Infinite dimensional representation of Lorentz group Given a multicomponent field $\Phi_a$ we have the transformation law
$$
\Phi_a(x) \rightarrow M_{ab}(\Lambda) \Phi_b(\Lambda^{-1} x)
$$
Where $M_{ab}$ is some finite dimensional representation of the Lorentz group. I want to understand the transformation of the coordinate $\Lambda^{-1} x$. In my notes it's claimed without proof that we can expand this coordinate transformation
$$
\psi(\Lambda^{-1}x) = e^{-i \omega_{\mu\nu} L^{\mu \nu}/2} \psi(x)
$$
where
$$
L^{\mu \nu} = i(x^\mu \partial^\nu - x^\nu \partial^\mu)
$$
I want to know how these generators $L^{\mu \nu}$ are derived ?
 A: I divided the answer into 3 sections, the actual derivation of the expression for $L^{\mu\nu}$ is in Part III but some useful equations are given or deduced before.
Part I
(Some comments about the comments, may contain judgement errors so feel free to correct)
I would just like to further dig into some aspects mentioned in the comments: concerning whether, in general, any element of the Lorentz Lie group can be expressed as an exponential of the corresponding Lorentz Lie algebra generators. It is indeed true that, since the Lorentz group is neither compact nor simply-connected, this may not be done in a naive way. In physics, it is customary to work with representations. One is often interested in irreducible representations of the Lorentz group. They may either by finite dimensional and non-unitary or infinite dimensional and unitary. But how to obtain such representations, if exponentiation fails?
For a Lorentz transformation $\Lambda$, the interest is to find the representations $\mathsf{D}(\Lambda)$ of the Lorentz group. It's easier to find the representation for the corresponding Lie algebra. This is done by working infinitesimaly, expanding around identity by a small deviation $\Lambda=1+\omega$. An infinitesimal Lorentz transformation is given by
$$\Lambda^\mu_{\phantom{\mu}\nu}\approx\delta^\mu_{\phantom{\mu}\nu}+\omega^\mu_{\phantom{\mu}\nu}\tag{1}\label{infinitezimallambda}$$
The parameters of the transformation
$$\omega_{\mu\nu}\equiv\eta_{\mu\rho}\omega^\rho_{\phantom{\rho}\nu}\tag{2}\label{omegamunu}$$
are antisymmetric $\omega_{\mu\nu}=-\omega_{\nu\mu}$ (this may be proven by using the definition relation for a Lorentz transformation, namely $\eta_{\rho\sigma}=\eta_{\mu\nu}\Lambda^\mu_{\phantom{\mu}\rho}\Lambda^\nu_{\phantom{\nu}\sigma}$). The representation of the corresponding Lie algebra is given by
$$\mathsf{D}(1+\omega)\approx1-\frac{\mathrm{i}}{2}\omega_{\mu\nu}M^{\mu\nu}\tag{3}\label{algebrarepr}$$
where the generators $M^{\mu\nu}$ of $\mathfrak{so}(1,3)$ obey the generic commutation relations (these may actually be proven but it's a little lengthy)
$$\left[M^{\mu\nu},M^{\rho\sigma}\right]=\mathrm{i}\left(\eta^{\nu\rho}M^{\mu\sigma}-\eta^{\mu\rho}M^{\nu\sigma}-\eta^{\nu\sigma}M^{\mu\rho}+\eta^{\mu\sigma}M^{\nu\rho}\right)\tag{4}\label{lorentzliealgebra}$$
Once you put your hand on a representation, by finding some particular $M^{\mu\nu}$ which obey \eqref{lorentzliealgebra}, you are then interested to obtain the associated Lie group representation. Since the Lorentz group is not simply connected, this can't be done via exponentiation. Nevertheless, the proper orthochronous Lorentz group $\mathsf{SO}(1,3)_+^\uparrow$ has $\mathsf{SL}(2,\mathbb{C})$ as a (double) covering group, which is simply connected. In general, for a group which is not connected, if there exists a universal covering group, it is simply connected and shares the same Lie algebra. Therefore, from a certain representation of the Lie algebra $\mathfrak{so}(1,3)_+^\uparrow$, which will be isomorphic to some representation of $\mathfrak{sl}(2,\mathbb{C})$, one can obtain the representation for the group via exponentiation. Therefore, \eqref{algebrarepr} gives through exponentiation
$$\mathsf{D}(\Lambda)=\mathsf{exp}\left\{-\frac{\mathrm{i}}{2}\omega_{\mu\nu}M^{\mu\nu}\right\}\tag{5}\label{dlambda}$$
but remember that this is the representation of the covering group $\mathsf{SL}(2,\mathbb{C})$ and not of the Lorentz group $\mathsf{SO}(1,3)$.
Maybe the following links can be useful for more detials: Representations of Lorentz Group and Why do we classify states under covering groups instead of the group itself?
Part II
(Some prerequisites, skip if already familiar)
In this part I will deduce a result which will be useful in answering your question but if you already know this, don't bother too much with these computations.
An arbitrary four-vector transforms under Lorentz as
$$V^\mu\rightarrow\Lambda^\mu_{\phantom{\mu}\nu}V^\nu$$
In this case, $\mathsf{D}(\Lambda)=\Lambda$. Infinitesimally, this translates via \eqref{algebrarepr} to
$$\delta V^\mu\approx -\frac{\mathrm{i}}{2}\omega_{\rho\sigma}\left(M^{\rho\sigma}\right)^\mu_{\phantom{\mu}\nu}V^\nu\tag{6}\label{deltavmu1}$$
whereas using \eqref{infinitezimallambda} yields
$$\delta V^\mu=\omega^\mu_{\phantom{\mu}\nu}V^\nu$$
Simple manipulations allows one to express
$$\omega^\mu_{\phantom{\mu}\nu}=\frac{1}{2}(\underbrace{\omega^\mu_{\phantom{\mu}\sigma}}_{\displaystyle \omega_{\rho\sigma}\eta^{\rho\mu}}\delta^\sigma_{\phantom{\sigma}\nu}+\underbrace{\omega^\mu_{\phantom{\mu}\rho}}_{\displaystyle\omega_{\sigma\rho}\eta^{\sigma\mu}}\delta^\rho_{\phantom{\rho}\nu})=\frac{1}{2}(\omega_{\rho\sigma}\eta^{\rho\mu}\delta^\sigma_{\phantom{\sigma}\nu}+\underbrace{\omega_{\sigma\rho}}_{\displaystyle -\omega_{\rho\sigma}}\eta^{\sigma\mu}\delta^\rho_{\phantom{\rho}\nu})$$
which then gives
$$\delta V^\mu=\frac{1}{2}\omega_{\rho\sigma}\left(\eta^{\rho\mu}\delta^\sigma_{\phantom{\sigma}\nu}-\eta^{\sigma\mu}\delta^\rho_{\phantom{\rho}\nu}\right)V^\nu\tag{7}\label{deltavmu2}$$
By direct comparison of \eqref{deltavmu1} with \eqref{deltavmu2} one obtains
$$\left(M^{\rho\sigma}\right)^\mu_{\phantom{\mu}\nu}=\mathrm{i}\left(\eta^{\rho\mu}\delta^\sigma_{\phantom{\sigma}\nu}-\eta^{\sigma\mu}\delta^\rho_{\phantom{\rho}\nu}\right)\tag{8}\label{mmunu}$$
which represents a finite dimensional representation for the Lorentz algebra. This result will turn out to be useful in the actual proof for your question.
Part III
(The actual answer: deducing the expression for $L^{\mu\nu}$ without pulling it out of the hat)
In a similar way as above, one may treat representations for other types of coordinate-independent entities, besides four-vectors (scalars, spinors, tensors, for example). Nevertheless, in field theory, the fields also depend on space-time coordinates. Therefore, when studying a Lorentz transformation upon a field (which may itself be a scalar, spinor, tensor and so on), the coordinate transformation must also be considered. Let's study the most simple case, a scalar field $\psi(x)$. Under a Lorentz transformation $x^\prime=\Lambda x$, it remains unchanged
$$\psi^\prime(x^\prime)=\psi(x)\quad\Leftrightarrow\quad \psi^\prime(x)=\psi(\Lambda^{-1}x) $$
When attempting to do infinitesimal variations of the fields, one finds that there are actually two possible manners of defining these small variations. The standard one
$$\delta\psi\equiv\psi^\prime(x^\prime)-\psi(x)=0$$
is obviously null for a scalar field. A second option is to consider variations at fixed space-time coordinate as
$$\delta_x\psi\equiv\psi^\prime(x)-\psi(x)=\psi^\prime(\Lambda^{-1}x)-\psi(x)\tag{9}\label{deltax}$$
In a sense, this measures not how the fields change under Lorentz but rather how their functional dependence of coordinates behaves under a Lorentz transformation.
For an infinitesimal Lorentz transformation $x^{\prime\mu}\approx x^\mu+\delta x^\mu$, this may be expanded as
$$\delta_x\psi\approx\psi^\prime(x^\prime-\delta x)-\psi(x)\approx \underbrace{\psi^\prime(x^\prime)}_{\displaystyle \psi(x)}-\delta x^\mu\partial_\mu\underbrace{\psi^\prime(x^\prime)}_{\displaystyle \psi(x)}-\psi(x)=-\delta x^\mu \partial_\mu \psi(x)$$
The relation \eqref{deltavmu1} expresses how a four-vector transforms under Lorentz and it may be used to further replace $\delta x^\mu$ in the above expression as
$$\delta_x\psi=+\frac{\mathrm{i}}{2}\omega_{\rho\sigma}\left(M^{\rho\sigma}\right)^\mu_{\phantom{\mu}\nu}x^\nu\partial_\mu \psi\stackrel{\eqref{mmunu}}=-\frac{\mathrm{i}}{2}\omega_{\rho\sigma}\mathrm{i}\left(\eta^{\sigma\mu}\delta^\rho_{\phantom{\rho}\nu}-\eta^{\rho\mu}\delta^\sigma_{\phantom{\sigma}\nu}\right)x^\nu\partial_\mu\psi$$
Finally, you may see how the operator $L^{\rho\sigma}$ is introduced
$$\delta_x\psi=-\frac{\mathrm{i}}{2}\omega_{\rho\sigma}\underbrace{\mathrm{i}(x^\rho\partial^\sigma-x^\sigma\partial^\rho)}_{\displaystyle \equiv L^{\rho\sigma}}\psi\tag{10}\label{deltaxl}$$
You can actually check that the operators $L^{\rho\sigma}$ defined in this way satisfy the commutation relations \eqref{lorentzliealgebra}, thus they are the generators of a Lorentz Lie algebra. The corresponding representation is an infinite dimensional one.
Collecting the results from \eqref{deltax} and \eqref{deltaxl}, the infinitesimal Lorentz transformation of the field is
$$\delta_x\psi=\psi^\prime(\Lambda^{-1}x)-\psi(x)=-\frac{\mathrm{i}}{2}\omega_{\mu\nu}L^{\mu\nu}\psi$$
and thus, the whole transformation becomes, using the same line of argumentation as for why it's possible to write the exponential from \eqref{dlambda}
$$\psi^\prime(\Lambda^{-1}x)=\mathsf{exp}\left\{-\frac{\mathrm{i}}{2}\omega_{\mu\nu}L^{\mu\nu}\right\}\psi(x)$$
Notice that this formula differs from yours by the prime from the first field (I also checked with the same result from Chapter 3.7.11 in Physics From Symmetry by J. Schwichtenberg and there it's also written with a prime, but $\Lambda$ instead of $\Lambda^{-1}$ since it uses a different convention)
If interested, you may further study Chapter 2.6 from M. Maggiore's Quantum Field Theory or briefly read an Appendix from G. Eichmann's notes, available at http://cftp.ist.utl.pt/~gernot.eichmann/2014-hadron-physics/hadron-app-2.pdf .
