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In QFT, we need to use infinite-dimensional representations of the Lorentz algebra, because all the non-trivial finite-dimension al representations are not unitary, and we need unitary representations so that $\langle\psi\vert\psi^\prime\rangle$ is invariant under Lorentz transformations (Schwartz, Quantum Field Theory, pag 110), $$\langle\psi\vert\psi^\prime\rangle \to \langle\psi\vert\Lambda^{\dagger}\Lambda\vert \psi^\prime\rangle = \langle\psi\vert\psi^\prime\rangle$$ if $\Lambda^{\dagger}\Lambda = I. $

Infinite-dimensional representations of the Lorentz algebra are achieved by means of differential operators (Maggiore, A Modern introduction to Quantum Field Theroy, pag 30, and Zee, Group Theory, pag 434), which act on the Hilbert space of square-integrable functions (which is infinite dimensional), instead of matrices acting on the $\Bbb R^{4}$ vector space.

So my understanding is that the goal of infinite-dimensional representations is to get Hermitian generators $J_i, K_i$ for the Lorentz Lie algebra, so that the Lorentz transformation $\Lambda=e^{i(\alpha_i J_i+\beta_i K_i)}$ is unitary ($\Lambda^{\dagger} = e^{-i(\alpha_i J_i+\beta_i K_i)})$.

In the Zee book on the group (page 434) we have the following definition for $K_1$, $$iK_1 = t\frac{\partial}{\partial x} + x\frac{\partial}{\partial t},$$
and later he states (page 436) that $iK_1$ is Hermitian.

To me, this is completely wrong, because if $iK_1$ were Hermitian, then $K_1$ would be anti-Hermitian, and then we would have missed the goal of using an infinite-dimensional representation to get Hermitian generators.

Also, explicit calculation shows that $K_1$ is Hermitian and then $iK_1$ is anti-hermitian:

  • The adjoint of an operator $A$ is defined by (Hassani, Mathematical Physics, page 61): $\langle \psi\vert A \vert \psi^\prime\rangle^{*} = \langle \psi^{\prime}\vert A^{\dagger}\vert \psi\rangle$.

So $A$ is Hermitian (i.e. self-adjoint) if $$\langle \psi\vert A \vert \psi^\prime\rangle^{*} = \langle \psi^{\prime}\vert A \vert \psi\rangle.$$

  • In the Hilbert space of square-integrable functions the inner product is defined by (below $x$ is a four-vector) $$\langle f \vert g \rangle = \int f(x)g(x)^* \, dx .$$

Integrating by parts and assuming as usual that $\psi(x),\psi^{\prime}(x) \to 0$ when $x\to boundary$, you get that $$\langle \psi\vert K_1 \vert \psi^\prime\rangle^{*} = \langle \psi^{\prime}\vert K_1 \vert \psi\rangle.$$

I am very reluctant to believe Zee is doing it wrong, so I am asking if my reasoning is correct or I am missing some crucial step.

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    $\begingroup$ There's a conventional issue where some people like their generators to be anti-Hermitian (i.e. anti-Hermitian $T$ generates the unitary $\mathrm{e}^{T}$) and some people like their generators to be Hermitian (i.e. Hermitian $T$ generates the unitary $\mathrm{e}^{\mathrm{i}T}$). Are you sure this isn't just a difference in convention? Comparing factors of $\mathrm{i}$ and $-1$ across different sources is often difficult because of this. $\endgroup$
    – ACuriousMind
    Nov 21, 2021 at 11:52
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    $\begingroup$ It doesn't seem merely a conventional issue, cause in the same pages he states that $J_i$ is hermitian, so $J_i$ is hermitian and $K_i$ anti-hermitian. If it were only convetional issue I would expect generators to be all hermitian, or all anti-hermitian. $\endgroup$
    – Andrea
    Nov 21, 2021 at 12:23
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    $\begingroup$ In that case, are you sure the representation Zee is really describing is the unitary representation on the space of states, and not the definitely not unitary representation on the fields? In the latter, the boosts are anti-unitary, see e.g. physics.stackexchange.com/q/669780/50583 $\endgroup$
    – ACuriousMind
    Nov 21, 2021 at 12:26
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    $\begingroup$ In that page he is using the generator defined thorough differential operator. My understanding is that the generators defined thorough differential operators are used just for the space of states. If it were the representation on the fields shouldn't he uses the usual matrix representation (which is not unitarian)? $\endgroup$
    – Andrea
    Nov 21, 2021 at 12:47
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    $\begingroup$ It does not make sense. $L^2$ is referred to spatial variables only. In that context $t$ is an external parameter $\partial_t$ is not an operator acting in the Hilbert space. The vector fiel $K_1$ you wrote is not the generator of the Lorentz symmetry in the Hilbert space. $\endgroup$ Nov 21, 2021 at 18:50

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I think the linked questions and WP article and books resolve the title question, which most responders identified as the heart of it, by pointing out quantum fields transforming in the finite-dimensional non-unitary irrep of the Lorentz group, while states/particles in the infinite-dimensional irrep of the same group, and how the two mesh/repackage and transition into each other.

I believe, however, the core of your question is the misunderstanding of the differential realization you are utilizing for the antihermitean boosts, $K_i$, e.g. $$ iK_1= t\partial_x + x\partial_t~ . $$ This is vector-field shorthand summarizing the action of the boost on the non-unitary 4D irreps of the Lorentz boost, namely 4-vectors $(t,x,y,z)$.

For simplicity, since you are only acting on the 2D subspace (t,x), let's truncate out (y,z), $$ \begin{pmatrix} t'\\x' \end{pmatrix}=e^{-\zeta \sigma_1} \begin{pmatrix} t\\x \end{pmatrix}= \begin{pmatrix} \cosh \zeta &-\sinh \zeta \\ -\sinh \zeta &\cosh \zeta \end{pmatrix}\begin{pmatrix} t\\x \end{pmatrix}, $$ where $\zeta$ is the rapidity and c=1. You note this transformation is definitely not unitary, which is just as well: it is only meant to preserve the interval $t^2-x^2$, and not a naive positive- definite Euclidean dot product.

The infinitesimal transformation is $$ (t',x')^T =(t,x)^T -\zeta (x,t)^T + O(\zeta^2). $$ So, for example, a transformed scalar field is $$ \phi(t',x',y',z')= \phi(t,x,y,z) -\zeta (x\partial_t +t\partial_x)\phi(t,x,y,z) +O(\zeta^2). $$ But, "deep down", you appreciate you are only transforming 4-vectors--not the paradigmatic infinite-dimensional vectors of elementary QM Hilbert space (where $i\partial_x$ truly translates to the familiar infinite-dimensional Heisenberg matrices).

For non-scalar fields, like the spinor field, for example, you further scramble the four components of the 4-spinor, again a finite not infinite-dimensional representation of the Lorentz group, through the likewise hermitian, not antihermitean, increment involving the γ matrix boost. I recall Schwartz's text, bottom of p 171, neatly illustrates how this perfectly fits with the Lorentz invariance of $\bar \psi \psi$!

So, the takeaway is that your differential realization, counterintuitively acts on finite-dimensional representations, and differs dramatically from the $L^2$ formalism you are focussing on (as well as the infinite-dimensional matrix irrep of the creation and annihilation operators describing particle states). I do not wish to get involved in the pedagogical glibness or inoptimalities of the texts you are discussing. (I have my personal grim opinions about one of them, half a century ago, before it was even written...).

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  • $\begingroup$ May I ask you which is in your opinion an good introductory book on QFT with a focus on group concepts? $\endgroup$
    – Andrea
    Nov 22, 2021 at 17:00
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    $\begingroup$ I fear you are right; Matt, too, promulgates nonsense on occasion. Still, he gets the point on spinors right, at the bottom of page 171, where non-unitarity is essential. Weinberg has a lot, but it is exceptionally formal and unfriendly reading. Choose from here. Wu-Ki Tung's group theory book, referred to above, is the friendliest. $\endgroup$ Nov 22, 2021 at 19:15
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    $\begingroup$ A compromise way to read Matt's statement with acceptance is to note that this action on the space of functions is reducible. That is, as shown above, action of the Lorentz group transforms the variables of the function, keeping you on the same function form. The action happens to be the same for any other function, which, in a sense, $\phi(t,x,y,z)$ doesn't see directly. The structure is an infinite direct product of 4D irreps. $\endgroup$ Nov 22, 2021 at 19:31
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    $\begingroup$ @Andrea The space of 4-vector valued functions is infinite-dimensional. The representation of the Lorentz (or Poincaré) group on this space is not unitary, and is "in spirit" just the finite-dimensional 4-vector representation as Cosmas explains here, but of course is technically infinite-dimensional because the space of functions is. Be careful: Just because every unitary representation must be infinite-dimensional, this doesn't mean every infinite-dimensional representation should be unitary. $\endgroup$
    – ACuriousMind
    Nov 22, 2021 at 23:03
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    $\begingroup$ Thanks to all the responders that helped me to have (hopefully) a better understanding. The message I takes is that that representation is actually a rep of the Lorentz algebra, technically is infinite-dimensional, but it's no irreducible. The irrep are the one provided as an example in Wi-Ki Tung $\endgroup$
    – Andrea
    Nov 23, 2021 at 12:09

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