Why are the odd point green functions in free field theory zero? I don't understand why the $(N=\mathrm{odd})$-point Green functions calculated in free field theory are identically zero. Is it because the Green functions are odd? If so, then how do I prove it? Is there any physical interpretation of the result?
 A: Let's try to analytically find out the n-point function (for a free theory). The partition function(al) with a source term $j(x)$ is,
$$Z[j] = \int [D\phi] \, \exp\{i\int dx \left(-\frac12 \phi(x)(\partial_x^2 + m^2 -i\epsilon) \phi(x) + j(x) \phi(x) \right)\} \,. $$
The extra $i\epsilon$ term is added to facilitate $t \rightarrow \infty$ limit. To complete the square you can use a shifted field $\phi' = \phi + \int dy \, \Delta_F(x-y) j(y)$, using the property of the Feynman propagator (or Green function) $\Delta_F$ the partition function gets simplified to 
$$Z[j] = Z[0] \exp\{-\frac12 \int \, dx \, dy \, j(x) \, \Delta_F(x-y) \, j(y) \} \,.  $$
The n-point function can be derived by differentiating the partition function $n$ times with respect to $j(x)$ and setting the source to zero in the end. For instance the 2-point function would be
$$\langle T(\phi(x) \phi(y) \, ) \rangle = \frac{1}{Z[j]} \frac{\delta^2 Z[j]}{i \delta j(x) i\delta j(y)} = i \Delta_F(x-y) + \text{ (terms linear in j)}$$
Since we set $j=0$, so the two point function simply becomes $i \Delta_F(x-y)$. However, you can see from the structure of the partition function the odd-point functions vanish because they are always linear in $j$. 
If you want to have a physical idea of why they vanish, then you can think in terms of diagrams. Because there is no interaction all the vertices vanish, you just have a bunch of (disconnected) lines joining pairs of points. For example a 4-point function would be 3 different terms, which are basically three different ways of connection two pairs of points. But now if you bring one more field inside the expectation you have to attach that extra (odd-th) leg to one of the lines, there by creating a vertex, however that's not allowed in a free theory so they have to vanish.
