= Solution
Use $c_j=(\gamma_j^A+i\gamma_j^B)/2$ and $c_j^\dagger=(\gamma_j^A-i\gamma_j^B)/2$. Moving operators from different modes past one another introduces a minus sign. Direct expansion gives
$$
\begin{aligned}
c_1^\dagger c_2+c_2^\dagger c_1
&=\frac i2(\gamma_1^A\gamma_2^B-\gamma_1^B\gamma_2^A),\\
c_1^\dagger c_2^\dagger+c_2c_1
&=-\frac i2(\gamma_1^A\gamma_2^B+\gamma_1^B\gamma_2^A).
\end{aligned}
$$
Thus, for real hopping and pairing parameters,
$$
H=\frac i2\left[(\Delta-t)\gamma_1^A\gamma_2^B+(t+\Delta)\gamma_1^B\gamma_2^A\right].
$$
At $\Delta=t\ne0$ this becomes
$$
\boxed{H=it\gamma_1^B\gamma_2^A}.
$$
Use $\tau$ for time to distinguish it from the hopping amplitude $t$. Each unused Majorana operator anticommutes with both factors in $H$, so commuting it through their product gives two minus signs. Hence $[H,\gamma_1^A]=[H,\gamma_2^B]=0$. The <Heisenberg picture> equation $dO/d\tau=(i/\hbar)[H,O]$ therefore gives
$$
\boxed{\gamma_1^A(\tau)=\gamma_1^A(0),\qquad \gamma_2^B(\tau)=\gamma_2^B(0)}.
$$
These are the two uncoupled <Majorana zero modes>.
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