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Why when we quantize EM field, whe quantize the vector potential $A^\mu$ obtaining vectorial particles (photons) like the elastic field (phonons) and we can't quantize directly the EM-field tensor $F^{\mu\nu}$? We in that situation should obtain tensorial particles of spin 2 like graviton..It's wrong? Why?

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Yes! Yes! Yes! It can't be wrong. It has to be an equally valid alternative. –  Marty Green May 28 '11 at 23:15
Possibly related: physics.stackexchange.com/q/9998/2451 –  Qmechanic May 29 '11 at 10:14
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2 Answers

up vote 8 down vote accepted

Just because $F^{\mu\nu}$ has two indices does not mean that it represents a spin-2 particle. Note that the metric $g^{\mu\nu}$ is a symmetric two indexed object while the EM field strength $F^{\mu\nu}$ is antisymmetric. In fact, the metric $g^{\mu\nu}$ is analogous to potential $A^\mu$ in EM and the field strength of gravity is the four indexed Riemann tensor $R_{\mu\nu\rho\lambda}$.

What spin the field represents depends on the symmetries of the indices and the field equations that it obeys. In particular, the physical degrees of freedom of a massless field of spin $(A/2,B/2)$ can be written in terms of dotted and undotted indices $G_{\alpha_1,\dots,\alpha_A,\dot\beta_1,\dots,\dot\beta_B}$ totally symmetric in the dotted and undotted indices. In order for it to be a representation of the Poincare group it must satisfy the supplementary conditions $$ \begin{align} \partial^{\gamma\dot\gamma}G_{\alpha_1,\dots,\alpha_{A-1}\gamma,\dot\beta_1,\dots,\dot\beta_B} &= 0 \\ \partial^{\gamma\dot\gamma}G_{\alpha_1,\dots,\alpha_{A},\dot\beta_1,\dots,\dot\beta_{B-1}\dot\gamma} &=0 \ . \end{align}$$ These supplementary conditions imply that each component of $G$ satisfies the Klein-Gordon equation. (The only exception is the scalar case where there are no supplementary conditions because there's no indices. In this case there's just the KG equation $\Box G = 0$.) It is also shown that these conditions imply that the field has a definite helicity of $(A-B)/2$ and there is only one degree of freedom in such a field. Physical fields are made of the sum of two such fields of opposite helicity

This is clearly the case for the EM field strength $F_{\mu\nu}$ when written in the form $F_{\alpha\beta\dot\alpha\dot\beta}=(\sigma^\mu)_{\alpha\dot\alpha}(\sigma^\nu)_{\beta\dot\beta}F_{\mu\nu}=2\varepsilon_{\alpha\beta}\bar{F}_{\dot\alpha\dot\beta} + 2\varepsilon_{\dot\alpha\dot\beta}F_{\alpha\beta}$, so the field strength decomposes into the sum of two massless fields carrying helicity $\pm1$.

There's a good description of the field representations of the Poincare group in section 1.8 of Ideas and Methods in Superspace and Supergravity. Section 1.8.3 deals with the massless representations applicable in the two cases raised in your question. Section 1.8.4 has the examples of massless scalar, spin-1/2, EM (spin-1), spin-3/2 and linearized gravity (spin-2).

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There's also a really nice argument I've heard from Prof Jim Gates that looks at the structure of a plane wave for fields of different spin. Moving to cylindrical coordinates in the direction that the wave is travelling makes the spin of the field manifest. If someone has a written copy of such an argument, I'd love to get a copy. –  Simon May 29 '11 at 1:54
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It's possible to construct a free quantum field on the basis of the EM field instead of on the basis of the EM potential, however if we do so then we cannot use minimal coupling to define an interaction between a spinor field and the EM field. Whatever alternative model one constructs would have to be empirically closely similar to QED. That's moderately killing, I believe, although I have been actively pursuing algebraic constructions in this general area.

For the EM field, the commutation relations for the creation and annihilation operators can be presented as $$[a_{\mu\alpha}(x),a_{\nu\beta}^\dagger(y)]=\int k_{[\mu}g_{\alpha][\nu}k_{\beta]} 2\pi\delta(k^2)\theta(k_0)e^{-ik\cdot (x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4},$$ antisymmetrizing on the $\mu,\alpha$ and $\nu,\beta$ components, in contrast to the massless Klein-Gordon free field, where there are no space-time indices, for which $$[a(x),a^\dagger(y)]=\int 2\pi\delta(k^2)\theta(k_0)e^{-ik\cdot (x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4}.$$ The first equation is what one obtains by taking the derivatives of the commutation relations of the EM potential. One doesn't see it much in the literature, AFAIK, although a closely related statement can be found in R. Menikoff and D. H. Sharp, J. Math. Phys. 18 (1977) 471.

Note, perhaps, the greater simplicity of the EM field commutation relations, which are already positive semi-definite on the space of test functions as stated here without requiring the Gupta-Bleuler or other machinery to ensure that there is no negative norm sector. It may be that this approach might be helpful in quantum optics, or in other Physics that does not implement interactions by minimal coupling, because of this simplification, however, again AFAIK, I do not believe it is much used. I would be somewhat surprised if it's often used, because the simplification is more conceptual than helpful for calculation, and there is always some pressure to use conventional notations and formalisms.

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