Hamiltonian with ext. vector potential - complex kinetic energy

in a given (TD)DFT code with an atomic basis set, i.e. $$\psi(\mathbf{r},t) = \sum_i c_i(t) \phi_i(\mathbf{r})$$ (where the non-on-site basis functions $\phi_i$ aren't necessarily orthogonal), the kinetic energy matrix is calculated via $$T_{ij}=\langle \phi_i|\hat{\mathbf{T}}|\phi_j\rangle \sim\langle\phi_i|\hat{\mathbf{p}}^2|\phi_j\rangle \sim -\langle\phi_i|\nabla^2|\phi_j\rangle.$$ Now in the case where an external time-dependent and spacial homogeneous vector potential $\mathbf{A}(t)$ is considered, the momentum operator reads $$\hat{\mathbf{p}}'=\hat{\mathbf{p}}+e\mathbf{A}$$ and the kinetic energy matrix elements will be like $$T_{ij}'\sim\langle\phi_i|-\nabla^2-2ie\mathbf{A}\cdot\nabla+e^2\mathbf{A}^2|\phi_j\rangle$$ since $\nabla\cdot\mathbf{A}=0$ is used.

Now my problem: obviously, there is a complex contribution to the kinetic energy that won't vanish. How can this be resolved, what am I missing?

Thanks and best regards!

• Why is this a problem? A Hermitian operator corresponds in quantum mechanics to a real observable of classical mechanics. And are you sure your basis functions are real? – akhmeteli Dec 21 '16 at 9:49

The operator $\vec{p} \vec{A}+\vec{A} \vec{p}$ is Hermitian which means that its eigenvalues are real. So if you evaluate the matrix element between identical states (it corresponds to the mean value of the observable) which are real in some basis, you will just get $0$. The same story as with the operator of momentum.
The basis functions $\phi_i$ (not eigenfunctions of the hamiltonian) are given real functions and the program gives you $\phi_i(\mathbf{r})$, $\nabla \phi_i(\mathbf{r})$ and so on. Then, the matrix elements are calculated via a numerical integration as defined above: $$T_{ij}'\sim-\int_V\phi_i(\mathbf{r})[\nabla^2\phi_j](\mathbf{r})-2ie\mathbf{A}\cdot\int_V\phi_i(\mathbf{r})[\nabla\phi_j](\mathbf{r})+e^2\mathbf{A}^2\int_V\phi_i(\mathbf{r})\phi_j(\mathbf{r})$$ So this is a complex number. Maybe I am missing the point in the sense, that, finally, the solution to the generalized eigenvalue problem, $$\left(\mathbf{H}-E\mathbf{S}\right)\mathbf{c}=0$$ must deliver real coefficients $c_{ij}$ while $\mathbf{H}$ can be complex (thus, the eigensolver has to be chosen appropriately)?