quantum mechanics - $pi, ~sigma$ - atomic transitions with respect to a quantization axis

In the absence of a magnetic field, how does one physically (i.e. perhaps in a thought expt) access $\Delta m = 0 ~~\textrm{or}~~ \pm1$ transitions since (as I understand it) the choice of quantization axis is arbitrary?

Answer

This is a confusion I've had as well. In trying to reconcile the confusion I also came across your post on PhysicsForums which has some more explication of the question: https://www.physicsforums.com/threads/photon-angular-momentum-and-magnetic-quantum-number-selection-rules.783055/page-2.

I'm not sure if you've found your answer already or what but I can give my input on how I think this should be interpreted.

First, obviously this confusion all stems from a choice of quantization axis so lets be clear about what we mean by that. I'll give the physical setup. We have a cloud of atoms. Let's consider a $J=0 \rightarrow J'=1$ transition. We also have a laser shooting light at those atoms. Say the laser is propagating in the $+z$ direction. Let's say the light is either right-hand circular (RHC), left-hand circular LHC, or linearly polarized. For RHC and LHC this definition is already unambiguous (Direction of propagation determines what RHC looks like for example). For linearly polarized light lets say that the light is linearly polarized in the $+x$ direction. Let's also say it is some simple superposition of RHC and LHC light. All of this amounts to fixing an arbitrary relative phase between the two definition of RHC and LHC light. Also, I will point out that RHC light carries angular momentum $+\hbar$ in the $+z$ direction and LHC light carries angular momentum $-\hbar$ in the $-z$ direction.

I will point out that the entire discussion above refers only to physical quantities. There has been NO CHOICE of quantization axis. There has, however, been a choice made about a coordinate axis. We will take the choice of coordinate axis as fixed with repsect to the physical apparatus so that won't change.

Notation about angular momentum states: For clarity, I will make sure to label my angular momentum states the following way: $|J,m\rangle_z$ refers to an atom which has total angular momentum $J$ (this is a quantization and coordinate axis independent statement) and angular momentum projection $m\hbar$ in the $+z$ direction. Note that that last statement is ALSO a quantization axis independent statement. It is a physical statement about the angular momentum carried by the atom. In other words, if $m\neq 0$ the atom isn't isotropic, it has an arrow attached to it. If you rotate the atom you change the physical system. Note that we can also talk about $|J,m\rangle_x$, a state which has angular momentum projection $m\hbar$ in the $+x$ direction.

However, and this is important, note that $|J,m\rangle_z \neq |J,m\rangle_x$. I think that people refer to the subscript I am using here as the quantization axis. So they would say that we have changed quantization axis from the LHS of the equation to the RHS. I think this talk of a quantization axis which can change mid-calculation is sort of confusing, but again, I'm pretty sure that the definition of $\sigma^+$, $\sigma^-$ and $\pi$ polarized light will rely on the choice of quantization axis so we need to talk about it.

Also, I’ll write out how to change from one set of angular momentum states to the other. $$|1,-1\rangle_x = \frac{1}{2}|1,+1\rangle_z -\frac{1}{\sqrt{2}}|1,0\rangle_z +\frac{1}{2}|1,-1\rangle_z$$ $$|1,0\rangle_x = -\frac{1}{\sqrt{2}}|1,+1\rangle_z +\frac{1}{\sqrt{2}}|1,-1\rangle_z$$ $$|1,+1\rangle_x = \frac{1}{2}|1,+1\rangle_z +\frac{1}{\sqrt{2}}|1,0\rangle_z +\frac{1}{2}|1,-1\rangle_z$$

Ok. I think we finally have the notation to tackle the problem.

1) Consider RHC polarized light. This light has $+\hbar$ angular momentum in the $+z$ direction. This means (by conservation of angular momentum) it will drive the atom from $|0,0\rangle$ (note this state doesn't need a coordinate subscript since it is rotationally symmetric) to $|1,+1\rangle_z$. We use this to define $\sigma^+$ polarization.

Definition of $\sigma^+$ polarization. FIRST choose a quantization axis. Say the direction of this axis is $\hat{n}$. We now define $\sigma^+$ polarized light to be light that drives a transition from $|J,m\rangle_{\hat{n}}\rightarrow |J',m'=m+1\rangle_{\hat{n}}$. If we really wanted to be pedantic maybe we could call this $\sigma^+_{\hat{n}}$ light to explicitly show that this definition of $\sigma^+$ light is dependant on the choice of quantization axis.

2) Now we consider linearly polarized light along the $+x$ direction. Say that the linearly polarized light vector looks like $\hat{\epsilon}_{x} = \frac{1}{\sqrt{2}}(-\hat{\epsilon}_+ +\hat{\epsilon}_-)$ (Hmm, I wonder why I'm choosing this particular linear combination*)

Let's consider what will happen if we think about it with respect to the $+z$ axis. Our photon is a superposition of RHC and LHC photons. We can say it is a superposition of $\sigma^+_z$ and $\sigma^-_z$ based on the definition above and angular momentum rules. This means that this photon (or superposition of photons depending on how we think about it will drive the following transition): $$|0,0\rangle \rightarrow \frac{1}{\sqrt{2}}(-|1,+1\rangle_z + |1,-1\rangle_z)$$ In other words, since the photon is in a superposition of $z$ angular momentum states it drives the atom into a superposition of $z$ angular momentum states. But notice what I've obviously set us up for. The state on the RHS of this equation is equal to $|1,0\rangle_x$ as written out above. Therefore we are also correct in saying that this linearly polarized photon drives: $$|0,0\rangle \rightarrow |1,0\rangle_x$$

We now have our definition for $\pi$ light.

Definition of $\pi$ light: First choose a quantization axis, $\hat{n}$. We define $\pi_{\hat{n}}$ light to be light which drives a transition from $|J,m\rangle_{\hat{n}}\rightarrow |J',m'=m+0\rangle_{\hat{n}}$

So I think to sum everything up: If we take a linear combination of $\sigma^+_z$ and $\sigma^-_z$ we end up with $\pi_x$ light**. That is, we DON'T end up with $\pi_z$ light which is I think a bit of the source of the misconception here. It's also a confusing concept and I haven't been able to find any sources explaining it in these terms so I would appreciate if anyone else can confirm that I'm explaining this the right way here.

*I'll point out again that this equation depends on the relative phases chosen in the definition of RHC and LHC light. If we change that relative phase then the linear polarization axis defined by this equation will no longer be the $+x$ axis, but rather some other axis and the rest of the analysis would change accordingly.

**If we take different linear combinations than the ones shown here we might get $\pi_y$ light or something else completely different. There's no linear combination of $\sigma^+_z$ and $\sigma^-_z$ that will give us $\pi_z$ light because they can never drive a transition to $|1,0\rangle_z$.