I am trying to compute the Hasse invariant of the quaternion algebra over $\mathbb{Q}_p$. I denote this algebra by $H$, and I'm assuming $p\equiv 3\pmod{4}$.
So, $\mathbb{Q}_p(i)$ is an unramified extension of $\mathbb{Q}_p$ and it splits $H$. The Frobenius element $\sigma$ in $Gal(\mathbb{Q}_p(i)/\mathbb{Q}_p)$ is (obvisouly) defined by $\sigma(i)=-i$.
The element $j$ of $H$ is an element such that $\sigma(x)=jxj^{-1}$ for every $x\in\mathbb{Q}_p(i)$.
So, the Hasse invariant of $H$ is $\operatorname{ord}(j) + \mathbb{Z}$. Now, $\operatorname{ord}$ is defined on $H$ by $\operatorname{ord}(\alpha)=\frac{1}{2}v_p(\alpha\cdot\overline{\alpha})$. In particular, $\operatorname{ord}(j)=\frac{1}{2}v_p(1)=0$.
So, the Hasse invariant of $H$ is $0$, but this is wrong! I would "guess" it should be $\frac{1}{2}$, and anyway, zero is already mapped to by $\mathbb{Q}_p$ itself.
What am I doing wrong?
Your $H$ is not a division algebra, it is isomorphic to $M_2(\mathbb{Q}_p)$. So, you got the Hasse invariant of $\mathbb{Q}_p$, which is indeed zero.
With $p=2$, we would actually get a division algebra.