\!\!\!\!\Box}\boldsymbol{=}\boldsymbol{\Box}_\nu\boldsymbol{=} \begin{bmatrix} \partial/\partial x^0\vphantom{\dfrac{a}{b}}\\ \partial/\partial x^1\vphantom{\dfrac{a}{b}}\\ \partial/\partial x^2\vphantom{\dfrac{a}{b}}\\ \partial/\partial x^3\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \partial_0\vphantom{\dfrac{a}{b}}\\ \partial_1\vphantom{\dfrac{a}{b}}\\ \partial_2\vphantom{\dfrac{a}{b}}\\ \partial_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \partial/\partial c\,t\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}}\\ \boldsymbol{\nabla}\vphantom{\dfrac{a}{b}}\\ \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-02a}\label{A-02a}\\ & \text{with transpose} \nonumber\\ & \boldsymbol{\Box\!\!\!\!\Box\!\!\!\!\Box}^{\boldsymbol{\top}}\boldsymbol{=} \begin{bmatrix} \partial_0 & \partial_1 & \partial_2 & \partial_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{=} \begin{bmatrix} \partial/\partial c\,t & \hphantom{x_1} & \boldsymbol{\nabla}^{\boldsymbol{\top}} & \hphantom{x_3} & \vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-02b}\label{A-02b} \end{align}

For their outer product we have \begin{align} \mathbf{W} & \boldsymbol{=}\mathbf{X}\boldsymbol{\times}\boldsymbol{\Box\!\!\!\!\!\Box}\boldsymbol{\equiv}\boldsymbol{\Box\!\!\!\!\!\Box}\,\mathbf{X}^{\boldsymbol{\top}}\boldsymbol{-}\mathbf{X}\,\boldsymbol{\Box\!\!\!\!\!\Box}^{\boldsymbol{\top}} \nonumber\\ &\boldsymbol{=} \begin{bmatrix} \partial_0\vphantom{\dfrac{a}{b}}\\ \partial_1\vphantom{\dfrac{a}{b}}\\ \partial_2\vphantom{\dfrac{a}{b}}\\ \partial_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x_0 & x_1 & x_2 & x_3 \vphantom{\dfrac{a}{b}} \end{bmatrix}\boldsymbol{-} \begin{bmatrix} x_0\vphantom{\dfrac{a}{b}}\\ x_1\vphantom{\dfrac{a}{b}}\\ x_2\vphantom{\dfrac{a}{b}}\\ x_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \partial_0 & \partial_1 & \partial_2 & \partial_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ & \boldsymbol{=} \begin{bmatrix} \partial_0x_0 & \partial_0x_1 & \partial_0x_2 & \partial_0x_3\vphantom{\dfrac{a}{b}}\\ \partial_1x_0 & \partial_1x_1 & \partial_1x_2 & \partial_1x_3\vphantom{\dfrac{a}{b}}\\ \partial_2x_0 & \partial_2x_1 & \partial_2x_2 & \partial_2x_3\vphantom{\dfrac{a}{b}}\\ \partial_3x_0 & \partial_3x_1 & \partial_3x_2 & \partial_3x_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{-} \begin{bmatrix} x_0\partial_0 & x_0\partial_1 & x_0\partial_2 & x_0\partial_3\vphantom{\dfrac{a}{b}}\\ x_1\partial_0 & x_1\partial_1 & x_1\partial_2 & x_1\partial_3\vphantom{\dfrac{a}{b}}\\ x_2\partial_0 & x_2\partial_1 & x_2\partial_2 & x_2\partial_3\vphantom{\dfrac{a}{b}}\\ x_3\partial_0 & x_3\partial_1 & x_3\partial_2 & x_3\partial_3\vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \nonumber\\ & \boldsymbol{=} \begin{bmatrix} 1\boldsymbol{+}x_0\partial_0 & \partial_0x_1 & \partial_0x_2 & \partial_0x_3\vphantom{\dfrac{a}{b}}\\ \partial_1x_0 & 1\boldsymbol{+}x_1\partial_1 & \partial_1x_2 & \partial_1x_3\vphantom{\dfrac{a}{b}}\\ \partial_2x_0 & \partial_2x_1 & 1\boldsymbol{+}x_2\partial_2 & \partial_2x_3\vphantom{\dfrac{a}{b}}\\ \partial_3x_0 & \partial_3x_1 & \partial_3x_2 & 1\boldsymbol{+}x_3\partial_3\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{-} \begin{bmatrix} x_0\partial_0 & x_0\partial_1 & x_0\partial_2 & x_0\partial_3\vphantom{\dfrac{a}{b}}\\ x_1\partial_0 & x_1\partial_1 & x_1\partial_2 & x_1\partial_3\vphantom{\dfrac{a}{b}}\\ x_2\partial_0 & x_2\partial_1 & x_2\partial_2 & x_2\partial_3\vphantom{\dfrac{a}{b}}\\ x_3\partial_0 & x_3\partial_1 & x_3\partial_2 & x_3\partial_3\vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \nonumber\\ & \boldsymbol{=}\mathbf{I}_{4\times 4}\boldsymbol{-}\mathbf{M} \tag{A-03}\label{A-03} \end{align} where $\mathbf{I}_{4\times 4}$ the identity matrix and $\mathbf{M}$ the $4\times4$ antisymmetric matrix of the question \begin{align} \mathbf{M} & \boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}\left(x_0\partial_1\boldsymbol{-}x_1\partial_0\right) & \hphantom{\boldsymbol{-}}\left(x_0\partial_2\boldsymbol{-}x_2\partial_0\right) & \hphantom{\boldsymbol{-}} \left(x_0\partial_3\boldsymbol{-}x_3\partial_0\right)\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\left(x_0\partial_1\boldsymbol{-}x_1\partial_0\right) & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\left(x_2\partial_1\boldsymbol{-}x_1\partial_2\right) & \hphantom{\boldsymbol{-}} \left(x_1\partial_3\boldsymbol{-}x_3\partial_1\right)\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\left(x_0\partial_2\boldsymbol{-}x_2\partial_0\right) & \hphantom{\boldsymbol{-}}\left(x_2\partial_1\boldsymbol{-}x_1\partial_2\right) & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}\left(x_3\partial_2\boldsymbol{-}x_2\partial_3\right) \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}\left(x_0\partial_3\boldsymbol{-}x_3\partial_0\right) & \boldsymbol{-} \left(x_1\partial_3\boldsymbol{-}x_3\partial_1\right) & \hphantom{\boldsymbol{-}}\left(x_3\partial_2\boldsymbol{-}x_2\partial_3\right) & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ & \boldsymbol{=} \begin{bmatrix} \hphantom{\boldsymbol{-}}0 & \hphantom{\boldsymbol{-}}K_1 & \hphantom{\boldsymbol{-}}K_2 & \hphantom{\boldsymbol{-}} K_3\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}K_1 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}J_3 & \hphantom{\boldsymbol{-}} J_2\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}K_2 & \hphantom{\boldsymbol{-}}J_3 & \hphantom{\boldsymbol{-}}0 & \boldsymbol{-}J_1 \vphantom{\dfrac{a}{b}}\\ \boldsymbol{-}K_3 & \boldsymbol{-} J_2 & \hphantom{\boldsymbol{-}}J_1 & \hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{A-04}\label{A-04} \end{align} Note that the scalar operators $L_k \boldsymbol{=} \boldsymbol{-}i\hbar J_k\,,k=1,2,3$ are the components of the orbital angular momentum which are hermitian, that's why the scalar operators $ J_k\,,k=1,2,3$ are antihermitian.

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The only thing you need to compute is the transpose of $M_{\mu\nu}$: $$M_{\mu\nu}^T = M_{\nu\mu} = x_\nu\partial_\mu - x_\mu\partial_\nu = -(x_\mu\partial_\nu-x_\nu\partial_\mu) = -M_{\mu\nu}.$$

So: $$ J_i^\dagger = -\frac{1}{2}(\epsilon_i^{jk})^\dagger (M_{jk})^\dagger,$$ but $\epsilon_i^{jk}$ is a real scalar so $(\epsilon_i^{jk})^\dagger = \epsilon_i^{jk}$, and $M_{jk}$ has real entries so $(M_{jk})^\dagger = (M_{jk})^T$ which we established is equal to $-M_{jk}$. So... $$ J_i^\dagger = +\frac{1}{2}\epsilon_i^{jk} M_{jk} = -J_i.$$

The $K_{0i}$ is trivial.

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  • $\begingroup$ Of course, $M_{\mu\nu}$ can also be shown to be antisymmetric without defaulting to a particular representation $\endgroup$ – Nihar Karve Jan 2 at 6:48
  • $\begingroup$ But ${\epsilon_{i}}^{jk}M_{jk}$ is not matrix multiplication? If I write it out explicitly, it will be something like $J_i = a + b + c$ like the addition of scalars. Unless the author mean $J_i = (J_1, J_2, J_3)$ is a matrix? $\endgroup$ – TaeNyFan Jan 2 at 7:57
  • $\begingroup$ no in index notation they are just entries of the matrix. So just numbers. They all commute too. $\endgroup$ – SuperCiocia Jan 2 at 8:01
  • $\begingroup$ I guess what I'm trying to ask is this: $J_1$ can be written explicitly as $J_1=x_3 \partial_2 - x_2 \partial_3$. Then $J_1$ is like a $1\times 1$ matrix so ${J_1}^\dagger = {(x_3 \partial_2)}^\dagger - {(x_2 \partial_3)}^\dagger$. How do we evaluate this? What is ${\partial_\mu}^\dagger$? $\endgroup$ – TaeNyFan Jan 2 at 8:10
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    $\begingroup$ No $J_i$ is a 1x3 Column vector, with each entry equal to what you said but times the Levi Civita symbol with $i=$ the row. $\endgroup$ – SuperCiocia Jan 2 at 8:12

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