Why the dipole interaction term in the Hamiltonian has all diagonal elements to be zero in the energy eigenbasis? I have been studying the semi-classical light matter interaction from the book, "Light matter interaction" by Weiner and Ho. They have defined the total Hamiltonian of a two level atom placed in an EM radiation as
$$\hat{H}=\hat{H_{0}}+\hat{V}(t)$$
where $H_0$ is the unperturbed Hamiltonian of the two level atom and $\hat{V}(t)$ is the dipole interaction term given by $\hat{V}(t)=\hat{\vec{d}}\cdot\vec{E}$. $\vec{d}$ is the dipole moment of the atom given by $\vec{d}=-e\vec{r}$. The eigenstates of $H_0$ are $|1\rangle$ and $|2\rangle$.
Now, it has been argued that since $V(t)$ has an odd parity with respect to $\vec{r}$, the diagonal terms
$V_{ii}=\langle i|\hat{V}|i \rangle=0$. However, I am not able to understand why this should be so.
Some thoughts
Without loss of generality, we can take the dipole moment to be $\hat{\vec{d}}=-e\hat{x} \mathbf{e_x}$ and the driving field $\vec{E}=E_0 cos(\omega t) \mathbf{e_n}$, so that $\hat{V}(t)=-e\hat{x} E_0 cos(\omega t) cos(\phi)$ where $\phi$ is the angle between $\mathbf{e_n}$ and $\mathbf{e_x}$. In such a case, we have
$$V_{ii}=\langle i|\hat{V}|i \rangle=e E_0 cos(\omega t)cos(\phi)\int_{-\infty}^{\infty}\phi^{*}_i(\vec{r}) x \phi_i(\vec{r}) d\vec{r}$$
If the wavefunction $\phi^{*}_i(\vec{r})$ has a definite parity(assumption 1), then indeed this integral is $0$.
In case of a spherically symmetric potential with no interaction between electrons in the atom, assumption 1 indeed holds. However, in the presence of interaction between electrons, I am not so sure if it will hold true!
Am I thinking about it the right way? Is there a way to prove this without assumption 1? Or is the assumption 1 always true?
 A: 
If the wavefunction $\phi^{*}_i(\vec{r})$ has a definite parity (assumption 1), then indeed this integral is $0$.

This is the only thing that's going on. We assume that the system is invariant under parity, and therefore that its eigenfunctions have definite parity and therefore that the eigenstates do not have a permanent dipole moment.
For some systems, this assumption can indeed break: notable examples are (the electronic states of) the water and ammonia molecules. If the assumption breaks, then the on-diagonal terms of the interaction potential do need to be included. This is inconvenient, and it makes everything more of a hassle, but it doesn't really introduce any qualitative changes to the physics, which is why it's rarely included unless it's explicitly necessary.
A: The reason is that such terms are usually "absorbed" in the main Hamiltonian, where they represent a small correction to the difference between the energy levels. Note that absorbing the diagonal terms to the Hamiltonian is a rather common procedure, by no means specific to the dipole approximation.
Example
E.g., for a two-level system with eigenstates $|1\rangle, |2\rangle$ we have
$$
H=H_0 + V(t) = \begin{bmatrix} E_1 & 0 \\ 0 & E_2\end{bmatrix} +
\begin{bmatrix} V_{11}(t) & V_{12}(t) \\ V_{21}(t) & V_{22}(t)\end{bmatrix} =
\begin{bmatrix} E_1+V_{11}(t) & 0 \\ 0 & E_2+V_{22}(t)\end{bmatrix} +
\begin{bmatrix} 0 & V_{12}(t) \\ V_{21}(t) & 0\end{bmatrix} = H' + V'(t)
$$
Different energy scales
If, e.g., we want to calculate the transition probability using the Fermi golden rule, we have
$$
w_{i\rightarrow f} =\frac{2\pi}{\hbar}|V_{if}|^2\delta(E_f-E_i\pm \hbar\omega)
$$
As one can see, the role of the non-diagonal and the diagonal elements of the eprturbation is different: the diagonal elements, absorbed into the energies $E_{i,f}$ adjust the energy conservation equality $E_f-E_i\pm \hbar\omega=0$, but this adjustment is small, since in most practical situations $$\left|\frac{V_{ij}}{\hbar\omega}\right|\ll 1, \left|\frac{V_{ij}}{E_2 - E_1}\right|\ll 1.$$
On the other hand, the non-diaginal elements, $V_{if}$, determine the rate of transitions, which cannot be neglected, since it is compared to zero (no transitions at all).
