I'm reading Timo Weigand notes for Gupta-Bleuler quantization of free EM field. .
On page 109, Author has made the following statements.
The Gupta-Bleuler condition for physical state is
$|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys} \leftrightarrow p^\mu\zeta_\mu = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.56)$$$|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys} \leftrightarrow p^\mu\zeta_\mu = 0 \tag{4.56}$$
since $p^2 = 0 $ for massless photon, such $\zeta^\mu$ can be decomposed as
$\zeta^\mu = \zeta^\mu_T + \zeta^\mu_S \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.57)$$$\zeta^\mu = \zeta^\mu_T + \zeta^\mu_S \tag{4.57}$$
with $\zeta_S = c\cdot p, \zeta^2_S = 0$ and $\vec{\zeta}_T\cdot\vec{p} = 0, \zeta_T^2<0$. So $|\vec{p},\zeta\rangle \in \cal{H}_\textrm{phys}$ can be written as
$|\vec{p},\zeta\rangle = |\vec{p},\zeta_T\rangle + |\vec{p},\zeta_S\rangle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4.58)$$$|\vec{p},\zeta\rangle = |\vec{p},\zeta_T\rangle + |\vec{p},\zeta_S\rangle\tag{4.58}$$
where $|\vec{p},\zeta_T\rangle$ describes 2 transverse DOF of positive norm. and $|\vec{p},\zeta_S\rangle$ describes 1 combined timelike and longitudinal DOF of zero norm.
I have following doubts regarding this :
- How is such decomposition of $\zeta^\mu$ made ?
How is such decomposition of $\zeta^\mu$ made?
- why do $|\vec{p},\zeta_T\rangle$ have positive norm and $|\vec{p},\zeta_S\rangle$ have zero norm ?
why do $|\vec{p},\zeta_T\rangle$ have positive norm and $|\vec{p},\zeta_S\rangle$ have zero norm?
- Why do the decomposed parts $\zeta^{\mu}_T$ and $\zeta^{\mu}_S$ follow the properties listed above the equation 4.58. ?
Why do the decomposed parts $\zeta^{\mu}_T$ and $\zeta^{\mu}_S$ follow the properties listed above the equation 4.58?
Please Help. Thanks.
