In the RNS formulation of superstring theory we have the action:
$$S=-\frac{T}{2}\int d^2\sigma(\partial_\alpha X^\mu \partial^\alpha X_\mu + \bar{\psi}^\mu \rho^\alpha \partial_\alpha \psi_\mu)$$
where $X^\mu$ is a worldsheet scalar and $\psi^\mu$ is a worldsheet spinor. We can Fourier decompose these two. For the open string in the Ramond sector we obtain modes $\alpha_n^\mu$ for $X^\mu$ and $d_n^\mu$ for $\psi^\mu$. We can relate the zero mode of $X$ to its momentum, i.e. $\alpha_0^\mu \sim p^\mu$.
My first question is, why can't we do something similar for $\psi^\mu$, so relating $d_0^\mu$ to the momentum of the spinor?
Secondly, I know that from the anticommutation relation $\{d_0^\mu,d_0^\nu\}=\eta^{\mu\nu}$ we deduce that the $d_0^\mu$ act as gamma matrices on the states. Why then does it disappear in the mass formula for the Ramond sector open string (in the light-cone gauge, equation 4.109 in Becker, Becker and Schwarz):
$$\alpha ' M^2 = \sum^\infty_{n=1}\alpha_{-n} \cdot \alpha_{n}+\sum^\infty_{n=1} n d_{-n} \cdot d_{n}$$
I would expect a term $\sim d_0\cdot d_0=\Gamma^\mu\Gamma_\mu=10I_{32}$ on the right. Is it absorbed in the mass?
Finally, if $M^2$ only depends on $\alpha_0^\mu$, why do we consider this to represent the total mass? Doesn't it miss a mass contribution from $\psi$? Would this mean the spinor is always massless? If so, why isn't $X$ massless?