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I am reading the textbook "Lectures on Quantum Field Theory", second edition by Ashok Das. On page 137, he talks about the non-unitarity of the finite-dimensional representation of the Lorentz group as I will explain now.

The generators of the Lorentz group in coordinate representation are given as: $$ M_{ij} = x_i \partial_j - x_j \partial_i \tag{4.26}$$ These generators follow the algebra: $$ [ M_{\mu \nu}, M_{\lambda \rho} ] = - \eta_{\mu \lambda}M_{\nu\rho} - \eta_{\nu \rho}M_{\mu\lambda} + \eta_{\mu \rho}M_{\nu\lambda} + \eta_{\nu \lambda}M_{\mu\rho} \tag{4.30}$$ We then define the rotation and boost operators respectively in terms of these generators as: $$ J_i = - \frac{1}{2}\epsilon_i^{\ jk}M_{jk},$$ $$ K_i = M_{0i} \tag{4.41}$$ It turns out that rotation and boost generators cannot both be hermitian simultaneously, namely: $$ J_i^\dagger = - J_i, \ \ K_i^\dagger = K_i \tag{4.45}$$

Then the author explains that this was supposed to happen because the Lorentz group is non-compact. Any finite-dimensional representation of a non-compact group must be non-unitary. However, we can get a unitary representation if we are willing to use infinite-dimensional representations. For example, author says, we can get both J and K to be hermitian if we define the generators with a factor of i: $$ M_{\mu \nu} = i ( x_\mu \partial_\nu - x_\nu \partial_\mu ) = M_{\mu \nu}^\dagger \tag{4.47}$$

Now, I can see that with this new definition, J and K both turn out to be Hermitian. But my question is, how is the representation given in equation (4.47) inifinite-dimensional? It is just the same as given in equation (4.26), with an additional i. What makes (4.26) finite-dimensional and (4.47) infinite-dimensional?

I am pretty sure my misunderstanding lies somewhere else, and I should be asking a different question. But I can't figure it out. So I have delineated my entire thought process here so that hopefully someone can find the mistake in my understanding here. Thanks.

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In 4.47, the representation is infinite dimensional. The Lorentz group acts on the infinite dimensional Hilbert space $H = L^2(\mathbb R^4)$. Intuitively, given a "wave function" $\psi\in H$, the Lorentz group acts by composition: $$ \Lambda\psi = x\to\psi(\Lambda^{-1}x) $$ The inverse is taken inside the argument so that it is a left action. Since a Lorentz transformation is metric preserving, it is of unit determinant, so by change of variables, the action is unitary.

From the unitary Lie group representation, you can deduce the hermitian Lie algebra representation, given by 4.47. There is always the usual caveat with domains and that technically the generators are self adjoint.

4.26 is just another convention for defining the generators. The issue is that without the extra factor $i$, the generators are not self adjoint on $H$. It is therefore more difficult to properly define their action using functional analysis, which is why it is more convenient to resort to the hermitian convention.

Usually, you start with the defining representation of the Lorentz group with $4\times4$ matrices, in which case: $$ \begin{align} J_1 &= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} & J_2 &= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} & J_3 &= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ K_1 &= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} & K_2 &= \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} & K_3 &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \end{align} $$ so that you indeed have: $$ \begin{align} J_i^\dagger &= -J_i & K_i^\dagger &= K_i \end{align} $$ However, starting from 4.26, you rather have: $$ \begin{align} J_i^\dagger &= -J_i & K_i^\dagger &= -K_i \end{align} $$ and from 4.47 $$ \begin{align} J_i^\dagger &= J_i & K_i^\dagger &= K_i \end{align} $$ Hope this helps.

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  • $\begingroup$ Thanks. This is what I understood from your answer: Generators in (4.26) are ALSO infinite-dimensional. But they are not hermitian. That's why we didn't get the same hermiticity for both J and K in (4.45). It seems that it's not enough just to naively pick up an infinite-dimensional representation. We must also be careful with the hermiticity of the generators in order to get an honest unitary representation. Is this correct @LPZ ? $\endgroup$
    – baba26
    Jun 30, 2023 at 13:46
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    $\begingroup$ Yes pretty much. Just to clarify, in infinite dimensional Hilbert spaces, it is usually easier to start from the group and then get to the generators in order to avoid domain issues (take for example the Stone-von Neumann theorem). Actually, from 4.26, equation 4.45 should rather read:$$K_i^\dagger = -K_i$$ $\endgroup$
    – LPZ
    Jun 30, 2023 at 14:57
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    $\begingroup$ You usually get 4.45 from finite dimensional representations like the defining representation of the Lorentz group, with $4\times 4$ matrices, which is often the usual starting point rather than 4.26. $\endgroup$
    – LPZ
    Jun 30, 2023 at 15:01

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