# Why is Hall response $(\sigma_{xy}-\sigma_{yx})/2$ rather than $\sigma_{xy}$?

In this PRL paper (and other works like the review article), the Hall response is defined as the antisymmetric part $$\sigma_H=(\sigma_{xy}-\sigma_{yx})/2$$ instead of $$\sigma_{xy}$$ itself.

What is the physical reason? Why not just $$\sigma_{xy}$$, which apparently describes the transverse response to a stimulus such as voltage?

The conductivity tensor $$\sigma$$ can be decomposed as the sum of a scalar part proportional to the identity matrix, an antisymmetric part, and a traceless symmetric part. As an explicit example in 2D, $$\pmatrix{a&b\\c&d} = \frac{a+d}{2}\pmatrix{1&0\\0&1} + \frac{b-c}{2}\pmatrix{0 & 1\\-1 & 0} + \pmatrix{\frac{a-d}{2}&\frac{b+c}{2}\\\frac{b+c}{2}&\frac{d-a}{2}} \tag{\star}$$

More generally, $$\sigma = \mathrm{Tr}(\sigma)\mathbb 1 + \frac{1}{2}\big(\sigma-\sigma^T\big) + \frac{1}{2}\big(\sigma+\sigma^T - 2\mathrm{Tr}(\sigma) \mathbb 1\big)$$

Unlike the other two terms, the traceless symmetric part singles out two special axes in $$\mathbb R^2$$ (its eigenspaces). This is certainly not unreasonable in a crystal, which already possesses special directions by virtue of the crystalline lattice, but in rotationally symmetric systems (such as a free electron gas) the traceless symmetric part must vanish.

With that in mind, the answer to your question is that the Hall response refers to the antisymmetric part of the conductivity tensor, not merely to its off-diagonal element(s). In a non-isotropic material, the conductivity tensor generically has nonzero off-diagonal elements by virtue of the traceless-symmetric part of the conductivity tensor; when we talk about the Hall response of the material, we want to subtract that contribution out.

Why [do we not define the Hall conductivity to be] just $$\sigma_{xy}$$, which apparently describes the transverse response to a stimulus such as voltage?

Because splitting the conductivity up into a longitudinal part and a transverse part is not a good way to understand how the material responds to an applied field.

You can think of it like this. The simplest possible type of (linear) material is one in which the current density is given by $$\mathbf J = \sigma_0 \mathbf E$$ for some scalar $$\sigma_0$$. More general materials have an "easy direction" (in which the conductivity is enhanced) and a perpendicular "hard direction" (in which the conductivity is reduced). These directions are generally determined by the crystal lattice. If we align our axes with these directions, the conductivity tensor becomes $$\pmatrix{\sigma_0 + \Delta & 0 \\ 0 & \sigma_0 - \Delta}$$ In particular, note that there is no intrinsically transverse response here - if I apply a field along the easy direction, I get a current along the easy direction, and the same is true for the hard direction. If I apply a field along some other direction I will get a transverse response, but only because $$E_x$$ and $$E_y$$ are scaled by different factors to produce $$J_x$$ and $$J_y$$.

The Hall effect is different - it is an intrinsically transverse response, in the sense that regardless of what direction I apply my field, the Hall conductivity will generate a current in the perpendicular direction. This manifests itself as an antisymmetric contribution to the conductivity, which becomes $$\sigma = \pmatrix{\sigma_0+\Delta & \sigma_H \\ -\sigma_H & \sigma_0-\Delta} = \sigma_0 \pmatrix{1&0\\0&1} + \sigma_H \pmatrix{0 & 1\\-1 & 0} + \Delta\pmatrix{1 & 0 \\ 0 &-1}$$

This is the form of the conductivity tensor when the axes are aligned with the special directions in the material. If we rotate our axes by an angle $$\theta$$, it's straightforward to show that the scalar and antisymmetric parts are actually invariant, while the third term becomes $$\Delta\pmatrix{\cos(2\theta) & -\sin(2\theta)\\ -\sin(2\theta) & -\cos(2\theta)}$$ As a result, the conductivity tensor in an arbitrary basis which is rotated by an angle $$\theta$$ with respect to the material's special axes is given by

$$\sigma = \sigma_0 \pmatrix{1&0\\0&1} + \sigma_H \pmatrix{0 & 1\\-1 & 0}$$ $$+\Delta\pmatrix{\cos(2\theta) & -\sin(2\theta)\\ -\sin(2\theta) & -\cos(2\theta)}$$

You can identify each term with the corresponding terms in $$(\star)$$.

The point is that the Hall response of a material refers to the intrinsically perpendicular response. Such a response can appear only in the absence of time-reversal symmetry, and has a different underlying mechanism than the other terms. It is therefore worthy of study in its own right.

By simply looking at $$\sigma_{xy}$$ all by itself, you would be muddling up the Hall response with the (time-reversal respecting) $$\Delta$$ term in a way which depends on your alignment convention - which is not a helpful way to approach the problem.

Do you know any 3D isotropic system that has Hall response? I'm asking because of the argument that rotation preserves the antisymmetric part, which seems to hold in general dimensions. (I suppose that one 2D example is just applying a magnetic field normal to the planar system.)

A 3D electron gas in an applied magnetic field has a Hall response in the plane perpendicular to $$\mathbf B$$.

Note that in 2D, an antisymmetric matrix has only one independent component. Under rotations, that component transforms like a scalar (i.e. it is invariant), but under reflections it gets an extra minus sign; as a result, it is sometimes referred to as a pseudoscalar.

In 3D, an antisymmetric matrix has three independent components: $$A=\pmatrix{0&c&-b\\-c&0&a\\b&-a&0}$$ We can arrange these components into a column: $$\mathbf a = \pmatrix{a\\b\\c}$$ The somewhat strange ordering of the components of $$A$$ is because using this convention, we see that $$A_{ij} = \epsilon_{ijk}a_k$$. One can show that under rotations, $$\mathbf a$$ transforms like a vector does; under reflections, it once again gets an extra minus sign. As a result, we sometimes call $$\mathbf a$$ a pseudovector.

Finally, note that the action of $$A$$ on a vector $$\mathbf v$$ can be expressed in terms of $$\mathbf a$$ as $$A\mathbf v = \mathbf a \times \mathbf b$$. Therefore, the antisymmetric part of the conductivity tensor can be thought of as a cross product with the associated pseudovector (which should evoke the flavor of the Lorentz force $$\mathbf v \times \mathbf B$$ ... perhaps the pseudovector $$\mathbf B$$ is better represented in matrix form?).

• Thank you. This is helpful information, but I don't see how it answers the question. It's effectively saying that Hall response is merely by definition the antisymmetric part. If so, my question is more about the physical reason for such? For instance, why is the traceless symmetric part excluded although it apparently gives some transverse response? Jan 2, 2023 at 6:01
• @xiaohuamao I have expanded my answer to discuss your follow-up and edit. Jan 2, 2023 at 6:49
• Very nice explanation, I have a doubt though, normally in topological insulators, if there is a gap then we find quantize d.c. Hall conductivity. Because of the gap we have an insulator and $\sigma_{0} = 0$. And also the TKNN formula or Kubo formula diverges for gapless cases. In a gapless case can we have both of $\sigma_0$ and $\sigma_H$ (might not be quantized) to be non-zero? Jan 2, 2023 at 7:49
• @Galilean Sure, a 2D electron gas in a small magnetic field has both $\sigma_0$ and $\sigma_H$ nonzero. Jan 2, 2023 at 7:54
• This intrinsically anti-symmetric nature of the conductivity tensor is an example of how Onsager's reciprocity relations fail in external magnetic field unless they are modified as per Casimir. Jan 2, 2023 at 11:12