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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?

Tainty
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  • But you're (typically) probing these transitions with light, which has a specific polarization axis, right? – Emilio Pisanty Jul 11 '16 at 13:41
  • Yes. Are you suggesting that the quantization axis related to the laser? But how do the atoms know which direction the laser is coming from and what polarization it possesses? – Tainty Jul 11 '16 at 13:54
  • The atoms are generally not sensitive to the direction of propagation of the laser (since propagation takes place on length scales of $\gtrsim400:\mathrm{nm}$, a thousand times bigger than the atom), but they do respond to the polarization of the laser. After all, the laser field is an oscillating electric field, along a given direction, right? – Emilio Pisanty Jul 11 '16 at 13:56
  • Yes, I agree that the atoms respond to the laser field. But I am confused with what then constitutes the quantization axis. I could send in light that is linearly polarized with respect to the laser, but depending on how the quantization axis is defined, that linearly polarized light could be treated as σ + and - polarization or π polarization and the transitions accessed (at least in theory) would be different. – Tainty Jul 11 '16 at 14:01

2 Answers2

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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 $|1,0\rangle_z$.

Julius
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Jagerber48
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  • Instead of using > use \rangle which renders $\rangle,;$ I've edited your other post, but as it has more such >, it's tedious for me to do so myself. –  Oct 02 '16 at 19:21
  • Thanks for the tip. I just used Microsoft word to make the replacement everywhere. I wish we I could just use \ket{}! – Jagerber48 Oct 02 '16 at 19:56
  • Yes, you can; you have to first use \newcommand to define the command \ket as \rangle; then use \ket generously. –  Oct 02 '16 at 20:09
  • @MAFIA36790 that's a potentially dangerous suggestion as explained in this meta post. – DanielSank Oct 03 '16 at 04:46
  • I know @Daniel; we are advised not to use this; but since OP demanded whether it's possible, I only showed him that it's possible and this can be done that way; but haven't told him to use it ;) Nevertheless, noted; should have warned him. –  Oct 03 '16 at 04:52
  • @jgerber. First up, thanks for that detailed response. I think your response is well summarized by the statement that "If we take a linear combination of σ+zσz+ and σ−zσz− we end up with πxπx light – Tainty Oct 19 '16 at 17:10
  • Ignore that response. I meant to say that I do understand that a linear combination of sigma+ (z) and sigma- (z) light does not equate to pi (z) light but rather either pi (x) or pi (y) light depending on the relative phase difference. But i suppose my main confusion arises from differentiating between which transitions (or atoms) are actually addressed for a particular light polarization. – Tainty Oct 19 '16 at 17:25
  • I think you have probably answered my question but allow me to clarify a few things just to be sure. Perhaps a better way of re-posing my question is as follows: Based on your earlier defined coordinate system, (and assuming there is no external magnetic field) suppose i send in pi (x) light, does this mean i will be inducing BOTH delta_m=0 transitions for atoms aligned in the x direction and delta_m = +-1 transitions for atoms aligned in the z direction? – Tainty Oct 19 '16 at 17:30
  • @Tainty So based on how you're asking your question you've already defined a coordinate system. This same coordinate system also applies to the atoms you are talking about. You can describe the angular momentum of the atoms in either the x basis or in the z basis. So we can make two statements. (1) $\pi_x$ light will drive $\Delta m_x = 0$ transitions. (2) $\pi_x$ light will drive $\Delta m_z = \pm 1$ transitions. The important point is that $m_x$ and $m_z$ are different (but related) quantum numbers in the same way the x basis is different from (but related to) the z basis. – Jagerber48 Oct 19 '16 at 17:39
  • So you statement that "i will be inducing BOTH delta_m=0 transitions for atoms aligned in the x direction and delta_m = +-1 transitions for atoms aligned in the z direction?" is sort of right, but it sounds like you're thinking of this as two separate processes when in fact those two statements correspond to one physical situation. It is related to the fact that the x basis is superpositions of the z basis and vice-versa (simultaneously for circular basis being superpositions of the linear basis and vice versa). – Jagerber48 Oct 19 '16 at 17:42
  • @jgerber But the transition probabilities for delta_m=0 transitions and delta_m = +-1 transitions are generally different. If these are one and the same thing as you point out (which does indeed seem correct), any analysis that follows should be able to account for/ capture that difference and have the transition probabilities eventually equate. – Tainty Oct 19 '16 at 17:53
  • I'll point out that three descriptions transform when you change between bases. The initial states, the final states, and the light field. So maybe you start out in $|1,+1\rangle_x$ and are driven to $|2,+1\rangle_x$ by $\pi_x$ light. But when you change basis you're now in a superposition of ground states and you're driving to a superposition of excited states with a superposition of light fields. All of these transformations are related by pauli matrices and the rates are related by Clebsch-Gordan coefficients. I suspect you could prove that the total rates come out the same. – Jagerber48 Oct 19 '16 at 22:52
  • After further thought, I've concluded that the rates do indeed come out to be the same. One way I think about it (which might be incorrect) is that the Clebsch Gordan coefficients can be treated as is, but the change of basis accounts for a modification in these coefficients which eventually leads to the rates being equal. Thanks a lot for your response. The 3 equations showing the change of angular momentum eigenstates were particularly helpful. Could you elaborate more on the steps involved in this basis change or point me to a good resource that could help me better understand this? – Tainty Oct 25 '16 at 18:18
  • That sounds right. It is a combination of the C-G coefficients and the change of basis that preserves the physical observables (such as population difference between ground and excited manifolds). Here is a reference with the pauli matrices. The way to think about this is that the matrices on these page are matrices written in the $z$ basis. So the eigenvectors of $\sigma_z$ are just $(1,0,0)$, $(0,1,0)$... However, the eigenvectors of $\sigma_x$ (states with definite $x$ projection of angular momentum) will be linear combinations of the $z$ eigenvectors. – Jagerber48 Oct 25 '16 at 20:19
  • These are precisely the linear combinations which appear in my answer. The components of the higher spin pauli matrices can be calculated by working out things like $\langle j,m| S_x |j,m'\rangle$ and recalling $S_x=\frac{S_+ + S_-}{2}$. These notes: http://atomoptics-nas.uoregon.edu/~dsteck/teaching/quantum-optics/ have been very helpful to me for these sorts of questions. see section 7.1. Here's the link I meant to add in the previous comment http://easyspin.org/documentation/spinoperators.html. – Jagerber48 Oct 25 '16 at 20:24
  • I have a similar answer to Jagerber48, but it seems to me that the meaning of z and x-axis in scenario 2 should be swapped. Namely, the photon is propagating along the x-axis, if it is linearly polarized, we should express it as a linear superposition of the two $\sigma_x$ components. Following calculations would then reveal that the photon is exciting transition $|0,0\rangle \rightarrow |1,0\rangle_z$, which makes sense as we have defined our transitions in the basis where $z$ is the quantization axis. – Corwinkko Aug 21 '22 at 09:41
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In reality, you need the quantization field to prepare your atom in a given spin state and have it stay in that state- at zero field, any small perturbation could change it. But in a thought experiment that is not a problem.

Still, you have to specify your atom as starting in some initial spin, and if it is starting as usual in an eigenstate of $S_z,$ you have to pick a direction for that $S_z$ axis. So the direction in the absence of an external magnetic field is just chosen by the initial conditions you set the atom in.

Of course, in the case of no field you could also initialize the atom in some superposition of spin states. For that matter, this is also true in the case with a field. But in the zero field case several of the transitions might be resonant at the same time, and you could imagine that interference between those possibilities makes the overall dynamics more complicated.

Rococo
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