Lorentz transform of contravariant 4-vector - one or two transformation matrices? I have a question regarding the Lorentz transformation of the position 4-vector. The contravariant 4-vector $x^{\mu}$ transforms infinitesimally like
$$x^{\mu}\rightarrow x^{\mu}-\frac{i}{2}\epsilon_{\nu\lambda}[J^{\nu\lambda}, x^{\nu}]=x^{\mu}+\epsilon_{\nu}^{\text{  }\mu}x^{\nu}$$
which leads to the following expression for the finite transform:
$$x^{\mu}\rightarrow \text{exp}\left(-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\right)\text{ }x^{\mu}\text{ }\text{exp}\left(\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}\right)=U^{-1}(\Lambda)x^{\mu}U(\Lambda)$$
My question is as follows: I often encounter expressions for the transform of this form:
$$x^{\mu}\rightarrow U^{-1}(\Lambda)x^{\mu}$$
without the final $U$-matrix. What is the difference between the $x$ sandwiched between two transform-matrices and the one with only one transform matrix in front? I suspect it has something to do with the fact that transformations often are written like $A=U^{-1}BU$, but I cannot seem to find an answer to this.
Anyone who can help me out here?
 A: In case of one  transformation matrix  we talk about a contravariant 4-vector represented by a $\,4\times1\,$ matrix that is by a one-column matrix
\begin{equation}
\mathbf X \boldsymbol{=}
\begin{bmatrix}
x_0 \\
x_1 \\
x_2 \\
x_3
\end{bmatrix}
 \boldsymbol{=}
\begin{bmatrix}
x_0 \\
\\
\mathbf x\\
\hphantom{=}
\end{bmatrix} 
\tag{01}\label{01}   
\end{equation}
But in case of two  transformation matrices may be we talk about a contravariant 4-vector represented by a $\,2\times2\,$ matrix, something like that
\begin{equation}
\mathrm X_{\boldsymbol{+}} \boldsymbol{=} 
\begin{bmatrix}
\hphantom{\boldsymbol{-}}x_0 \boldsymbol{-} x_3 &  \:\boldsymbol{-} x_1\boldsymbol{+}ix_2 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\
\:\boldsymbol{-}x_1\boldsymbol{-}ix_2 & \hphantom{-}x_0 \boldsymbol{+} x_3 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}
\end{bmatrix}\boldsymbol{=} x_0\mathrm I_{4}\boldsymbol{-}\mathbf x\boldsymbol{\cdot}\boldsymbol{\sigma} 
\tag{02}\label{02}   
\end{equation}
or that
\begin{equation}
\mathrm X_{\boldsymbol{-}} \boldsymbol{=} 
\begin{bmatrix}
\hphantom{\boldsymbol{-}}x_0 \boldsymbol{+} x_3 &  \:\hphantom{\boldsymbol{-}}x_1\boldsymbol{-}ix_2 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\
\:\hphantom{\boldsymbol{-}}x_1\boldsymbol{+}ix_2 & \hphantom{-}x_0 \boldsymbol{-} x_3 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}
\end{bmatrix}\boldsymbol{=} x_0\mathrm I_{4}\boldsymbol{+}\mathbf x\boldsymbol{\cdot}\boldsymbol{\sigma} 
\tag{03}\label{03}   
\end{equation}
that is(1)
\begin{equation}
\mathrm X_{\boldsymbol{\pm}} \boldsymbol{=} 
\begin{bmatrix}
\hphantom{\boldsymbol{-}}x_0 \boldsymbol{\mp} x_3 &  \:\boldsymbol{\mp} x_1\boldsymbol{\pm}ix_2 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\
\:\boldsymbol{\mp}x_1\boldsymbol{\mp}ix_2 & \hphantom{-}x_0 \boldsymbol{\pm} x_3 \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}
\end{bmatrix}\boldsymbol{=} x_0\mathrm I_{4}\boldsymbol{\mp}\mathbf x\boldsymbol{\cdot}\boldsymbol{\sigma} 
\tag{04}\label{04}   
\end{equation}
Observe that
\begin{align}
\det\mathrm X_{\boldsymbol{\pm}} &\boldsymbol{=}x^2_0\boldsymbol{-}\left(x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3\right) 
\tag{05a}\label{05a}\\
\mathrm{trace}\mathrm X_{\boldsymbol{\pm}} & \boldsymbol{=}2x_0
\tag{05b}\label{05b}
\end{align}
A Lorentz transformation keeps $\,\det\mathrm X_{\boldsymbol{\pm}}\, $ invariant while a rotation keeps $\,\mathrm{trace}\mathrm X_{\boldsymbol{\pm}}\,$ invariant.
Also under a reflection through the origin
\begin{equation}
\mathrm X_{\boldsymbol{+}} \boldsymbol{\longleftrightarrow}\mathrm X_{\boldsymbol{-}} 
\tag{06}\label{06}   
\end{equation}
Note that  this 4-dimensional case is the analog of the 3-dimensional one. If $\mathbf x \in \mathbb R^3$ is represented by a $\,3\times 1\,$ matrix
\begin{equation}
\mathbf x \boldsymbol{=}
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
x_2 \vphantom{\dfrac{a}{b}}\\
x_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{07}\label{07}   
\end{equation}
then a pure rotation is represented by a $\,3\times 3\,$ real matrix
\begin{equation}
\mathbf x' \boldsymbol{=}
\begin{bmatrix}
x'_1 \vphantom{\dfrac{a}{b}}\\
x'_2 \vphantom{\dfrac{a}{b}}\\
x'_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
R_{11} & R_{12} & R_{13} \vphantom{\dfrac{a}{b}}\\ 
R_{21} & R_{22} & R_{23} \vphantom{\dfrac{a}{b}}\\ 
R_{31} & R_{32} & R_{33} \vphantom{\dfrac{a}{b}} 
\end{bmatrix}
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
x_2 \vphantom{\dfrac{a}{b}}\\
x_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}\mathcal R\,\mathbf x
\tag{08}\label{08} 
\end{equation}
But if this $\mathbf x \in \mathbb R^3$ is represented by an hermitian traceless $\,2\times 2\,$ matrix
\begin{align}
\mathrm X & \boldsymbol{=}
\begin{bmatrix}
x_3 & x_1\boldsymbol{-}ix_2 \vphantom{\dfrac{a}{b}}\\ 
x_1\boldsymbol{+}ix_2 & \boldsymbol{-}x_3  \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}x_1
\begin{bmatrix}
0 &  \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\dfrac{a}{b}}\\
1 &  \!\!\hphantom{\boldsymbol{-}}0\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{+}x_2
\begin{bmatrix}
0 & \!\!\boldsymbol{-} i \vphantom{\dfrac{a}{b}}\\
i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{+}x_3
\begin{bmatrix}
1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\dfrac{a}{b}}\\
0 & \!\!\boldsymbol{-} 1\vphantom{\dfrac{a}{b}}
\end{bmatrix}
\nonumber\\
& \boldsymbol{=} x_1\,\sigma_1\boldsymbol{+}x_2\,\sigma_2\boldsymbol{+}x_3\,\sigma_3\boldsymbol{=} \mathbf x\boldsymbol{\cdot}\boldsymbol{\sigma}  
\tag{09}\label{09}   
\end{align}
then the rotation is represented by
\begin{equation}
\mathrm X' \boldsymbol{=}
\mathrm U \,\mathrm X\,\mathrm U^{\boldsymbol{*}}
\tag{10}\label{10}   
\end{equation}
where $\,\mathrm U \in SU(2)\,$ a special unitary $\,2\times 2\,$ matrix.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
(1)
Reference : $'$Concepts of Particle Physics, Vol.II$'$ by K.Gottfried-V.F.Weisskopf, Appendix III THE DIRAC FIELD.

A: You are confusing two different transfomations. The correct statement is the one in  @nwolijin's  comment
$$
U(\Lambda)^{-1} \hat x^\mu U(\Lambda)= {\Lambda^\mu}_\nu \hat x^\nu
$$
here
$$
U(\Lambda) =exp\left\{ \frac 12 \omega_{\mu\nu}J^{\mu\nu}\right\}
$$
is acting on the infinite-dimensional Hilbert space of functions $\phi(x)$ in which the opertor  $\hat x^\mu$ acts by
$$
\hat x^\mu: \phi(x) \mapsto x^\mu \phi(x).
$$
The ${\Lambda^\mu}_\nu$ is a $d$-by-$d$ finite Lorentz transform matrix acting on the finite-dimensional Minkowski space in which $x^\mu$ are the numerical components of a $d$-vector.
A good excercie to see if you have understood this is to to verify that
$$
U(\Lambda_2)^{-1} U(\Lambda_1)^{-1} \hat x^\mu U(\Lambda_1)U(\Lambda_2)= {\Lambda^\mu_1}_\sigma {\Lambda^\sigma_2}_\nu\hat x^\nu, 
$$
so that $U(\Lambda_1)U(\Lambda_2)= U(\Lambda_1\Lambda_2)$. If you are confused by the difference between $x^\mu$ and $\hat x^\mu$ you  will get the $\Lambda$ matrices in the wrong order.
