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I almost solved the problem Equivalence of Bogoliubov-de Gennes Hamiltonian for nanowire. In the next steps I used the notation by arXiv:0707.1692:

$$ \Psi^{\dagger} = \left(\left(\psi_{\uparrow}^{\dagger}, \psi_{\downarrow}^{\dagger}\right), \left(\psi_{\downarrow}, -\psi_{\uparrow}\right)\right) $$


$$ \Psi = \left(\left(\psi_{\uparrow}, \psi_{\downarrow}\right), \left(\psi_{\downarrow}^{\dagger}, -\psi_{\uparrow}^{\dagger}\right)\right)^{T}\text{.} $$

I'm trying to show that the Hamiltonian for a nanowire with proximity-induced superconductivity

$$ \hat{H} = \int dx \text{ } \left[\sum_{\sigma\epsilon\{\uparrow,\downarrow\}}\psi_{\sigma}^{\dagger}\left(\xi_{p} + \alpha p\sigma_{y} + B\sigma_{z}\right)\psi_{\sigma} + \Delta\left(\psi_{\downarrow}^{\dagger}\psi_{\uparrow}^{\dagger} + \psi_{\uparrow}\psi_{\downarrow}\right)\right]\text{,} $$

can be written as

$$ \hat{H} = \frac{1}{2}\int dx \text{ } \Psi^{\dagger}\mathcal{H}\Psi $$

with $\mathcal{H} = \xi_{p} 1\otimes \tau_{z} + \alpha p \sigma_{y}\otimes\tau_{z} + B\sigma_{z}\otimes 1 + \Delta 1\otimes\tau_{x}$ (here $\tau_{i}$ are the Pauli matrix for the particle-hole space and $\otimes$ means the Kronecker product).

Here I calculate as example the first und third term of $\Psi^{\dagger}\mathcal{H}\Psi$.

$$ \tau_{z}\Psi = \left(\left(\psi_{\uparrow}, \psi_{\downarrow}\right), -\left(\psi_{\downarrow}^{\dagger}, -\psi_{\uparrow}^{\dagger}\right)\right)^{T} = \left(\left(\psi_{\uparrow}, \psi_{\downarrow}\right), \left(-\psi_{\downarrow}^{\dagger}, \psi_{\uparrow}^{\dagger}\right)\right)^{T} $$

$$ \Rightarrow \left(\left(\psi_{\uparrow}^{\dagger}, \psi_{\downarrow}^{\dagger}\right), \left(\psi_{\downarrow}, -\psi_{\uparrow}\right)\right)\xi_{p}\left(\left(\psi_{\uparrow}, \psi_{\downarrow}\right), \left(-\psi_{\downarrow}^{\dagger}, \psi_{\uparrow}^{\dagger}\right)\right)^{T} = \left(\psi_{\uparrow}^{\dagger}, \psi_{\downarrow}^{\dagger}\right)\xi_{p}\left(\psi_{\uparrow}, \psi_{\downarrow}\right)^{T} + \left(\psi_{\downarrow}, -\psi_{\uparrow}\right)\xi_{p}\left(-\psi_{\downarrow}^{\dagger}, \psi_{\uparrow}^{\dagger}\right)^{T} = \psi_{\uparrow}^{\dagger}\xi_{p}\psi_{\uparrow} + \psi_{\downarrow}^{\dagger}\xi_{p}\psi_{\downarrow} - \psi_{\downarrow}\xi_{p}\psi_{\downarrow}^{\dagger} -\psi_{\uparrow}\xi_{p}\psi_{\uparrow}^{\dagger} $$

Now I use the anticommutatorrelation $\{\psi_{\sigma}, \psi^{\dagger}_{\sigma^{\prime}}\} = \delta_{\sigma,\sigma^{\prime}} \Leftrightarrow \psi_{\sigma}\psi^{\dagger}_{\sigma} = 1 - \psi_{\sigma}^{\dagger}\psi_{\sigma}$

$$ \Leftrightarrow 2\psi_{\uparrow}^{\dagger}\xi_{p}\psi_{\uparrow} + 2\psi_{\downarrow}^{\dagger}\xi_{p}\psi_{\downarrow} - 2\xi_{p} $$

However, the term $-2\xi_{p}$ here are wrong.

For the third term I obtain

$$ \psi_{\uparrow}^{\dagger}B\sigma_{z}\psi_{\uparrow} + \psi_{\downarrow}^{\dagger}B\sigma_{z}\psi_{\downarrow} + \psi_{\downarrow}B\sigma_{z}\psi_{\downarrow}^{\dagger} +\psi_{\uparrow}B\sigma_{z}\psi_{\uparrow}^{\dagger} = \psi_{\uparrow}^{\dagger}B\sigma_{z}\psi_{\uparrow} + \psi_{\downarrow}^{\dagger}B\sigma_{z}\psi_{\downarrow} - \psi_{\downarrow}^{\dagger}B\sigma_{z}\psi_{\downarrow} -\psi_{\uparrow}^{\dagger}B\sigma_{z}\psi_{\uparrow} + 2B\sigma_{z} = 2B\sigma_{z} $$

Does anybody see my mistake?

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2 Answers 2

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I define $H\sim\psi_{\alpha}^{\dagger}\xi_{\alpha\beta}\psi_{\beta}$, where I forget the sum/integrals and all these boring staff. I also define $\xi_{\alpha\beta}\equiv\boldsymbol{\xi}\cdot\boldsymbol{\sigma}=\xi_{0}+\xi_{x}\sigma_{x}+\xi_{y}\sigma_{y}+\xi_{z}\sigma_{z}$ to have the most generic one-body Hamiltonian written in a compact form. The one body Hamiltonian then reads, in matrix notation $$ H\sim\left(\begin{array}{cc} \psi_{\uparrow}^{\dagger} & \psi{}_{\downarrow}^{\dagger}\end{array}\right)\left(\begin{array}{cc} \xi_{0}+\xi_{z} & \xi_{x}-\mathbf{i}\xi_{y}\\ \xi_{x}+\mathbf{i}\xi_{y} & \xi_{0}-\xi_{z} \end{array}\right)\left(\begin{array}{c} \psi_{\uparrow}\\ \psi_{\downarrow} \end{array}\right) $$ as can be easily check.

One now wants to add the particle-hole double space (Nambu space). One uses that (the anti-commutation relation) $$ \psi_{\alpha}^{\dagger}\xi_{\alpha\beta}\psi_{\beta}=-\xi_{\alpha\beta}\psi_{\beta}\psi{}_{\alpha}^{\dagger}+\delta_{\alpha\beta}\xi_{\alpha\beta}=-\psi_{\beta}\left(\xi_{\alpha\beta}\right)^{T}\psi{}_{\alpha}^{\dagger}+\delta_{\alpha\beta}\xi_{\alpha\beta} $$ and you get the unavoidable trace over the one-body energy. This nevertheless renormalizes your energy in a standard way, and one usually drops off this extra term. We thus get $$ H\sim\dfrac{1}{2}\left(\begin{array}{cccc} \psi{}_{\uparrow}^{\dagger} & \psi{}_{\downarrow}^{\dagger} & \psi_{\uparrow} & \psi_{\downarrow}\end{array}\right)\left(\begin{array}{cc} \boldsymbol{\xi}\cdot\boldsymbol{\sigma} & -\mathbf{i}\sigma_{y}\Delta\\ \mathbf{i}\sigma_{y}\Delta & -\left(\boldsymbol{\xi}\cdot\boldsymbol{\sigma}\right)^{T} \end{array}\right)\left(\begin{array}{c} \psi_{\uparrow}\\ \psi_{\downarrow}\\ \psi{}_{\uparrow}^{\dagger}\\ \psi{}_{\downarrow}^{\dagger} \end{array}\right)-\dfrac{1}{2}\text{Tr}\left\{ \boldsymbol{\xi}\cdot\boldsymbol{\sigma}\right\} $$ in a mixed notation (block matrix in the middle, full vectors on the edge). Note the only important thing here that $\left(\boldsymbol{\xi}\cdot\boldsymbol{\sigma}\right)^{T}=\left(\boldsymbol{\xi}\cdot\boldsymbol{\sigma}\right)^{\ast}$ and only the $\sigma_{y}$ component changes sign (look at your $\alpha p\sigma_{y}\tau_{z}$ term in the BdG Hamiltonian)

Your ordering convention is found by an obvious change of basis from mine. Then you choose a representation for the tensor product and you're done. One more time, you can not avoid the final trace term, but most of the people forget to discuss it. It has almmost no role, except when you want to describe some effects related to the phase transition of superconductivity (for instance, to correctly write the free energy, you need it).

One more thing: the Hamiltonian you gave is a bit famous at the moment for hosting Majorana fermions. If you diagonalise the spin-part, you end up with a $p$-wave effective superconductivity at low energy.

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That is the perfect answer. I want to rewrite the Hamiltonian as a tensor product (here Kronecker product)

$$ H \sim \left(\begin{array}{cc} \boldsymbol{\xi}\cdot\boldsymbol{\sigma} & -\mathbf{i}\sigma_{y}\Delta\\ \mathbf{i}\sigma_{y}\Delta & -\left(\boldsymbol{\xi}\cdot\boldsymbol{\sigma}\right)^{T} \end{array}\right) = \xi_{0}1\otimes\tau_{z} + \xi_{z}\sigma_{z}\otimes\tau_{z} + \xi_{y}1\otimes\sigma_{y} + \Delta\sigma_{y}\otimes\tau_{y}\text{.} $$

My problem now is spin-orbit coupling term $\xi_{y}1\otimes\sigma_{y}$ that here is different as in the literature. Is this on the physical aspect okay? I known that the applied magnetic field in the Zeeman field term must perpendicular to spin-orbit field, what in that case is fulfilled.

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Sorry for this late answer. You should ask for precision as comment, not as answer... which are... well, for answer :-) I'm not sure I understand your last statement. Usually, the only thing you have to keep is the symmetry, whatever its representation. So, as long as you have some Zeeman effect along the $z$-axis (whatever it means, here it means you have a term $h\sigma_{z}$) and a spin-orbit coupling along a perpendicular axis (here something like $vp\sigma_{x,y}$, with $v$ a velocity to keep the dimension of energy) you're fine. Have fun. – FraSchelle Sep 25 '13 at 4:20

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