General form of the matrix element of a Lorentz invariant rank two tensor operator? Why can the matrix element of a traceless, symmetric tensor $Q_{\alpha \beta}$ that does not commute with Lorentz transformations, for a one particle state be written as
$$ \langle p | Q_{\alpha \beta} | p\rangle  = p_\alpha p_\beta - \frac{1}{4} \eta_{\alpha \beta} p^2  \quad ?$$
This is claimed by E. Witten in his "INTRODUCTION TO SUPERSYMMETRY" and he argues this is the case because of Lorentz invariance.
 A: In a relativistic quantum theory there is a unitary operator $U$ for every Lorentz transformation $\Lambda$. If the states $|p\rangle$ represent scalar particles, then their action under the unitary operators is $U_\Lambda |p\rangle = |\Lambda p \rangle$ where $(\Lambda p)^\mu = \Lambda^\mu_{\,\,\nu}p^\nu$. A tensor operator is an operator such that for any Lorentz transformation $U_\Lambda Q_{\mu\nu} U^†_\Lambda = \Lambda_{\mu}^{\,\,\alpha}\Lambda_{\nu}^{\,\,\beta}Q_{\alpha\beta}$. These are the standard definitions and properties of a relativistic quantum theory. A traceless Lorentz tensor is one that satisfies $Q^\alpha_{\,\,\alpha} = 0$. A symmetric Lorentz tensor is one that satisfies $Q_{\alpha\beta} = Q_{\beta\alpha}$.
This then puts constraints on the forms of the matrix elements of $Q$. The matrix element $\langle p |Q_{\mu\nu}|p\rangle = q_{\mu\nu}(p)$ is a complex function of four-momentum $p$. Since the operator $U_\Lambda$ is unitary for every operator we must have
$\langle p |Q_{\mu\nu}|p\rangle = \langle p |U_\Lambda^† U_\Lambda Q_{\mu\nu}U^†_\Lambda U_\Lambda |p\rangle$ which implies that 


*

*$q_{\mu\nu}(p) = \Lambda_{\mu}^{\,\,\alpha}\Lambda_\nu^{\,\,\beta}q_{\alpha\beta}(\Lambda p)$. 

*$q_{\mu\nu}(p) = q_{\nu\mu}(p)$

*$q^{\mu}_{\,\,\mu}(p)=0$ for all $p$


Note that $q_{\mu\nu}(p)=f(p^2)\eta_{\mu\nu}$ satisfies conditions 1 and 2 because $p^2 = (\Lambda p)^2$ and $\Lambda_{\mu}^{\,\,\alpha}\Lambda_\nu^{\,\,\beta}\eta_{\alpha\beta}=\eta_{\mu\nu}$. This term by itself doesn't satisfy condition 3 though.
Also, $q_{\mu\nu}(p)=g(p^2)p_\mu p_\nu$ satisfies conditions 1 and 2 because of how the four momentum $p$ transforms.
Some thought tells us that there are no other symmetric Lorentz tensors available to us to build terms that satisfy the first two conditions, at least as long as we don't introduce any other momenta.
In order to satisfy condition 3 we can combine the two terms. The trace of the first term is $f(p^2)\eta^{\mu}_{\,\,\mu} = 4f(p^2)$ and the trace of the second term is $g(p^2)p^\mu p_\mu = p^2 g(p^2)$. So then our matrix element must have the form
$q_{\mu\nu}(p) = q(p^2)\left(p_\mu p_\nu - \frac{1}{4}p^2\eta_{\mu\nu}\right)$ where $q$ is some arbitrary function of $p^2$.
If the states do not represent scalar particles then the argument must change because the matrix element transforms in a different way. For traceless symmetric tensor operators with additional nice properties there may be ways to further constrain the function $q(p^2)$, but we would need to know those properties.
