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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    $\begingroup$ 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. $\endgroup$
    – user8817
    Commented Aug 31, 2013 at 17:27
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    $\begingroup$ related: physics.stackexchange.com/q/29766 $\endgroup$
    – user4552
    Commented Aug 31, 2013 at 20:36
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    $\begingroup$ 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. $\endgroup$
    – user4552
    Commented Sep 1, 2013 at 0:01
  • $\begingroup$ I'm recently reading vol.I of Weinberg's book and I like it's explanations. Is this question is answered explicitly or implicitly "but exactly", somewhere in it? @acuri $\endgroup$
    – moshtaba
    Commented May 17, 2019 at 1:15

7 Answers 7


Massless particles with spin do not have a "$S_z = 0$" state because they actually do not have spin like massive particles do. They have helicity, which is the value of the projection of the spin operator onto the momentum operator. The reason for this is the representation theory of the group of spacetime symmetry, the Poincaré group.

To understand this, we must first recall that "spin" is the number that labels irreducible representations of $\mathrm{SU}(2)$, the double cover of the rotation group $\mathrm{SO}(3)$. But, in relativistic quantum field theory, which is the theory needed to describe photons, this rotation group is not the spacetime symmetry group we need to represent. Instead, we must seek representations of the identity connected component of the Poincaré group $\mathrm{SO}(1,3)\rtimes\mathbb{R}^4$, i.e. of the proper orthochronous Lorentz transformation together with translations.

Now, for the finite-dimensional representations of the Lorentz group, we're lucky in that there is an "accidental" equivalence of algebra representations of $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\times\mathfrak{su}(2)$, allowing us to label the finite-dimensional representations in which classical relativistic fields transform by pairs of half-integers $(s_1,s_2)$ where $s_i\in\frac{1}{2}\mathbb{Z}$ labels a single $\mathfrak{su}(2)$ representation. The actual rotation algebra sits diagonally in this $\mathfrak{su}(2)\times\mathfrak{su}(2)$, so the physical spin of such a representation is $s_1+s_2$. This determines the classical spin associated to a field.

As so often, the quantum theory makes things more complicated: Wigner's theorem implies that we must now seek unitary representations of the Poincaré group on our Hilbert space of states. Except for the trivial representation corresponding to the vacuum, none of the finite-dimensional representations is unitary (essentially because the Poincaré group is non-compact and doesn't have compact normal subgroups). So we must turn to infinite-dimensional representations, and here we do not have the equivalence between $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\times\mathfrak{su}(2)$. The techniques exploited to realize this equivalence explicitly rely on finite-dimensionality of the representation. In particular, there is no such isomorphism as $\mathrm{SO}(1,3)\cong\mathrm{SU}(2)\times\mathrm{SU}(2)$, regardless of how often you will read similar claims in physics books. For more on this issue, see e.g. this answer by Qmechanic.

It turns out that classifying the unitary representations is not so simple a task. The full classification is called Wigner's classification, and it turns out that to construct irreducible unitary representations, it is relevant to look at the little group corresponding to the momentum of a particle - the subgroup of the Lorentz group which leaves the momentum of the particle invariant. For a massive particle, this is $\mathrm{SO}(3)$, and it turns out we can label the unitary representation also with our familiar spin $s$.

But for a massless particle, the momentum $(p,-p,0,0)$ is not invariant under $\mathrm{SO}(3)$, but under a group called $\mathrm{ISO}(2)$ or $\mathrm{SE}(2)$, which is essentially $\mathrm{SO}(2)$ with translations. Being Abelian, $\mathrm{SO}(2)$ has only one-dimensional irreducible representations, labeled by a single number $h$, which turns physically out to be the eigenvalue of helicity. There are more general cases for $\mathrm{ISO}(2)$, called the continuous spin representations (CSR), but they have so far not been physically relevant.

Now, this single number $h$ flips its sign under parity, so for particles associated to classical fields with non-zero spin, we must take both the $h$ and the $-h$ representations. And that's it - massless particles of helicity $h$ have the $h\oplus -h$ representation on their space of states, not a spin representation of $\mathrm{SO}(3)$. Evaluation of the actual spin operator shows that the our classical idea of spin coincides with the number $h$.

Therefore, without having said anything about the photon or the electromagnetic field in particular, we know that massless particles of non-zero spin come with two degrees of freedom. This is completely general, and at the heart of the argument that all massless vector bosons are gauge bosons:

We know that a generic vector field has three d.o.f. - the independent field components that transform into each other under Lorentz transformation, hence three independent sets of creation and annihiliation operators that transform into each other, hence we expect three distinct kinds of particle states.

But the two d.o.f. of a massless spin-1 particle don't match with this - so one of the d.o.f. of a massless vector field must be "fake". The way d.o.f.s of fields are "fake" is by the field being a gauge field and there being 1 d.o.f. in the freedom to choose a gauge. The story of the quantization of gauge theory - even in the Abelian case of electromagnetism - is subtle, and you are right to not blindly accept the argument that the two classical polarizations of the gauge field - the longitudinal one is eliminated by gauge symmetry - become distinct kinds of particle states in the quantum theory: The decoupling of the states one would naively associate to the longitudinal modes is ensured by the Ward identities, and not at all obvious a priori.

It is by this that the properties of being a gauge boson and of not having a $S_z = 0$ and of being massless are all interrelated: Being one of these things immediately also forces the other two. In this answer, I regarded "being massless" as the fundamental property, since this shows "no $S_z=0$" without assuming anything more specific about the field - in particular, without restricting to gauge fields or electromagnetism a priori.

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    $\begingroup$ Can you comment on how this relates to photon absorption? You can have an atom in an $S_z=0$ state absorb a photon and transition to an $S_z=±1$ state, so the photon 'spin' does couple to the mechanical angular momentum degrees of freedom. How does that process look like in terms of the group representations in your answer? $\endgroup$ Commented Jul 1, 2016 at 10:43
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    $\begingroup$ @EmilioPisanty: Bound states like atoms are horribly difficult to deal with in quantum field theory, so I think one usually models that in the non-relativistic QM regime and just imposes that $S_z = 0$ doesn't exist for a photon by hand. You can, however, have an electron in QFT have absorb or emit a photon in the presence of a nucleus - that's (reverse) bremsstrahlung, but there you don't have $S_z$ levels for the electron, the intrinsic helicity of the photon just goes into angular momentum of the electron. $\endgroup$
    – ACuriousMind
    Commented Jul 1, 2016 at 10:57
  • $\begingroup$ Hard it may be, but interesting it remains. In the usual non-relativistic QM, you think of photon absorption as going from one state of global well-defined $J_z=+1$ to another such state. Here you say that the photon's state isn't in that representation - so what's going on there? $\endgroup$ Commented Jul 1, 2016 at 11:01
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    $\begingroup$ @EmilioPisanty: With "horribly difficult" I mean that I think that is an unsolved problem. Interacting QFTs have generically unknown spaces of states - no one knows how to write the state of an atom, so we can't really ask how the interaction looks in QFT. I'm not sure what the best way to do this is - that probably makes for an interesting new question in itself. $\endgroup$
    – ACuriousMind
    Commented Jul 1, 2016 at 11:14
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    $\begingroup$ This is very nice, but if the goal of this question was to provide an intuitive understanding for people not familiar with QFT (a tall order, to be sure), I doubt it will be successful. $\endgroup$
    – Rococo
    Commented Jul 2, 2016 at 19:06

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.

  • $\begingroup$ 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! $\endgroup$
    – Todd R
    Commented Dec 13, 2012 at 1:50
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    $\begingroup$ 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 ? $\endgroup$
    – user4552
    Commented Aug 31, 2013 at 20:50
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    $\begingroup$ This sounds like a reasonable answer but I'm completely lost at the claim that for spin 1 an $S_z=0$ state would be symmetric w.r.t. rotations around e.g. the $x$ axis. It's certainly not what happens for massive particles; normally, such a state would be characterized by complete symmetry (i.e. returns to itself, and not to a dephased version of itself like circular polarizations do) upon a rotation about the axis of propagation. I don't doubt that the (non-)existence of a rest frame can be seen to come in, but this doesn't feel like it. $\endgroup$ Commented Jul 1, 2016 at 0:36
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    $\begingroup$ I don't understand this argument - it seems to imply that for a massive spin-1 particle, you can only have $S_z = 0$ in the rest frame. But that's not right - the correct statement is that you can only have $S_z = 0$ if there exists a rest frame. $\endgroup$
    – tparker
    Commented Jul 5, 2016 at 12:41
  • $\begingroup$ @BenCrowell I think Rennie’s argument applies as well to graviton. For a massless particle, the direction of momentum breaks the rotational symmetry (down to the little group). Now only rotating around this axis survives. Therefore the helicity can only be ±2. $\endgroup$ Commented Sep 28, 2017 at 4:42

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 $$ A^{\mu}(x) = \int{\frac{1}{\sqrt{p^0}}A^{\mu}({\bf{p}})e^{-ip.x}}d^3{\bf{p}} + \textrm{c.c.}$$ If the particle is moving in the z direction, then its momentum is $$ p^{\mu} = (p^0, 0, 0, p^3)$$ and the Lorenz 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 $$ A^{\mu}(x) = \int{\frac{1}{\sqrt{p^0}}a_{3}({\bf{p}})\epsilon_3^{\mu}e^{-ip.x}}d^3{\bf{p}} + \textrm{c.c.}$$ 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.


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 left- and right-circularly polarized photons, respectively. 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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    $\begingroup$ 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. $\endgroup$
    – Todd R
    Commented Dec 12, 2012 at 6:11
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    $\begingroup$ @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. $\endgroup$
    – KDN
    Commented Dec 12, 2012 at 13:54

In Quantum Field Theory the one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles.

The Casimirs

If we have an irreducible representation of a group then Schur's Lemma says that an operator that commutes with all generators, a Casimir Operator, must be a multiple of the identity. Then applying this operator to any state of the representation gives the same eigenvalue (sometimes also called Casimir). We use the eigenvalues of different representations to label them. This is exaclty what we do in Quantum Mechanics when we use the Casimir $J^2$ and they eigenvalues $j$ to label irreducible representations of the angular momenum algebra.

The Poincare Group has two Casimir Operators, $P_\mu P^\mu$ and $W_\mu W^\mu$, where $P^\mu$ is the momentum generator and $$W^\mu=-\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}J_{\nu\sigma}P_\rho,$$ is the Pauli-Lubanski vector. The $J^{\mu\nu}$ are the Lorentz Group generator. We can assume therefore that we have two labels for the irreducible representations of the Poincare Group.

We write the one particle states as $$|p,\sigma\rangle,$$ where $p$ is the four momentum and $\sigma$ is the other label to be determined. The eigenvalues of $P_\mu P^\mu$ are $m^2$, the square mass of the particle. This gives rise to an infinite dimensional representation whose states are labeled by four momentum $p$. So we are left to find the irreducible representations of the homogenous Lorentz Group. However we have to consider the massive and massless cases separately.

The Little Group

Let us first pick up a particular four momentum $k$. We write a general Lorentz group transformation as $$\Lambda=L(\Lambda p)W(\Lambda,p)L^{-1}(p),$$ where $L(p)$ is the boost relating $k$ and $p$, $$L(p)k=p,$$ $$W(\Lambda,p)\equiv L^{-1}(\Lambda p)\Lambda L(p),$$ is the so-called Wigner rotation and the $L^{-1}$ denote the inverse transformation. These elements form the so-called Little Group which leaves the rest frame momentum $k$ invariant, $$W(\Lambda,p)k=k.$$ Acting with $\Lambda$ on a state $|p,\sigma\rangle$, $$\Lambda |p,\sigma\rangle=L(\Lambda p)W(\Lambda,p)|k,\sigma\rangle,$$ and noticing the resulting state must have four momentum $\Lambda p$ and be in a linear combination of states with the unknown label $\sigma$ we conclude that the $W(\Lambda,p)$ act on the unknown label $\sigma$. Therefore knowing the irreducible representation of the Little Group is what we need to know the irreducible representations of the Poincare Group.

Massive Particles

In this case we can go to the rest frame, $p^\mu=(m,0,0,0)\equiv k^\mu$. We see that the Little Group leaving $k^\mu=(m,0,0,0)$ can be the rotation group in three dimensions, $SO(3)$, or even the more general $SU(2)$ which is a double cover of $SO(3)$. For the later case we know (standard Quantum Mechanics) that their irreducible representations are labeled by the spin $j=0,1/2,1,3/2,...$ and the total number of states for a given spin is $2j+1$.

Massless Particles

There is no rest frame so we choose $P^\mu=(k,0,0,k)$. The Little Group leaving $k$ invariant is the Euclidean group in two dimensions $ISO(2)$ which consists of two translations and rotations in the $x^1x^2$ plane. The two translation generators give rise to another continuous eingenvalue $\theta$ but it is an experimental fact that there is no particle with $\theta\neq 0$. So we only need to consider the plane rotations. These rotations (about the $x^3$ axis) form the Abelian group $SO(2)$ whose elements are $e^{i\phi \vec J\cdot\vec e_3}$. Each representation of this group has only one state, and they are labeled by integers $$h\equiv \vec J\cdot\vec e_3,$$ which we will call helicity. A massless particle in principle has one possible value of the helicity $h$ but from its definition the helicity is a pseudo-scalar. For a massless particle interacting through a parity conserving interaction we have to assign the two representations $h$ and $-h$ to represent the particle. That is why the phtoton has helicity $+1$ and $-1$ and the graviton has helicity $+2$ and $-2$.

  • $\begingroup$ About the last equation you wrote " they are labeled by integers... which we will call helicity". but I think the main problem is why this "integer" couldn't be zero? (corresponding to trivial but whoever allowed representation) $\endgroup$
    – moshtaba
    Commented May 17, 2019 at 0:55

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.


According to quantum electrodynamics, the most accurately verified theory in physics, a photon is a single-particle excitation of the free quantum electromagnetic field. More formally, it is a state of the free electromagnetic field which is an eigenstate of the photon number operator with eigenvalue 1.

The single-particle Hilbert space of the photon carries a unitary irreducible massless spin 1 representation of the extended Poincare group. In the massless case, the vector representation (which is an irreducible spin 1 representation in the massive case) is reducible and decomposes into an irreducible scalar representation on the longitudinal modes and an irreducible representation on the transversal modes; the latter is the photon representation.

In momentum space, longitudinal modes have a vector potential $A(p)$ parallel to the 3-momentum $p$, and transversal modes have a vector potential $A(p)$ orthogonal to $p$ (typically split into two linear or circular polarization modes). The lack of longitudinal modes in the irreducible representation accounts for the lack of $S_z=0$ states of photons propagating in $z$-direction (i.e., with momentum parallel to $(0,0,1)^T$).

The most general single photon states have the form $|A\rangle = \int \frac{dp^3}{2p_0} A(p)|p\rangle$, where $|p\rangle$ is a single particle state with definite 3-momentum $p$, $p_0=|p|$ is the corresponding photon energy divided by $c$, and the photon amplitude $A(p)$ is a polarization 3-vector orthogonal to $p$. Thus a general photon is a superposition of monochromatic waves with arbitrary polarizations, frequencies and directions.

The photon amplitude $A(p)$ can be regarded as the photon's wave function in momentum space. Since photons are not localizable (though they are localizable approximately), there is no photon wave function in coordinate space with a probability interpretation of being localized at a position.

The Fourier transform of $A(p)$ is the so-called analytic signal $A^{(+)}(x)$. By adding its complex conjugate one gets a real 3-vector potential $A(x)$. In terms of this, the mass zero and transversality conditions together translate into the free Maxwell equations written in vector potential form. Extending the 3-vector potential to a 4-vector potential by adding a vanishing 0-component and allowing for gauge transformations brings the conditions into the covariant 4-dimensional form of the free Maxwell equations in the Lorentz gauge, $$\nabla \cdot \nabla A(x) = 0,~~~~\nabla \cdot A(x) = 0.$$ In particular, a single photon has precisely the same degrees of freedom as a classical vacuum radiation field.

[Added July 6] Note that photons couple through matter only through the conserved charge current $j(x)$. Conservation of charge means that $\partial \cdot j(x)=0$. Therefore integration by part implies that in the matter - photon interaction $\int dx~ j(x)\cdot A(x)$, the longitudinal part of $A(x)$ is irrelevant as the term doesn't change when one adds to $A(x)$ a longitudinal term $\partial V(x)$ with scalar $V$. This also shows that massless vector potentials and gauge invariance go hand in hand. Also note that the Coulomb part of the electromagnetic field is not represented by physical photons. (It can be viewed in terms of virtual photons; these don't form a causal representation of the Poincare group but have all possible 4-momenta including the tachyonic ones and all possible spin 1 states.)


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