Galois cohomology H^1(Q_p, Z_p(2)) = 0?  For Tate twists Z_p(2), which is defined by the projective limit of 
\mu_{p^m}(2) over all m>0, I would like to calculate H^1(Q_p, Z_p(2)). 
I guess this is zero, but cannot prove it. 
Is it possible to calculate and prove H^1(Q_p, \mu_{p^m}(2)) = 0 for each m> 0? 
Just teach me, please. 
Pierre MATSUMI 
 A: Explicitly we have $H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) = \begin{cases} \mathbb{Z}_p \oplus \mathbb{Z}/p\mathbb{Z} & \text{if } p \le 3 \newline \mathbb{Z}_p & \text{if } p > 3.\end{cases}$
This follows from the Remark following Prop. 7.3.10 of Neukirch et. al (same book as in Timo's answer): 
$$H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) \cong H^1(\mathbb{Q}_p,\mathbb{Q}_p/\mathbb{Z}_p(-1))^\vee \overset{7.3.10}{\cong} (\mathbb{Q}_p/\mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z})^\vee = \mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z}$$ where $\vee$ denotes the Pontryagin dual and $n=w_p^2$ is the maximal $p$-power such the the degree of $\mathbb{Q}_p(\mu_n) \mid \mathbb{Q}_p$ divides $2$. 
The same argument can be used to compute $H^1(K,\mathbb{Z}_p(i))$ for all finite extensions $K \mid \mathbb{Q}_p$ and all $i \in \mathbb{Z}$. 
A: Have you tried the Hochschild-Serre spectral sequence $H^p(\mathbf{Q}_p(\mu_{p^m})/\mathbf{Q}_p, H^q(\mathbf{Q}_p(\mu_{p^m}), \mu_{p^m}^{\otimes 2})) \Rightarrow H^{p+q}(\mathbf{Q}_p, \mu_{p^m}^{\otimes 2})$ and the exact sequence of lower terms $0 \to E_2^{1,0} \to E^1 \to E_2^{0,1} \to E_2^{2,0} \to E^2$?
See also Neukirch, Schmidt, Wingberg, Cohomology of Number Fields, Corollary (7.3.8). This reduces the determination of $H^1$ to that of $H^0$, which is trivial, and of $H^2$, which can be treated using the dualising module $\mu$.
