I am looking at this energy dispersion relationship that has the form

$$ \varepsilon = -\alpha k_y + \sqrt{\beta k_y^2 + \gamma k_x^2 } $$

Here is my conundrum - I am able to find the minima of the energy dispersion (happens to be at (0,0)) and using the definition

$$ [M_{ij}]^{-1} = \left(\frac{1}{\hbar^2}{\frac{\partial^2 \varepsilon}{\partial k_{i}\partial k_{j}}}\right)\bigg|_{k_x=0, k_y =0} $$

I encounter singularities in finding the effective mass.

So I took a different approach and tried to find the Taylor expansion of the second term, but I believe this is also a futile effort. Any advice on how to proceed?


1 Answer 1


The effective mass can be nulled out in specific case (like in graphene, some k values will lead to an M$_{ij}$=0). It can also reach infinite values in specific case (see this free article: https://aip.scitation.org/doi/abs/10.1063/1.3488833#:~:text=An%20electron%20rotating%20under%20a,to%20an%20azimuthal%20electric%20field.) It can also be negative (one can see this looking at the dispersion/energy curves if I recall correctly).

In other words, I think your answer that M$_{i}^{-1}$=0 (since you said you obtain infinite M$_{ij}$, I assumed that you get this result as well) is plausible if the minima is at k$_{x}$=k$_{y}$=0.

However, after a quick plot of your energy function (I might have done the plot a bit fast), I'm not sure the minimum is at k$_{x}$ and k$_{y}$=0, but at something closer to k$_{x}$=0 and k$_{y}$<0 (depending on the size of $\alpha$,$\beta$ and $\gamma$ the minima could move).

Hope it helps out, cheers


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