Feynman rules with helicity states. 
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*Whenever Feynman rules are stated they are always without any mention of the helicities - this I find to be very confusing. How does one introduce and account for that? 

*Is there an intuitive/simple argument for why massless particles should have "helicities" (and not polarizations) and they can only be of the form $\pm\text{ some positive integer}$? (..i have seen some very detailed arguments for that which hinge on the representation theory for the little group of massless particles and various other topological considerations - i am here looking for some "quick" explanation for that..) 

*Is there some reason why polarized gluon scattering amplitudes at the tree-level can somehow "obviously" be written down? Like for example, consider a process where two positive helicity gluons of momenta $p_1$ and $p_2$ scatter into two negative helicity gluons of momenta $p_3$ and $p_4$ then at tree level the scattering amplitude is, 
$A(p_1^+,p_2^+,p_3^-,p_4^-)= \frac{ig^2}{4p_1.p_2} \epsilon_2^+ \epsilon_3^-(-2p_3.\epsilon_4^-)(-2p_2.\epsilon_1^+)$
where $\epsilon^{\pm}_i$ is the polarization of the $i^{th}$ particle.
I have at places seen this expression being almost directly written down.
Is the above somehow obvious? 
 A: The argument for the first question goes as follows:
Consider the Pauli-Lubanski vector $ W_{\mu} = \epsilon_{\mu\nu\rho\sigma}P^{\nu}M^{\rho\sigma}$. 
Where $P^{\mu}$ are the momenta and $M^{\mu\nu}$ are the Lorentz generators. 
(The norm of this vector is a Poincare group casimir but this fact will not be needed for the argument.)
By symmetry considerations We have $W_{\mu} P^{\mu} = 0$. Now, in the case of a massless particle, a vector orthogonal to a light-like vector must be proportional to it (easy exercise). Thus
$ W^{\mu} = h P^{\mu}$, ($ h = const.$). Now, the zero component of the Pauli-Lubanski vector is given by:
$ W_{0} = \epsilon_{0\nu\rho\sigma}P^{\mu}M^{\mu\nu} = \epsilon_{abc}P^{a}M^{bc} = \mathbf{P}.\mathbf{J}$, (where the summation after the second equality is on the spatial indices only, and $\mathbf{J}$ are the rotation generators ).
Therefore the proportionality constant
$ h = \frac{W^{0}}{P^{0}}= \frac{\mathbf{P}.\mathbf{J}}{|\mathbf{P}|}$
is the helicity.
Now, on the quantum level, if we rotate by an angle of $2 \pi$ around the momentum axis, the wave function acquires a phase of:
$exp(2 \pi i\frac{\mathbf{P}}{|\mathbf{P}|}.\mathbf{J}) = exp(2 \pi i h)$.
This factor should be $\pm 1$ according to the particle statistics thus $h$ must be half integer.
As for the second question, a very powerful method to construct the gluon amplitudes  is by the twistor approach.
Please see the following article by N.P. Nair for a clear exposition. 
Update:
This update refers to the questions asked by user6818 in the comments:
For simplicity I'll consider the case of a photon and not gluons. 
The strategy of the solution is based on the explicit construction of the angular momentum and spin of a free photon field (which depend on the polarization vectors) and showing that the above relations are satisfied for the photon field.
The photon momentum and the angular momentum densities can be obtained via the Noether theorem from the photon Lagrangian. Alternatively, it is well known that the photon linear momentum is given by the Poynting vector (proportional to) $\vec{E}\times\vec{B}$, 
and it is not difficult to convince onself that the total angular momentum density is (proportional to) $\vec{x}\times \vec{E}\times\vec{B}$. 
Now, the total angular momentum can be decomposed into angular and spin angular momenta (please see K.T. Hecht: quantum mechanics (page 584 equation 16))
$\vec{J} = \int d^3x (\vec{x}\times \vec{E}\times\vec{B}) =\int d^3x (\vec{E}\times\vec{A} + \sum_{i=1}^3 E_j \vec{x} \times \vec{\nabla} A_j )$
The first term on the right hand side can be interpreted as the spin and the second as the orbital angular momentum as it is proportional to the position.
Now, Neither the spin nor the orbital angular momentum densities are gauge invariant (only their sum is). But, one can argue that the total orbital angular momentum is zero because the position averages to zero, thus the total spin: 
$ \vec{S} =\int d^3x (\vec{E}\times\vec{A})$
is gauge invariant:
Now, we can obseve that in canonical quantization: $[A_j, E_k] = i \delta_{jk}$, we get $[S_j, S_k] = 2i \epsilon_{jkl} S_l$. Which are the angular momentum commutation relations apart from the factor 2.
Now, by substituting the plane wave solution:
$A_k = \sum_{k,m=1,2} a_{km} \vec{\epsilon_m}(k) exp(i(\vec{k}.\vec{x}-|k|t)) +h.c.$
(The condition $\vec{\epsilon_m}(k).\vec{k} = 0$, is just a consequence of the vanishing of the sources).
We obtain:
$\vec{S} = \sum_{k,m=1,2}(-1){m} a^\dagger_{km}a_{km} \hat{k} = \sum_{k}(n_1-n_2)\hat{k}$
(where $n_1$, $n_2$ are the numbers of right and left circularly polarized photons). Thus for a single free photon, the total spin, thus the total angular momentum are aligned along or opposite to the momentum, which is the same result stated in the first part of the answer.
Secondly, the photon total spin operators exist and transform (up to a factor of two) as spin 1/2 angular momentum operators.
A: 
Whenever Feynman rules are stated they are always without any mention of the helicities - this I find to be very confusing. How does one introduce and account for that?

In QFT you can represent the state of a gauge quanta by its momentum and helicity.  You can also do it in gauge dependent way by specifying the momentum and a polarization vector $\epsilon_\mu(p)$.  This is null $\epsilon(p)^2 = 0$ and is subject to a gauge equivalence $\epsilon_\mu(p) \sim \epsilon_\mu(p) + p_\mu$.  When you compute a scattering amplitude by using Feynman rules the way you describe the states of the external particles is by using polarization vectors.  This is standard textbook material so I don't see what's confusing you.

Is there an intuitive/simple argument for why massless particles should have "helicities" (and not polarizations) and they can only be of the form ± some positive integer? (..i have seen some very detailed arguments for that which hinge on the representation theory for the little group of massless particles and various other topological considerations - i am here looking for some "quick" explanation for that..)

Mathematically, particles are in correspondence with irreducible representations of the Lorentz group.  The representation theory of the Lorentz group is a bit delicate because it is non-compact.  But the non-compactness is easy to understand physically: it's because one can boost by an arbitrary amount.  So let's forget about the boosts and look only at the Lorentz transformations which preserve the direction of momentum.  You should be able to show that these transformations form a $SO(2) \simeq U(1)$ subgroup.  They act on the states by multiplying them by a phase.  The charge under this $U(1)$ group is just the helicity.

Is there some reason why polarized gluon scattering amplitudes at the tree-level can somehow "obviously" be written down?

What do you mean by obviously?  It's easy to write it down at tree level, it's a typical QFT exercise in using Feynman rules.  And by the way, it seems to me the formula you wrote down can't be right since it breaks gauge invariance.  It should be invariant under $\epsilon_\mu(p) \sim \epsilon_\mu(p) + p_\mu$.
