Yambo Community Forum

Dear users and developers:

When solving the BSE in Bloch space, we know that yambo uses single electron-hole transitions as the basis to build BSE Hamiltonian.

So, for a spin-polarized semiconductor, will the single electron-hole transitions with flipped spin (e.g., the hole is from a spin-up valence band and the electron is from a spin-down conducting band) be included in the basis when constructing BSE Hamiltonian ?

From the theory of BSE, the inclusion of these spin-flipped electron-hole transitions is quite natural, and at sometime it's also necessary (e.g., for the description of singlet-triplet splitting). But by checking the total number of eigenvalues in "exc_E_sorted" file created by "ypp -e s", it seems that yambo constructs and diagonalizes BSE Hamiltonian for spin-up and spin-down subspace respectively, and the transitions between different spin are not included……

I will be very grateful that if anyone can give me a final confirmation.

Thank you !

Dear Zeyu Jiang,
spin flip terms are not included in the Hamiltonian as they are not dipole allowed.
Please note that singlet triplet splitting can be calculated by calculating separately singlet and triplet excitations.
Triplet excitations are calculated by omitting the exchange term in the excitonic Hamiltonian. See e.g. Rohlfing and Louie PRB 62, 4927 (2000).
It is also possible to consider non collinear spin, by calculating a ground state including spin orbs coupling and the BSE will be solved in a spinorial space.

Best,

Daniele

Daniele Varsano wrote:Dear Zeyu Jiang,
spin flip terms are not included in the Hamiltonian as they are not dipole allowed.
Please note that singlet triplet splitting can be calculated by calculating separately singlet and triplet excitations.
Triplet excitations are calculated by omitting the exchange term in the excitonic Hamiltonian. See e.g. Rohlfing and Louie PRB 62, 4927 (2000).
It is also possible to consider non collinear spin, by calculating a ground state including spin orbs coupling and the BSE will be solved in a spinorial space.

Best,

Daniele

OK, I think I get it, thank you very much~