I'm attempting the calculate the Khovanov homology of the unknot using the figure eight diagram of the unknot with exactly one crossing going from top left to bottom right as shown below.
I also include the convention I'll be using for the resolutions of the crossings.
I calculate that the resulting chain complex given by this diagram is $$\cdots\to 0\to V\otimes V\stackrel{m}{\to} V\to0\to\cdots$$ where $V$ is the graded abelian group $\mathbb{Z}_{(1)}\oplus\mathbb{Z}_{(-1)}$ (I'll drop the gradings from here as I don't think they play any part in my eventual misunderstanding - I could be wrong) generated by $v_+$ and $v_-$ and $m$ is the map which acts on generators by $$\begin{array}{rcl}m(v_-\otimes v_-)&=&0\\ m(v_+\otimes v_-)&=&v_-\\ m(v_-\otimes v_+)&=&v_-\\ m(v_+\otimes v_+)&=&v_+.\end{array}$$
Here we see that the kernal of $m$ is freely generated by $v_-\otimes v_-$ and $v_-\otimes v_+ - v_+\otimes v_-$ and so we get that $H^{Kh}_0\cong\mathbb{Z}\oplus\mathbb{Z}$, and $m$ is surjective so $H^{Kh}_1$ is trivial.
Now consider the mirror image of the above diagram which is still isotopic to the unknot, and with the same choice of crossing resolutions. Because of our choice of resolutions, we now get the chain complex $$\cdots\to 0\to V\stackrel{\Delta}{\to} V\otimes V\to0\to\cdots$$ where $\Delta$ is the map which acts on generators by $$\begin{array}{rcl}\Delta(v_-)&=&v_-\otimes v_-\\ \Delta(v_+)&=&v_+\otimes v_- + v_-\otimes v_+.\end{array}$$
Here we see that the kernal of $\Delta$ is trivial so $H^{Kh}_0$ is trivial, but $\mbox{Im}\Delta\cong \mathbb{Z}\oplus\mathbb{Z}$ and so $H^{Kh}_1\cong\mathbb{Z}\oplus\mathbb{Z}$.
It seems that I'm missing something here as either I've inadvertently shifted homological degrees somewhere where I should not have (or haven't where I should have), or I need to maybe dualise the chain complex for some reason when taking the mirror image of our knot. Which of my calculations above is incorrect (or both?) and am I not understanding some crucial step in the definition of the Khovanov homology of a knot which has lead to this mistake?
You're probably forgetting to renormalize the grading based on the number of $+$ and $-$ crossings in the diagram.