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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?

Frobenius
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TaeNyFan
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2 Answers2

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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.

Frobenius
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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.

SuperCiocia
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  • Of course, $M_{\mu\nu}$ can also be shown to be antisymmetric without defaulting to a particular representation – Nihar Karve Jan 02 '21 at 06:48
  • 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? – TaeNyFan Jan 02 '21 at 07:57
  • no in index notation they are just entries of the matrix. So just numbers. They all commute too. – SuperCiocia Jan 02 '21 at 08:01
  • 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 ${\partial_\mu}^\dagger$? – TaeNyFan Jan 02 '21 at 08:10
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    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. – SuperCiocia Jan 02 '21 at 08:12
  • For example, the angular momentum component operator $J_x = i\hbar z\partial_y - i\hbar y\partial_z$. Can't we say we take the hermitian of $J_x$? The author said that $J_i $are the angular momentum operators, so he should mean taking the hermitian of each component, instead of the vector operator $\vec{J}$. – TaeNyFan Jan 02 '21 at 08:20
  • I think we are getting into a rabbit hole here. First, the $J_i$ used in the question is not strictly speaking the angular momentum (physically), as it's not Hermitian. The author makes this point in the picture you attached. Second, to prove the hermiticity of the $J_x$ that you write in the comments one has to show that $\langle J_x f|i\rangle = \langle f| J_x i\rangle$. Because it is not a matrix operator. As example of how it's done is here. However, – SuperCiocia Jan 02 '21 at 23:21
  • one can see that $J_x$ must be hermitian as it is composed from Hermitian operators $\mathbf{r}$ and $\mathbf{p}$. How would you prove $\hat{\mathbf{p}} = \frac{\hbar}{\mathrm{i}}\nabla$ is Hermitian? – SuperCiocia Jan 02 '21 at 23:26