Here we solve the Hamiltonian Eq. (50) through a perturbative approach. As the
Hamiltonian due to the spin-orbit terms are generally much smaller than the
Hamiltonian corresponding to free electron gas, we rewrite Eq. (50) as

$\displaystyle\frac{{\cal\tilde{H}}}{\hbar\omega_{c}}$
$\displaystyle=\overset{{\cal
H}_{0}/\hbar\omega_{c}}{\overbrace{a^{\dagger}a+\frac{1}{2}+\frac{\tilde{\Delta}}{2}\sigma_{z}}}+\overset{{\cal
V}/\hbar\omega_{c}}{\overbrace{\gamma\left(a^{\dagger}+a\right)\sigma_{x}+i\delta\left(a-a^{\dagger}\right)\sigma_{y}}}$
$\displaystyle={\cal H}_{0}/\hbar\omega_{c}+{\cal V}/\hbar\omega_{c}.$

with corresponding unperturbed Hamiltonian and perturbation, ${\cal H}_{0}$
and ${\cal V}$, respectively. Using now the Schrieffer–Wolff transformation
Schrieffer and Wolff (1966); Sergey Bravyi (2011), defined by $e^{S}$, with
the constraint ${\cal V}+\left[{\cal S},{\cal H}_{0}\right]=0$, we obtain an
effective Hamiltonian given by ${\cal H}_{eff}={\cal
H}_{0}+\frac{1}{2}\left[{\cal S},{\cal V}\right]+{\cal O}\left({\cal
V}^{3}\right)$. For our system we find ${\cal S}={\cal S}_{\gamma}+{\cal
S}_{\delta}$, with

$\displaystyle{\cal S}_{\gamma}$
$\displaystyle=-\frac{\gamma}{1-\tilde{\Delta}^{2}}\left\\{a\left(\sigma_{x}+i\tilde{\Delta}\sigma_{y}\right)-a^{\dagger}\left(\sigma_{x}-i\tilde{\Delta}\sigma_{y}\right)\right\\},$
(139) $\displaystyle{\cal S}_{\delta}$
$\displaystyle=-\frac{i\delta}{1-\tilde{\Delta}^{2}}\left\\{a\left(\sigma_{y}-i\tilde{\Delta}\sigma_{x}\right)+a^{\dagger}\left(\sigma_{y}+i\tilde{\Delta}\sigma_{x}\right)\right\\},$
(140)

yielding

$\displaystyle\frac{\tilde{{\cal H}}_{eff}}{\hbar\omega_{c}}$
$\displaystyle=\frac{1}{2}\left(1+\tilde{\Delta}\sigma_{z}\right)-\Omega-\Lambda\sigma_{z}+\left(1-2\Lambda\sigma_{z}\right)a^{\dagger}a$
$\displaystyle+\Gamma\left(aa+a^{\dagger}a^{\dagger}\right)\sigma_{z},$ (141)

with

$\displaystyle\Omega$
$\displaystyle=\frac{\left(\gamma^{2}+\delta^{2}\right)+2\delta\gamma\tilde{\Delta}}{1-\tilde{\Delta}^{2}},$
(142) $\displaystyle\Lambda$
$\displaystyle=\frac{\left(\gamma^{2}+\delta^{2}\right)\tilde{\Delta}+2\delta\gamma}{1-\tilde{\Delta}^{2}},$
(143) $\displaystyle\varGamma$
$\displaystyle={\frac{\delta^{2}-\gamma^{2}}{1-\tilde{\Delta}^{2}}\tilde{\Delta}}.$
(144)

The Hamiltonian Eq. (141) can be rewritten in the Bogoliubov-de Gennes form as

$\displaystyle\frac{\tilde{{\cal H}}_{eff}}{\hbar\omega_{c}}$
$\displaystyle=\frac{1}{2}(1+\tilde{\Delta}\sigma_{z})-\Omega-\Lambda\sigma_{z}-\frac{1}{2}\left(1-2\Lambda\sigma_{z}\right)$
$\displaystyle+\frac{1}{2}\left(a^{\dagger}\quad
a\right)\left[\begin{array}[]{cc}1-2\Lambda\sigma_{z}&{2\Gamma\sigma_{z}}\\\
{2\Gamma\sigma_{z}}&1-2\Lambda\sigma_{z}\end{array}\right]\left(\begin{array}[]{c}a\\\
a^{\dagger}\end{array}\right),$ (149)

which can be diagonalized by a $2\times 2$ Bogoliubov-de Gennes
transformation, and reads

$\frac{\tilde{{\cal
H}}_{eff}}{\hbar\omega_{c}}=\frac{1}{2}(1+\tilde{\Delta}\sigma_{z})-\Omega-\Lambda\sigma_{z}-\frac{1}{2}\left(1-2\Lambda\sigma_{z}\right)+\frac{1}{2}\left(\tilde{a}^{\dagger}\quad\tilde{a}\right)\left[\begin{array}[]{cc}\sqrt{(1-2\Lambda\sigma_{z})^{2}-4\Gamma^{2}}&0\\\
0&\sqrt{(1-2\Lambda\sigma_{z})^{2}-4\Gamma^{2}}\end{array}\right]\left(\begin{array}[]{c}\tilde{a}\\\
\tilde{a}^{\dagger}\end{array}\right),$ (150)

with the diagonal operators $\tilde{a}$ and $\tilde{a}^{\dagger}$. For most
semiconductors, we have $\Omega,\Lambda,\Gamma\ll 1$. By neglecting the fourth
order or higher spin-orbit terms, i.e., $\delta^{i}\gamma^{j}$ with $i+j\geq
4$, we obtain

$\displaystyle\frac{\tilde{{\cal
H}}_{eff}}{\hbar\omega_{c}}=\frac{\tilde{\Delta}}{2}\sigma_{z}+\left|1-2\Lambda\sigma_{z}\right|\left(\tilde{a}^{\dagger}\tilde{a}+\frac{1}{2}\right)-\Omega$
(151)

with energies

$\frac{\varepsilon_{l,s}}{\hbar\omega_{c}}=\frac{s}{2}\tilde{\Delta}-\Omega+\left|1-2\Lambda
s\right|\left(l+\frac{1}{2}\right).$ (152)

For $1-2\Lambda>0$ we obtain

$\frac{\varepsilon_{l,s}}{\hbar\omega_{c}}=\left(l+\frac{1}{2}+\tilde{\Delta}\frac{s}{2}\right)-2s\Lambda\left(l+\frac{1}{2}\right)-\Omega,$
(153)

which is Eq. (53) in the main text.

##  Appendix F Approximations leading to Eqs. (58) and (59)

Starting from Eq. (57) one can obtain the the $F$-function by inverting the
energy levels to obtain $l$, for each value of $s$. The resulting equations
are

$\displaystyle{\cal F}_{+}$
$\displaystyle=\frac{\varepsilon}{\hbar\omega_{c}}-\frac{1}{2}+\left|\Lambda\right|\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)+\frac{1}{4}\frac{\Lambda}{\left|\Lambda\right|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\sqrt{\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2}+8\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left[\frac{\varepsilon}{\hbar\omega_{c}}\left|\Lambda\right|+\frac{\left|\Lambda\right|^{2}}{2}\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)+\frac{1}{2}\left(\,\Omega\frac{\Lambda}{\left|\Lambda\right|}-\Lambda\right)\right]}$
$\displaystyle-\frac{1}{4}\frac{\Lambda}{\left|\Lambda\right|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\sqrt{\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2}+8\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left[\frac{\varepsilon}{\hbar\omega_{c}}\left|\Lambda\right|+\frac{\left|\Lambda\right|^{2}}{2}\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)-\frac{1}{2}\left(\,\Omega\frac{\Lambda}{\left|\Lambda\right|}-\Lambda\right)\right]}$
(154) $\displaystyle{\cal F}_{-}$
$\displaystyle=-\frac{1}{2}\frac{\Lambda}{\left|\Lambda\right|}+\frac{1}{4}\frac{\Lambda}{\left|\Lambda\right|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\sqrt{\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2}+8\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left[\frac{\varepsilon}{\hbar\omega_{c}}\left|\Lambda\right|+\frac{\left|\Lambda\right|^{2}}{2}\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)+\frac{1}{2}\left(\,\Omega\frac{\Lambda}{\left|\Lambda\right|}-\Lambda\right)\right]}$
$\displaystyle+\frac{1}{4}\frac{\Lambda}{\left|\Lambda\right|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\sqrt{\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2}+8\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left[\frac{\varepsilon}{\hbar\omega_{c}}\left|\Lambda\right|+\frac{\left|\Lambda\right|^{2}}{2}\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)-\frac{1}{2}\left(\,\Omega\frac{\Lambda}{\left|\Lambda\right|}-\Lambda\right)\right]}$
(155)

We will further simplify these equations by approximating Eqs. (154) and (155)
up to second order in the spin-orbit parameters $\Lambda$ and $\Omega$ (or
fourth order in $\gamma$ and $\delta$). Accordingly, we rewrite these
equations as

$\displaystyle{\cal F}_{+}$
$\displaystyle=\frac{\varepsilon}{\hbar\omega_{c}}-\frac{1}{2}+|\Lambda|\left(1-\frac{\Lambda}{|\Lambda|}\tilde{\Delta}\right)$
$\displaystyle+\frac{1}{4}\frac{\Lambda}{|\Lambda|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\left(\sqrt{A+B}-\sqrt{A-B}\right),$
(156) $\displaystyle{\cal F}_{-}$
$\displaystyle=-\frac{1}{2}\frac{\Lambda}{|\Lambda|}+\frac{1}{4}\frac{\Lambda}{|\Lambda|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\left(\sqrt{A+B}+\sqrt{A-B}\right),$
(157)

where $A=A_{0}+A_{1}+A_{2}$ and $B=B_{1}$, with

$\displaystyle A_{0}$
$\displaystyle=\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2},$
(158) $\displaystyle A_{1}$
$\displaystyle=8\frac{\varepsilon}{\hbar\omega_{c}}\left|\Lambda\right|\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right),$
(159) $\displaystyle A_{2}$
$\displaystyle=4\left|\Lambda\right|^{2}\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2},$
(160) $\displaystyle B_{1}$
$\displaystyle=4\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left(\,\Omega\frac{\Lambda}{\left|\Lambda\right|}-\Lambda\right).$
(161)

Here, the nominal values of the subindices of $A_{i}$ and $B_{j}$ indicate
their order on the spin-orbit terms $\Lambda$ and $\Omega$. Accordingly, we
expand the square roots of Eqs. (156) and (157) and keep only terms up to
second order in either $\Lambda$ or $\Omega$, yielding

$\displaystyle\sqrt{A+B}+\sqrt{A-B}$ $\displaystyle\approx
2\sqrt{A_{0}+A_{1}}\left(1+\frac{1}{2}\frac{A_{2}}{A_{0}+A_{1}}\right),$
$\displaystyle=2\sqrt{A_{0}+A_{1}+A_{2}}$ (162)
$\displaystyle\sqrt{A+B}-\sqrt{A-B}$
$\displaystyle\approx\thinspace\frac{B_{1}}{\sqrt{A_{0}}}.$ (163)

As a consequence, we can finally write

$\displaystyle{\cal F}_{+}$
$\displaystyle=\frac{\varepsilon}{\hbar\omega_{c}}-\frac{1}{2}+\Omega-\Lambda\tilde{\Delta}$
(164) $\displaystyle{\cal F}_{-}$
$\displaystyle=-\frac{1}{2}\frac{\Lambda}{|\Lambda|}+\frac{1}{2}\frac{\Lambda}{|\Lambda|}\frac{1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}}{\left|1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right|}\times\sqrt{\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)^{2}+8\left|\Lambda\right|\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\left[\frac{\varepsilon}{\hbar\omega_{c}}+\frac{1}{2}\left|\Lambda\right|\left(1-\frac{\Lambda}{\left|\Lambda\right|}\tilde{\Delta}\right)\right]},$
(165)

which are Eqs. (58) and (59), respectively.

##  Appendix G Temperature dependence of the normalized differential
resistivity

In this section we derive the general temperature dependence of the normalized
differential magnetoresistivity in Eq. (24) for the systems studied in this
work.

$\displaystyle\delta\rho_{xx}(B)$ $\displaystyle=2\sum_{l=1}^{\infty}\int
d\varepsilon\tilde{L}_{\Gamma}\left(l\frac{\Gamma}{\hbar\omega_{c}}\right)\left(-\frac{df_{0}(\varepsilon)}{d\varepsilon}\right)$
$\displaystyle\times\cos(2\pi l\mathcal{F}_{-})\cos(2\pi l\mathcal{F}_{+}).$
(166)

At $T=0$ K, we have
$-df_{0}(\varepsilon)/d\varepsilon\rightarrow\delta(\varepsilon-\varepsilon_{F})$,
which simplifies Eq. (166) to

$\delta\rho_{xx}(B)=2\sum_{l=1}^{\infty}\tilde{L}_{\Gamma}\left(l\frac{\Gamma}{\hbar\omega_{c}}\right)\left.\cos(2\pi
l\mathcal{F}_{-})\cos(2\pi
l\mathcal{F}_{+})\right|_{\varepsilon=\varepsilon_{F}},$ (167)

being obviously temperature independent. When the temperature is finite but
small, i.e., $k_{B}T\ll\mu\sim\varepsilon_{F}$, we have a temperature
dependent $\delta\rho_{xx}(B)$. We now analyze the relevant case for low-
density semiconductors, but with
$\varepsilon_{F}\gg\varepsilon_{R},\varepsilon_{D}$, and high number of
populated Landau levels, i.e., $\varepsilon_{F}/\hbar\omega_{c}\gg 1$. With
these conditions, all the different cases analyzed in this manuscript present
${\cal F}_{\pm}$-functions constant or linearly dependent on the energy, so we
write here, ${\cal F}_{\pm}\propto\varepsilon+cte$, see, for example Eqs.
(29), (44), (45), (58), and (59). Using $2\pi l{\cal
F}_{+}=2\Lambda_{+}^{l}\varepsilon+\phi_{+}^{l}$, ${2\pi l{\cal
F}_{-}=2\Lambda_{-}^{l}\varepsilon+\phi_{-}^{l}}$, with $\phi_{\pm}^{l}$
properly defined by comparison with these equations, and assuming an energy-
independent Dingle factor (only true for Lorentzian broadening.), we need to
calculate integrals of the following form,

$\int_{0}^{\infty}d\varepsilon\left(-\frac{\partial
f^{0}}{\partial\varepsilon}\right)\cos\left(2\Lambda_{+}^{l}\varepsilon+\phi_{+}^{l}\right)\cos\left(2\Lambda_{-}^{l}\varepsilon+\phi_{-}^{l}\right)=\int_{-\frac{\mu}{2k_{B}T}}^{\infty}dx\frac{\cos\left(4\Lambda_{+}^{l}k_{B}Tx+2\Lambda_{+}^{l}\mu+\phi_{+}^{l}\right)\cos\left(4\Lambda_{-}^{l}k_{B}Tx+2\Lambda_{-}^{l}\mu+\phi_{-}^{l}\right)}{2\sinh^{2}x},$
(168)

where we have introduced the dimensioness quantity
$x=\frac{\varepsilon-\mu}{2k_{B}T}$. For $\mu\gg k_{B}T$, we obtain

$\displaystyle\int_{0}^{\infty}d\varepsilon\left(-\frac{\partial
f^{0}}{\partial\varepsilon}\right)\cos\left(2\Lambda_{+}^{l}\varepsilon+\phi_{+}^{l}\right)\cos\left(2\Lambda_{-}^{l}\varepsilon+\phi_{-}^{l}\right)=\pi
k_{B}T\left\\{\frac{\left(\Lambda_{+}^{l}-\Lambda_{-}^{l}\right)\cos\left[2\mu\left(\Lambda_{+}^{l}-\Lambda_{-}^{l}\right)+\phi_{+}^{l}-\phi_{-}^{l}\right]}{\sinh\left[2\pi
k_{B}T\left(\Lambda_{+}^{l}-\Lambda_{-}^{l}\right)\right]}\right.$ (169)
$\displaystyle+\left.\frac{\left(\Lambda_{+}^{l}+\Lambda_{-}^{l}\right)\cos\left[2\mu\left(\Lambda_{+}^{l}+\Lambda_{-}^{l}\right)+\phi_{+}^{l}+\phi_{-}^{l}\right]}{\sinh\left[2\pi
k_{B}T\left(\Lambda_{+}^{l}+\Lambda_{-}^{l}\right)\right]}\right\\}.$ (170)

using

$\displaystyle\int_{-\infty}^{\infty}dx\frac{\cos(2\lambda_{1}+a_{1})\cos(2\lambda_{2}+a_{2})}{\cosh{x}^{2}}$
$\displaystyle=\frac{\pi(\lambda_{1}-\lambda_{2})\cos(a_{1}-a_{2})}{\sinh\pi(\lambda_{1}-\lambda_{2})}+\frac{\pi(\lambda_{1}+\lambda_{2})\cos(a_{1}+a_{2})}{\sinh\pi(\lambda_{1}+\lambda_{2})}$

For the cases treated in this work, $\Lambda_{+}^{l}\gg\Lambda_{-}^{l}$ holds,
and we obtain

$\displaystyle\int_{0}^{\infty}d\varepsilon\left(-\frac{\partial
f^{0}}{\partial\varepsilon}\right)\cos\left(2\Lambda_{+}^{l}\varepsilon+\phi_{+}^{l}\right)\cos\left(2\Lambda_{-}^{l}\varepsilon+\phi_{-}^{l}\right)$
$\displaystyle\approx{\cal
A}_{l}(T)\cos\left(2\mu\Lambda_{+}^{l}+\phi_{+}^{l}\right)\cos\left(2\mu\Lambda_{-}^{l}+\phi_{-}^{l}\right),$

with

${\cal A}_{l}(T)=\frac{2\pi k_{B}T\Lambda_{+}^{l}}{\sinh\left(2\pi
k_{B}T\Lambda_{+}^{l}\right)},$ (171)

for the temperature dependent coefficient for the SdH oscillation. For all the
cases investigated in this work, we have $\Lambda_{+}=\pi l/\hbar\omega_{c}$,
yielding Eq. (25) in the main text,

${\cal
A}_{l}\left(T\right)=\frac{2\pi^{2}lk_{B}T/\hbar\omega_{c}}{\sinh\left(2\pi^{2}lk_{B}T/\hbar\omega_{c}\right)}.$
(172)

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