- #1
DeathbyGreen
- 83
- 15
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}<br />
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
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}<br />
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