Spin Sums & Conservation of Angular Momentum External fermions and bosons have momentum and spin (polarization) degrees of freedom.
E.g. decay rate of $t\rightarrow W^{+}+\bar{b}$ with an unpolarized t beam.
Peskin & Schröder sum over t, $\bar{b}$ spin and W polarization independently. So one can use spin and polarization rules. 
I understand that one sums over these spins but why independently ($\Sigma_{s,s',\lambda}$)? Aren't there any constraints from conservation of total angular momentum? 
E.g. angular momentum balance in rest frame of t: $1/2 \rightarrow ??? =! 1/2$. How is angular momentum conservation enforced? Let's say t & $\bar{b}$ spin are fixed why is the W's polarization arbitrary?
 A: [I know the question is already more than one year old, but just in case someone, like me yesterday, looks for the answer - here it comes]
The conservation law is implicitly imposed by the form of polarization vectors/spinors, so indeed you can sum over all polarizations - the non-physical combinations will automatically vanish.
To see it working, let us consider a theory with a massive vector X and two fermions: $\psi_1$ and $\psi_2$. Let the Lagrangian contain a vertex combining all of them and let $m_1 > m_2 + m_X$. Let us consider the decay:
$$\psi_1 \to X + \psi_2$$
with particles' momenta given by:
$$p_1=(m_1,0,0,0)\;,\quad
p_2=(E_2,0,0,p)\;,\quad
p_X=(E_X,0,0,-p)\;.$$
What happens if, for example, the $\psi_1$ and $\psi_2$ quarks are polarized oppositely and projection of $X$'s spin onto a given axis is 0?
The matrix element, up to a constant, reads
$$\mathcal{M}^{(-\,,\;0\,,\;+)}=\bar u_2^{(+)}(p_2)\;\gamma_\mu\;u_1^{(-)}(p_1)\cdot\epsilon^\mu(p_X\,,\;s=0)\;.$$
Inserting polarization vectors/spinors in their explicit form, we obtain (up to normalization)
\begin{align}
\mathcal{M}^{(-\,,\;0\,,\;+)}
%
&=\left(\begin{array}{c}
\left(\begin{array}{c}1\\0\end{array}\right)\\
\frac{\vec\sigma\cdot\vec p_2}{E_2+m_2}\left(\begin{array}{c}1\\0\end{array}\right)
\end{array}\right)^\dagger
\cdot\gamma_0\cdot\gamma_\mu\cdot
\left(\begin{array}{c}
\left(\begin{array}{c}0\\1\end{array}\right)\\
\frac{\vec\sigma\cdot\vec p_1}{E_1+m_1}\left(\begin{array}{c}0\\1\end{array}\right)
\end{array}\right)
\cdot\left(\frac{q}{m_X}\,,\;0\,,\;0\,,\;\frac{E_X}{m_X}\right)^\mu\\
%
&=\left(1\,,\;0\,,\;\frac{p}{E_2+m_2}\,,\;0\right)
\cdot\gamma_0\cdot\gamma_\mu\cdot
\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)
\cdot\left(\frac{q}{m_X}\,,\;0\,,\;0\,,\;\frac{E_X}{m_X}\right)^\mu\\
%
&=\left(1\,,\;0\,,\;\frac{p}{E_2+m_2}\,,\;0\right)
\cdot\gamma_0\cdot\gamma_0\cdot
\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)
\cdot\frac{q}{m_X}\\
&\quad-\left(1\,,\;0\,,\;\frac{p}{E_2+m_2}\,,\;0\right)
\cdot\gamma_0\cdot\gamma_3\cdot
\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)
\cdot\frac{E_X}{m_X}\;.
\end{align}
If you now express the gamma matrices explicitly (in Weyl reprezentation):
$$\gamma_0=\left(\begin{array}{cccc}
0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0
\end{array}\right)\;,\qquad
\gamma_3=\left(\begin{array}{cccc}
0&0&1&0\\0&0&0&-1\\-1&0&0&0\\0&1&0&0
\end{array}\right)
$$
you will see that
$$\mathcal{M}^{(-\,,\;0\,,\;+)}=0$$
as it should since spin is not conserved: $-\frac12 \neq 0 + \frac12$.
