Im brooding about this now for a long time and I dont get why this equals 0. Its an excerpt from the book " R.Loudon – The Quantum Theory of Light " (Link to the full Book). This excerpt is on Page 50 and its about the interaction of light/electromagnetic wave with a two level System (here an atom with Z electrons and two possible energy states). $\hat{H}_I$ is the interaction Hamiltonian ($\hat{H}_I=e\sum_{i=1}^{Z}\vec{r}_i\cdot\vec{E}_0cos(wt)$ with $\vec{E}_0cos(wt)$ being the electric field of the light and $\vec{r}_i$ being the vectors to the electrons (the nucleus is in the origin of the coordinate system)). And finally $1$ and $2$ represent the states of the atom being in the first or rather in the second energy state.
Now I hope someone can explain to me why this needs to be zero. I really dont get it. And I'm also not sure what "Its matrix elements diagonal in the atomic states must vanish" means
Best Answer
In the basis given by the two states $|1\rangle, |2\rangle$, the Hamiltonian is a 2 x 2 matrix, with element $(i,j)$ given by the braket $\langle i| \mathcal{H} |j\rangle$. Thus, the diagonal of the matrix is the elements $\langle 1| \mathcal{H} |1\rangle$ and $\langle 2| \mathcal{H} |2\rangle$.
Bra-ket notation can be rewritten as an integral, as the author points out in Eq. (2.1.12). If we for simplicity assume one dimension only, the position representation of the braket is
$$\langle 1| \mathcal{H} |1\rangle = \int_{-\infty}^{\infty} dx\ \psi^*_1(x) \mathcal{H}(x) \psi_1(x). $$
They claim that the integrand is odd in $x$. If that is the case (e.g. if $\mathcal{H}(x)$ is odd and $\psi^*_1(x)$ and $\psi_1(x)$ have the same parity), the symmetric bounds on the integral mean it is zero.