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How to write electron hole Hamiltonian into quasi-boson?

  1. Jul 23, 2015 #1
    V Chernyak, Wei Min Zhang, S Mukamel, J Chem Phys Vol. 109, 9587
    (can download 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.
     
  2. jcsd
  3. Jul 24, 2015 #2
    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)

    [tex]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}[/tex]

    Eq.(B.1)

    [tex]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}[/tex].

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

    the definition of exciton operator

    Eq.(B.4)


    [tex]B^\dagger_{m_1,m_2}=a^\dagger_{m_1}b^\dagger_{m_2}[/tex]

    [tex]B_{m_1,m_2}=b_{m_2}a_{m_1}[/tex]


    Eq.(B.5)


    [tex]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}[/tex]



    Eq.(B6)

    [tex]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}][/tex].
     
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