= Solution
\b[Relaxation-time model.] The <Drude model> treats mobile charge carriers as independent particles accelerated by the <Lorentz force> between scattering events. Collisions randomize the drift <momentum> toward local equilibrium and are represented by a uniform, field-independent <momentum relaxation time>. For the present calculation the <carrier density>, isotropic <mass> and <relaxation time> are fixed, the fields are uniform and stationary, and quantum orbital quantization and intercarrier correlations are neglected. The steady <drift velocity> obeys
$$
0=q(\mathbf E+\mathbf v\times\mathbf B)-\frac m\tau\mathbf v.
$$
Here $n$ is the number of carriers of that species per unit volume, $q$ their \b[signed] charge, $m>0$ their inertial or isotropic band <mass>, $\tau$ the <momentum relaxation time>, and $\sigma_0$ their single-species <electrical conductivity> at zero field.
\b[One-carrier <tensor>.] With $\mathbf B=B\hat{\mathbf z}$ and $\mathbf J=nq\mathbf v$, solving for the <electric field> gives
$$
\mathbf E=\frac{m}{nq^2\tau}\mathbf J-\frac1{nq}\mathbf J\times\mathbf B.
$$
Since $(\mathbf J\times\mathbf B)_x=BJ_y$ and $(\mathbf J\times\mathbf B)_y=-BJ_x$,
$$
E_x=\frac{m}{nq^2\tau}J_x-\frac{B}{nq}J_y,\qquad E_y=\frac{B}{nq}J_x+\frac{m}{nq^2\tau}J_y.
$$
Thus the <resistivity tensor>, defined by $\mathbf E=\boldsymbol\rho\mathbf J$, is
$$
\boxed{\boldsymbol\rho_q=\begin{pmatrix}\sigma_0^{-1}&-B/(nq)\\B/(nq)&\sigma_0^{-1}\end{pmatrix},\qquad \sigma_0=\frac{nq^2\tau}{m}.}
$$
Its off-diagonal signs follow from the stated axes and signed $q$. For zero transverse <electric current>, $E_y/J_x=B/(nq)$ is the Hall <electrical resistivity>; reversing charge reverses it. The single-species longitudinal <electrical resistivity> is independent of $B$ in this ideal isotropic model.
\b[Adding <Electrons> and <electron holes>.] Let both species have density $n$ and charges $q=\pm e$, with $e>0$. The stipulated equal zero-field <electrical conductivity> means equal positive <charge carrier mobility> $\mu=\sigma_0/(ne)$, even if their masses and <momentum relaxation times> separately differ. Put $r=1/\sigma_0$ and $h=B/(ne)$. Inverting each <electrical resistivity> matrix gives
$$
\boldsymbol\sigma_+=\frac1{r^2+h^2}\begin{pmatrix}r&h\\-h&r\end{pmatrix},\qquad \boldsymbol\sigma_-=\frac1{r^2+h^2}\begin{pmatrix}r&-h\\h&r\end{pmatrix}.
$$
Both species experience the same <electric field>, so their <electric current> densities add. Therefore it is their \b[<conductivity tensors>], rather than their <resistivity tensors>, that must be summed. The Hall parts cancel and
$$
\boldsymbol\sigma_{\rm tot}=\frac{2\sigma_0}{1+(\mu B)^2}\mathbf I_2,\qquad \boxed{\boldsymbol\rho_{\rm tot}=\frac{1+(\mu B)^2}{2\sigma_0}\mathbf I_2.}
$$
This is <compensated two-carrier magnetoresistance>. Although the net Hall field vanishes, each carrier species is bent in its own direction; maintaining the total longitudinal <electric current> requires an increasing longitudinal field. Equal carrier numbers alone do not ensure zero Hall response if <charge carrier mobilities> differ; both equalities given here are used.
\b[Large-field limit and quantum oscillations.] Expanding the exact diagonal entry,
$$
\rho_{xx}=\rho_{yy}=\frac1{2\sigma_0}+\frac{\sigma_0B^2}{2n^2e^2}.
$$
For $\mu B\gg1$,
$$
\boxed{\rho_{xx}\sim\rho_{yy}\sim\frac{\sigma_0B^2}{2n^2e^2},\qquad B\gg\frac{ne}{\sigma_0}=\frac{m}{e\tau_{\rm tr}}.}
$$
Here the last equality applies to either species with its own <mass> and transport time; the stipulated <charge carrier mobilities> give the same threshold. This is the classical strong-field condition $\omega_c\tau_{\rm tr}\gg1$: carriers turn substantially before their drift <momentum> relaxes.
Observable <quantum oscillations> additionally require sufficiently sharp <Landau levels> and a sufficiently cold <Fermi-Dirac distribution>, approximately
$$
\omega_c\tau_q\gtrsim1,\qquad \hbar\omega_c\gtrsim k_BT.
$$
When $\tau_q\simeq\tau_{\rm tr}$, the disorder field scale is comparable to the onset scale for large classical <magnetoresistance>. The thermal condition can demand a much larger field, and when $\tau_q\ll\tau_{\rm tr}$ even the disorder criterion is stronger. The quantitative <damping> factors in Q2(c) show that these are crossover scales rather than sharp visibility boundaries. A large smooth $B^2$ <magnetoresistance> therefore does not itself demonstrate <Landau level>; it can exist while thermal smearing suppresses oscillations. Conversely, the asymptotic condition $\omega_c\tau_{\rm tr}\gg1$ is stronger than merely reaching unity, so there is no universal ordering of the two experimental onset fields without temperatures and lifetime information.
\b[Scaling function.] The material's actual zero-field longitudinal <electrical resistivity> is $\rho_{\rm tot}(0)=1/(2\sigma_0)$. Dividing the exact answer by this value gives
$$
\boxed{\frac{\rho_{xx}(B)}{\rho_{xx}(0)}=1+\left(\frac{B\sigma_0}{ne}\right)^2=F(B\sigma_0),\qquad F(x)=1+\frac{x^2}{n^2e^2}.}
$$
The assumption that $n$ stays constant is essential to eliminating further temperature-dependent parameters. $B\sigma_0$ is a dimensionful scaling variable; $B\sigma_0/(ne)=\mu B$ is its dimensionless version. The corresponding fractional increase is $[\rho(B)-\rho(0)]/\rho(0)=(\mu B)^2$.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2016-compensated-transport.png]
{title=Original equal-mobility two-carrier curves collapse under zero-field resistivity scaling}
\b[What the supplied data test.] In the original PDF's left panel, the field dependence is strongly <temperature> dependent. In the right panel, plotting the fractional <magnetoresistance> against the squared rescaled field makes the displayed \b[125, 150, 175, 200 and 225 K curves collapse approximately onto one straight line], over the field intervals where they overlap. This supports the proposed one-parameter scaling and, on that range, the quadratic field law. The \b[100 K curve does not collapse]: it bends away and has a different slope in the rescaled plot. The indicated scaling is therefore supported for the displayed \b[125–225 K] data, not for the entire displayed <temperature> range. The graph alone does not locate a sharp transition between 100 and 125 K or establish behavior above 225 K.
This empirical collapse is <Kohler's rule>. It is consistent with a common temperature-dependent <charge carrier mobility> or relaxation scale and a fixed carrier population; it is not a unique proof of exact compensation or of one microscopic scattering mechanism. A change in relative scattering times, carrier population or Fermi-surface structure could spoil the collapse, and this figure alone does not select the cause of the 100 K departure.
There is a factor-of-two normalization convention in the printed text. The $\sigma_0$ introduced for \b[each] species gives $\rho_{\rm tot}(0)=1/(2\sigma_0)$, whereas the final paragraph calls $1/\sigma_0$ “$\rho_0$.” If $\rho_0$ denotes the experimentally measured zero-field <electrical resistivity>, the consistent slope is
$$
\frac{\rho(B)-\rho_0}{\rho_0}=\frac1{4n^2e^2}\left(\frac B{\rho_0}\right)^2,\qquad \rho_0=\rho_{\rm tot}(0).
$$
If one instead keeps the literal single-species definition $\rho_s=1/\sigma_0$, the same model has $[\rho(B)-\rho_s]/\rho_s=-1/2+[B/\rho_s]^2/(2n^2e^2)$. The choice rescales and offsets the axes by temperature-independent constants, so it does not change which <temperature> curves collapse, but it must be resolved before inferring a <carrier density> from the slope.
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