Let's have the solution for vector boson Lagrangian in form of 4-vector field: $$ A_{\mu } (x) = \int \sum_{n = 1}^{3} e^{n}_{\mu}(\mathbf p) \left( a_{n}(\mathbf {p})e^{-ipx} + b_{n}^{+} (\mathbf p )e^{ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} }, $$ $$A^{+}_{\mu } (x) = \int \sum_{n = 1}^{3}e^{n}_{\mu}(\mathbf p) \left( a^{+}_{n}(\mathbf {p})e^{ipx} + b_{n} (\mathbf p )e^{-ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} }, $$ where $e^{n}_{\mu}$ are the components of 3 4-vectors, which are called polarization vectors. There are some properties of these vectors: $$ e_{\mu}^{n}e_{\mu}^{l} = -\delta^{nl}, \quad \partial^{\mu}e^{n}_{\mu} = 0, \quad e_{\mu}^{n}e_{\nu}^{n} = -\left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right). $$ The first two are obviously, but I have the question about the third. It is equivalent to the transverse projection operator $(\partial^{\perp })^{\mu \nu}$ relative to $\partial_{\mu}$ space. So it realizes Lorentz gauge in form of $$ \partial^{\mu}A^{\perp}_{\mu} = 0, \quad A^{\perp}_{\mu} = \left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right)A^{\nu}, $$ which is need for decreasing the number of components $A_{\mu}$ as vector form of representation $\left(\frac{1}{2}, \frac{1}{2} \right)$ of the Lorentz group by one (according to the number of components of spin-1 field).

I don't understand how to interprete this property. Can it be interpreted as matrice of dot product of 4 3-vectors $e_{\mu}$, which makes one of component of $A_{\mu}$ dependent of anothers three?

  • $\begingroup$ I find your question extremely confusing. What can you be possibly asking about? What does it mean to "interpret this property" if "the property" actually isn't a property but an equation or a definition of an object? How can an equation with no dot products be interpreted as a matrix of dot products? Or can "what" be interpreted as that? Moreover, the equations above are meant to completely isolate the 4 components of the vector field, treat them separately, and decide in the right basis which of them are physical and which of them are not. Where is any dependence in this isolated treatment? $\endgroup$ – Luboš Motl Sep 19 '13 at 17:50
  • $\begingroup$ @LubošMotl . The third "property" is the result of first two. But I want to know how it happened that it linking $e_{\mu}^{n}e_{\nu}^{n}$ (which can be interpreted as dot product of 4 3-vectors, so I don't understand, why you din't see dot product) with transverse projection operator in space of $\partial^{\mu}$. The equations for $A_{\mu}, A^{*}_{\mu}$ are solutions of Klein-Gordon equation for each components with transverse condition, so we can forget about the condition, because the $e_{\mu}^{n}$ vectors determined by the two first properties promotes to satisfying it identically. $\endgroup$ – user8817 Sep 19 '13 at 18:01

In a sum on polarizations, like $\sum_\lambda~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)$, there is a fundamental difference if you are considering all the polarizations, or only the physical polarizations .

If you take all polarizations, the sum is equals to $g_{\mu\nu}$, and it is in fact a normalizations of the $e_{\mu}^{\lambda}(k)$. This sum is non-physical.

$$\sum_{all ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-g_{\mu\nu} \tag{1}$$

If you consider only the physical polarizations, this sum is physical, and you will get the pole of the propagator, which is a physical quantity too :

$$\sum_{physical ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})\tag{2}$$

The propagator here is :

$$D_{\mu\nu}(k) = \frac {-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})}{k^2-m^2} \tag{3}$$

  • $\begingroup$ Thank you. Also, is it connected with my explanation? And if not, is my explanation correctly at all? $\endgroup$ – user8817 Sep 19 '13 at 19:32
  • $\begingroup$ I am afraid that no one understands your explanation, @PhysiXxx, so getting the usual wisdom about these polarization vectors is the maximum you may hope for. $\endgroup$ – Luboš Motl Sep 20 '13 at 5:04
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    $\begingroup$ @PhysiXxx : It is true that your question is not completely clear. I tried to present the important physical point as clear as possible. Of course, you have to kill one unphysical freedom degree. And you can see directly that this is represented in the expression of the physical propagator, which is directly linked to the sum on physical polarizations. You may verify that the equation $(2)$ is compatible, on shell (so with $k^2=m^2$), with $k^\mu e_{\mu}^{\lambda}(k)=0$ $\endgroup$ – Trimok Sep 20 '13 at 7:25

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