# The definition of a $\pi$ polarized photon?

I am looking at the definition of $\sigma^\pm$ and $\pi$ polarized photons (in the context of atomic transitions), however I have seem to come across two (both seen in numerous sources surrounding the same context):

Definition 1

$\sigma^+$ is a left circularly polarized photon, $\sigma^-$ is right circularly polarized photon and $\pi$ is a linear combination of both. (see e.g.here )

Definition 2

w.r.t a particular axis (let us take the $z$-axis) a $\sigma^+$ photon is one with $j_z=m\hbar$, a $\sigma^-$ photon is one with $j_z=-m\hbar$ and a $\pi$ photon is one with $j_z=0$. (see e.g. here)

Am I correct in saying that these two definitions are not equivalent (since in the second case $\sigma^+$, $\sigma^-$ and $\pi$ seem to be forming eigenstates of a Hermitian operator)? If they are not equivalent why are both used (or have I misinterpreted one), if the are equivalent please can you explain why?

• They look the same to me. Can you be more specific with your concerns? – garyp Aug 25 '16 at 13:09
• There is no $S_z=0$ state for photons. Your definition 2 does not make sense to me. Whenever you compare such apparently different definition, please give their sources and contexts. It may be that one is just wrong, but it may also just be that the two are talking about two different notions of "photon state". – ACuriousMind Aug 25 '16 at 13:19
• @garyp: In definition 1, the $\pi$-polarization is a sum of $\sigma^+$ and $\sigma^-$. If in definition 2 saying "with $j_z=m\hbar$" means "is an eigenstate of $j_z$ with that eigenvalue", then the $\pi$ state is linearly independent from $\sigma^\pm$ in definition 2. – ACuriousMind Aug 25 '16 at 13:21
• @ACuriousMind I have added an example source for both definitions. – Quantum spaghettification Aug 25 '16 at 14:03
• You linked the sources exactly the wrong way around. You need to look at what the two different sources are doing: The one with definition 1 gives a general, relativistic description of photon states, while the second talks about a very specific non-relativistic case of RF photons relative to a given $\vec B$-field axis. That the two use the same name for states in totally different contexts is unfortunate, but the situations are effectively incomparable. What answer do you want except: "Yes, they're not equivalent, because the situations described are not equivalent"? – ACuriousMind Aug 25 '16 at 14:10

I think these definitions (which I explain more in my answer here $\pi, ~\sigma$ - atomic transitions with respect to a quantization axis) might help answer your question:

Notation for angular momentum states: I'll label angular momentum states with the axis along which they are defined. So for example $|1,+1\rangle_z$ is a state which has a projection of 1 unit of angular momentum in the $+z$ direction. Whereas $|1,-1\rangle_x$ refers to a state which has a projection of -1 unit of angular momentum in the $+x$ direction. Note very importantly that $|1,+1\rangle_z \neq |1,+1\rangle_x$. See the linked question for more details.

Definition of $\sigma^+$ light:

First pick a quantization axis direction. Call this direction $\hat{n}$. $\sigma^+_\hat{n}$ light is light which drives a transition from $|J,m\rangle_\hat{n} \rightarrow |J',m'=m+1\rangle_{\hat{n}}$

Definition of $\pi$ light:

First pick a quantization axis direction. Call this direction $\hat{n}$. $\pi_\hat{n}$ light is light which drives a transition from $|J,m\rangle_\hat{n} \rightarrow |J',m'=m+0\rangle_{\hat{n}}$

Note this definition depends on the choice of quantization axis. Note also that they make no reference to whether the light is circularly or linearly polarized. I think you will find these to be the most general definitions which can usually be applied to statements people make about photons. To refer specifically to the definitions you gave:

Definition 1: Consider a left circular photon propagating in the $+z$ direction. This photon has $-\hbar$ angular momentum in the $+z$ direction. This means it can drive a transition from $|1,+1\rangle_+z\rightarrow |1,\rangle_{+z}$ for example. So by my definition this would be a $\sigma^-_{+z}$ photon which conflicts with the definition you gave. However, If I choose the opposite direction as my definition of angular momentum FOR MY ATOMS we see that this photon can drive a transition from $|1,-1\rangle_{-z} \rightarrow |1,0\rangle_{-z}$ meaning it is a $\sigma^+_{-z}$ photon, corresponding with the definition you gave.

Definition 2: This one looks like it kind of corresponds with my definition. I should point out the $m$'s are unnecessary since a single photon can only carry 1 unit of angular momentum. Also, I will point out again (and as others have pointed out) that if the light is travelling in the $+z$ direction it is impossible for it to be $\pi_z$ light. However, that doesn't mean $\pi_z$ light is impossible. I think this caveat makes this second definition a bit confusing.

In summary, the definitions of $\sigma^+$, $\sigma^-$ and $\pi$ light are not intrinsic to either the light itself, or the atom itself, but rather to the interaction between light and the atom. The definitions also depend on the choice of quantization axis.