Decay of a scalar Particle-Symmetry factors Consider the Lagrangian
$$\mathcal{L} = \dfrac{1}{2} (\partial_{\mu}\phi_{1})^2 + \dfrac{1}{2}(\partial_{\mu}\chi)^2 - \dfrac{M_1^2}{2}\phi_{1}^2 -\dfrac{M^2_\chi}{2} \chi^2 - \dfrac{\mu_\chi}{2} \phi_1\chi^2 $$
where $\phi_1$ and $\chi$ are real scalar fields.
The $\phi_1$ particle decays into the 2 identical $\chi$ particles.
Firstly,
When drawing the feynman diagram for the decay, should the amplitude of the vertex be -2$i\mu_{\chi}$ or just -$i\mu_{\chi}$ because the $\chi$ particles are identical? Or will this be taken care by the $\dfrac{1}{2}$ out front the interaction term? and I get only -$i\mu_{\chi}$
I know that the amplitude should have been -2$i\mu_{\chi}$ if there was no half in front of the coupling constant $\mu_{\chi}$, the 2 is because of the identical particles, but does that mean now that I have the half out front, the amplitude should be -$i\mu_{\chi}$ only?
Secondly,
When calculating the decay width, when I integrate d$\Pi_{LIPS}$, the integral over the solid angle d$\Omega$, should it be $4\pi$ or just $2\pi$ because of the identical particles? or will this again be taken care with the half out front the interaction term? and the solid angle should be $4\pi$ instead of $2\pi$
 A: To get the right vertex factor for the Feynman diagram, we use the rule: for every field $\phi$ that appears $n_\phi$ times in a given term, we should divide the term by $n_\phi!$ to leave the remaining factor as the actual vertex factor. Since $\phi_1$ appears once and $\chi$ twice in the final term, the normalisation of a $1/2$ is correct, and the vertex factor is just $-i\mu_\chi$.
You say you already know what should happen "if there was no half out front". It should be straightforward to convert between normalisations. Just think of $\mu_\chi/2$ as $\mu'$. Now there is no half out front, and so the Feynman rule is $-2i\mu'$, which is simply $-i\mu_\chi$.
The second 'symmetry factor', from the phase space integral, should be thought of separately, even if both factors have their roots in particle indistinguishability. The factor in the Feynman diagram comes from the combinatorics of how the fields in the S-matrix can act on the initial and final states, annihilating or creating particles. The factor in the phase space integral comes from considering which initial and final states we consider distinct, and which we consider identical.
In particular, a state with the first $\phi_1$ particle having momentum $\mathbf{p}$ and the second having momentum $\mathbf{q}$ is the same as the state with the first $\phi_1$ particle having momentum $\mathbf{q}$ and the second having momentum $\mathbf{p}$, by Bose statistics. In our case, in the rest frame of the $\chi$ particle, the two momenta must be equal and opposite, and so the phase space integral reduces merely to an integral over the direction $\mathbf{n}$ of the first one. But the state with the first particle leaving in direction $\mathbf{n}$ is identical to the state with the first particle leaving in direction $-\mathbf{n}$, so if we insist on integrating over all directions (read: solid angles), we need an accompanying factor of $1/2$.
