Group representation acting on operators (QFT)

Question

I have found in many texts the following statement:

Let $T_g$ be a representation of a group (of transformations, e.g. rotations, translations, Lorentz transformations ) acting on a given Hilbert space H. Then $T_g$ acts on the states of $H$ as $T_g|\varphi\rangle$ and on operators $T_g^{-1} A T_g$ where $A$ is any linear operator in H.

Starting from the expression for operators, and taking the infinitesimal generators F of the representation, the expression can be written as $(1-\delta\omega F)A(1+\delta\omega F)$ and from this many commutation relations are derived

The question is why the expression for operators is $T_g^{-1} A T_g$ and not the other way around $T_g A T_g^{-1}$?. I can not find any plausible explanation.

May be is so simple that I can no see it. Hope someone can help me.

Answer 1

It's a matter of convention (basically the active v. passive distinction as mentioned in a comment by user jabirali). The punchline is that if the states transform as $|\psi\rangle \to T_g|\psi\rangle$, then operators should transform as $A\to T_gAT_g^{-1}$, and vice versa.

To see why, suppose that states are taken to transform as $|\psi\rangle\to T_g|\psi\rangle$, and consider a state of the form $A|\phi\rangle$ for some state $|\phi\rangle$. On one hand, the full state $A|\phi\rangle$ should transform as follows: \begin{align} A|\phi\rangle\to T_gA|\phi\rangle \end{align} On the other hand, we could attempt to transform $A|\phi\rangle$ by transforming $A$ and $|\phi\rangle$ separately in some way. Let's call $A_g$ the transformed operator, then we would have \begin{align} A|\phi\rangle \to A_gT_g|\phi\rangle. \end{align} Now we could ask what $A_g$ needs to be for these two procedures to agree, namely so that \begin{align} A_gT_g|\phi\rangle = T_gA|\phi\rangle \end{align} This holds for all $|\phi\rangle$ provided \begin{align} A_gT_g = T_gA \end{align} which implies \begin{align} A_g = T_gAT_g^{-1} \end{align} as desired.

See, for example, Peskin and Schroeder p. 59 where $U(\Lambda)$ implements Lorents transformations on states, and in eq. 3.108, the operator-valued fields transform as $U\psi(x)U^{-1}$.

Answer 2

Whenever the Hilbert space of a representation $\pi$ of a $G$-algebra admits a true strongly continuous unitary representation of the group $G$, then this representation, say $u$ is covariant w.r.t. $\pi$ in the sense that $$u_g\pi(a)u_g^* = \pi(\alpha_g(a)),$$ where $a$ is any element of the algebra, $g$ is any element of the group and $\alpha$ is the action of $G$ on the algebra. In order for all the states in the same superselection state (here $\pi$ is assumed to be $\alpha$-regular, hence irreducible) to give coherent "outcomes", the representation $u$ must transform them according to $\xi\mapsto u_g\xi$.