Question about perturbation theory and even and odd wavefunctions I was solving a question about perturbation theory and I came across something my teacher didn't mention and I  can't seem to understand it. In the question there is an external electric field on a H-atom. I can neglect hyperfine structure and lamb shift. The atom-field interaction term in the Hamiltonian is 
$$−\vec{d}\cdot \vec{E} = e z E$$
where $\vec{d}$ is the dipole moment of the atom.
I have to find the first order correction to the energy level of the ground state using perturbation theory. In the answers they say that the wave function is even while the perturbation is odd so the resultant energy shift is 0.
I don't really understand this. Why is the perturbation odd and why is the wave function even? And why is the resultant energy shift then 0? Thanks!
 A: The first order perturbation correction is just the matrix element of the perturbation in respect to the wave function of the state of interest:
$$E^{(1)} = \int dxdydz \psi(x,y,z)^*\hat{V}\psi(x,y,z).$$
The wave function of a bound state can be taken to be real. In an atom some of these functions are even and other are odd, i.e.
$$\psi(-x, -y, -z) = \pm\psi(x,y,z).$$
This is frequently referred to as the state parity.
The perturbation is odd:
$$V(x,y,z) = ezE = -e(-z)E = -V(-x, -y, -z).$$
Therefore
$$E^{(1)} = \int dxdydz \psi(x,y,z)^*\hat{V}(x,y,z)\psi(x,y,z) =\\
\int dxdydz \psi(-x,-y,-z)^*\hat{V}(-x,-y,-z)\psi(-x,-y,-z)=\\
-\int dxdydz \psi(x,y,z)^*\hat{V}(x,y,z)\psi(x,y,z) = -E^{(1)},
$$
That is $E^{(1)} = -E^{(1)} =0$.
Note that this is the case for both even and odd states, since $\psi(x,y,z)^*\psi(x,y,z) = |\psi(x,y,z)|^2$ is even for either parity. This also has an important consequence that the transitions are possible only between the states of different parity - this may surface in calculating the second order perturbative correction (the proof is very similar).
