1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Apr 7 '16 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 Apr 7 '16 at 11:54
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Apr 8 '16 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 Apr 11 '16 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 Apr 11 '16 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.