# Is there any atom which is dia-electric?

Take an atom. Suppose we impose some magnetic field on it. For some atoms, the energy increases---this is a phenomenon of diamagnetism.

The question is, how about an electric field? Can the energy of the atom increase when the electric field is turned on?

Put in a different way, can the electric field induce a dipole anti-parallel to it? If not, why can a magnetic field?

• The phenomenon is called "paraelectric," not "dia-electric."
– Buzz
Commented May 22, 2016 at 14:10

## 2 Answers

No, there can't be atoms like that, at least not in the real world.

In the magnetic case, diamagnetism means that the magnetic susceptibility may be negative (so the permeability may be lower or higher than in the vacuum, the magnetic susceptibility may have both signs).

But in the electric case, the electric susceptibility is always positive, and the permittivity of a material is always greater than the permittivity of the vacuum.

The asymmetry arises because atoms are full of charged particles that move in the expected direction. On the other hand, they don't contain any magnetic monopoles, just dipoles, and their behavior is harder to predict, if I omit the detailed explanations of paramagnetism, diamagnetism etc.

• I do not doubt your conclusion. But, it seems that the argument is very hollow. Commented May 22, 2016 at 12:42

Possibly it can be proven in this way. Let $H_0$ denote the unperturbed Hamiltonian, and let $D_z = \sum_{i=1}^N z_i$ denote the $z$-component of the electric dipole.

Define

$$H (f) = H_0 - f D_z .$$

Let $E_g(f)$ be the ground state energy of $H(f)$. First, apparently, $E_g$ is an even function of $f$.

Let $|\psi(f)\rangle$ be the ground state of $H(f)$. We have

$$E_g(0) = \langle \psi(0)|H(0)|\psi(0)\rangle =\frac{1}{2} \langle \psi(0)|H(f)|\psi(0)\rangle + \frac{1}{2} \langle \psi(0)|H(-f)|\psi(0)\rangle \\ \geq \frac{1}{2} \langle \psi(f)|H(f)|\psi(f)\rangle + \frac{1}{2} \langle \psi(-f)|H(-f)|\psi(-f)\rangle \\ = \frac{1}{2} E_g(f) + \frac{1}{2} E_g (-f) = E_g(f) . \\$$

QED.

Anyway, the point is that, the ground state energy as a function of $f$ is a concave function.

• Sorry, how does the fact that it's an electric dipole and not a magnetic dipole enters? It doesn't seem to enter, so you may give the same proof for the magnetic field as well, and that conclusion is wrong. The question is whether the function may be concave or convex or just one of them and you don't give a valid proof of an answer. Commented May 22, 2016 at 14:17
• Is this salad of complicated symbols generated by AI? Looks like nonsense. Commented Jun 20 at 17:41
• So where does the $\geq$ statement come from? Seems to be that you assume the desired conclusion there. And you didn't appeal to any properties of $d$ or $\psi$. Commented Jun 20 at 20:57