In the MSCKF paper proposed for visual-inertial odometry, the authors use an error-state extended Kalman filter. When describing the error quaternion, they state that:
for the quaternion a different error definition is employed. In particular, if $\hat{\bar{q}}$ is the estimated value of the quaternion $\bar{q}$, then the orientation error is described by the error quaternion $δ\bar{q}$, which is defined by the relation $\bar{q} = δ\bar{q}⊗\hat{\bar{q}}$. In this expression, the symbol ⊗ denotes quaternion multiplication. The error quaternion is $$δ\bar{q} \approx [\frac{1}{2}δ\theta^T 1]^T$$
However, I don't understand how they come up to that conclusion. I'm also struggling to understand how to calculate $δ\theta$. I've read this question that is similar: How to compute the orientation error between two 3D coordinate frames? but I'm still confuse about it! What does $δ\theta$ contain? It's a 3x1 vector right?
It should be taken as a definition of $\delta\theta$. The quaterion $q$ is the exponential of a purely imaginary quaternion $v$ representing the axis of rotation: $q = e^{v/2}$. The factor of $1/2$ makes so that $|v|$ is exactly the angle of rotation. If we have some error $\delta\theta$ in $v$ which is small (and we ignore the commutator $[v, \delta\theta] = 2v\times \delta\theta$) then $$ q_{\mathrm{approx}} = e^{v/2 + \delta\theta/2} \approx e^{v/2}e^{\delta\theta/2} = q_{\mathrm{true}}e^{\delta\theta/2} \approx q_{\mathrm{true}}(1 + \delta\theta/2). $$ This motivates us to define $\delta\theta$ via $$ \bar q_{\mathrm{true}}q_{\mathrm{approx}} = 1 + \delta\theta/2. $$ The RHS here is the same as their awkward notation $$ [\tfrac12\delta\theta^T\quad 1]^T. $$