Supercharges for N=1 supersymmetry of a super Hamiltonian I currently have determined a super Hamiltonian for a non-relativistic electron in an external electromagnetic field using an appropriate Lagrangian and use of the Euler-Lagrange equations:
$$H=\frac{1}{2m}\left(\overrightarrow{p}-\frac{q}{c}\overrightarrow{A}\right)^2+q\phi$$
I then considered an electron in a purely magnetic field finding that there is an additional interaction such that
$$H=\frac{1}{2m}\left(\overrightarrow{p}+\frac{q}{c}\overrightarrow{A}\right)^2+\frac{g}{2}\frac{e\hbar}{2mC}\overrightarrow{B}\cdot{\overrightarrow{\sigma}}.$$
By setting $g=2$ I can analyse the Hamiltonian in $\mathcal N=1$ supersymmetry by considering the anti-commutation relations between supercharges and the Hamiltonian:
$$H=2Q_i^2=\{Q_i,Q_i\}$$
where  $[H,Q_i]=0$. My struggle is that this closely follows the Fred Cooper book on Supersymmetric Quantum mechanics and he achieves a supercharge as shown below 
$$Q_i=\frac{1}{\sqrt{4m}}\left(\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}\right)\cdot{\overrightarrow{\sigma}}.$$
I am lost as to how this is found by simple factorisation. I am unsure, how is this done?
 A: If you are unsure of how to verify it, start with
$$Q^2=(p+A)^i\sigma_i\,(p+A)^j\sigma_j,$$
where I'm summing over both indices and I'm getting rid of the constants for simplicity. This splits into a term where both $p$ are acting on the wavefunction, and a term where the left $p$ is acting on the right $A$.
$$=(p+A)^i(p+A)^j\sigma_i\sigma_j-i\partial^iA^j \sigma_i\sigma_j$$
Now use the identity $\sigma_i\sigma_j=\delta_{ij}+i\epsilon_{ijk}\sigma_k.$ The first term in $Q^2$ is symmetric in $i,j$ so only the Kronecker delta matters, in the second term the Kronecker delta leads to the divergence of $A$ which vanishes in the Coulomb gauge, so only the Levi-Civita symbol matters.
$$=(p+A)^2+\epsilon_{ijk}\partial^iA^j \sigma_k=(p+A)^2+(\nabla\times \vec A)\cdot\vec\sigma=(p+A)^2+\vec B\cdot\vec\sigma.$$
If you are wondering how they figured out $Q$ in the first place, I imagine they were aware of the identity $\vec{V}\cdot{\vec\sigma}\,\vec{V}\cdot{\vec\sigma}=V^2$, for an ordinary (non-operator) $\vec V$, so they just looked at the kinetic term and tried out an operator $Q=(\vec p+\vec A)\cdot \vec\sigma$. The maybe unexpected thing is this also gives you the magnetic field term with $g=2$.
