Diagonalization of a hamiltonian for a quantum wire with proximity-induced superconductivity I'm trying to diagonalize the Hamiltonian for a 1D wire with proximity-induced superconductivity. In the case without a superconductor it's all fine. However, with  a superconductor I don't get the correct result for the energy spectrum of the Hamiltonian in

Tudor D. Stanescu and Sumanta Tewari. “Majorana fermions in semiconductor nanowires: fundamentals, modeling, and experiment.” Journal of Physics: Condensed Matter 25, no. 23 (2013): 233201. arXiv:1302.5433 [cond-mat.supr-con].

given by
$$
H = \eta_{k}\tau_{z} + B\sigma_{x} + \alpha k\sigma_{y}\tau_{z} + \Delta\tau_{x}
$$
Here $\sigma$ and $\tau$ are the Pauli matrices for the spin and particle-hole space.
Now the correct result is: $E^{2}_{k} = \Delta^{2} + \eta_{k}^{2} + B^{2} + \left(\alpha k\right)^{2} \pm 2\sqrt{B^{2}\Delta^{2} + \eta^{2}_{k}B^{2} + \eta^{2}_{k}\left(\alpha k\right)^{2}}$
Now, my problem is that I don't know how I can bring the Hamiltonian in the correct matrix form for the calculation of the eigenvalues. If I try it with the upper Hamiltonian I have completely wrong results for the energy spectrum. I believe my mistake is the interpretation of the Pauli matrices $\tau$ but I don't know how I can write the Hamiltonian in the form to get the correct eigenvalues.
 A: Here is a cute little trick I've often found pretty handy: just keep squaring your matrices until they're diagonal! In this case you're going to have to make use of the standard identities of Pauli matrices $$\left\{ \sigma_{i},\sigma_{j}\right\} =2\delta_{ij}$$ You also need to make use of the fact that the different “species” of Pauli matrices, $\sigma$ and $\tau$, won’t “see” each other. In other words, when you’re working through the algebra, Pauli matrices of different species can pass through each other as if they were scalars.
Anyways, the given Hamiltonian is $$H = \eta_{k}\tau_{z}+B\sigma_{x}+\alpha k\sigma_{y}\tau_{z}+\Delta\tau_{x}$$ As I mentioned above, we first square it: $$H^2 = \eta_{k}^{2}\tau_{z}^{2}+B^{2}\sigma_{x}^{2}+\left(\alpha k\right)^{2}\sigma_{y}^{2}\tau_{z}^{2}+\Delta^{2}\tau_{x}^{2}+2B\eta_{k}\tau_{z}\sigma_{x}+2\alpha k\eta_{k}\sigma_{y}\tau_{z}^{2}+2\Delta B\sigma_{x}\tau_{x}+\Delta\eta_{k}\left\{ \tau_{z},\tau_{x}\right\} +\alpha kB\left\{ \sigma_{x},\sigma_{y}\right\} \tau_{z}+\alpha k\Delta\sigma_{y}\left\{ \tau_{z},\tau_{x}\right\}$$ Now, using the anticommutator identity for either species of Pauli matrices, the above expression simplifies to $$H^2 = \eta_{k}^{2}+B^{2}+\left(\alpha k\right)^{2}+\Delta^{2}+2B\eta_{k}\tau_{z}\sigma_{x}+2\alpha k\eta_{k}\sigma_{y}+2\Delta B\sigma_{x}\tau_{x}$$ For reasons that will become obvious shortly, we rearrange the above expression in the following way and square it $$\left(H^{2}-\eta_{k}^{2}-B^{2}-\left(\alpha k\right)^{2}-\Delta^{2}\right)^{2}=\left(2B\eta_{k}\tau_{z}\sigma_{x}+2\alpha k\eta_{k}\sigma_{y}+2\Delta B\sigma_{x}\tau_{x}\right)^{2}$$ Expanding out that further we get $$\left(H^{2}-\eta_{k}^{2}-B^{2}-\left(\alpha k\right)^{2}-\Delta^{2}\right)^{2} = 4B^{2}\eta_{k}^{2}\tau_{z}^{2}\sigma_{x}^{2}+4\left(\alpha k\right)^{2}\eta_{k}^{2}\sigma_{y}^{2}+4\Delta^{2}B^{2}\sigma_{x}^{2}\tau_{x}^{2}+4\alpha kB\eta_{k}^{2}\tau_{z}\left\{ \sigma_{x},\sigma_{y}\right\} +4\Delta B^{2}\eta_{k}\sigma_{x}^{2}\left\{ \tau_{x},\tau_{z}\right\} +4\alpha k\eta_{k}\Delta B\left\{ \sigma_{x},\sigma_{y}\right\} \tau_{x}$$ Once again, using the anticommutators identities we get $$\left(H^{2}-\eta_{k}^{2}-B^{2}-\left(\alpha k\right)^{2}-\Delta^{2}\right)^{2}=4B^{2}\eta_{k}^{2}+4\left(\alpha k\right)^{2}\eta_{k}^{2}+4\Delta^{2}B^{2}$$ Note that the above expression contains only diagonal matrices; we have effectively diagonalized the Hamiltonian. This can be made more explicit by writing $$H^{2}=\left[\eta_{k}^{2}+B^{2}+\left(\alpha k\right)^{2}+\Delta^{2}\pm2\sqrt{B^{2}\eta_{k}^{2}+\left(\alpha k\right)^{2}\eta_{k}^{2}+\Delta^{2}B^{2}}\right]\mathbb{I}_{4\times4}$$ Now, from the above expression, it's not hard to figure out that $$E_{k}^{2}=\eta_{k}^{2}+B^{2}+\left(\alpha k\right)^{2}+\Delta^{2}\pm2\sqrt{B^{2}\eta_{k}^{2}+\left(\alpha k\right)^{2}\eta_{k}^{2}+\Delta^{2}B^{2}}$$ Forgive me if you’re wondering: “with all this algebra, how is that a trick?” Well, this was a tough example. But this trick is pretty general whenever your Hamiltonian consists of matrices (or their tensor products) which satisfy the Clifford algebra. I’m sure you can pick a much simpler example where this trick will really be a trick. For example, you can check the Hamiltonian in equation (51) of:

Martin Leijnse and Karsten Flensberg. “[Introduction to topological superconductivity and Majorana fermions.][1]” Semiconductor Science and Technology 27, no. 12 (2012): 124003. ([arXiv][2])

where you can simply compute the eigenvalues (equation (52)) in your head using this trick.
