# Group representation acting on operators (QFT)

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.

• I think the difference between $T^{-1}AT$ and $TAT^{-1}$ is the difference between active and passive transformations, i.e. whether you mathematically transform the actual physical system, or just transform the coordinates used to describe the system. – jabirali Jan 14 '15 at 0:34
• Many thanks, jabirali. I'd rather see all the texts using the same convention. Best Regards. – edel Jan 14 '15 at 9:23
• I am adding this link on the subject of active vs passive transformations: [link] (physics.stackexchange.com/questions/51994/…) – edel Jan 14 '15 at 17:08
• Another link on the subject. Yes, my question was also originated while reading Srednicki's text: [link] (physics.stackexchange.com/questions/62000/…) – edel Jan 14 '15 at 17:09
• Another link on the subject that uses the first form: [link] (physics.stackexchange.com/questions/113639/…). According to what have been said, is this an example of the active transformation convention? – edel Jan 15 '15 at 11:02

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 $U\psi(x)U^{-1}$.

• Many thanks for your answer, joshphysics. It is a very neat derivation. The comment of jaribali clarifies the difference between the two forms. Unfortunately this and other conventions (e.g. Minkowski metric signature) vary from text to text, and this makes our life harder. Best Regards. – edel Jan 14 '15 at 9:21
• @edel Sure thing. Yes it's quite irritating that there isn't notational uniformity in the literature. – joshphysics Jan 14 '15 at 17:37

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 $\xi\mapsto u_g\xi$.

• Hi, There is something in your answer that calls for more – edel Jan 14 '15 at 23:43
• @edel what exactly? – Phoenix87 Jan 14 '15 at 23:48
• It seems to provide a solid mathematical ground for this discussion. Is there any text where I can find more about the statement and a proof?. On the other hand, if $u_g$ is unitary then $u_g^* = u_g^{-1}$ and we are left again with the second form. Regards – edel Jan 14 '15 at 23:53
• Bogoliubov et al., General Principles of Quantum Field Theory. – Phoenix87 Jan 14 '15 at 23:57
• @Phoenix87 I just checked out your book recommendation. What a remarkable text! I'm quite sad I'd never seen it until now but glad I came across you comment. – joshphysics Jan 15 '15 at 21:37