Gauge Freedom

In the previous section, it was implied that a gauge freedom exists with regard to the zeroth order occupied orbitals. In order to clarify this issue, consider the generalised Kohn-Sham equations, equation 3.26. These differ from the Kohn-Sham equations as usually presented

$$H\vert\Psi_{\alpha}^{(0)}\rangle = \epsilon_{\alpha}\vert\Psi_{\alpha}^{(0)}\rangle.$$ (3.35)

Defining a $(N\times N)$ unitary matrix such that

$$U^{-1}_{\alpha\beta} = U^{\dagger}_{\alpha\beta}$$ (3.36)

then if another set of occupied orbitals is defined through a unitary transformation

$$\vert\Phi^{\prime}_{\alpha}\rangle = \sum_{\gamma=1}^{N} U_{\gamma\alpha}\vert\Phi_{\gamma}\rangle$$ (3.37)

it is clear that both total electronic energies and densities remain the same, whilst the orthonormality condition is still satisfied. This is the result of a $U(N)$ gauge freedom. If one takes the occupied orbitals to be the solution of the Kohn-Sham equations, as will be done for the purposes of this work, then this is equivalent to stating that

$$\Lambda^{(0)}_{\beta\alpha} = \delta_{\beta\alpha}\epsilon^{(0)}_{\alpha}$$ (3.38)

and

$$U_{\beta\alpha} = \delta_{\beta\alpha}.$$ (3.39)

However, this only applies to the zeroth order wavefunctions. It is necessary to consider the issue of gauge freedom with regard to the higher order wavefunctions. It is most convenient to consider the orthonormalisation condition, which may be written as

$$\sum_{j=0}^{i}\langle \Phi_{\alpha}^{(i)}\vert\Phi_{\beta}^{(i-j)}\rangle = 0$$ (3.40)

which implies, to first order, $i=1$

$$\langle \Phi_{\alpha}^{(0)}\vert\Phi_{\beta}^{(1)}\rangle + \langle \Phi_{\alpha}^{(1))}\vert\Phi_{\beta}^{(0)}\rangle = 0.$$ (3.41)

This though only fixes the real part of the above scalar product; the imaginary part is not fixed, and thus one may impose the stronger condition that

$$\langle \Phi_{\alpha}^{(0)}\vert\Phi_{\beta}^{(i)}\rangle - \langle \Phi_{\alpha}^{(i)}\vert\Phi_{\beta}^{(0)}\rangle = 0.$$ (3.42)

This defines the parallel transport gauge, and for the first order wavefunctions yields

$$\langle \Phi_{\alpha}^{(0)}\vert\Phi^{(1)}_{\beta} \rangle = 0$$ (3.43)

which determines the projection of the first order wavefunction on the valence manifold. As can be seen, the parallel transport gauge ensures that the first order wavefunction is completely orthogonal to the valence manifold. It should be noted that this is not the only gauge possible; it is sometimes useful, if one is interested in obtaining derivatives of the Kohn-Sham eigenenergies, to demand that the Lagrange multiplier matrix is diagonal at all orders. This is the ``diagonal'' gauge [71], but it is only mentioned in passing here.