# What does the non-Lorentz indices $\lambda$ of the polarization vector $\boldsymbol{\epsilon}_\lambda$ count?

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).

• 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 – SRS Dec 29 '17 at 7:00
• 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 $\lambda$. – SRS Dec 29 '17 at 7:02
• @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. – Sean E. Lake Dec 29 '17 at 7:16
• 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 – SRS Dec 29 '17 at 8:12