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The effective mass of charge carriers is defined as $$ \left(\frac{1}{m}\right)_{ij} = \frac{1}{\hbar^2}\frac{\partial^2E }{\partial k_i \partial k_j} $$

Considering that at the CBM we have a parabola. Let's say the equation is of the form $$ E(k_i, k_j) = ak_i^2 + bk_j^2 $$ where a and b are constants. Now in this case all the values where we have $ \frac{\partial^2E }{\partial k_i \partial k_j}_{i\neq j}$ will be equalent to 0. In other words all the off diagonal terms will be 0 and we will have a diagonal matrix for effective mass tensor. And in such a scenerio we can say that the effective mass along our basis vectors is 100% correct. So for this case eigenvectors and basis vectors would overlap.

My main question is if I calculate the effective mass at the CBM along three prependicular directions will the off diagonal elements be 0 or not?? I believe this is what has been done in this paper. Or do I have to calculate the effective mass tensor and then determine the eigen vectors.

Furthermore if anyone could recommend a good text for effective mass tensors basics that would be really appreciated.

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No, in general the off-diagonal parts of the tensor are not zero. They are only zero if you express the tensor in the basis of its eigenvectors. The safest way to do the calculation is always to calculate the tensor, and determine the eigenvectors.

I wrote a more general discussion about the effective mass tensor on the Matter Modeling SE site here, but I will answer your specific concerns here.

2D example

Let us consider your 2D example, i.e.

$$ E(k_x,k_y) = ak_x^2 + bk_y^2\tag{1} $$ and let us assume that $a\neq b$ (since otherwise it is degenerate, so any vector would be an eigenvector). If we set $k_y=0$ and sample along $k_x$ then we see: $$ E(k_x,0) = ak_x^2, \tag{2} $$ so we can indeed infer $a$ and, hence, the effective mass is $\frac{1}{a}$ in this direction (in atomic units).

Similarly, sampling along $k_y$ would give $$ E(0,k_y) = bk_y^2, \tag{3} $$ which also tells us the value of $b$, and hence $\frac{1}{b}$ is the effective mass in this direction. However, this only works because we know what the eigenvectors are in this case, and have sampled along them. In general, we do not know the eigenvectors, and so this approach will not work.

To illustrate the problem, let us consider sampling along the alternative orthogonal directions: $$ \vec{k_1}=\frac{1}{\sqrt{2}}\left(\begin{array}{c} k\\ k \end{array}\right) ~\mathrm{and} ~\vec{k_2}=\frac{1}{\sqrt{2}}\left(\begin{array}{c} k\\ -k \end{array}\right) $$ Now we have, $$ E(\vec{k_1}) = E\left(\frac{k}{\sqrt{2}},\frac{k}{\sqrt{2}}\right) =a\left(\frac{k^2}{2}\right)+b\left(\frac{k^2}{2}\right)= \frac{1}{2}\left(a+b\right)k^2; \tag{4} $$ thus, sampling along this direction only would lead us to suppose the effective mass is $\frac{2}{\left(a+b\right)}$.

Similarly, sampling in the $\vec{k_2}$ direction leads us to $$ E(\vec{k_2}) = E\left(\frac{k}{\sqrt{2}},-\frac{k}{\sqrt{2}}\right) =a\left(\frac{k^2}{2}\right)+b\left(\frac{k^2}{2}\right)= \frac{1}{2}\left(a+b\right)k^2, \tag{5} $$ which also gives an apparent effective mass of $\frac{2}{\left(a+b\right)}$.

Alternative viewpoint

Another way of looking at this, is to first write down the general parabolic energy form: $$ E(k_x,k_y) = ak_x^2 + bk_y^2 + ck_xk_y, \tag{6} $$ where we now have the extra "cross term". Let us sample this along the original $(k_x,0)$ and $(0,k_y)$ directions.

If we set $k_y$ to zero and sample by varying $k_x$, we obtain: $$ E(k_x,0) = ak_x^2, \tag{7} $$ which is exactly the same as expression (2), even though this is not the same parabola, and $a$ is not the reciprocal of the effective mass eigenvalue.

Similarly, setting $k_x=0$ and sampling along $k_y$ leads to $$ E(0,k_y) = bk_y^2, \tag{8} $$ which is exactly the same as expression (3), even though $b$ is not a reciprocal eigenvalue of the effective mass tensor either.

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  • $\begingroup$ thanks for taking out the time and for the amazing explanation $\endgroup$
    – Chan
    Aug 27, 2022 at 5:29

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