What are the unitary operators for various transformation?

Question

Transformations, at least in lagrangian-symmetries context, are usualy described as uintary operators. I dont understand what are these operators exactly.

For example, let's look at the Lorentz transform of 4-vector. All the books state the next equation,

$U^{-1}(\Lambda) A^\mu U(\Lambda) = {\Lambda^\mu}_\nu A^\nu$.

No, I perfectly understand the right-hand side. I know to write $\Lambda$, whether in infinitesimal or finite form. What I dont understand is what is going on on the left-hand side. Can we actually write $U(\Lambda)$ or is it just an abstract way to say that we expect unitary operator? Is it a differential operator (like $x_\mu\partial_\nu-x_\nu\partial_\mu$)? Is it the representation of the actual operator?

One of the reasons it's difficault for me to understand it, is the next observation - on the RHS I see that the components of A "mix", when transformed, but on the LHS there's only a single component, so nothing can mix it there.

~ thanks a lot ~

Answer 1

It is a requirement one imposes on physical representations of the algebra of observables, known as $\alpha$-regularity, where $\alpha$ denotes the action of a certain group over the algebra. In more concrete terms, the group involved here is the Lorentz group (but more generally is the Poincaré group). Under certain hypotheses (see Wigner 1931 and Bargmann 1954) projective representations of the group can be lifted to genuine (unitary) representations $U$ that satisfy to the covariance relation with the chosen representation $\pi$ of the algebra, viz. $$U(g)\pi(A)U(g)^* = \pi(\alpha_g(A))$$ for any element $g$ of the group and any element $A$ of the algebra. If $A$ is a 4-vector in the tangent space to Minkowski space-time at a point, then its transformation law is $\alpha_\Lambda(A) = \Lambda A$, so that at the level of the representation this is implemented by a unitary $U(\Lambda)$ such that $$U(\Lambda)\pi(A)U(\Lambda)^* = \pi(\Lambda A).$$ The mixing of indices in this case is irrelevant in the sense that the unitary operator is taking care of that through covariance.

Answer 2

The 4-vector $A^{\mu}$ is an operator which acts on a Hilbert space with states $|a\rangle$. These things are called tensor operators - see chapter 4 of Howard Georgi's book Lie Algebras in Particle Physics. So, they have a matrix representation $\langle a|A^{\mu}|b\rangle$ which I'll write as $A^{\mu a}_{\ \ \ b}=\langle a|A^{\mu}|b\rangle$ to emphasize the labels that are usually suppressed. The complete object $A^{\mu a}_{\ \ \ b}$ is assumed to transform trivially under the Lorentz group, \begin{equation} [U(g)A]^{\mu a}_{\ \ \ b}=A^{\mu a}_{\ \ \ b} \end{equation} Let's transform the complete object on the LHS. \begin{equation} [U(g)A]^{\mu a}_{\ \ \ b}=\Lambda^{\mu}_{\ \nu}[U(g)]^{a}_{\ c}[U(g^{-T})]_{b}^{\ d}A^{\nu c}_{\ \ \ d}=\Lambda^{\mu}_{\ \nu}[U(g)]^{a}_{\ c}[U(g^{-1})]^{d}_{\ b}A^{\nu c}_{\ \ \ d}=A^{\mu a}_{\ \ \ b} \end{equation} Here, the superscript $T$ denotes matrix transpose. $[U(g)]^{a}_{\ b}=\langle a|U(g)|b\rangle$ are the matrices of the unitary rep of the Lorentz group and $\Lambda(g)^{\mu}_{\ \nu}$ are the matrices of the defining rep carried on the space of 4-vectors. Using the rightmost equality, move the two unitary matrices onto the RHS by multiplying by their respective inverses. \begin{equation} [U(g^{-1})]^{r}_{\ a}\Lambda^{\mu}_{\ \nu}[U(g)]^{a}_{\ c}[U(g^{-1})]^{d}_{\ b}A^{\nu c}_{\ \ \ d}[U(g)]^{b}_{\ s}=[U(g^{-1})]^{r}_{\ a}A^{\mu a}_{\ \ \ b}[U(g)]^{b}_{\ s} \end{equation} \begin{equation} \delta^{r}_{\ c}\Lambda^{\mu}_{\ \nu}\delta^{d}_{\ s}A^{\nu c}_{\ \ \ d}=[U(g^{-1})]^{r}_{\ a}A^{\mu a}_{\ \ \ b}[U(g)]^{b}_{\ s} \end{equation} \begin{equation} \Lambda^{\mu}_{\ \nu}A^{\nu r}_{\ \ \ s}=[U(g^{-1})]^{r}_{\ a}A^{\mu a}_{\ \ \ b}[U(g)]^{b}_{\ s} \end{equation} Now suppress the operator labels $r,s,a,b$, \begin{equation} \Lambda^{\mu}_{\ \nu}A^{\nu}=U(g^{-1})A^{\mu}U(g) \end{equation} and we recover the way a tensor operator transforms as in the question.