The Fourier mode expansion of the free electromagnetic field in radiation gauge is given by $$\textbf{A}(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2\omega_\textbf{p}}}\sum\limits_{\lambda=1,2}[\boldsymbol{\epsilon}_\lambda a_{\textbf{p},\lambda}e^{-ip\cdot x}+\boldsymbol{\epsilon}^{*}_\lambda a_{\textbf{p},\lambda}e^{+ip\cdot x}].$$

What does $\lambda$ count? As I understand, it doesn't count the $x,y,z$ components of $\boldsymbol{\epsilon}$ because those are counted by the spatial Lorentz indices $i=1,2,3$ in $\boldsymbol{\epsilon}_\lambda=\{\epsilon_\lambda^i\}=(\epsilon_\lambda^1,\epsilon_\lambda^2,\epsilon_\lambda^3)$. Which clearly shows that $\lambda$ doesn't count the spatial components of $\boldsymbol{\epsilon}$.

On the other hand, the relation $\boldsymbol{\epsilon}_{\lambda}\cdot\textbf{p}=0$ implies that 2 of the 3 spatial components of $\boldsymbol{\epsilon}_{\lambda}$ i.e., $\epsilon^1_\lambda, \epsilon^2_\lambda, \epsilon^3_\lambda$ will be independent.

Therefore, I do not understand where does the restriction $\lambda=1,2$ come from? It appears to me that there is a restriction on the components of a given $\boldsymbol{\epsilon}_{\lambda}$ vector.

Am I misinterpreting something?


$\lambda$ counts the number of independent polarizations of a photon. Note that the polarization tensor is a 4-vector $\epsilon^\mu$. In Coulomb gauge, $A^0 = 0$ so that $\epsilon^0 = 0$. Thus, a generic polarization tensor in Coulomb gauge takes the form $\epsilon^\mu = (0,\epsilon^i)$ and is therefore given by 3 variables. Further, these three variables are not all independent but are constrained by the condition $$ \epsilon^i p_i = 0 \, . $$ This is 1 equation for 3 variables. We can therefore solve for one of the variables in terms of the other 2. Thus, in total, there are 2 independent solutions to the equation above which we label as $\epsilon_\lambda^i$ with $\lambda = 1 ,2$.

For example, if you wish, you can solve for $\epsilon^3$ in terms of $\epsilon^1$ and $\epsilon^2$ and a generic solution to the constraint takes the form $$ \epsilon^\mu = (0,\epsilon^1,\epsilon^2, - \frac{p^1 \epsilon^1 + p^2 \epsilon^2 }{ p^3 } ) $$ Then, the two independent polarizations can be found by choosing $(\epsilon^1,\epsilon^2)=(1,0)$ and $(0,1)$. For instance $$ \epsilon^\mu_{\lambda=1} = (0, 1 , 0 , - \frac{p^1 }{ p^3 } ) \, , \qquad \epsilon^\mu_{\lambda=2} = (0, 0 , 1 , - \frac{p^2 }{ p^3 } ) $$ The above choice of independent polarizations are the linear polarizations.

Another set of two independent polarizations can be found by choosing $(\epsilon^1,\epsilon^2)=(1,i)$ and $(1,-i)$. These are called circular polarizations.


That $\lambda$ counts the number of independent polarization states available to the photon. Real photons exist only in the solenoidal part of the vector potential, $\mathbf{A}$. Since there are two independent degrees of freedom at any point in a solenoidal field, you get two possible values for $\lambda$. You could get three possible values for $\lambda$ if the photon had mass, making the longitudinal (irrotational) part of $\mathbf{A}$ physical, but that wouldn't be gauge invariant.

It's easier to see what's going on mode space ($\mathbf{k}$-space), where the integral and sum in the question already are. There, $\lambda$ just labels two vectors orthogonal to the radial unit vector. One choice for them would be $\boldsymbol{\epsilon}_1 = \hat{\theta}$ and $\boldsymbol{\epsilon}_1 = \hat{\phi}$, the unit vectors traditionally chosen to be orthogonal to a radial unit vector in a spherical coordinate system. In physical space this only makes sense when you work with two physical points, a source and observer, instead of one. When you do that, this implies that the vector potential will be perpendicular to the line of sight connecting the two points.

What about $\phi=A^0$ you ask? Well, that field isn't actually a dynamical field because it's time derivative doesn't appear in the Lagrangian. Thus, it plays a role more like a Lagrange multiplier constraint field than anything else. So, $A^\mu$ has only 2 degrees of freedom that behave like particles, one that acts like a Lagrange multiplier ($\phi$), and another that cannot take any particular value due to gauge invariance (up to a linear integral, $\nabla \cdot \mathbf{A}$).

For details on how to quantize the electromagnetic field correctly (handling both gauge fixing constraints, and the constraint equation you get from varying the Lagrangian with respect to $\phi$), I recommend Weinberg's "Quantum Theory of Fields" Vol. 1 and 2 since he goes over both the canonical and Lagrangian formalisms thoroughly (even if the notation is a little cumbersome).

  • $\begingroup$ My question is not about $A^0$. I'm confusing between the indices i and $\lambda$ that $\epsilon$ carry. If the question is unclear, I'll try to modify it. @SeanE.Lake $\endgroup$ – SRS Dec 29 '17 at 7:00
  • $\begingroup$ The point of my confusion is that epsilon carry two types of indices and all I can see that the index i can be constrained to have two values not $\lambda$. $\endgroup$ – SRS Dec 29 '17 at 7:02
  • $\begingroup$ @SRS The key piece of information is that $\hat{p}$ is like $\hat{r}$ - it doesn't always point in the same physical direction. So there are two polarization vectors, and they map to a Euclidean vector space with Cartesian basis vectors, so that index needs to take three values even if not all elements are independent. $\endgroup$ – Sean E. Lake Dec 29 '17 at 7:16
  • $\begingroup$ All I can see from the constraint $\boldsymbol{\epsilon}_\lambda\cdot\textbf{p}=0$ is that the components of a $\boldsymbol{\epsilon}_{\lambda}$ (for a fixed $\lambda$) are not all independent. But is there a constraint which tells us that there are two independent polarization vectors? If we had a constraint of the form $\sum\limits_{\lambda}\boldsymbol{\epsilon}_\lambda\cdot\textbf{p}=0$, then, we could have said that there are two independent $\boldsymbol{\epsilon}$ vectors, $\boldsymbol{\epsilon}_1$ and $\boldsymbol{\epsilon}_2$, labelled by $\lambda=1,2$ respectively. @SeanE.Lake $\endgroup$ – SRS Dec 29 '17 at 8:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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