Monday, 19 October 2020

homework and exercises - Anticommutatorrelation in Bogoliubov-de Gennes Hamiltonian


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:


Ψ=((ψ,ψ),(ψ,ψ))


and



Ψ=((ψ,ψ),(ψ,ψ))T.


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


ˆH=dx [σϵ{,}ψσ(ξp+αpσy+Bσz)ψσ+Δ(ψψ+ψψ)],


can be written as


ˆH=12dx ΨHΨ


with H=ξp1τz+αpσyτz+Bσz1+Δ1τx (here τi are the Pauli matrix for the particle-hole space and means the Kronecker product).


Here I calculate as example the first und third term of ΨHΨ.


τzΨ=((ψ,ψ),(ψ,ψ))T=((ψ,ψ),(ψ,ψ))T


((ψ,ψ),(ψ,ψ))ξp((ψ,ψ),(ψ,ψ))T=(ψ,ψ)ξp(ψ,ψ)T+(ψ,ψ)ξp(ψ,ψ)T=ψξpψ+ψξpψψξpψψξpψ


Now I use the anticommutatorrelation {ψσ,ψσ}=δσ,σψσψσ=1ψσψσ



2ψξpψ+2ψξpψ2ξp


However, the term 2ξp here are wrong.


For the third term I obtain


ψBσzψ+ψBσzψ+ψBσzψ+ψBσzψ=ψBσzψ+ψBσzψψBσzψψBσzψ+2Bσz=2Bσz


Does anybody see my mistake?



Answer



I define Hψαξαβψβ, where I forget the sum/integrals and all these boring staff. I also define ξαβξσ=ξ0+ξxσx+ξyσy+ξzσz to have the most generic one-body Hamiltonian written in a compact form. The one body Hamiltonian then reads, in matrix notation H(ψψ)(ξ0+ξzξxiξyξx+iξyξ0ξz)(ψψ)

as can be easily check.


One now wants to add the particle-hole double space (Nambu space). One uses that (the anti-commutation relation) ψαξαβψβ=ξαβψβψα+δαβξαβ=ψβ(ξαβ)Tψα+δαβξαβ

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 H12(ψψψψ)(ξσiσyΔiσyΔ(ξσ)T)(ψψψψ)12Tr{ξσ}
in a mixed notation (block matrix in the middle, full vectors on the edge). Note the only important thing here that (ξσ)T=(ξσ) and only the σy component changes sign (look at your αpσyτz 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.



No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...