Problem with the conservation law in Feynman diagram I have some problem with the virtual particle process as discussed below.
We have some interactions that have real photon as produced particle. We consider a tree level Feynman diagram for it and the interaction is shown in the picture here:

(direction of time is from left to right)
Here, $u$ can be any particle obeying the usual conservation laws. 
My problem is that, if we have conservation of four momentum in upper vertex and we go to $C.M.$-frame of $uu$(real $u$ and virtual $u$) in that vertex, then the momentum of real photon should be zero and this is impossible!
I know that virtual particles don't respect any particular physical relation but I can't see how this would solve the problem. Since, we can always go to the $C.M.$ of $uu$ and it should not depend on whether we have real or virtual particle. 
Or is it that we can't go to $C.M.$ if we have virtual particle? What could be the reason for this?
 A: There is no center of momentum frame for those two particles. This question is equivalent to asking "how can a photon have momentum in its own rest frame?" And just like in that case, the answer is simply "it doesn't have a rest frame."
We can use a concrete example to show this. Consider the two particles have equal mass and equal but opposite momentum, so in the lab frame they have four-momenta $(E,p)$ and $(E,-p)$, and the outgoing photons are collinear with the incoming particles, with four-momenta $(q,q)$ and $(q,-q)$, where $q^2=E^2-p^2$.
Then the virtual particle has a four-momentum of $(q-E,q-p)$, by conservation of momentum. If you boost the top particle and the virtual particle to a new frame, the momenta are given by:
$$p^\prime=\gamma(p-\beta E) $$
$$p_{\text{virtual}}^\prime=\gamma(-p+q+\beta E-\beta q)$$
$$p_{\text{real}}^\prime=\gamma(p-\beta E)$$
The center of momentum frame for these two particles is, by definition, one where the momenta sum to zero:
$$p_{\text{virtual}}^\prime+p_{\text{real}}^\prime=\gamma(q-\beta q)=0$$
$\gamma\ne 0$, so the only valid solution is $\beta=1$. But of course that is not a valid frame in special relativity.
The key takeaway here is that not every set of particles necessarily has a center of momentum frame. Specifically, a center of momentum frame exists if and only if the four-momentum of the system is time-like.
A: The outgoing photons do not necessarily have zero momentum. 
Let $p$, $q$ be the 4-momenta of the incoming $u$ spinors, and let $p'$ and $q'$ be the 4-momenta of the outgoing photons. 
Then, conservation of momenta at the top vertex merely fixes the 4-momentum of the propagator as $p - p'$ (depending on your convention of the direction of the momentum). 
However, conservation of momentum imposed on the system as a whole asserts that 
$$p+ q = p ' + q ' . $$
The CoM frame is chosen when the three momenta $\mathbf{p} + \mathbf{q} = 0$. 
So all we have is that 
$$ p + q  = \left(2 \sqrt{|\mathbf{p}|^2 + m ^2}, \mathbf{0}\right) = p ' + q ' $$
where $m$ is the mass of both spinors. This doesn't give us enough constraints to fix $\mathbf{p'}$ as zero. 
