I have been working on trying to calculate the effective mass in different directions by fitting a parabola to the VBM and CBM and using the fitted parameters to calculate $\frac{d^2E}{dk^2}$. And then we evaluate the value of effective mass using this. First we start by calculating the band structure along the high symmetry points and we use this in order to identify the location of CBM and VBM.

Once we know the exact location of the k point, we plan on the performing variation along different directions in order to get $m^*_{hkl}$, but I am a bit confused about how do I choose the said [hkl] direction. Furthermore how is the [hkl] direction even realized in the reciprocal space, I do know that the [hkl] direction is perpendicular wrt the (hkl) plane in real space. But how would I vary the coordinates in the k space. Like let's say my high symmetry point lies at (0,0.5,0) How should I do the small variations that would span some space.

Two possible ways to do this that I can think of -

  1. We do the variation along (x,y,z) directions as is defined by Quantum Espresso. But I am not sure how would we explain it, because we are also uncertain about the miller indices we have to assign to it.
  2. We would define longitudinal and transverse directions, longitudinal direction is the direction joining Gamma - and the high symmetry point which has CBM or VBM. The transverse will be simply perpendicular to this. This has been done here and here. But how would we define directions for Gamma point


  • $\begingroup$ In order to do the band calculations, you (or the software) did the variation between the high symmetry points. You only need to fit the resulting bands. If you want a new path, you have to recalculate the bands again with the new path. $\endgroup$
    – Camps
    Aug 19, 2022 at 11:57
  • $\begingroup$ @Camps I do understand that but considering that the eff mass is an anisotropic quantity it will vary differently in different directions. Due to which we will have to calculate the band structure again with different k points, it is just that this time we will vary the coordinates in a particular direction around the k point where the CBM or the VBM is located. My question is how does one decide which direction we will have to vary it in. Also we plan on calculating the eff mass of conductivity $\endgroup$
    – Chan
    Aug 19, 2022 at 15:54
  • $\begingroup$ I think that this depend on the experimental data or use of your material. $\endgroup$
    – Camps
    Aug 20, 2022 at 18:57
  • 1
    $\begingroup$ @Camps If I simply determined the effective mass along the direction of band structure(by simply fitting to the calculated band structures along the high symmetry points), the could still be used as an indicator of the materials possible applications in photovoltaic, right ?? Furthermore, I think this is what most papers do when they calculate the effective mass. $\endgroup$
    – Chan
    Aug 21, 2022 at 6:15
  • 1
    $\begingroup$ Lots of good discussion here! $\endgroup$ Aug 25, 2022 at 14:19

1 Answer 1


The effective mass of a particle is not a scalar, but a rank-2 tensor, defined by, $$ \left(\frac{1}{m}\right)_{ij} = \frac{1}{\hbar^2}\frac{d^2E}{dk_idk_j}, \tag{1} $$ where $i$ and $j$ label reciprocal-space directions, $E$ is the band energy and $k$ is a reciprocal-space vector.

The eigenvalues of $\mathrm{m}$ are the effective masses, and the eigenvectors are the corresponding directions; these directions give the basis in which the effective mass tensor is diagonal, and so form a natural basis in which to specify the effective masses.

This means that, in general, you should calculate all nine elements of the tensor (although it's symmetric, so only six are independent) and diagonalise it to find the eigenvectors and eigenvalues. This is more work than a single direction, but it means that you don't have to worry about "getting the direction wrong" because the natural directions will be calculated for you.

Isotropic systems

If your system is isotropic, then all of the effective masses are the same. In this case, the tensor will be diagonal in any coordinate system and you can just talk about a single scalar as "the effective mass". An obvious example of this is free space, where we write equations such as, $$ \vec{F} = m\vec{a}, \tag{2} $$ (where $\vec{F}$ is a force, and $\vec{a}$ an acceleration) and we often forget that $m$ is actually of rank 2 (not 0).

Calculating the effective mass tensor

In principle, it's straightforward to calculate the effective mass tensor by finite difference of either the band energy directly, or its reciprocal-space gradient, $\nabla_kE$. However, in practice the units of $k$-points can be rather confusing, both in terms of formulating the mathematics correctly, and implementing it in your ab initio modelling program of choice.

When k-points are specified in fractional coordinates, they are fractions of the reciprocal-lattice vectors; e.g. a k-point specified as:

$$ \vec{k} = (0.2 , 0.4, 0.5), \tag{3} $$ actually means the vector $$ \vec{k} = 0.2\vec{A}^\ast + 0.4\vec{B}^\ast + 0.5\vec{C}^\ast, \tag{4} $$ where $\vec{A}^\ast$, $\vec{B}^\ast$ and $\vec{C}^\ast$ are the reciprocal-lattice vectors.

When we calculate the effective mass tensor by finite-difference, it's common to change the fractional coordinates of $\vec{k}$, but the derivative is defined in terms of the change in actual reciprocal-space coordinates, so you need to convert the $\delta\vec{k}$ into a reciprocal-distance by multiplying by the reciprocal-lattice vectors.

Why mass is a rank-2 tensor

It's easy to state that mass is a rank-2 tensor, but it may not be clear why this is so. Newton's 2nd Law tells us that force, $\vec{F}$, and acceleration $\vec{a}$ are linearly related, and that mass is the constant of proportionality.

Scalars and linear relationships

We are trained from an early age to think about mass as a scalar (rank-0 tensor) quantity, such that

$$ \vec{F} = m\vec{a} \Rightarrow F_i = ma_i, \tag{5} $$ i.e. the $i$th component of the force vector is $m$ times the $i$th component of the acceleration.

Thus, when mass is a scalar, the force and acceleration are always parallel. We are used to this physically, as we expect an object to accelerate in the direction in which we pull it.

General linear relationships

An expression where $F_i$ depends only on $a_i$ is not the most general linear relationship. There is no fundamental reason why, for example, the 1st component of the force should not be linearly related to the 3rd component of the acceleration, $$ F_1 = m_{13} a_3,\tag{6} $$ (for some constant $m_{13}$) or even a combination of all the components of $\vec{a}$, $$ F_1 = m_{11} a_1 + m_{12} a_2 + m_{13} a_3 = \sum_j m_{1j}a_j,\tag{7} $$ with the additional constants $m_{11}$ and $m_{12}$ as well. The same holds for $F_2$ and $F_3$, and so we can write a more general expression, $$ F_i = \sum_{j} m_{ij}a_j.\tag{8} $$ which is a general linear relationship between all of the components of the force and all of the components of the acceleration. If we look at this expression mathematically, we see that it is exactly the same as the matrix-vector product in component form, $$ F_i = \left(\mathrm{M}\vec{a}\right)_i,\tag{9} $$ where $\matrix{M}$ is a rank-2 tensor (expressed as a matrix in any given coordinate system).

Physical meaning

Since mass is a rank-2 tensor, then there can be cases where we exert a force on an object and it accelerates in a different direction (not parallel to the force). Empirically, we might observe such an effect if we pull a wheeled vehicle at an angle not parallel to the direction of the wheels; the vehicle will preferentially accelerate in the direction the wheels are pointing in, rather than parallel to the force.

The effective mass of a particle arises due to interactions between the particle and the material, and if the material is not isotropic then there is no reason to expect that the mass will be isotropic either. In the everyday case of electrons flowing in a wire, the electrons will flow down the wire even if the electric field is not parallel to the wire.

This is why it's risky to try to determine the direction of propagation first, and then compute the effective mass (energy curvature) in this direction. You will always get a curvature, and so you'll have "an effective mass" for that direction, but unless your chosen direction is an eigenvector of the effective mass tensor, your results doesn't mean what you think it means, because to make the particle travel along that direction would require a force in a different direction, that you haven't calculated!

  • $\begingroup$ Since it was getting a little long for the comments, I moved the discussion here to chat $\endgroup$
    – Tyberius
    Aug 29, 2022 at 14:00

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