(1) In the canonical quantization of the free electromagnetic field, the Coulomb gauge condition $$A^0=0,~~ \nabla\cdot\textbf{A}=0\tag{1}$$ implies that the polarization vector $\epsilon^\mu$ satisfies $$\epsilon^0=0,~~\boldsymbol{\epsilon}\cdot\hat{\textbf{p}}=0\tag{2}$$ which says that the electromagnetic field has two independent transverse states of polarization.

(2) From the representation theory of Poincare group, it is known that for photons $\textbf{S}\cdot\hat{\textbf{p}}$ has eigenvalues $h=\pm 1$ where $\textbf{S}$ denotes the spin operator.

Both the descriptions above make it clear that the electromagnetic field has two independent degrees of freedom. The description (1) says the electromagnetic field has two independent states of polarization and the description (2) says that it has two independent helicity states.


$\bullet$ Does it mean states of polarization are identical to states of helicity?

$\bullet$ Is there is a unique one-to-one correspondence between the states of helicity and the independent states of polarization? In that case, $h=+1$ corresponds to which polarization and $h=-1$ corresponds to which? How can such a correspondence, if exists, be understood?

A similar question was asked here.

  • $\begingroup$ Polarization vectors are the photon wave-functions (in momentum space) corresponding to states of a given helicity. $\endgroup$
    – Prahar
    Commented Dec 30, 2017 at 13:14
  • $\begingroup$ because the photon has zero mass and spin one, there are only two possible states +1 to the direction of motion and -1 to the direction of motion.look at the figures in this link to see the connection between the polarization of a light wave and the spin of the photon en.wikipedia.org/wiki/Spin_angular_momentum_of_light $\endgroup$
    – anna v
    Commented Dec 30, 2017 at 15:08

2 Answers 2


States of definite helicity are states of definite spin measured along a particular axis. You can, in principle, use any axis to define your spin eigen-basis, it's just not commonly done because the result isn't Lorentz invariant and you have to be careful about what spins are forbidden by the lack of a helicity 0 state. It can be shown that there is a unique one to one mapping between the choice of spin basis and the polarization state basis. The polarization vectors, $\boldsymbol{\epsilon}_\lambda$, handle that mapping (the $\lambda$ indices exist in spin/polarization space, and the spatial index exists in physical space).

  • 1
    $\begingroup$ "It can be shown that there is a unique one to one mapping between the choice of spin basis and the polarization state basis. " Can you show this or suggest a reference? @SeanE.Lake $\endgroup$
    – SRS
    Commented Jun 12, 2019 at 19:17
  • $\begingroup$ The polarization of a wave is set by a linear combination of the $\mathbf{\epsilon}_r$, right? The index $r$ lives in helicity space, and the vector components of each $\mathbf{\epsilon}$ live in physical/polarization space. In other words, you assume this to be the case when you use $\mathbf{\epsilon}_r$. @SRS $\endgroup$ Commented Jun 12, 2019 at 19:52

In the radiation gauge, the 3-vector potential has the most general Fourier mode expansion given by $$\textbf{A}(\textbf{x})=\int\frac{d^3\textbf{p}}{(2\pi)^3\sqrt{2E_{\textbf{p}}}}\sum\limits_{r=1}^{2}[\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}+\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{1}$$ Using the definition of spin operator $$S^{ij}=\int d^3\textbf{x}:(A^i\partial_0 A^j-A^j\partial_0 A^i):$$ and Eq.(1), one obtains after performing the integral over space $$S^{ij}=i\int\frac{d^3\textbf{p}}{(2\pi)^3}\sum\limits_{r,s}[\epsilon_r^i(\textbf{p})\epsilon_s^{j*}(\textbf{p})-\epsilon_s^{i*}(\textbf{p})\epsilon_r^j(\textbf{p})]a^\dagger_{\textbf{p},r}a_{\textbf{p},s}.$$ The action of the spin operator $S^{ij}$, as obtained in Eq. (2), on a one-particle state $a^\dagger_{\textbf{k},m}|0\rangle$, one finds, $$S^{ij}a^\dagger_{\textbf{k},m}|0\rangle=i\sum\limits_{s=1}^{2}\Bigg[\epsilon_m^i(\textbf{p})\epsilon_s^{j*}(\textbf{k})-\epsilon_s^{i*}(\textbf{k})\epsilon_m^j(\textbf{k})\Bigg]a^\dagger_{\textbf{k},s}|0\rangle.$$

Let us choose $\textbf{k}=(0,0,k)$ so that the helicity is measured by $\textbf{S}\cdot\hat{\textbf{k}}=S^3=S^{12}$. We choose, $\boldsymbol{\epsilon}_{1}(\textbf{k})=1/\sqrt{2}(1,i,0)$ and $\boldsymbol{\epsilon}_{2}(\textbf{k})=1/\sqrt{2}(1,-i,0)$. Therefore, $$S^3a^\dagger_{\textbf{k},1}|0\rangle=(+1)a^\dagger_{\textbf{k},1}|0\rangle,\\ S^3a^\dagger_{\textbf{k},2}|0\rangle=(-1)a^\dagger_{\textbf{k},2}|0\rangle.$$

Conclusion A one-particle state with right circular polarization corresponds to helicity $+1$, and the one-particle state with left-circular polarization corresponds to a state with helicity $-1$.


The electric field is given by $$\textbf{E}(\textbf{x})=-\frac{\partial\textbf{A}}{\partial t}=(-i)\int\frac{d^3\textbf{p}}{(2\pi)^3}\sqrt{\frac{E_\textbf{p}}{2}}\sum\limits_{r=1}^{2}[\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}-\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{2}$$ and

$$\textbf{B}(\textbf{x})=\nabla\times\textbf{A}=(i)\int\frac{d^3\textbf{p}}{(2\pi)^3}\sqrt{\frac{E_\textbf{p}}{2}}\sum\limits_{r=1}^{2}[\hat{\textbf{p}}\times\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}-\hat{\textbf{p}}\times\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{3}$$ where I used the fact that $E_{\textbf{p}}=|\textbf{p}|$. Using $\hat{\textbf{p}}=(0,0,1)$, it should be easily checked that the operators $\textbf{E}\pm i\textbf{B}$ acting on the vacuum $|0\rangle$ respectively creates one-particle states of photon with right-circular and left-circular states of polarization. I'm lazy to work out the algebra.

Reference A Modern Introduction to Quantum Field Theory-Michele Maggiore.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.