# Why is the $S_{z} =0$ state forbidden for photons?

If photons are spin-1 bosons, then doesn't quantum mechanics imply that the allowed values for the z-component of spin (in units of $\hbar$) are -1, 0, and 1?

Why then in practice do we only use the $\pm 1$ states?

I have been told that this is directly related to the two polarizations of the photon. This seems to be more of a classical argument however, arising from the fact that Maxwell's equations do not permit longitudinal EM waves in a vacuum.

I have also heard that it is related to the fact that photons have no rest mass, although I understand far less of this reasoning.

What I'm looking for are elaborations on these two arguments (if I have them correct), and perhaps an argument as to how these two are equivalent (if one exists).

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physics.stackexchange.com/q/73911 . For photon we use helicity, not spin, because massless representations of the Poincare group can't be characterized by the spin. –  PhysiXxx Aug 31 '13 at 17:27
–  Ben Crowell Aug 31 '13 at 20:36
Note that the converse of this is much easier to prove: if there is a constraint on $s_z$, then the particle must be massless. This is because if the particle were massive, we could go into its rest frame, and in that frame there would be no preferred axis to use in defining the constraint on the spin. –  Ben Crowell Sep 1 '13 at 0:01

I can't improve on KDN's answer, but given Todd's comments this is an attempt to rephrase KDN's answer in layman's terms.

A system is only in an eigenstate of spin around an axis if a rotation about the axis doesn't change the system. Take $z$ to be the direction of travel, then for a spin 1 system the $S_z$ = 0 state would be symmetric to a rotation about an axis normal to the direction of travel. But this can only be the case if the momentum is zero i.e. in the rest frame. If the system has a non-zero momentum any rotation will change the direction of the momentum so it won't leave the system unchanged.

For a massive particle we can always find a rest frame, but for a massless particle there is no rest frame and therefore it is impossible to find a spin eigenfunction about any axis other than along the direction of travel. This applies to all massless particles e.g. gravitons also have only two spin states.

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This was very helpful. I was tired of hearing the "classical" explanation of polarizations resulting from Maxwell's equations, and am not familiar enough with field theory yet to fully grasp much of the mathematics in the other posts (although as an exposure to the ideas and notation of field theory they are very helpful in their own right). This argument based on symmetry and the basic ideas of invariance was very helpful and gives me a groundwork to understand some of the technical details from other posts. Thank you @John Rennie and everyone else who posted! –  Todd R Dec 13 '12 at 1:50
This is a nice shot at a minimal conceptual explanation, but I think the simplification goes a little too far and becomes incorrect. In particular, this argument would seem to show that gravitons have helicities -2, -1, +1, and +2, whereas in fact they only have -2 and +2. Since the argument appears to be either incorrect or incomplete in the case of gravitons, I have doubts about it in the case of photons. Can it be mapped somehow onto the standard treatment in terms of the little group, as described in Arnold Neumaier's answer to this question physics.stackexchange.com/q/29766/4552 ? –  Ben Crowell Aug 31 '13 at 20:50

The absence of the $S_z=0$ spin projection is related to masslessness of the photon. Because the photon is massless, it propagates at the speed of light and has no rest-frame time evolution. This removes one of the allowed polarization states that would be present for massive bosons. Solving the eigenvalue problem for the spin operator S gives eigenvalues of $S_z=\pm\hbar,0$, where the normalized eigenvectors, given in $(x,y,z)$ cartesian notation, correspond to eigenvectors $\frac{1}{\sqrt{2}}(1,i,0)$ (for $+\hbar$), $\frac{1}{\sqrt{2}}(1,-i,0)$ (for $-\hbar$) and (0,0,1) (for 0).
The first two eigenvectors represent propagating right- and left-circularly polarized photons. The third eigenvector represents a non-propagating field. The photon that does not propagate, being massless, has no energy at all.

There is some validity to the notion of $S_z=0$ virtual photons however.

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Thank you for the response @KDN but this is actually the sort of answer that I have been getting frustrated with. What I am interested in is your remark that because there is no rest frame it removes one of the allowed polarizations. Why is this exactly? Also, it's not clear to me why $S_z = 0$ represents and non-propagating field, and not simply a propagating field with no angular momentum. If possible, clarification would be appreciated. Thank you. –  Todd R Dec 12 '12 at 6:11
@juanrga gives a very complete answer why this polarization state vanishes. Without gauge fixing, more polarization states at first appear possible, but these just represent spurious degrees of freedom. These non-propagating polarization states do exist, in a mathematical sense, but they are not observable (even in the mathematical sense, i.e., an observable quantity cannot be constructed for these states). The Wikipedia article on the Gupta-Bleuler formalism (en.wikipedia.org/wiki/Gupta%E2%80%93Bleuler_formalism) does a good job of addressing this issue in not-too-complex terms. –  KDN Dec 12 '12 at 13:54

The answers from KDN and John Rennie are right - I'll just try to illustrate how it works:

The components of a massless spin 1 field satisfy $$\Box^2 A_{\mu}(x) = 0$$ Traditionally we perform the expansion in momentum variables $$A^{\mu}(x) = \int{\frac{1}{\sqrt{p^0}}A^{\mu}({\bf{p}})e^{-ip.x}}d^3{\bf{p}} + comp\ conj$$ If the particle is moving in the z direction, then its momentum is $$p^{\mu} = (p^0, 0, 0, p^0)$$ and the Lorentz condition $\partial_{\mu}A^{\mu}=0$ which, on the momentum space variables looks like $$p_{\mu}A^{\mu}({\bf{p}})=0$$ now becomes $$p^0A^{0}({\bf{p}})-p^3A^{0}({\bf{p}})=0$$ and so we see that $$A^{0}({\bf{p}}) = A^{3}({\bf{p}})$$ So we can express $A^{\mu}({\bf{p}})$ in terms of polarization vectors $$A^{\mu}({\bf{p}}) = \sum\limits_{\lambda}a_{\lambda}({\bf{p}})\epsilon^{\mu}_{\lambda}$$ where the three polarization vectors look like $$\epsilon^{\mu}_{1}=(0, 1, 0, 0)$$ $$\epsilon^{\mu}_2=(0, 0, 1, 0)$$ $$\epsilon^{\mu}_{3}=(1, 0, 0, 1)$$ If you now take the special case of a wave with just the third polarization $$A^{\mu}(x) = \int{\frac{1}{\sqrt{p^0}}a_{3}({\bf{p}})\epsilon_3^{\mu}e^{-ip.x}}d^3{\bf{p}} + comp\ conj$$ and you now compute the ${\bf{E}}$ and ${\bf{B}}$ fields, then the special form of $\epsilon_3^{\mu}$ ensures you get zero. Hence the polarization in the direction of propagation does nothing to contribute to the field.

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Applying a covariant quantization scheme on the free electromagnetic field $A^{\mu}$ one can show the existence of one-photon states described by momentum $k$ and one of four possible polarization states. Those four polarizations states correspond to the four possible values of spin -1,0,0,+1. Those correspond to transversal (2), longitudinal (1), and scalar photons (1).

However, this is obtained from assuming that the four states are truly independent, when are not. By imposing the Lorentz condition (or some other equivalent as Gupta Bleuler condition) one obtains that longitudinal and scalar photons are linearly dependent for each value of momentum

$$[a_3(k) - a_0(k)] |\Psi \rangle = 0$$

Here the $a_0$ and $a_3$ are destruction operators for scalar and longitudinal photons, respectively. It is easy to show that the above combination implies that longitudinal and scalar photons do not contribute to field observables. Thus the expectation value for the energy of the electromagnetic field only involves transversal photons

$$\langle \Psi | H | \Psi \rangle = \langle \Psi | \sum_k \sum_{r=1}^2 \hbar \omega_k a_r^\dagger(k) a_r(k)] |\Psi \rangle$$

As a consequence, only transverse photons can be observed as free particles associated to the electromagnetic field.

However, scalar and longitudinal photons play an important role in presence of charges. In my opinion the most simple and direct way to understand why is to use the photon propagator $D^{\mu\nu}(k)$. Again this depends on four polarization states. The interpretation of the transverse photon contribution $D_T^{\mu\nu}(k)$ is direct, whereas the contributions of longitudinal and scalar cannot be physically interpreted by separate. However, they can be reorganized in linear combinations $D_C^{\mu\nu}(k)$ and $D_R^{\mu\nu}(k)$ that allow a simple physical interpretation

$$D^{\mu\nu}(k) = D_T^{\mu\nu}(k) + D_C^{\mu\nu}(k) + D_R^{\mu\nu}(k)$$

The first term is the usual radiation contribution and involves transversal photons. The second term is the usual Coulomb term and involves a mixture of scalar and longitudinal photons. The remaining term, also involving a mixture of scalar and longitudinal photons, is unobservable (it can be shown that its contribution to scattering is zero).

Note that although the Coulomb interaction emerges as an exchange of scalar and longitudinal photons, those photons are not observable. They do not appear in initial and final states of scattering processes (only transverse photons do), but are virtual particles in intermediate states.

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