Question 2.7: The polarization vectors of a massive spin-1 particle with f......
The polarization vectors of a massive spin-1 particle with fourmomentum p^{\mu} and helicity λ are denoted by \varepsilon^{\mu}(p;\lambda). It holds that
\varepsilon^{\mu^{\ast}}(p;\lambda)\varepsilon_{\mu}(p;\lambda^{\prime})=\varepsilon_{\mu}^{\ast}(p;\lambda)\varepsilon^{\mu}(p;\lambda^{\prime})=-\delta_{\lambda\lambda^{\prime}}\ . (1)
The minus sign occurs because these vectors are spacelike. Owing to Lorentz covariance, the sum over the polarization states
\sum\limits_{\lambda}\varepsilon_{\mu}^{\ast}(p;\lambda)\varepsilon_{\nu}(p;\lambda)\equiv\eta_{\mu\nu}(p) (2)
has to be of the form
\eta_{\mu\nu}(p)=A~g_{\mu\nu}+B p_{\mu}p_{\nu}~. (3)
Find arguments for this fact and determine the constants A and B. Make use of scalar multiplications by p^{\mu},\,p^{\nu}, and g^{uv}. Note that g^{uv} g_{uv} =4.
The condition \partial_{\mu}\phi^{\mu}=0 for the wave function of a spin-1 particle leads to p^{\mu}\varepsilon_{\mu}(p;\lambda)=0. Here p^{\mu} and \varepsilon^{\mu}(p;\lambda) are a system of four linearly independent, orthogonal vectors; i.e., any four-vector a^{\mu} can be represented as a linear combination
a^{\mu}=a_{p}p^{\mu}+\sum\limits_{\lambda}a_{\lambda}\varepsilon^{\mu}(p;\lambda)\,\,\,.\, (4)
Now we evaluate
a^{\mu}\eta_{\mu\nu}(p)=\sum\limits_{\lambda}\Biggl[a_{p}p^{\mu}\varepsilon_{\mu}^{*}(p;\lambda)+\sum\limits_{\lambda^{\prime}}a_{\lambda^{\prime}}\varepsilon_{\mu}^{*}(p;\lambda)\varepsilon^{\mu}(p;\lambda^{\prime})\Biggr]\varepsilon_{\nu}(p;\lambda)\\ =\sum\limits_{\lambda}\Biggl[a_{p}\cdot0+\sum\limits_{\lambda^{\prime}}(-\delta_{\lambda\lambda^{\prime}})a_{\lambda^{\prime}}\Biggr]\varepsilon_{\nu}(p;\lambda)\\=-\sum\limits_{\lambda}a_{\lambda}\varepsilon_{\nu}(p;\lambda)\,\,\,. (5)
\varepsilon_{\mu}(p;\lambda) and shows that -\eta_{\mu\nu}(p) eliminates the part of a given four-vector that is proportional to p^{\mu}. Therefore \eta_{\mu\nu}(p) can only depend on p^{\mu},\,\mathrm{i.e.}\,,\,a_{\nu}^{\perp}(p)= -a^{\mu}\eta_{\mu\nu}(p). Since the \varepsilon_{\mu}(p;\lambda) are four-vectors, the polarization sum \eta_{\mu\nu} transforms like a second-rank tensor. Any symmetric tensor of second rank that is built by a four-vector p^{\mu} is of the general form
\eta_{\mu\nu}(p)=A(p^{2})g_{\mu\nu}+B(p^{2})p_{\mu}p_{\nu}\ . (6)
There are no other possibilities, because the only Lorentz-covariant quantities available are g_{\mu\nu},\,p_{\mu}, and p^{2}. Here we have p^{2}=M^{2}=\mathrm{const.},and therefore A and B must be constants. Multiplying the polarization sum by p^{\mu} yields
p^{\mu}\eta_{\mu\nu}=\sum\limits_{\lambda}p\cdot\varepsilon^{\ast}(p;\lambda)\varepsilon_{\nu}(p;\lambda)=0\\ \rightarrow p^{\mu}(A\ g_{\mu\nu}+B p_{\mu}p_{\nu})=(A+B p^{2})p_{\nu}=0\rightarrow B=-\frac{A}{M^{2}}\ . (7)
Hence the polarization sum assumes the form A(g_{\mu\nu}-p_{\mu}\,p_{\nu}/M^{2}). A contraction with g^{\mu\nu} leads to
g^{\mu\nu}\eta_{\mu\nu}(p)=\eta^{\mu}{}_{\mu}(p)=\sum\limits_{\lambda}\varepsilon^{\mu^{\star}}(p;\lambda)\varepsilon_{\mu}(p;\lambda)=\sum\limits_{\lambda}(-\delta_{\lambda\lambda})=-3 (8)
\rightarrow\eta^{\mu}{}_{\mu}(p)=A~g^{\mu}{}_{\mu}+B p^{2}=A\left(g^{\mu}{}_{\mu}-\frac{p^{2}}{M^{2}}\right) =A(4-1)=3A→ A = −1 . (9)
The final result is then
\sum\limits_{\lambda}\varepsilon_{\mu}^{\ast}(p;\lambda)\varepsilon_{\nu}(p;\lambda)=-\left(g_{\mu\nu}-\frac{p_{\mu}\,p_{\nu}}{M^{2}}\right)\ . (10)