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I am wondering how I can see that the Hartree-Fock orbitals of a periodic crystal obey the Bloch theorem? My problem is that the Hartree-Fock Hamiltonian does not have the form $-\frac 1 2 \vec\nabla^2 + V(\vec r)$ but also includes the Coulomb and exchange terms: $$ H_{\text{HF}}\,\psi_i(\vec r) = -\frac 1 2 \vec\nabla^2\psi_i(\vec r) + V(\vec r)\psi_i(\vec r) + \sum_j\int\text d^3 r' \frac{\psi_j^*(\vec r')\psi_j(\vec r')}{|\vec r - \vec r'|}\psi_i(\vec r) - \sum_j\int\text d^3 r' \frac{\psi_j^*(\vec r')\psi_i(\vec r')}{|\vec r - \vec r'|}\psi_j(\vec r) $$ Here $V(\vec r)$ is the periodic potential of the atoms such that $V(\vec r) = V(\vec r + \vec R)$ for any lattice vector $\vec R$ of the crystal.

So my idea is that it should be enough to show that $[H_{\text{HF}}, T_{\vec R}] = 0$ where $T_{\vec R} = \text e^{\vec R \vec\nabla}$ is the translation operator. If this is true then $H_{\text{HF}}$ and $T_{\vec R}$ have the same eigenfunctions which have the form $$ \psi_i(\vec r ) = \text e^{\text i \vec k_i \vec r}u_i(\vec r)\; $$ where $u_i(\vec r) = u_i(\vec r+\vec R)$ has the same periodicity as the crystal.

However, the problem is that the Coulomb and exchange operator are composed by the orbitals. So I do not really understand how to compute $[H_{\text{HF}}, T_{\vec R}]$ ?

Or is there a better way to show that the Hartree-Fock orbitals of a crystal are Bloch waves?

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    $\begingroup$ I don't believe HF orbitals are necessarily Bloch waves. When Hartree-Fock is applied in practice, it is commonly assumed that the orbitals can be written as a linear combination of Bloch waves (e.g. eq 72 here). $\endgroup$
    – lemon
    Commented Apr 7, 2016 at 10:41
  • $\begingroup$ Hm, ok but for which physical system is the Bloch theorem applicable? If the Hamiltonian has to be of the form $-\frac 1 2 \vec \nabla^2+V(\vec r)$ then electron-electron interaction can never be considered. Even for the Hartree-Fock or the DFT approximation it can not be applied... $\endgroup$
    – thyme
    Commented Apr 7, 2016 at 11:54

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Please keep in mind lemon's observation as to the correct nature of Block waves. This being said, the Hartree-Fock Hamiltonian is a general ansatz for localized orbitals, it applies equally well to periodic lattices and to non-periodic molecular systems. For a periodic lattice however, translation symmetry imposes that localized HF orbitals are (ideally) translated copies of reference orbitals. Then the HF Hamiltonian becomes invariant under translations as well, $$ H({\vec r}) = H({\vec r} + {\vec R}) $$ and this implies, for any wavefunction $\psi({\vec r})$, $$ T_{\vec R}H({\vec r})\psi({\vec r}) = H({\vec r} + {\vec R}) \psi({\vec r} + {\vec R}) = H({\vec r}) \psi({\vec r} + {\vec R}) = H({\vec r}) T_{\vec R} \psi({\vec r}) $$ So $[T_{\vec R}, H({\vec r})] = 0$ and the Bloch theorem applies. See Eq.(8.8) of Ashcroft and Mermin, Solid State Physics.

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  • $\begingroup$ Thanks for your answer. What do you mean with "For a periodic lattice translation symmetry imposes that localized HF orbitals are translated copies of reference orbitals"? Does that mean that the Hartree-Fock orbitals are calculated only for one unit cell? My problem is that I intuitively accept that we get Bloch orbitals but I can not see it in a mathematical rigorous way. Could you please show me how you find $H(\vec r) = H(\vec r + \vec R)$? If you use $H_{\text{HF}}$ of my question this is simply not true. $\endgroup$
    – thyme
    Commented Apr 8, 2016 at 15:34
  • $\begingroup$ Sorry for answering late. The HF Hamiltonian is after all a self-consistent approximation to the total crystal Hamiltonian, and as such is expected to carry its defining symmetries, including of course translation. With this in mind, if $\psi_i({\vec r})$ is any localized HF solution (not translationally invariant), then translation symmetry requires that its translations are also valid HF solutions. $\endgroup$
    – udrv
    Commented Apr 11, 2016 at 7:00
  • $\begingroup$ If you take this into account for the sums defining the Coulomb and exchange terms, you can show that the HF Hamiltonian is indeed self-consistent wrt translational symmetry. Hence $H({\vec r}) = H({\vec r} + {\vec R})$. $\endgroup$
    – udrv
    Commented Apr 11, 2016 at 7:00

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