Show Energy-momentum operator transforms as a tensor under Lorentz transformations

Question

I know, from my professor notes, that a general field operator can transform under Lorentz as $$U(\Lambda)\hat{\mathcal{O}}^r(x)U^\dagger(\Lambda)=\hat{\cal{O}}^{r'} {M(\Lambda)_{r'}}^r$$ where $M(\Lambda_1\Lambda_2)=M(\Lambda_1)M(\Lambda_2)$. Where does this relation come from?

Then I have to show that $$U(\Lambda)\hat{\tau}_{\mu\nu}U^\dagger(\Lambda)=\hat{\tau}_{\mu'\nu'} {\Lambda^{\mu'}}_{\mu}{\Lambda^{\nu'}}_\nu$$ namely, this transforms as a tensor.

I started at the infinitesimal level where $${\Lambda^\mu}_\nu=\delta^\mu_\nu-\omega^\mu_\nu$$ and $$U(\Lambda)=\mathbf{1}-\frac{1}{2}i\omega_{\mu\nu} \hat{M}^{\mu\nu}$$ I don't know how to proceed with this calculation. I tried substituting these two relation for the infinitesimal level but I can't reach any good result.

Answer 1

In regards to your first question, that is, in fact, a definition. To appreciate it, let's recall what a representation is. Let $G$ be some group. A representation of $G$ is a pair $(V,D)$ where $V$ is some vector space and where $D:{G}\to{\rm GL}(V)$ is a map associating to every group element $g$ one linear transformation $D(g)$ satisfying the condition: $$D(g_1)D(g_2)=D(g_1g_2).$$

The idea of a representation is simple. A group, by itself, is one abstract entity which encodes a set of elements with a composition law. A representation is a way in which these objects can actually be realized as concrete transformations acting linearly on a vector space. As you may imagine for the same group (i.e., the same underlying abstract pattern of composition) there are many actual concrete realizations. That is the case for the Lorentz group where we can find scalar, vector, tensor and spinor representations.

Now, let $V$ be one vector space carrying a representation $D:{\rm SO}(1,3)\to {\rm GL}(V)$ of the Lorentz group ${\rm SO}(1,3)$. A quantum field transforming in the representation $R$ is any set of operators $\Phi^a(x)$ where $a=1,\dots, \dim V$, such that under the unitary representation $U(\Lambda)$ of the Lorentz group on the theory's Hilbert space, the condition $$U(\Lambda)\Phi^a(x)U(\Lambda)^{\dagger}=D(\Lambda^{-1})^a_{\phantom a b}\Phi^b(\Lambda x)\tag{1}$$ holds true. Here $D(\Lambda)^a_{\phantom a b}$ is the matrix representative of the transformation $D(\Lambda)$ in some basis chosen for the representation space $V$.

Regarding your second question, it's not very clear the context (you want to show that a specific stress-energy tensor transforms as it should or a general one starting from some definition?). In any case, I'll give you one example. A massless scalar field has stress-energy tensor $$T_{\mu\nu}=\partial_\mu \phi^\ast \partial_\nu \phi-\dfrac{\eta_{\mu\nu}}{2}\partial_\sigma\phi^\ast \partial^\sigma \phi.\tag{2}$$

Moreover, a scalar field operator obeys, by definition $$U(\Lambda)\phi(x)U(\Lambda)^\dagger = \phi(\Lambda x).\tag{3}$$

In that case you can derive how $\partial_\mu \phi(x)$ transforms by taking a derivative of (3). Then you can construct $U(\Lambda)T_{\mu\nu}(x)U(\Lambda)^{\dagger}$ and use the transformation you derive for $\partial_\mu \phi(x)$ to evaluate it.