photon propagator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:22:50Z http://mathoverflow.net/feeds/question/7357 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7357/photon-propagator photon propagator Dan 2009-12-01T03:30:18Z 2012-10-01T12:16:10Z <p>I am reading Zee's book "QFT in a nutshell". I have a question on the photon propagator computation. For a massive photon, consider the Lagrangian $L = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2}m^2A_\mu A^\mu + A_\mu J^\mu$, then the path integral is $Z = \int dx ~L = \int dx ~{ \frac{1}{2}A_\mu[(\partial^2 + m^2)g^{\mu \nu} - \partial^\mu \partial^\nu]A_\nu + A_\mu J^\mu }$. From this we get that the photon propagator $D_{\mu \nu}$ satisfies $[(\partial^2 + m^2)g^{\mu \nu} -\partial^\mu \partial^\nu ] D_{\nu \lambda}(x) = \delta^\mu_\lambda \delta^{(4)}(x)$, and solving this, $$D_{\nu \lambda}(k) = \frac{-g_{\nu \lambda} + k_\nu k_\lambda/m^2}{k^2 - m^2}.$$ </p> <p>I can not see why the numerator has a term $k_\nu k_\lambda/m^2$. Any ideas?</p> http://mathoverflow.net/questions/7357/photon-propagator/7369#7369 Answer by David Bar Moshe for photon propagator David Bar Moshe 2009-12-01T04:28:08Z 2009-12-01T04:28:08Z <p>This is just because of the algebraic matrix inversion in the momentum space. It can be seen from the identity:</p> <p>(g_mu_nu - k_mu * k_nu/(k^2-m^2)) *(g_nu_lambda - k_nu k_lambda/m^2) = g_mu_lambda.</p> <p>This is a special case of a general result in linear algebra: The inverse of a matrix prportional to a weighted sum of a unit matrix and a Hermitian matrix of unit rank is also proportional to a (generally different) weighted sum of the unit matrix and the Hermitian unit rank matrix.</p> http://mathoverflow.net/questions/7357/photon-propagator/7372#7372 Answer by Theo Johnson-Freyd for photon propagator Theo Johnson-Freyd 2009-12-01T04:43:16Z 2009-12-01T04:43:16Z <p>Just multiply it out in Fourier, where $\partial = ik$:</p> <p>$$\bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] \frac{ -g_{\nu\lambda} + m^{-2} k_\nu k_\lambda }{k^2 - m^2} = \frac1{k^2 - m^2} \bigl( - (-k^2 + m^2) \delta^\mu_\lambda + (-k^2 + m^2) m^{-2} k^\mu k_\lambda - k^\mu k_\lambda + k^\mu k^2 k_\lambda m^{-2} \bigr) = \delta^\mu_\lambda$$</p> <p>which is a function of $k$. But converting back to position space, $1(k) = \delta(x)$.</p> <p>This proves that $D$ is a solution. To be <em>the</em> solution, you usually have to impose boundary conditions, etc. In this case, there are no solutions to $\bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] f_\nu = 0$: the corresponding equation in Fourier is $(-k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu = 0$, and contracting with $g_{\mu \nu}$ gives $0 = d(-k^2 + m^2) + k^2 = dm^2 - (d-1)k^2$, where $d$ is the dimension of spacetime, so $k^2 = \frac{d}{d-1}m^2$, but $-\frac1{d-1}m^2 g^{\mu\nu} + k^\mu k^\nu$ cannot equal $0$, as $k^\mu k^\nu$ cannot be an invertible matrix. So $D$ is the only solution.</p> http://mathoverflow.net/questions/7357/photon-propagator/108477#108477 Answer by Jimiras for photon propagator Jimiras 2012-09-30T15:12:27Z 2012-09-30T15:12:27Z <p>Sir Theo Johnson-Freyd had not answer "how" the kνkλ/m2 term came up in numerator. If plugging directly the answer which A.Zee gave into the differential equation, it just confirm the massive photon propagator has that form.</p> http://mathoverflow.net/questions/7357/photon-propagator/108541#108541 Answer by Jeff Harvey for photon propagator Jeff Harvey 2012-10-01T12:16:10Z 2012-10-01T12:16:10Z <p>Theo and David are perfectly correct. To add a bit more of a physical explanation which might help with the why part, a massive spin one particle has 3 physical degrees of freedom so there must be some condition on the four components $A_\mu$. The equation of motion for $A_\mu$ is equivalent to saying that each component of $A_\mu$ satisfies the massive Klein-Gordon equation and that in addition $\partial^\mu A_\mu=0$. This latter condition in momentum space implies that $k^\mu D_{\mu \nu}(k)=0$. So one can understand the $1/(k^2-m^2)$ from each component obeying the massive KG equation and the factor in the numerator as ensuring that $k^\mu D_{\mu \nu}(k)=0$.</p>