5
$\begingroup$

Just as a photon's polarization determines its spin angular momentum (or more accurately its helicity), its wavefront can carry orbital angular momentum (OAM) much like an electron in an atom.

Following this Phys.SE question, Selection rules for atomic emission: why doesn't the photon carry orbital angular momentum, it is clear that since atoms are so small (<0.1 nm), they don't really "feel" the OAM of light. This is because the wavefront of optical light carrying OAM changes on the scale of many hundreds of nanometers for optical light, much larger than the atom. Thus, the atom shouldn't really be able to tell the difference between an optical photon carrying non-zero OAM and one with zero OAM.

The logic of the above question is quite sensible in my opinion, but the issue I am having is: where does the OAM go when an atom happens to absorb a photon with non-zero OAM?

I would assume that OAM is conserved in the absorption of the photon and goes into center-of-mass atomic OAM, but is that necessarily true? And how do we reconcile the atom's photon-acquired OAM with the fact that the field is locally identical to an ordinary photon with zero OAM on the length-scale of atoms?

Edit: I want to emphasize that I am only considering optical (visible) light, which is much larger in wavelength than the atom. Additionally, I am thinking of the general case of an atom not located precisely in the central vortex of the OAM beam, but on one of its side lobes.

$\endgroup$
1
$\begingroup$

This is going to be a bit of a half baked answer but here it is.

The usual picture for light with orbital angular momentum is light in a Laguerre-Gaussian Mode mode with non-zero azimuthal index. This sort of field has phase wrapping azimuthally around the propagation axis.

We suppose without loss of generality that the atom is on the propagation axis of the beam*

As I point out in the linked post and as you emphasize in the question, typically when we think about interactions between atoms and light we work in the dipole approximation meaning we assume the electric field is homogeneous across the whole volume of the atom. In the case of light with angular momentum we can see this is not the case. Consider the simplest Laguerre-Gaussian mode with a non-zero phase wrapping.

Brightness indicates intensity, color indicates optical phase

If the atom is in the center of this picture we can clearly see that the dipole approximation is NOT satisfied. This is because the phase of light on the right half of the atom is opposite to the on the left side of the atom. This means no matter what we do with, for example, the scaling of the beam, the atom will never be experiencing a homogeneous field. This means the dipole approximation fails.

When the dipole approximation fails all of the "usual" selection rules that you learn in intro classes are thrown out the window and you need to use higher order selection rules.

I don't know enough about all the details of these selections rules but you can see that this pattern could impart angular momentum to the electron around the nucleus.

And that is my answer to your question. Just like how spin angular momentum of light imparts angular momentum to the electron (atomic "internal" angular momentum) orbital angular momentum of light also imparts momentum to the angular momentum of the electron.

Note that you state

it is clear that since atoms are so small (<0.1 nm), they don't really "feel" the OAM of light.

And you drew this conclusion from my answer to the other question. This conclusion is incorrect. It is possible to have optical fields which vary on atomic length scales (Like the one I have shown above). It just requires using tightly focused or very high order optical fields which need to be carefully aligned so that their regions of large gradients are overlapped with the atom. For example, if the Laguerre-Gaussian mode depicted above is made to be a large beam (waist of a few mm for example) and is offset so that the atom is like in the middle of the red part then the field will look homogeneous to the atom. It is only when the atom is at the central vortex that it strongly experiences the field inhomogeneity.

All of that said you may be interested in Spin-orbital-angular-momentum coupled Bose-Einstein condensates. Note that here the light is not coupled to a single atom but rather to a quantum gas of many atoms in the BEC state. In this case the orbital angular momentum of the light is transferred into the orbital angular momentum of the BEC.

*We can always choose a basis for the optical field which is centered on the atom. If beam is displaced from the atom then in this particular basis it will be composed of a large number of high order modes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A few comments: I agree the dipole approximation fails if the atom is located in the dead center, but I was imagining a more realistic scenario where the atom was not located at the center but off to the side. I also agree that if the wavelength of the photon is short enough, it's wavefront size can become similar to the atom and things make sense, which is why I focused on optical light in my question. $\endgroup$ – KF Gauss Nov 6 '19 at 13:04
  • $\begingroup$ I think the answer lies within your final footnote about off-axis atoms! Angular momentum to me should be conserved regardless of where the atom is (though the exact value is dependent on the com frame). How does the atom manage to sense the OAM in the off-axis case, and where exactly is that OAM going? $\endgroup$ – KF Gauss Nov 6 '19 at 13:07
  • $\begingroup$ For the sake of concreteness, I am thinking about light tuned at the Hydrogen $s\rightarrow p$ transition with light carrying OAM of $m=1$ with the hydrogen located outside of the central vortex. By angular momentum conservation, it would naively seem that the s to p transition is impossible because the light has an angular momentum mismatch. Yet when i think of the situation classically, the off-axis atom should absorb the resonant light just fine without any knowledge about the OAM. $\endgroup$ – KF Gauss Nov 6 '19 at 13:22
  • $\begingroup$ @KFGauss yes some of your insight is correct. I might expand the footnote as part of this answer or as another answer since that is where you’re question lies. For now I’ll say the following. In your situation with an atom off axis from an m=1 optical mode the following would happen. Everything would be in a superposition state. If we adjust the axis to be aligned with the atom instead of the beam we’d see the optical mode is a superposition of modes with different angular momenta. This would include a big fraction of m=0 spatial mode. The atom would absorb from this low order spatial mode. $\endgroup$ – jgerber Nov 6 '19 at 16:31
  • $\begingroup$ However, the atom would not absorb light from the higher order spatial mode components. This means after the atom the light will be a superposition of modes, but now the m=0 mode will have less amplitude. The atom will be in a superposition of having absorbed and not absorbed a photon. Furthermore, the light and atom will be entangled. That is, if the atom is found in the excited state Then there will be no light since it was absorbed. If the atom is found in the ground state there will still be light, but less light than if there were no atom in the m=0 spatial component of light. $\endgroup$ – jgerber Nov 6 '19 at 16:36
1
$\begingroup$

After reviewing the comments I believe KF Gauss is correct in their statement that the atom picks ups angular momentum with respect to the light propagation axis. See Eq. 5.448 in Quantum and Atom Optics by Steck regarding the mechanical force on an atom by an optical field.

$$ \langle \boldsymbol{F}\rangle = \frac{i\hbar|\Omega(\boldsymbol{r})|^2}{4\left(\frac{\Gamma}{2} - i \Delta\right)(1+s(\boldsymbol{r}))}\left(\nabla \log\left(|\Omega(\boldsymbol{r})|\right) - i \nabla \phi(\boldsymbol{r}) \right) + c.c. $$

Here $\Omega(\boldsymbol{r}) = |\Omega(\boldsymbol{r})|e^{i\phi(\boldsymbol{r})}$ is the complex spatially dependent Rabi frequency. The square magnitude is proportional to the local field intensity as well as some atomic structure parameters and the phase is the phase of the optical field. $\Gamma$ is the atomic spontaneous emission decay rate from whatever excited states are considered for the atomic transition, a two-level approximation is appropriate so that $\Gamma$ is decay from the excited state. $\Delta$ is the detuning between the light field and the atomic transition under consideration. $s(\boldsymbol{r})$ is the atomic transition saturation parameter.

$$ s(\boldsymbol{r}) = \frac{|\Omega(\boldsymbol{r})|^2}{2\left[\left(\frac{\Gamma}{2}\right)^2 + \Delta^2\right]} $$

The first term is the dipole force which says that there is a force proportional to gradient of the intensity of the field. The second term is the radiation pressure force which is a force proportional to the gradient of the phase of the optical field.

Because the phase of an optical field with orbital angular momentum wraps azimuthally (at least away from the center), an atom place in the region of high intensity would feel an azimuthal force, thus impartin orbital angular momentum to the atom.

Note that the formula for the force above is derived from the optical Bloch equations including spontaneous emission. Thus the force arises as a result of the atom absorbing light and spontaneously emitting it in a random direction. I bring this up because it's not entirely obvious to me what would happen if we considered absorption alone.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Would you mind defining the terms in your expression to keep the answer self-contained? $\endgroup$ – KF Gauss Nov 9 '19 at 21:01
  • $\begingroup$ @KFGauss Ok, I've done it. $\endgroup$ – jgerber Nov 9 '19 at 21:12
1
$\begingroup$

The key to understanding how orbital angular momentum (OAM) can be conserved when a photon that carries OAM is absorbed by a small particle is to understand OAM itself a little better. Note that OAM in a paraxial light beam is always defined with respect to the propagation axis of that beam. If one would pick some random axis pointing in some random direction and displaced away from the beam by some random distance, then the OAM would be different. This is because the linear momentum of the beam now contributes significantly to that amount.

Now back to the particle that absorbs a photon with OAM. As you pointed out, the probability for such an absorption is very small, on the order of $$ \text{probability}=\left(\frac{d}{\lambda}\right)^{2\ell} , $$ where $d$ is the size of the particle and $\ell$ is the topological charge of the optical mode (an integer proportional to the OAM). However, this expression assumes that the particle sits in the centre of the beam, at the vortex, where the light would have been able to apply some torque on the particle. However, the intensity of the light is almost zero. That's why the probability for absorption is so small. It also assumed the mechanism for the absorption is govern by the dipole moment of the particle. If the quadrupole moment is involved, the probability for absorption is proportional to the gradient of the optical field and thus becomes much larger.

But what if the particle sits off-axis in the beam? The probability for absorption (via the dipole moment) becomes much larger, because the light intensity is much larger. This time, there is no vortex that can apply a torque to the particle. So what happens to the OAM? This is where the picture of the linear momentum that I explained before becomes important. The particle would pick up a linear momentum from the photon. In this region, the wave front of the optical field is tilted. So the local propagation vector that optical field has in this region would give the particle a linear momentum that has a slight transverse component. It would be projected along a straight path that is twisted with respect to the propagation axis of the optical beam. When one computes the OAM due to this linear momentum with respect to the original propagation axis of the optical beam, one finds that it gives exactly the amount of OAM that was carried by the photon that was absorbed.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you quantify and/or provide a diagram for your last paragraph? The linear momentum of light $\hbar k$ doesn't seem to be relevant (or at least it's not obvious to me) since it contributes to a perpendicular angular momentum relative to the OAM. Is it just the case that an off-axis atom absorbing the OAM light gets a velocity perpendicular to the original OAM such that $r \cdot mv = \hbar l$ with $\mathbf{v} \perp \mathbf{L}$. So OAM is conserved by just adding a small transverse velocity to the atom? $\endgroup$ – KF Gauss Nov 9 '19 at 20:58
  • $\begingroup$ Also, do you have similar expressions for the absorption probability for the quadrupole (or higher) terms? $\endgroup$ – KF Gauss Nov 9 '19 at 20:59
  • $\begingroup$ See the slight edits. Yes, the OAM is conserved thanks to a slight transverse velocity/momentum component due to the tilt in the wavefront of the optical field. I also added a bit about the quadrupole interaction being proportional to the gradient of the optical field. $\endgroup$ – flippiefanus Nov 10 '19 at 4:17
-1
$\begingroup$

where does the OAM go when an atom happens to absorb a photon with non-zero OAM

It is simple : the photon can raise an electron to a higher energy level only if all the quantum numbers are conserved,including angular momentum. It is not enough to have the energy difference ( within the width of the energy level).

If the photon is head on with the atom, its trajectory passes through the center of mass, it will have zero angular momentum, which is the usual visualization of exciting energy levels. If its angular momentum is close (within the width of the level) to an integer number of h_bar, and an energy level exists that can absorb its energy ( within the width) the photon will be absorbed, other wise it will scatter off. The atom will be at the appropriate l, so the photon's angular momentum becomes the atom's, ( until the atom de-excites the electron falling to a lower energy level)

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Unlike a normal Gaussian beam, Orbital angular momentum carrying light definitely has non-zero Angular momentum if the atom is in the center. See the image in jgerber's answer for reference. You seem to be saying all the momentum is conserved in the normal sense, how do we reconcile this with the fact that on the atomic scale the light field is practically constant outside the vortex region? $\endgroup$ – KF Gauss Nov 6 '19 at 13:17
  • $\begingroup$ @KFGauss Look at the sentence I am quoting . The question is about photons, single quantum mechanical particles. , and atoms are single quantum mechanical particles. Light emerges from a confluence of photons, but photons are not light. See my answer here for links on how light emerges from a zillion of photons. also the graphic may help physics.stackexchange.com/questions/243641/… $\endgroup$ – anna v Nov 6 '19 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.