Polarization vector basis in Peskin & Schroeder

Question

I am studying chapter 16.1 of Peskin & Schroeder and I am trying to understand how the chosen polarization vector basis works. It is given by the following:

$$ \epsilon_i^T\cdot\epsilon_j^{*T}=-\delta_{ij} $$ where $\epsilon^T$ are the transverse polarization vectors while the longitudinal/timelike polarization states are given by a lightlike linear combination of $$ \epsilon_\mu^+(k)=\left(\frac{k^0}{\sqrt{2}|\vec{k}|},\frac{\vec{k}}{\sqrt{2}|\vec{k}|}\right)\;\;\;\;\;\;\;\;\;\;\epsilon_\mu^-(k)=\left(\frac{k^0}{\sqrt{2}|\vec{k}|},-\frac{\vec{k}}{\sqrt{2}|\vec{k}|}\right) $$ Basically, I am trying to calculate $$ g_{\mu\nu}=\epsilon_\mu^-\epsilon_\nu^{+*}+\epsilon_\mu^+\epsilon_\nu^{-*}-\sum_i\epsilon_{i\mu}^T\epsilon_{i\nu}^{T*} $$ I am missing an important part in the calculation. What is the form of the transverse polarization states? I could use the relation given but for example will there be a minus sign difference between $ \epsilon_{10}^T\epsilon_{10}^{T*} $ and $ \epsilon_{11}^T\epsilon_{11}^{T*} $ ?

Essentially, what would be the form of $\epsilon_{i\mu}^T\epsilon_{j\nu}^{T*}$?

Answer 1

The transverse polarisation vectors are of the form $\varepsilon^{\text{T}}(0,\boldsymbol{p})$ with $\boldsymbol{k}\cdot \boldsymbol{p}=0$ and so $$\varepsilon^\pm \cdot \varepsilon^{\text{T}} \propto (k^0, \pm \boldsymbol{k})\cdot (0,\boldsymbol{p}) = \mp\boldsymbol{k} \cdot \boldsymbol{p}=0$$.