1. Aug 8, 2013

### Kontilera

Hello!
I have worked with Green's functions in electrodynamics and have now started reading qft.
First I encountered the spin-0 propagator,
$$D(x-y) = \int \frac{d^4 k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2 -m^2}.$$
This seems not so new.. We ahve a blow up around the mass-shell and the wave propagates through spacetime as a planewave. Our Greenfunction is a bilocal function over spacetime.

Now the problem is the spin-1 field:
$$D_{\nu \lambda} = \frac{-g_{\nu \lambda} + k_\nu k_\lambda /m^2}{k^2 - m^2}.$$
Here the function is not dependent on which spacetime points the emission and absorption take places (right?). How should I interpret the lorentz indices? (What does the tensor ''eat'' and what does it ''spit out''?)

Last edited: Aug 8, 2013
2. Aug 8, 2013

### RGevo

I'm not quite sure I understand the question.

The propagator indices will be contracted at one of the vertices. So for example, if my propagator is a photon in e+e- to q qbar I will have a gamma_nu at one point in space.

I.e the e+e- annihilation vertice.

Then photon propagates to the next space time point gamma_mu (where the q qbar is created).

The photon propagator contracts with one of these space points so I end up with the connected piece. Which now is something with gamma_nu gamma^nu. Which fully contracts.

A similar thing happens if I have a fermion propagator (instead of spin 1), in this case the space time vertices are contracted with 2 photon external lines.

I hope this is partially useful!

3. Aug 8, 2013

### fzero

You've written the first Green function in the spatial representation, whereas the 2nd one is written in the momentum space representation. These representations are Fourier transforms of one another, as you should be able to work out from the relationship between the representations of the fields, i.e., for a scalar

$$\phi(x) = \int \frac{d^4 k}{(2\pi)^4}e^{ikx} \phi(k),$$

where the factors of $2\pi$ and the sign in the exponent are some conventions that you might want to check with your textbook. Then the relationship between the different representations of propagator is obtained by defining the propagator in an appropriate way in terms of $\langle 0 | [\phi(x),\phi(y)]|0\rangle$. There is a discussion on wikipedia and no doubt somewhere in your text.

In momentum space, the scalar Green function is

$$D(k) = \frac{1}{k^2 -m^2},$$

while the spatial representation of the spin 1 propagator will come out to be something like

$$D_{\nu \lambda}(x-y) = \int \frac{d^4 k}{(2\pi)^4}e^{ik(x-y)}\frac{-g_{\nu \lambda} + k_\nu k_\lambda /m^2}{k^2 - m^2}.$$

4. Aug 8, 2013

### king vitamin

There are many equivalent ways to derive the propagators, each with its own way of illuminating why they take the form they have. Since you mentioned being familiar with propagators from classical EM, it might be easiest to understand the propagator simply as the Green's function of the equation of motion. From the Proca lagrangian

$$\mathcal{L} = - \frac{1}{4}F_{\mu \nu}F^{\mu \nu} + \frac{1}{2} m^2 A_{\mu}A^{\mu}$$

we find the equation of motion from Euler-Lagrange:

$$\left[ (\partial^2 + m^2) g_{\mu \nu} - \partial_{\mu} \partial_{\nu} \right] A^{\mu} = 0$$

so the Green's function must satisfy

$$\left[ (\partial^2 + m^2) g_{\mu \nu} - \partial_{\mu} \partial_{\nu} \right] D^{\nu \rho}(x - y) = \delta_{\mu}^{\rho} \delta^3(x - y).$$

This is essentially where the tensor structure comes in. The dynamics are diagonalized by defining the fourier transform of D in the usual way

$$\left[ (- k^2 + m^2) g_{\mu \nu} + k_{\mu} k_{\nu} \right] D^{\nu \rho}(k) = \delta_{\mu}^{\rho}.$$

The propagator you gave in the OP is the solution to this equation. Physically, the reason for the nontrivial tensor structure is that the spin-1 vector field is constrained by its representation in the Lorentz group. The equations of motion I gave above imply that each component satisfies the Klein-Gordon equation $(\partial^2+m^2)A^{\mu} = 0$, but we could just stick four scalar fields on top of each other and they would also satisfy this constraint. A (massive) spin-1 field actually has only 3 degrees of freedom, so we need to project a degree of freedom out. This is done by the equations of motion above, which also imply $\partial_{\mu}A^{\mu} = 0$.

5. Aug 9, 2013

### vanhees71

One should, however, emphasize that the Fourier integral in #1 is undefined. It is utmost important to be aware that this integral only makes sense as soon as a prescription of how to circumwent the poles of the propagator in the complex $p^0$ is given. The various possibilities to run around the poles leads to different propagators, and only the physics application you are considering can tell you which propagator you need. E.g., for vacuum-QFT perturbation theory you need the time-ordered propagator (which in the vacuum is identical with the Feynman propagator). There you get the correct prescription by adding an infinitesimal part in the denominator:
$$D(p)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$
This tells you, which pole is inside the contour you get by closing the integration path in either the upper or lower $p^0$ half-plane for $t_x-t_y<0$ or $t_x-t_y>0$ respectively when the usual sign convention of the Fourier transform is chosen, i.e.,
$$D(x-y)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \frac{\exp[-\mathrm{i} p \cdot(x-y)]}{p^2-m^2+\mathrm{i} 0^+}.$$
I used the "west-coast convention" of the metrix $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$.

6. Aug 9, 2013

### andrien

As others have pointed out,that your first propagator is in coordinate representation and second one is in momentum space.By the way,no indices on spin 0 green function represents it's scalar nature.The vertex for the diagram does not contain any index.While with spin 1 the two vertex which will be joined by that boson line will have index ,the propagator between these two points must come up with two indices in order for whole matrix element to be lorentz invariant(in very simple case).It does not matter who is emitting that boson and who is absorbing.It is already taken care when one writes the two point green function.

7. Aug 9, 2013

### Kontilera

Thanks guys! This reallly helped.