Lorentz transformation of Hilbert space operators

Question

I want to address a confusion that has arisen in certain questions of mine on this site, which you can find "here" and "here".

My confusion is related to the transformation under Lorentz of an operator $O$ defined on the Hilbert space of states $\mathcal{H}$, $O\in \mathcal{B(H)}$. Because Lorentz transformation are a symmetry of space-time, by "Wigner's theorem", they are represented on $\mathcal{H}$ by a unitary or anti-unitary operator. Consider now how the operator $O$ transform under such a unitary representation of Lorentz; if my understanding is correct, "this post" gives the transformation as

\begin{equation}\tag{1} O\mapsto U(\Lambda)OU^\dagger(\Lambda), \end{equation}

where $U(\Lambda)\equiv U(g(\Lambda,0)),\:g\in ISO(3,1),$ is the unitary representation of $\Lambda$ on $\mathcal{H}$. Let us now consider a quantized vector field $\hat{A}\,^{\mu}(x)$. One should then have, in view of (1), that

\begin{equation}\tag{2} U(\Lambda)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda) = \Lambda^\mu_{\,\nu}A^\nu(x). \end{equation}

Using (2), I want to verify

\begin{equation}\tag{3} U(\Lambda_1\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_1\Lambda_2)= (\Lambda_1 \Lambda_2)^\mu_{\,\nu}\hat{A}\,^{\nu}(x), \end{equation}

which should hold due to $U(\Lambda)$ being a representation of the Lorentz group. Let me now introduce the following identities which I will use later on;

\begin{equation}\tag{4} U^\dagger(\Lambda_1\Lambda_2)=U(g^{-1}(\Lambda_1\Lambda_2,0))=U(g(\Lambda_2^{-1}\Lambda_1^{-1},0)) = U(\Lambda_2^{-1})U(\Lambda_1^{-1}) = U^\dagger(\Lambda_2)U^\dagger(\Lambda_1). \end{equation}

Using (2) and (4) into the LHS of (3) one obtains

\begin{equation}\tag{5} U(\Lambda_1\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_1\Lambda_2)= U(\Lambda_1)\left(U(\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_2)\right)U^\dagger(\Lambda_1). \end{equation}

Using (2) for the term inside the parenthesis

\begin{equation}\tag{6} U(\Lambda_1\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_1\Lambda_2)= U(\Lambda_1)\Lambda^{\mu}_{2\,\nu}\hat{A}\,^{\nu}(x)U^\dagger(\Lambda_1). \end{equation}

$\Lambda^{\mu}_{2\,\nu}$ are numbers, so we can commute them past $U(\Lambda_1)$, and obtain

\begin{equation}\tag{7} U(\Lambda_1\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_1\Lambda_2)=\Lambda^{\mu}_{2\,\nu}\left(U(\Lambda_1)\hat{A}\,^{\nu}(x)U^\dagger(\Lambda_1)\right) \end{equation}

Finally, using (2) for the term in parenthesis, one gets

\begin{equation}\tag{8} U(\Lambda_1\Lambda_2)\hat{A}\,^{\mu}(x')U^\dagger(\Lambda_1\Lambda_2) = (\Lambda_2\Lambda_1)^{\mu}_{\,\sigma}\hat{A}\,^{\sigma}(x), \end{equation}

where I used that $\Lambda^\mu_{2\,\nu}\Lambda^\nu_{1\,\sigma}=(\Lambda_2\Lambda_1)^\mu_{\,\sigma}$.

This result doesn't match (3), thus it would seem it is incorrect. When doing the same computation with the transformation (1) rewritten as $U^\dagger(\Lambda)OU(\Lambda)$, I obtain a coherent result with the representation properties. Here is my question; how to correctly transform an operator acting on $\mathcal{H}$ under the Lorentz group? Should it be (1), or the modified version I just gave above?

Answer 1

Since the question is about general transformation properties of operators in Hilbert space, it would be easier to consider a simpler case of a vector operator in ordinary quantum mechanics. Consider for example the momentum operator $ \mathbf{\hat P} = (\hat P_x,\hat P_y,\hat P_z)$, and define a rotated momentum operator $ \mathbf{\hat P'}$ by applying some rotation matrix $O$:

$$ \mathbf{\hat P'} = O \mathbf{\hat P} .$$

Let $|\mathbf{p}\rangle$ denote eigenstates of $ \mathbf{\hat P}$ with eigenvalues $\mathbf{p}$, namely $\mathbf{\hat P}|\mathbf{p}\rangle = \mathbf{p}|\mathbf{p}\rangle$ (which should be understood as three equations, one for each momentum component). Now we have

$$ \mathbf{\hat P'} |\mathbf{p}\rangle = O \mathbf{\hat P} |\mathbf{p}\rangle = O \mathbf{p}|\mathbf{p}\rangle \equiv \mathbf{p}'|\mathbf{p}\rangle$$

Where we defined $\mathbf{p}' = O \mathbf{p}$. In other words $|\mathbf{p}\rangle$ is also an eigenstate of $ \mathbf{\hat P'}$ but with different eigenvalues $\mathbf{p}'$. We can therefore define a unitary operator that sends $|\mathbf{p}\rangle$ to $|\mathbf{p}'\rangle$, that is $U |\mathbf{p}\rangle=|\mathbf{p}'\rangle$ so we can write

$$\mathbf{\hat P'} U^\dagger |\mathbf{p'}\rangle=U^\dagger\mathbf{p}'|\mathbf{p'}\rangle$$

or equivalently

$$U \mathbf{\hat P'} U^\dagger |\mathbf{p'}\rangle=\mathbf{p}'|\mathbf{p'}\rangle$$

which implies that $U \mathbf{\hat P'} U^\dagger = \mathbf{\hat P}$ (since the above equation holds for all $\mathbf{p'}$) and therefore $\mathbf{\hat P'} = U^\dagger\mathbf{\hat P}U$.

Now we can repeat exactly the same calculation with a rotation that is composed from two successive rotations, $O = O_1 O_2$. The unitary operator is defined by $U |\mathbf{p}\rangle=|O_1O_2\mathbf{p}\rangle = U_1|O_2\mathbf{p}\rangle = U_1 U_2 |\mathbf{p}\rangle$ (where $U_{1,2}$ are the corresponding operators for rotations $O_{1,2}$), i.e. it is a representation of the rotations group.

Note that whether we call the unitary transformation $U$ or $U^\dagger$ is of course entirely a matter of choice. Had we defined it the other way around, then $U$ would have been the dual representation.