# [Physics] Why is effective mass a tensor

condensed-matterfield-theorylagrangian-formalismmass

So I came across the effective mass concept for solids the other day. It was mentioned that the effective mass is a tensor and may have different values in different directions. However, this is stark contrast to ordinary mass which is direction independent (as far as I know). So how do we physically explain this direction dependence of electrons (holes) mass inside a solid? Is it called "mass" just because it has the dimensions of mass? Or is it just a mathematical tool without any physical significance?

Effective mass $$m_{\ast}$$ is just a constant that shows up in the dispersion relation $$\epsilon(\mathbf{k})$$ of an energy band. Consider a one-dimensional band with dispersion $$\epsilon(k)$$. Expanding near a minimum of $$\epsilon$$ \begin{align} \epsilon(k) \approx \epsilon_0 + \frac{\hbar^2 k^2}{2m_{\ast}} \end{align} It's defined by analogy to the quantum mechanical energy of a free particle of mass $$m$$, \begin{align} E(k) = \frac{\hbar^2 k^2}{2m}. \end{align} For example, consider a one-dimensional tight-binding model with dispersion \begin{align} \epsilon(k) = -2 t\cos(ka). \end{align} Taylor expanding $$\epsilon$$ near $$k=0$$ we get \begin{align} \epsilon(k)\approx -2t + tk^2 a^2. \end{align} so that the effective mass is \begin{align} m_{\ast}= \frac{\hbar^2}{2ta^2}. \end{align} In higher dimensions, the dispersion relation can be more complicated. If the dispersion is isotropic (the same in every direction), then \begin{align} \epsilon(\mathbf{k}) = \epsilon_0 + \frac{\hbar^2 |\mathbf{k}|^2}{2m_{\ast}}. \end{align} However, you could have a dispersion where the constants in front of different components of $$\mathbf{k}$$ are different, e.g. \begin{align} \epsilon(\mathbf{k}) = \frac{\hbar^2}{2}\left(\frac{k_x^2}{m_x} + \frac{k_y^2}{m_y}\right) \end{align} Such a dispersion could arise from, e.g., a tight-binding model in which the hopping parameters or lattice constants in each direction are different. You could even have a dispersion with cross-terms between the components of $$\mathbf{k}$$, e.g. \begin{align} \epsilon(\mathbf{k}) = \frac{\hbar^2}{2}\left(\frac{k_x^2}{m_1} + \frac{k_x k_y}{m_2} + \frac{k_y^2}{m_3}\right) \end{align} In general, the dispersion (to quadratic order, and dropping the constant) will be \begin{align} \epsilon(\mathbf{k}) = \frac{\hbar^2}{2}\sum_{a,b}h_{ab} k_a k_b \end{align} for a tensor $$h_{ab}$$ — the inverse of the effective mass tensor.