Suppose we have the following:
$E(\vec{k}) = E_0 - A(\cos(k_x a) + \cos(k_y a) + \cos(k_z a))$, a tight binding energy.
I know that the effective mass is given by $m^{*} = \hbar^2 [\frac{\partial^2 E}{\partial k_i \partial k_j}]^{-1}$ (where $i$ and $j$ denote k-space vector directions) which is just the curvature of the band.
In the current case as I have cosines of $k_x,k_y,k_z$
Does this means that I have three effective masses?
$$m^{*}_{xx} = -\hbar^2 \frac{1}{Aa^2 \cos(k_x a)}$$
$$m^{*}_{yy} = -\hbar^2 \frac{1}{Aa^2 \cos(k_y a)}$$
$$m^{*}_{zz} = -\hbar^2 \frac{1}{Aa^2 \cos(k_z a)}$$
Could we not come up with a single expression for the effective mass as the energy band is isotropic?