Lorentz transformation of momentum operator

Question

I encountered an element of my QFT lecture notes related to representations of the Poincaré group which is not clear to me.

Consider a representation of the Poincaré group $ISO(3,1)$ on the Hilbert space of states $\mathcal{H}$, such that for any $g(\Lambda,a)\in ISO(3,1)$, it can be represented by a unitary operator $U(g(\Lambda,a))\equiv U(\Lambda,a)$, and $U(g(\Lambda,0))\equiv U(\Lambda)$.

Then consider the Hermitian operator $P^\mu$ under Lorentz transformation;

\begin{equation}\tag{1} U^\dagger(\Lambda)P^\mu U(\Lambda)=\Lambda^\mu_{\:\nu} P^\nu, \end{equation}

where a summation over $\nu$ is intended. Here is my question; why does the action of a Lorentz transformation when represented on a Hilbert space takes the form given by (1)?

My current understanding is that we are considering an adjoint representation, hence the form. But I am not entirely convinced; when looking at the "Wikipedia definition" of an adjoint representation, it is indeed given by $\Psi_g(h)=ghg^{-1}$, for $g,h$ in a Lie group. So (1) seems to be inverted with respect to the definition I just gave. But this definition is the adjoint action for elements of a Lie group, while $P^\mu$ is an element of a Lie algebra if I am correct, thus it doesn't make much sense to consider it.

Answer 1

You are right about everything you say. For the inverted order of the elements in the adjoint representation, zou can however just rename the elements. Since $g^{-1}$ is also an element of the group you can just rename $g':=g^{-1}$ and then write the adjoint representation as $\Psi_{g'}(h)=g'hg'^{-1}=g^{-1}hg$, which shows that it really is a matter of definition. For your other question regarding the definition of the adjoint representation: while it is true that the adjoint representation is an action of the group on itself, you can take the pushforward of this group representation to get a representation of the corresponding Lie algebra, which is also called the adjoint representation, but now it's the action of the Lie group acting on its own Lie algebra. This justifies the validity of the left hand side of equation (1). Concerning the right hand side, since the adjoint (algebra) representation acts on the Lie algebra, it gives us again an element of the Lie algebra. This element can then again be represented as a linear combination of the generators of the Lie algebra, hence the right hand side. I hope this makes more sense now.

Answer 2

You have to have the inverse $U(\Lambda)$ on the left $$ U(\Lambda)^{-1} P^\mu U(\Lambda) = {\Lambda^\mu}_\nu P^\nu, $$ if you want to be consistent with the representation property $$ \Lambda_2\Lambda_1\leftrightarrow U(\Lambda_2\Lambda_1)= U(\Lambda_2)U(\Lambda_1). $$ To see this work out the composition of two maps $$ U(\Lambda_2 \Lambda_1)^{-1} P^\mu \,U(\Lambda_2 \Lambda_1)\\ = U( \Lambda_1)^{-1} U(\Lambda_2)^{-1} P^\mu \,U(\Lambda_2)U( \Lambda_1)\\ = U( \Lambda_1)^{-1} ({\Lambda_2^\mu}_\nu P^\nu) U( \Lambda_1)\\ ={\Lambda_2^\mu}_\nu U( \Lambda_1)^{-1} P^\nu U( \Lambda_1)\\ ={\Lambda_2^\mu}_\nu {\Lambda_1^\nu}_\lambda P^\lambda\\ = {(\Lambda_2 \Lambda_1)^\mu}_\lambda P^\lambda $$ In passing from the third to fourth line we have remembered that the numbers ${\Lambda_1^\mu}_\nu$ can be commuted past $U(\Lambda)$ because $U(\Lambda)$ is linear map.

For the Adjoint representation of a Lie group on its Lie algebra, we usually write $$ {\rm Ad}(g)X_i\equiv gX_i g^{-1}= X_j {[{\rm Ad}(g)]^j}_i $$ in which we write the Adjoint matrix elements on the right as in $$ A|n\rangle= |m\rangle \langle m|A|n\rangle. $$ This is also consistent with the represenation property.