# $\pi$, $\sigma^\pm$ components with no magnetic field?

This question stems from The definition of a $\pi$ polarized photon?

In the presence of a magnetic field we can define a $\sigma^-$ polarized photon as one with $j_z=-m\hbar$, a $\sigma^+$ one with $j_z=m\hbar$ and a $\pi$ photon one with $j_z=0$.

A $\sigma^+$ ($\sigma^-$) photon traveling along the $z$ direction (i.e. parallel to the magnetic field) appears as a left (right) circularly polarized wave. But the same photons traveling in the $x$ or $y$ direction appear as linear (see here). A $\pi$ photon has a polarization parallel to the magnetic field and can't propagate along the magnetic field.

From this it is clear that when observing in the $x$ direction you will only see linearly polarized photons and when in the $z$ direction only circularly polarized photons.

Is the same true when there is no magnetic field present or does something change in the analysis meaning we can see linearly polarized photons and circularly polarized photons in both directions? If so what and if not why not?

• Duplicate of physics.stackexchange.com/q/267457 – Rococo Aug 26 '16 at 19:35
• @Rococo Although this question is related mine, I don't think it explicitly answers the question of that I have posed here. – Quantum spaghettification Aug 27 '16 at 3:46
• @Rococo that said I do think the principals described in your answer to this question could be adapted to answer this question. – Quantum spaghettification Aug 27 '16 at 4:50

There's several things going on here, and a good bit of misconceptions. The concepts of a $\sigma^\pm$ or a $\pi$ transition are always relative to a given quantization axis. If there is an external magnetic field, then that is implicitly taken as the quantization axis; if there isn't one then it needs to be explicitly set.

It is also important to note that there is no explicit need for the quantization axis to coincide with the external magnetic field. Doing it helps simplify the calculation both mathematically and conceptually, but ultimately it is nothing more than a choice of basis for a vector space and all the physics is unchanged if you change to some awkward basis.

What does happen, however, if you analyze a situation with an external field in some other axis, is that the at-rest Zeeman effect hamiltonian is no longer diagonal so even in the absence of a radiation field you will get Zeeman-induced "transitions" between your basis states, because they are not eigenstates of the atom-only hamiltonian.

OK, so let's get into some of your misconceptions.

A $\sigma^+$ ($\sigma^-$) photon traveling along the $z$ direction (i.e. parallel to the magnetic field) appears as a left (right) circularly polarized wave. But the same photons traveling in the $x$ or $y$ direction appear as linear.

This is only a vague nod in the direction of being correct. There are no $\sigma$ or $\pi$ photons, there are $\sigma$ and $\pi$ transitions, and you normally tend not to define both at the same time. Once you get into the language, then you can extend the term to say "$\pi$ photon" and mean "photon polarized so that it will effect a $\pi$ transition", but the photon itself has linear or circular polarization.

This is why the statement above is just completely wrong. If you have a circularly polarized photon and you change the axis it's propagating on, to be the "same" photon it still needs to be circularly polarized. Conversely, a $\sigma^\pm$ transition can be caused by a $z$-propagating circularly polarized photon as well as by $x$-propagating $y$-polarized photons. See the difference?

Coming now to the crux of the question,

From this it is clear that when observing in the $x$ direction you will only see linearly polarized photons and when in the $z$ direction only circularly polarized photons.

this is roughly true, but you should note that whenever you can see left- as well as right-circular photons you can also see superpositions of the two, which are generally linear polarizations. Similarly, if you observe from the $x$ axis you can see both $y$-polarized (linear) photons coming from $\sigma$ transitions and $z$-polarized photons coming from $\pi$ transitions, but you can also observe their superpositions, which include circular polarizations.

...or, at least, the argument above would work if you were superposing photons at the same frequency. In the presence of a magnetic field, however, the levels are shifted by different amounts depending on the magnetic quantum number $m$ (if you set the quantization axis along the magnetic field), which means that if you observe along the $x$ axis then the $y$-polarized and $z$-polarized components will be at different frequencies; the same holds for the two circular components observed along the quantization axis. You can still observe superpositions of these components, but being at different frequencies they no longer combine into a single component with different polarization.

Is the same true when there is no magnetic field present or does something change in the analysis meaning we can see linearly polarized photons and circularly polarized photons in both directions? If so what and if not why not?

If there is no magnetic field, then there is no splitting between the levels, and all the radiation is at the same frequency and it can be combined as you wish. This therefore comes in two flavours:

• If you're trying to use an external field to stimulate a transition up from an $S$ to a $P$ state, then any polarization will work, and the state of the polarization will determine which state within the $P$ manifold will get populated.

Circular polarizations will populate $m=\pm1$ states about a quantization axis along the propagation direction; linear polarizations will populate $m=0$ states about a quantization axis along the polarization direction.

You're not, of course, beholden to this analysis: you can put your quantization axis wherever you want, and the analysis of which states do or don't get populated will be equivalent, though it may be harder if you choose an awkward basis.

• On the other hand, if you already have an excited $P$ state and you're looking for its emission, then the polarization and propagation direction of the emission will depend on the state, pretty much as described in your book if you start with an $m=0$ or $m=\pm1$ state.

And, again, it's important to note that there are other, funkier states, like superpositions of $m=0$ and $m=\pm1$ states, and those will emit in different directions. Naturally, these are some of the states that come into play if you analyze an arbitrary state in an arbitrary frame.

Either way, in the absence of a natural quantization axis which plays a role in the hamiltonian, the analysis is exactly the same in any frame of reference.

• Would it be appropriate to view this in terms of a frequency-operator, $\hat W$ and polarization operator $\hat Y$ acting on photon states. In the case of a quantization axis along the propagation axis we have two basis vectors $\psi_r=\left|\omega_1, x+iy\right>$ (a $\sigma^+$ photon) and $\psi_l=\left|\omega_2, x-iy\right>$ (a $\sigma^-$ photon). Any photon propogating along this axis must be in a state $A\left|\omega_1, x+iy\right>+B\left|\omega_2, x-iy\right>$... – Quantum spaghettification Sep 16 '16 at 14:23
• In the case of a magnetic field $\omega_1 \ne \omega_2$ and thus on measuring the frequency and polarization the photon lies in one of pure states $\psi_r$ or $\psi_l$. When no magnetic field is present, however $\omega_1=\omega_2$ and on measuring the frequency and polarization the state remains in the state $A\left|\omega_1, x+iy\right>+B\left|\omega_2, x-iy\right>$ which may be e.g. linearly polarize. The same would hold for quantization axis perpendicular to the direction of propgation with minor alterations. Is this a good way to think about it or not? – Quantum spaghettification Sep 16 '16 at 14:23
• @Quantumspaghettification Yeah, that's mostly OK, though note that in principle you can have four relevant basis states, $|\omega_i, e_j⟩$, with $\omega_i$ and $e_j$ independent frequencies and polarizations, and all four states will come in if you have a nonzero magnetic field on a separate direction. – Emilio Pisanty Sep 16 '16 at 14:41

Firstly, the electromagnetic field has only two degrees of freedom. This is a consequency of the Ward identity, which follows from gauge invariance. These defrees of freedom give components of the vector field that are perpendicular to the direction of propagation. (In superconductors, where gauge invariance is broken, the photon becomes massive and thus has a component along the direction of propagation.)

These two degrees of freedom can be expressed as helicity eigenstates. Note that one cannot measure the spin of an electromagnetic field, because one cannot be in a rest-frame of the field. So instead, we need to consider the helicity, which means that we consider a projection of the spin in the direction of propagation.

Say the field propagates in the $z$-direction, then the relevant component of the angular momentum operator is $J_z$. If one restricts this matrix operator to the transverse directions, it looks like the $\sigma_y$ Pauli matrix $$\sigma_y = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] .$$ The eigenstates of this matrix are the two circular polarization states, denoted by $\sigma^{\pm}$. Therefore, these are the helicity eigenstates. However, one can form any linear combination of these states. Therefore one can form any arbitrary polarization state (such as the linear polarization states), but they would not be helicity eigenstates.

If the photons were to propagate in a different direction, then we need to rotate the whole setup. Unless we also rotate out reference frame, the mathematics would come out to look different, but in the end the helicity eigenstates in the new direction would again be the circular polarization states.

One cannot have the same photon that would have propagated in the $z$-direction, now proagating in the $x$-direction, because the polarization vector of the photon must always be perpendicular to the propagation direction. So circular polarization for $z$-directed propagation does not become linear polarization for $x$-directed propagation.

Another thing, if the photon has been prepared in a particular circular polarization state and we measure in some linear polarization basis then there is a nonzero probability that we will detect the photon. This is because circular polarization can be seen as a superposition of any two orthogonal linear polarization states, and visa versa.

There would always be a magnetic field associated with a propagating electromagnetic field. When we use polarizers or waveplates to prepare a photon in a particular polarization states, then the magnetic field is directly related to the electric field that is associated with that polarization state, via Maxwell's equations.