This is a question regarding the spin-1 massive field commutator $[A_i(\mathbf{x},t),\Pi_j(\mathbf{y},t)]$, where $\Pi$ is the conjugate field and $A^\mu$ is the four-potential. My result was, $$[A_i(\mathbf{x},t),\Pi_j(\mathbf{y},t)]=-i \left(\delta_{ij}-\frac{\nabla_i \nabla_j}{m^2}\right)\delta^3(\mathbf{x}-\mathbf{y}).$$ One of my textbooks says it should be equal to, $[A_i(\mathbf{x},t),\Pi_j(\mathbf{y},t)]=-ig_i^j\delta^3(\mathbf{x}-\mathbf{y})$. But where does the second term go? Another textbook replaces the $m^2$ with $\nabla^2$ for some reason. I am using the Lorenz gauge: $\partial_\mu A^\mu=0$.

Here's the university textbook:

enter image description here


A massless spin $j=1$ field has $$ [A_i(\mathbf{x},t),\Pi_j(\mathbf{y},t)]=-i \left(\delta_{ij}+\frac{\nabla_i \nabla_j}{{\color{red}\nabla^2}}\right)\delta^3(\mathbf{x}-\mathbf{y}). $$ while a massive spin $j=1$ field has $$ [A_i(\mathbf{x},t),\Pi_j(\mathbf{y},t)]=-i \left(\delta_{ij}-\frac{\nabla_i \nabla_j}{{\color{red}m^2}}\right)\delta^3(\mathbf{x}-\mathbf{y}). $$

Finally, a (massless or massive) Stückelberg field, in the 't Hooft-Feynman gauge $\xi=1$ has $$ [A_\mu(\mathbf{x},t),\Pi_\nu(\mathbf{y},t)]=-i \eta_{\mu\nu}\delta^3(\mathbf{x}-\mathbf{y}). $$

The first case is sometimes said to correspond to the Coulomb gauge (even though one could argue that we are not picking any gauge in particular, but just using the polarisation vectors as dictated by the properties of the massless representations of the Poincaré Group), while the second one is sometimes said to correspond to the unitary gauge ($\xi\to\infty$) of a Stückelberg field . This is also known as a Proca field.

  • $\begingroup$ Another good answer! So, as far as I understand, a $\xi=1$ gauge removes the $\frac{m^2(\partial_\mu A^\mu)^2}{2}$ term from the Lagrangian? $\endgroup$ – Yuri Kotsar May 3 '17 at 22:21
  • $\begingroup$ Sorry, I meant $m^2A_\mu^* A^\mu$. But I guess I have to read more, so I'll close this for now. $\endgroup$ – Yuri Kotsar May 4 '17 at 9:01

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.