Photon angular momentum and magnetic quantum number selection rules

In summary: The quantization axis is set in the laboratory frame, and is not related to the direction of propagation of the photon.
  • #1
Tainty
27
1
For an atom, the single photon electric dipole selection rules for the magnetic quantum number require that delta_m = -1, 0 or +1.

As I understand, the physical explanation for this set of selection rules is usually related to the conservation of the projection of the angular momentum on the quantization axis.

This leads to delta_m = + or – 1 for right or left circularly polarized light and delta_m=0 for linearly polarized light.

My question then is, what exactly do we mean by the quantization axis? Does it need to be the direction of photon propagation? On one hand, it seems that it should be, since the projection of the spin angular momentum of a photon is either +1 or -1 on the direction of propagation (also defined as its helicity). But on the other, we also know that a photon cannot have zero angular momentum in its direction of propagation.

On a related note, if we think of a linearly polarized photon as having equal probability of being left and right circularly polarized, why isn’t delta_m = + or -1 allowed for linearly polarized light?

Can someone point out the errors in my argument? Thanks!
 
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  • #2
Tainty said:
My question then is, what exactly do we mean by the quantization axis? Does it need to be the direction of photon propagation? On one hand, it seems that it should be, since the projection of the spin angular momentum of a photon is either +1 or -1 on the direction of propagation (also defined as its helicity). But on the other, we also know that a photon cannot have zero angular momentum in its direction of propagation.
The quantization axis is set in the laboratory frame, and is not related to the direction of propagation of the photon.

Tainty said:
On a related note, if we think of a linearly polarized photon as having equal probability of being left and right circularly polarized, why isn’t delta_m = + or -1 allowed for linearly polarized light?
The confusion comes from the fact that the polarization considered here is, as I said above, not related to the direction of propagation of the photon. The idea is that instead of using as a polarization basis for the photon the three cartesian axes defined in the lab frame, ##\hat{\mathbf{e}}_x##, ##\hat{\mathbf{e}}_y##, and ##\hat{\mathbf{e}}_z## (all unit vectors), we use linear combinations in x and y. This gives the three elements of the polarization basis, ##\sigma^+##, ##\sigma^-##, ##\pi##:
$$
\begin{align*}
\sigma^+ &= \frac{1}{\sqrt{2}} \left( \hat{\mathbf{e}}_x + i \hat{\mathbf{e}}_y \right) & \sigma^- &= \frac{1}{\sqrt{2}} \left( \hat{\mathbf{e}}_x - i \hat{\mathbf{e}}_y \right) & \pi &= \hat{\mathbf{e}}_z
\end{align*}
$$
For example, a ##\pi## photon will have a linear polarization along z, regardless of its direction of propagation (which has to be in the xy plane). If you have a photon traveling in the +z direction and polarized linearly along the x axis, then it is a combination of ##\sigma^+## and ##\sigma^-##.

Usually, when discussing the polarization of a photon, it is assumed that we are talking about its polarization in the plane perpendicular to its direction of propagation (the plane in which the electric field oscillates). In that case, it can be seen to be left-handed circular polarized, ##\sigma_L##, or right-handed circular polarized, ##\sigma_R##, or a linear combination of the two resulting in elliptic or linear polarization. So it is important not to confuse ##\sigma_L## and ##\sigma_R## with ##\sigma^+## and ##\sigma^-##.

Let me give you a stark example of this. If you take a ##\sigma_R## photon traveling in the +z direction, it will be ##\sigma^+## in the lab frame, and its absorption will lead to ##\Delta m = +1##. If the same ##\sigma_R## photon is traveling in the -z direction, it will then be ##\sigma^-## in the lab frame, and the atom will undergo a ##\Delta m = -1## transition.
 
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  • #3
Thank you DrClaude.

What I am understanding from your response is that the polarization of the photon is independent of the choice of polarization basis (ie coordinate system). And that the quantization axis is really an arbitrary choice in the lab frame. I am able to accept that and it is what I’ve been believing to be the case all along.

But on the other hand, isn’t the projection of angular momentum of a circularly polarized photon (i.e. sigma_R and sigma_L based on your notation) only equal to +-hbar in its direction of propagation? And it is this “+-hbar” of momentum that is usually cited as the reason for delta_m=+-1 selection rules (where m is the magnetic quantum number) for circularly polarized light.

So in essence what I am unable to reconcile is that even though the choice of quantization axis is evidently arbitrary, the physical reasoning behind the selection rules seem to require that this axis should be direction of propagation.

Could you enlighten me further?
 
  • #4
Tainty said:
But on the other hand, isn’t the projection of angular momentum of a circularly polarized photon (i.e. sigma_R and sigma_L based on your notation) only equal to +-hbar in its direction of propagation?
Correct. But you have to be careful at what "circularly polarized" means here, that is, in what basis set it is circularly polarized. Take for example an ensemble of atoms prepared in an excited ##l'=1##, ##m'=+1## level, which can only decay to the ground ##l=0## level, and thus ##m=0## and ##\Delta m = -1##. All atoms will thus emit ##\sigma^+## photons (remember that in emission, ##\Delta m = \pm1## corresponds to ##\sigma^{\mp}##). The lab frame is used for the quantization axis of the atom, with +z chosen as pointing upward. Now an experimentalist takes a photon detector with a polarizer in front to find out the polarization of the photons emitted. If they put the detector along the z axis, they will find that the photons are all circularly polarized ##\sigma_R## if the detector is put above the cloud of atoms, and ##\sigma_L## if the detector is put below the cloud of atoms. Elsewhere, the polarization will be elliptical (and linear in some special places). To summarize, in the lab frame the photons are all ##\sigma^+## circularly polarized, but in the frame of the photon, it is most likely elliptically polarized.

Tainty said:
And it is this “+-hbar” of momentum that is usually cited as the reason for delta_m=+-1 selection rules (where m is the magnetic quantum number) for circularly polarized light.
That is correct when "circularly polarized light" refers to the lab frame. But if in my example above you take a laser with a circular polarizer in front, such that you have ##\sigma_R## photons, and aim it at an arbitrary angle to a cloud of ground state atoms, it will induce all three transitions, ##\Delta m = 0, \pm1##, in the atoms (and put atoms in superpositions of ##l'=1, m'=+1, 0, -1##), because the light is not circularly polarized in the frame of the atom.
 
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  • #5
What you point out is very interesting DrClaude, thanks again!

So what I'm learning is that the magnetic quantum number selection rules are intricately related to the basis of both the lab and atom. Makes a lot of sense to me.

Is it possible to align the lab frame with that of the atom? I presume the answer would be yes, because processes like optical pumping, where an atom can be prepared in a particular m (magnetic quantum number) state (just like in your example above) have been experimentally conducted for a long time. Or am I missing something again?

But suppose in theory that it is possible that we can align both frames of reference, i.e. the lab frame is the same as that of the atom. What happens then when we tune onto a resonance of an atom with linearly polarized light? Are we able to simultaneously excite the Δm=±1 transitions in addition to the Δm=0 transition? As you can see, my real confusion is with the m selection rules and in particular why m cannot also be +-1 for linearly polarized light (of course assuming that we can somehow align both the atom and lab frame).
 
  • #6
I would consider first a simple case as the transition from an s state into p states and use px, py and pz instead of the complex m eigenstates. It is quite clear that a transition into px can only happen when the light is polarized along x direction and correspondingly for the other directions.
 
  • #7
Hi DrDu, thank you for joining the discussion.

I think I can understand the m selection rules as far as circularly polarized light is concerned. I am just getting confused with the physical interpretation of linearly polarized light, since one should be able to think of it as a combination of left and right circularly polarized light. So why then isn't Δm=±1 allowed?
 
  • #8
Of course m=+-1 are allowed for linearly polarized light if the polarization vector lues in the xy-plane.
 
  • #9
Tainty said:
Is it possible to align the lab frame with that of the atom? I presume the answer would be yes, because processes like optical pumping, where an atom can be prepared in a particular m (magnetic quantum number) state (just like in your example above) have been experimentally conducted for a long time. Or am I missing something again?
I'm not sure that aligning the lab fram with the atom's frame means. When an atom is in a defined ##m## state, it is with respect to a quantization axis that is external to the atom. But it is certainly possible to polarize atoms (see post #4), which can indeed be achieved by optical pumping.

Tainty said:
As you can see, my real confusion is with the m selection rules and in particular why m cannot also be +-1 for linearly polarized light
I'll add to what DrDu answered. The statement doesn't mean that ##\Delta m = 0## for any photon that has gone through a linear polarizer, since that linear polarization is in the plane perpendicular to the photon's direction of propagation. The statement means that in the lab basis for the photon polarization, ##\pi##, ##\sigma^+##, ##\sigma^-##, only ##\pi## photons can induce ##\Delta m = 0##. It simply turns out that in that basis, ##\pi## is the only linear polarization, the other two being circular. That's why some people say call that "linearly polarized light." To be honest, I find that this is an abuse of language and leads to confusion, of which you are the unfortunate victim. If people kept referring only to ##\pi##, ##\sigma^+##, and ##\sigma^-## photons, I think it would be easier to understand.
 
  • #10
DrDu and especially DrClaude, that was really helpful. Everything is beginning to fall into place now.

Two clarifications:

1. Am i right that the notion of a π, σ+ or σ- photon is defined by us based on the polarization basis that we prescribe? In other words, if i shine a laser with σ+ light onto an atom so that i induce a Δm=1 transition, I can move this laser around (say by any angle, maybe even 180deg), prescribe another basis, and still induce the same Δm=1 transition?

2. And just to remove any ambiguity, I am thinking of all this in the absence of any Zeeman splitting such that the m energy levels are degenerate.
 
  • #11
No, Claude pointed already out that the polarisations of the light are relative to the chosen z-axis for quantization. So moving the laser around will change it's polarisation.
 
  • #12
DrDu, but isn't the z-axis an arbitrary choice? Are you saying that there is an absolute z-axis? Just to be clear, when i say move the laser around, I'm just trying to clarify that the transition i induce has got nothing to do with where the laser was positioned to begin with.
 
  • #13
The m values refer to the axis chosen. However you are right that moving the laser and the axis won't have observable effects, at least as long as the system is isotropic.
 
  • #14
Thanks a lot, I think I got it.

Upon reflection, I think my main error was that even though a linearly polarized photon can be thought of as a combination a left and right circularly polarized photon, a π photon cannot be made up of a combination of a σ+ and σ- photon.

Really appreciate all of your help!
 
  • #15
Hello again, it seems like a i got a little too far ahead of myself.

After further thought, i have another query. Suppose i shine linearly polarized photons onto an ensemble of atoms, if i use this linear polarization to define my quantization axis, then these photons would induce a Δm=0 transition. However, had i chosen my quantization axis differently, these linearly polarized photons could very well have been expressed as a combination of σ+ and σ-, which would instead yield transitions of Δm=±1.

So in a physical experiment, how does an experimentalist know with certainty which transitions are being targeted?
 
  • #16
When you are changing the axis, also the labels m of the orbitals change. E.g. the orbital ##p_0=p_z## becomes upon a 90 deg rotation around the y-axis the orbital ##p_x=\frac{1}{\sqrt{2}} (p_{+1} + p_{-1})##, so what appears as a ##\Delta m=0## transition from s to ##p_0## becomes a transition with ##\Delta m=\pm 1## when you chose a rotated quantization axis.
 
  • #17
DrDu, am I understanding correctly that you are in agreement with me that the nature of transition depends on the choice of quantization axis?

To elaborate further, what I've appreciated most from both you and DrClaude's responses is the fact that π, σ+ and σ- photons (and not the actual light polarization) lead to Δm=0,1 and -1 transitions respectively.

However, what relates these π, σ+ and σ- photons to the Δm selection rules is the choice of quantization axis. Since this choice is arbitrary relative to the actual light polarization (i.e. linear, left or right circular), my question is how does someone in the lab with a laser design an experiment to access a particular Δm transition?

It seems to me that the quantization axis needs to be well defined, for example by imposing an external magnetic field.
 
  • #18
Tainty said:
However, what relates these π, σ+ and σ- photons to the Δm selection rules is the choice of quantization axis. Since this choice is arbitrary relative to the actual light polarization (i.e. linear, left or right circular), my question is how does someone in the lab with a laser design an experiment to access a particular Δm transition?

It seems to me that the quantization axis needs to be well defined, for example by imposing an external magnetic field.
You don't need an external magnetic field. Take a laser beam with right-handed polarization (##\sigma_R##), decide arbitrarily to say that the laser propagates along the +z direction, and then atoms absorbing from the laser will undergo ##\Delta m = +1## transitions, simple as that.
 
  • #19
Hmm DrClaude, I think you probably know the answer to what I'm getting at, but analogous to what i mentioned in post#15, if this choice is arbitrary, does this mean that the transitions that are accessed are arbitrary too?

To be more specific, in your example, if i had decided arbitrarily instead that the laser was propagating in the -z direction, then wouldn't the Δm=-1 transition be accessed? And of course one could easily generalize this even further.
 
  • #20
Tainty said:
Hmm DrClaude, I think you probably know the answer to what I'm getting at, but analogous to what i mentioned in post#15, if this choice is arbitrary, does this mean that the transitions that are accessed are arbitrary too?
In a sense, yes. The atom itself does not have its own preferred reference frame. When you say an atom in a ##m=+1## state, it is with respect to an external quantization axis. Change the lab axes, and the atom, which is still in the exact same quantum state, can now be seen as being in a superposition of different ##m## states.
Tainty said:
To be more specific, in your example, if i had decided arbitrarily instead that the laser was propagating in the -z direction, then wouldn't the Δm=-1 transition be accessed?
That is correct.
 
  • #21
Ok thanks, i think that clarifies things for me.

Am I right that an external magnetic field (for e.g. the Earth's magnetic field) would help to define the quantization axis? Do you happen to know if such a external field is necessary in optical pumping?
 
  • #22
I am coming back to my post #6. I think you should try to get a visual understanding of how the spherical harmonics involved look like and when and why transition dipole moment matrix elements between them vanish.
 
  • #23
For a atom (e.g., for simplicity a hydrogen atom) without any external fields, the choice of the quantization axis for angular momentum is arbitrary, as is the choice of a (inertial) reference frame in terms of spherical coordinates. Usually one chooses the polar axis of the spatial coordinate system as quantization axis for angular momentum. This is convention. Since in the most simple case, i.e., neglecting interactions between the magnetic moments of proton and electron, the energy eigenvalues are degenerate by the "magnetic quantum number" [itex]m[/itex], i.e., the eigenvalue of the component of the angular momentum operator in direction of the arbitrary quantization axis. That's due to rotational symmetry of the Coulomb potential, and thus the total Hamiltonian of the atom.

In case of the hydrogen atom, described by the pure Coulomb potential, you have an additional "accidental symmetry", enhancing the full symmetry group from the rotational SO(3) to SO(4), and thus you have further degeneracy of the energy eigenvalues, which in this case only depend on the main quantum number [itex]n[/itex].

If you switch on an external magnetic field, it is customary to choose its direction as both the polar axis of the spatial coordinate system and the quantization axis for angular momentum. The rotational symmetry is now broken and thus the energy values split, i.e., the degeneracy is (at least partially) lifted (Zeeman Effect).
 
  • #24
I would like to resume this discussion if possible.
Based on my understanding and what has been mentioned above, the selection rules are largely dictated by the choice of the quantization axis. And in the absence of any external field, the choice of quantization axis is totally arbitrary due to spherical symmetry. But the lack of a unique quantization axis is confusing for me particularly when I think of conducting a physical experiment with the objective of isolating particular transitions.

To be more specific, let me provide an example.
Suppose i have an ensemble of atoms in a cell that I would like to optically pump. I define my quantization axis as the z-axis and have this axis coincide with the direction of propagation of light from a laser. Further suppose that this light is linearly polarized (along the y axis). Based on my defined quantization axis, this linearly polarized light can be decomposed into σ+ and σ- light and therefore one would say that Δm=±1 transitions would be accessed. However, had I defined the y-axis as my quantization axis, this linearly polarized or π polarized light would drive a Δm=0 transition instead.

Since the atom does not have any knowledge of my choice of quantization axis, how does one know which transitions are accessed? Or more specifically, how does one know the final state(s) of the atoms? Suppose the theoretically calculated linestrengths for a particular transition vary depending on whether Δm=0 or ±1, (i.e. σ or π polarized light), how would one then demonstrate (through an experiment) this difference in linestrengths?

Please let me know if my question makes sense. Thanks for your help!
 
  • #25
If there is no preferred reference in your system you have full rotational symmetry, and the choice of the quantization axis of angular momentum is thus irrelevant. It will just determine your frame of reference, and both non-relativistic as well as relativistic QT are invariant under rotations. So the selection rules for em. transitions are independent on this choice of axis.

If there's a preferred axis, e.g., particular at presence of an external magnetic field, it's physically sensible to choose this axis as the quantization axis. First of all there may be still a symmetry under rotations around this axis left (if there's no other preferred direction in the problem that's necessarily the case), and it is useful to use a basis in Hilbert space with the corresponding quantization axis.
 
  • #26
Hi Vanhees71, thanks for your response. I've been taking some time to look into this a little more but I still can't reconcile certain things that you've said.

vanhees71 said:
So the selection rules for em. transitions are independent on this choice of axis.
Could you elaborate on this?

Perhaps a schematic would better illustrate my point.
If you refer to my attachment, consider linearly polarized light (red arrow) from a laser propagating along the y direction. If I choose the z axis as the quantization axis, then the light is π polarized and Δm=0 transitions are accessed. However, if i had chosen the y-axis as the quantization axis, then the light would be σ polarized and could be decomposed or treated as a linear superposition of σ+ and σ- light, in which case Δm = ±1 transitions are accessed.

Further suppose i consider a case where I am pumping from a J=1, mJ=0 initial level to an excited J=1 level. The mJ=0 to mJ=0 transition (dotted grey arrow) is forbidden whereas the mJ=0 to ±1 transitions are allowed with a probability of 1/6. This seems to suggest that my (arbitrary) choice of quantization axis can decide whether the photon is absorbed, i.e. if i choose the z-axis as my quantization axis, then the transition is unallowed but had i chosen the quantization axis to be the y-axis, then the transition probability would be finite. But surely this cannot be the case.

Could you or someone point out the fallacy in my argument?
Schematic.jpg
 
  • #27
The m = 0 state with z as the quantization axis is not the same state as the m = 0 state with y as the quantization axis.
 
  • #28
DrClaude, thanks for your response.

The m=0 state maps out a circle in the xy plane for the case where z is the quantization axis and the m=0 state with y as the quantization axis maps out a circle in the xz plane. Is this correct?

My assumption is that regardless of choice of quantization axis, the physical outcomes (say in an expt) should be equivalent. In other words, if I define z as my quantization axis, then shining linearly polarized (in this case equivalent to π polarized) light from a laser should not induce any transitions since the m=0 -> m=0 transition is forbidden. That is to say, even if i had chosen y as my quantization axis for the analysis, the outcome (i.e. of no light being absorbed) should still be the same. Is this assumption correct?

I wonder then if my problem lies with the fact that as I changed the quantization axis of the atom in my analysis, the quantization axis of the light remained unchanged and therefore the analysis was somehow inconsistent.
 
  • #29
What do you mean by "quantization axis of the light"? You only need to consider two polarization states as a basis. The most natural ones are to use helicity eigenstates (i.e., circular polarized waves). The helicity is the projection of the angular momentum to the direction of momentum (propagation direction of the em. wave).
 
  • #30
Hi vanhees71, thanks for pointing out my incorrect terminology. Perhaps propagation direction of the light as opposed to quantization axis of the light would be more appropriate.
 
  • #31
I believe I have gained a much better understanding of the selection rules and how they are defined with and without an external field. I would like to thank all of you for your time in explaining these concepts.

I have a relevant question regarding optical pumping that perhaps you might be able to help me with (If it is deemed necessary, I can post this in a separate thread). Consider an atom with a simplified two-level system: ground state: J=½ (mJ =±½) and excited state J'=½ (mJ'=±½).

In an idealized situation with no external field (in other words, let us neglect the Earth's field), suppose I shine left circularly polarized (LCP) CW laser light at an ensemble of these atoms. (Here it is assumed that the wavelength of the light is resonant with the energy difference between the two levels.) Further suppose that I define my quantization axis as along (i.e. // to) the propagation axis of the light.

This light will induce Δm=+1 (i.e. σ+)transitions, and eventually drive all the atoms into the J=½, mJ=½ ground state after multiple absorption and relaxation cycles (i.e. optical pumping). This suggests that after pumping with this LCP laser light for some time, the atoms will become transparent to the light, since atoms in the J=½, mJ=½ ground state cannot absorb this light. This resulting change in opaqueness or transparency of the atoms over time can be easily monitored with a light-detection device like a photodiode or photomultiplier.

My question is: is there any fundamental reason against observing such an outcome in a physical experiment? Is there a requirement for an external field to be applied? I have come across a few threads online which suggest that a magnetic field is required.

Furthermore, in a real-life configuration, does the Earth's field affect these outcomes?
 

1. What is photon angular momentum and why is it important?

Photon angular momentum refers to the amount of rotational energy carried by a photon, which is a fundamental particle of light. It is important because it helps us understand the behavior and properties of light, such as its polarization and interactions with matter.

2. How is photon angular momentum quantized?

Photon angular momentum is quantized because it can only exist in discrete values, which are determined by the photon's frequency and Planck's constant. This means that the angular momentum of a photon can only change in specific increments, rather than continuously.

3. What are the selection rules for photon angular momentum and magnetic quantum number?

The selection rules for photon angular momentum and magnetic quantum number refer to the restrictions on the possible values that these quantities can take during a photon's interaction with matter. These rules are based on conservation laws and symmetry principles, and they help determine the allowed transitions and emission spectra of atoms and molecules.

4. How do the selection rules affect the emission spectrum of an atom?

The selection rules for photon angular momentum and magnetic quantum number determine which transitions are allowed for an atom during emission. This means that only certain wavelengths of light will be emitted, resulting in a unique emission spectrum for each element. By studying these spectra, we can identify the elements present in a sample.

5. Can the selection rules be violated?

While the selection rules for photon angular momentum and magnetic quantum number are generally followed, there are some cases where they can be violated. For example, in certain rare atomic processes, the rules may be temporarily broken due to the presence of external electric or magnetic fields. However, these violations are typically small and do not significantly affect the overall behavior of light and matter.

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