Nambu Spinor Notation in Kitaev spinless p-wave model

[DeathbyGreen]

Hey all! Thanks for reading. I'm currently following along in some reading and had some trouble with re-writing a Hamiltonian in Bogluibov-de Gennes form using Nambu notation (Nambu spinors). Here is the low down:

Say we have a Hamiltonian:
$$\frac{1}{2} \sum_{i=1}^{N} c_{i}^{\dagger} D c_{i} + \frac{1}{2}\sum_{i=1}^{N-1}c^{\dagger}_{i+1}Tc_{i} + c^{\dagger}_{i}T^{\dagger}c_{i+1}$$

where we have
$$D = (\frac{\hbar^2}{ma^2} - \mu)\tau_{z}$$

and
$$T = (-\frac{\hbar^2}{2ma^2}\tau_{z} - \frac{i\Delta}{2a}\tau_{x})$$

The tau are Pauli matrices. We are to be able to write the Hamiltonian as a 2Nx2N matrix (N being the number of particles), by defining Nambu spinors

$$\tilde{c} = (c_{1}, c_{2},...,c_{N})^T$$

which is of length 2N since each c_{i} is a 2 spinor. Finally, we can simplify the Hamiltonian to
$$H = \frac{1}{2} \tilde{c^{\dagger}}H\tilde{c}$$

Where H is a tridiagonal matrix consisting of T's and D's. I understand how the two are equivalent, but how could I just look at a Hamiltonian and tell if I could simplify it using Nambu spinors? When using the Bogluibov quasiparticles we see the same type of thing to get a BdG Hamiltonian. I can check the equivalence by working backwards, but how could I start with the original H in my statement and rearrange it to ultimately get the Nambu form? Hopefully I'm being clear enough :P

 

[DrDu]

When you say "We are to be able to write the Hamiltonian as a 2Nx2N matrix (N being the number of particles), by defining Nambu spinors" you are simply forming a vector out of Nambu spinors, not defining them. Your starting Hamiltonian seems to be already defined in terms of Nambu spinors. I fear I don't quite get your point.

 

[DeathbyGreen]

Yes sorry, I was worried my question wouldn't be clear. I think it is a pretty simple procedure, but I just don't see how I can look at the first Hamiltonian and know that I can re-write it as a vector of Nambu spinors. It is already written in terms of them, but I don't know how the tridiagonal matrix form is so apparent. Is there some kind of matrix transformation that is commonly used that I'm just not aware of?

 

[DeathbyGreen]

I think I just figured it out...I just re-wrote each individual term in matrix notation and added them, then got the tridiagonal thing. :D Thanks!