this solvability condition determines the spectrum of the system.

In the absence of a magnetic field ($\Delta_Z=0$), the ABS spectrum is given by solutions to the transcendental equation \[Eq. \] $$\label{ABSnoB}
\begin{aligned}
0&=(\alpha^2+v_F^2)\Omega_\Delta\cos\chi-2E\alpha v_F\sin\chi \\
    &-(v_F^2-\alpha^2)\Omega_\Delta\cos(2\bar k_{F}L),
\end{aligned}$$ where $\chi=2EL/\alpha$ and $\bar k_F=(k_{F1}+k_{F2})/2$; $k_{F1}=2k_{so}+\mu/\alpha$ and $k_{F2}=\mu/\alpha$ are the two Fermi momenta of the Rashba spectrum. We note that Eq.  reproduces the known ABS spectrum $\Omega_\Delta\cos\chi=E\sin\chi$ in the limit of no Fermi velocity mismatch between the normal and superconducting regions ($\alpha=v_F$) [@deGennes:1963]. Fermi velocity mismatch acts as an effective sharp potential barrier at the interface ($x=0$) that introduces Fabry-Perot-like oscillations described by $\cos(2\bar k_FL)$. The number of ABSs lying within the gap is determined by the length of the quantum dot, as the spacing between ABS levels is given by the energy scale $\alpha/L$. Thus, for $\alpha/L\ll\Delta$, there are many subgap states, while for $\alpha/L\gg\Delta$, there is a single subgap state. To see that there is always at least one subgap ABS [@olesia2], let us examine the limit $L/\alpha\to0$, where the ABS energy should be near the gap edge. Expanding $E=\Delta-\delta E$ (with $0<\delta E\ll\Delta$) and solving Eq.  gives $$\label{ABSnoBsol}
\sqrt{\frac{\delta E}{\Delta}}=\frac{\sqrt{2}v_F\alpha\chi_\Delta}{\alpha^2[1+\cos(2\bar k_{F}L)]+v_F^2[1-\cos(2\bar k_FL)]},$$ where $\chi_\Delta=2\Delta L/\alpha$. As the right-hand side 