# How to write electron hole Hamiltonian into quasi-boson?

1. Jul 23, 2015

### PRB147

V Chernyak, Wei Min Zhang, S Mukamel, J Chem Phys Vol. 109, 9587
Eq.(2.2), Eq. (B1) Eq.(B4)-(B6).
When I substitue Eq.(B4)-(B6) into Eq.(2.2), I can not recover Eq.(B1).
Who can give me a reference or hint on
how to write electron hole Hamiltonian into quasi-boson operator?
I think it is not possible to write sole electron or sole hole Hamiltonian in Eq.(2.2)
into quasi-boson operator.
Eq.(2.2) describes a two-band (one conduction one-valence semiconductor) system in real space,
taking account of electron-electron, hole-hole, electron-hole interactions.

2. Jul 24, 2015

### PRB147

V Chernyak, Wei Min Zhang, S Mukamel, J Chem Phys Vol. 109, 9587 (can be freely downloaded here http://mukamel.ps.uci.edu/publications/pdfs/347.pdf ) Eq.(2.2), Eq. (B1) Eq.(B4)-(B6). When I substitue Eq.(B4)-(B6) into Eq.(2.2), I can not recover Eq.(B1). Who can give me a reference or hint on how to write electron hole Hamiltonian into quasi-boson operator? I think it is not possible to write sole electron or sole hole Hamiltonian in Eq.(2.2) into quasi-boson operator. Eq.(2.2) describes a two-band (one conduction one-valence semiconductor) system in real space, taking account of electron-electron, hole-hole, electron-hole interactions.

The equations listed as follows:

Eq.(2.2)

$$H_e=\sum\limits_{m_1,m_2;n_1,n_2} h_{m_1,m_2;n_1,n_2} B^\dagger_{m_1,m_2}B_{n_1,n_2}+U_{m_1,m_2;n_1,n_2;k_1,k_2;l_1,l_2}B^\dagger_{m_1,m_2}B^\dagger_{n_1,n_2}B_{k_1,k_2}B_{l_1,l_2}$$

Eq.(B.1)

$$H_e=\sum\limits_{m_1,n_1}t^{(1)}_{m_1,n_1}a^\dagger_{m_1}a_{n_1}+t^{(2)}_{m_2,n_2}a_{m_2}a_{n_2}+\frac{1}{2}\sum\limits_{m_1,n_1,k_1,l_1}V^{(1)}_{m_1,n_1,k_1,l_1}a^\dagger_{m_1}a^\dagger_{n_1}a_{k_1}a_{l_1} +\frac{1}{2}\sum\limits_{m_2,n_2,k_2,l_2}V^{(2)}_{m_2,n_2,k_2,l_2}a^\dagger_{m_2}a^\dagger_{n_2}a_{k_2}a_{l_2} +\frac{1}{2}\sum\limits_{m_1,n_2,k_2,l_1}W_{m_1,n_2,k_2,l_1}a^\dagger_{m_2}b^\dagger_{n_2}b_{k_2}a_{l_2}$$.

where $$a_{n_1}~(a^\dagger_{n_1})$$ is electron annihilation (creation) operator, $$b_{n_2}~(b^\dagger_{n_1})$$ is hole annihilation (creation) operator.

the definition of exciton operator

Eq.(B.4)

$$B^\dagger_{m_1,m_2}=a^\dagger_{m_1}b^\dagger_{m_2}$$

$$B_{m_1,m_2}=b_{m_2}a_{m_1}$$

Eq.(B.5)

$$h_{m_1,m_2;n_1,n_2}=t^{(1)}_{m_1,n_1}\delta_{m_2,n_2}+\delta_{m_1,n_1}t^{(2)}_{m_2,n_2} +W_{m_1m_2n_1n_2}$$

Eq.(B6)

$$U_{m_1,m_2;n_1,n_2;k_1,k_2;l_1,l_2}=-\frac{1}{4}[ t^{(1)}_{m_1,k_1}\delta_{m_2,k_2} \delta_{n_1,l_1} \delta_{n_2,l_2} +\delta_{m_1,k_1}t^{(2)}_{m_2,k_2}\delta_{n_1,l_1}\delta_{n_2,l_2} +\delta_{m_1,k_1}\delta_{m_2,k_2}t^{(1)}_{n_1,l_1} \delta_{n_2,l_2} +\delta_{m_1,k_1}\delta_{m_2,k_2}\delta_{n_1,l_1} t^{(2)}_{n_2,l_2}] +\frac{1}{4}[V^{(1)}_{m_1,n_1,k_1,l_1}\delta_{m_2,k_2}\delta_{n_2,l_2} + \delta_{m_1,k_1}\delta_{n_1,l_1}V^{(2)}_{m_2,n_2,k_2,l_2}]$$.