# About diagonal element of dipole

1. Nov 5, 2014

### KFC

Hi all,
I am reading a book on atom interaction with light. It starts with the 2 level system. The book said the diagonal elements of dipole operator is zero, i.e. $\langle g|x|g\rangle = \langle e|x|e\rangle = 0$. I don't understand why this is true but after searching online and there gives two reasons

1) the atom has no permanent dipole
2) the symmetry

https://www.physicsforums.com/threa...s-of-electric-dipole-element-are-zero.706659/

It explains the second reason. But is it always true that any atom will have such symmetry? or does it require the atom is in gas state?If so, why is that?

Also, why $\langle g|x|g\rangle = \langle e|x|e\rangle = 0$ implies no permanent dipole.

2. Nov 6, 2014

### Staff: Mentor

You can certainly polarize an atom. But because the Hamiltonian of an isolated atom is invariant under rotations, eigenstates have an overall symmetry.

The electric dipole moment of an ensemble of charges is defined as
$$\vec{\mu} = \sum_{i} q_i \vec{r}_i$$
Transposed to quantum mechanics, you need to calculate the expectation value of the right-hand-side over the state your are considering, hence $\vec{\mu} \propto \langle \psi | \hat{r} | \psi \rangle$.

3. Nov 6, 2014

### DrDu

You have to argue using parity, not rotational symmetry. E.g. in atomic nitrogen, one electron occupies a p orbital, so the wavefunction doesn't have full rotational symmetry and there will be a preferred direction.
However, due to parity, all eigenstates are either even or odd under inversion of coordinates $P \psi(r)=\psi(-r)=\pm \psi(r)$ and any expectation value of $\vec{\mu}$ vanishes as $\langle \psi(r)|\vec{\mu}|\psi(r)\rangle= \langle \psi(r)|PP\vec{\mu}PP|\psi(r)\rangle=-\langle \psi(r)|\vec{\mu}|\psi(r)\rangle$. However, parity is not an exact symmetry of nature and there are some high precision experiments which try to detect tiny dipole moments in ions or atoms due to parity violating effects .

4. Nov 6, 2014

### Staff: Mentor

Indeed. My mistake.

5. Nov 6, 2014

### KFC

Thanks for the explanation. But I still have two questions. Is this argument always true for any atom at any state? From your analysis, it seems that no matter what parity the state $|\psi\rangle$ has, $\langle\psi|r|\psi\rangle$ always be odd, so $\int dr^3 \langle\psi|r|\psi\rangle = 0$?

I read some other text, they said this is only true for atoms in gas state, for solid state, it doesn't hold?

The last question, again, why $\langle g|x|g\rangle$ and $\langle e|x|e\rangle$ called permanent dipole? I know DrClaude explained that but what I am asking is why they are "permanent". If they are not equal to zero, so what is physical significance of $\langle g|x|g\rangle$ and $\langle e|x|e\rangle$ separately?

Thanks.

6. Nov 7, 2014

### DrDu

Yes, generally, this statement holds only true in the gas phase. In a solid or molecule, the surrounding may break parity invariance. I.e. the orbitals get deformed by interaction with the surrounding atoms so that they are no longer invariant wrt inversion.
The states e and g are eigenstates of the hamiltonian. Hence time evolution changes them only by multiplication with a phase factor which drops out when you form expectation values. Hence these matrix elements are time independent or permanent.

7. Nov 7, 2014

### KFC

Thanks a lot. It makes sense to me.

May I ask one more question on this problem. In the text on classical E&M, it reads that electric dipole moment energy in a field $E(\vec{r}, t)$ should be

$-\vec{d}\cdot E(\vec{r}, t)$

So should the Hamiltonian be $H =-\vec{d}\cdot E(\vec{r}, t)$ ? In the text, it said the Hamiltonian is given under the dipole approximation. I find a material online about dipole approximation, they simply put that since $exp(ikr) = 1+ ikr + \cdots$, when we use only the first term, we have $H =-\vec{d}\cdot E(\vec{r}, t)$

It is really confusing! why we have to use $exp(ikr)$ and how does it connect to the Hamiltonian of the dipole moment?

8. Nov 8, 2014

### DrDu

The dipole approximation is an approximation for the hamiltonian of extended charge distributions who have not only a dipole moment but also higher moments, like the quadrupole moment. The coupling to higher multipole moments can be neglected, when the wavelength 2 pi/k is much larger than the dimensions of the charge distributions.

9. Nov 8, 2014

### KFC

I read this reasoning online. But what confusing me is what 'moment' really is, or what's the physical meaning for quadrupole,etc.? Why the moment always comes in "pair" so there are dipole quadrupole, octupole but nothing in between?

I still have a hard time to understand the statement "The coupling to higher multipole moments can be neglected, when the wavelength 2 pi/k is much larger than the dimensions of the charge distributions." So does it mean if you have a monochromatic laser which we consider it is a plane wave, if the wavelength is so long such that the size of the atoms is totally emerged in the wave and the wave just looks like a flat constant background, we can ignore the quadrupole and higher one so only consider the dipole is enough? What happens if the laser is not monochromatic or we don't use laser beam or the charged objected is so big so we should consider higher multipole moments?

I learn something called multipole expansion in a text about classical E&M, there they expand the inverse distance from the charge to the field point (1/r) in taylor expansion. They said the first term of the expansion is the monopole, the second term is the dipole, etc. Is this really the same thing you mentioned above? If yes, that what confusing me is how "2 pi/k is much larger than the dimensions of the charge distributions" comes into this analysis?

10. Nov 8, 2014

### DrDu

Yes, the multipole expansion in classical mechanics is the same we are talking about here.
The interaction term in the hamiltonian can be written as $-d_i E_i(0)-Q_{ij} \nabla_i E_j(0) -\ldots$ (summation over doubly occuring indices is understood). Note that I didn't check the signs and eventual pre-factors. Q is of the order $qr^2$, where q is the charge of the system and r the extension of the charge distribution, while $\nabla E$ is of the order $E_0/\lambda$. d is of order $qr$ and $E_0=E(0)$ so the quadrupolar term is smaller by about $r/\lambda$.

It is not important whether the laser beam is monochromatic or not. In the optical range, considering the dipole moment is usually enough, if not a transition of interest is dipole forbidden.
For nuclear transitions, the situation is different, as the wavelength of gamma radiation is comparable to the size of the nucleus.

11. Nov 8, 2014

### KFC

Thanks for the explanation.