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Note that those fermions has integer spin while the bosons has semi-integer spins, the opposite of the spin-statistic, characterizing them as ghosts (if you assume that your theory obey spin-statistic …

$L_1^{(m)}$ commutes with the $bc$-ghosts. Furthermore, $c_1$ and $c_{-1}$ anti-commute, so $c_1c_{-1}|0;k\rangle=-c_{-1}c_{1}|0;k\rangle=0$, and the firth term annihilate the state as well. …

If $|p\rangle$ is an eigenstate of $p^\mu$,there is no reason to make this state a ghost. Ghost would appear if you have polarization in this states. Is better to see this in a second quantization. No …

The opposite statistic of the ghosts will lead to an erasing of the wrong polarizations that you introduce in order to work in a covariant manner. … All this holds because the ghosts has the opposite statistics of the gauge parameters. …

Also, it is better to have the ghosts as fermions since they will shows up in loops with the opposite sign of the pure gauge loops, erasing them. … In this formalism of bosonic ghosts is not clear how this cancellation will be happening, but of course will be happen since there is no inconsistency over here. …

Ghosts insertions are related to moduli (b-ghosts) and fixing vertex operators (c-ghosts). …

The supercurrent is fixed by the superconformal algebra, which closes after adding the energy-momentum tensor $T_{B}$: $$ T_F(z)T_F(0)\sim\frac{2c}{3z^3}+\frac{2}{z}T_{B}(0)\,, $$ $$ T_{B}(z)T_F(0)\si …