7
$\begingroup$

Can we write down a generic expression for propagator for any arbitrary spin? At the least, about the ultraviolet behavior of these propagator. Especially, I would like to know whether there is a specific dependence on spin of the field.

$\endgroup$
  • $\begingroup$ @HolgerFiedler I meant a generic quantum field of arbitrary spin. Let us also suppose that it is massive. $\endgroup$ – Truthseeker Aug 14 '16 at 10:34
6
$\begingroup$

It is possible to write an expression for the propagator of a massive spin $j$ particle whose ultraviolet limit can be shown to vary as $q^{-2+2j}$.

To see why this holds one has to find a natural generalization of Pauli matrices for $(j,0)$ and $(0,j)$ and of gamma matrices for the one where you deal with parity conserving interaction which involves fields transforming as $(j,0)\oplus (0,j)$ representation the details to which can be found in the reference mentioned.

Momentum space propagator of a particle of spin $j$ (eqn. 5.13 of reference which need not be mentioned in the context) has the structure involving a matrix $\Pi(q)$ written in terms of $2j+1$ dimensional matrix of finite Lorentz transformation $D^j[L (\bf {p})]$ where $L (\bf{ p})$ is lorentz boost (eqn. 4.3 of reference) in the numerator. Explicit calculation of matrix $\Pi(q)$ is given in the appendix of the reference.

In ultraviolet limit, the propagator behaves as $\Pi(q)/q^2$ where the value of monomial $\Pi(q)$ is given in table 1 of the reference. From the table it is easily seen that the monomial’s leading term in ultraviolet region goes as $q^0,q,q^2,q^3$ and so on for respectively $0,1/2,1,3/2$ spin particle and hence contributes a factor of $q^{2j}$ and in total propagator behaves as $q^{-2+2j}$ for a spin $j$ particle.

The case of $(j,0)\oplus (0,j)$ yields similar behavior for massive particle. This analysis actually breaks down for massless particles because of the non semi-simple structure of the little group associated with $m=0$ case and there is not a definite relation between massless particles propagators and the spin as photon and graviton both will behave as $j=0$ in the above relation.

S. Weinberg , Feynman rules for any spin. Phys. Rev. 133,1964.

$\endgroup$

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.