# Si conductivity depenence on crystal orientation?

scivet
As I know, electrical conductivity is a second rank tensor. So in a cubic structure, there is no orientation dependence of electrical conductivity (isotropic).
But how come Si has higher electron conducitivity along [001] and hole conductivity along [110]?

tejas777
Silicon has a diamond lattice structure, not cubic. The bandstructure of Silicon is anisotropic. Therefore, the effective mass is a tensor can be found from the bandstructure as seen in Eq. (3) in:

http://141.213.232.243/bitstream/2027.42/69866/2/APPLAB-67-20-2966-1.pdf

The above reference talks about holes, but you can also find information for effective masses of electrons in different directions here:

The conductivity is given by:

$\sigma = q \mu_i n$

where ##i = e,h## subscripts represent quantities related to electrons and holes respectively, ##q## is the electron charge, ##\mu_i## and ##n## are the mobilities and charge densities of the respective carriers. The mobility is given by:

$\mu = \frac{q \tau}{m_{eff}}$

where ##\tau## is the mean free scattering time. Therefore, conductivity does depend on crystal orientation.

scivet

tejas, you said Si is not a cubic.
But the wikipedia says, "Diamond cubic is in the Fd3m space group, which follows the face-centered cubic bravais lattice.".
So is SI really not a cubic?
I think the Diamond is not a Bravais lattice but we can convert it to the FCC Bravais lattice by combining (0,0,0) and (1/4,1/4,1/4).
What's the origin of the anisotropy of their energy band in Si?
Is there any other hidden assumption in the statement the 2nd rank tensor physical properties must be isotropic?

Thank you.

Silicon has a diamond lattice structure, not cubic. The bandstructure of Silicon is anisotropic. Therefore, the effective mass is a tensor can be found from the bandstructure as seen in Eq. (3) in:

http://141.213.232.243/bitstream/2027.42/69866/2/APPLAB-67-20-2966-1.pdf

The above reference talks about holes, but you can also find information for effective masses of electrons in different directions here:

The conductivity is given by:

$\sigma = q \mu_i n$

where ##i = e,h## subscripts represent quantities related to electrons and holes respectively, ##q## is the electron charge, ##\mu_i## and ##n## are the mobilities and charge densities of the respective carriers. The mobility is given by:

$\mu = \frac{q \tau}{m_{eff}}$

where ##\tau## is the mean free scattering time. Therefore, conductivity does depend on crystal orientation.

harrylin
[..] So is SI really not a cubic?[..]
It's "diamond-cubic" - which is indeed not cubic!

The most densely packed crystal planes are [1,1,1]; anisotropic chemical etching follows those planes.

M Quack
The diamond structure is most definitely cubic. It is not densly packed. As scivet pointed out, it is FCC with a two-atom basis. There are about as many different cubic crystal structures as there is sand on the sea shore. Most of them have not only a basis of several atoms, but several elements. Pyrite, Fluorite, Zincblende, Rock salt, etc. They are all cubic but quite different from each other. You can find a limited list of common structures with pictures here:
http://cst-www.nrl.navy.mil/lattice/spcgrp/cubic.html [Broken]

The assumption in the usual formula $\vec{j} = \sigma \cdot \vec{E}$ is that the current density is proportional to the electric field E. The conductivity σ is a rank-2 tensor and under cubic symmetry that is isotropic.

For the conductivity to be asymmetric you need non-linear terms, e.g. 3rd order.

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scivet
All the FCC lattices have the most densel packed planes on [1,1,1].
Is the anisotropic ethcing rate related to the diamond structure rather than the FCC structure?

It's "diamond-cubic" - which is indeed not cubic!

The most densely packed crystal planes are [1,1,1]; anisotropic chemical etching follows those planes.

scivet
The crystal system of diamond is classified into cubic (Refer to Nye's book, "Physical properties of crystals", page 107, the 1985 edition, section thermal expansion coefficient).
In his book, Nye says diamond has isotropic thermal expansion coefficients (Chapter VI, table 6). The thermal expansion coefficient is a 2nd rank tensor.
Could you explain this?

So the basic questions is

How do some crystals which are classified into cubic have anisotropic 2nd rank tensor physical properties (such as conductivity, thermal expansion coefficient, dielectric susceptibility...)?

In Nye's book, all the 2nd rank tensor properties are isotropic as below.

Fluorite, Sodium chloride have isotropic magnetic susceptibility
(see chapter III, table 4).
Cesium chloride, Sodium chloride have isotropic dielectric constants (see chapter IV, table 5).
Copper, diamond, sodium chloride have isotropic thermal expansion coefficients. (see chapter VI, table 6).

http://quantumwise.com/documents/tutorials/latest/SiliconOptical/index.html/chap.Si.Band.html

In the above website, the caption of Fig. 2 says,
"Figure 2: Real and imaginary parts of the diagonal components of the dielectric constant. Since silicon has cubic symmetry the dielectric constant is isotropic".

Again, the dielectric constant is a 2nd rank tensor.

The diamond structure is most definitely cubic. It is not densly packed. As scivet pointed out, it is FCC with a two-atom basis. There are about as many different cubic crystal structures as there is sand on the sea shore. Most of them have not only a basis of several atoms, but several elements. Pyrite, Fluorite, Zincblende, Rock salt, etc. They are all cubic but quite different from each other. You can find a limited list of common structures with pictures here:
http://cst-www.nrl.navy.mil/lattice/spcgrp/cubic.html [Broken]

The assumption in the usual formula $\vec{j} = \sigma \cdot \vec{E}$ is that the current density is proportional to the electric field E. The conductivity σ is a rank-2 tensor and under cubic symmetry that is isotropic.

For the conductivity to be asymmetric you need non-linear terms, e.g. 3rd order.

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M Quack
A rank-2 tensor can be written as 3x3 matrix. The tensors has to be invariant under the crystal's symmetry operations. For cubic symmetry, that would be the permutation of the (a,b,c) axes as they are equivalent, and a 2-fold (180 deg) rotation about the faces of the cube.

If we write this down we get

$M=\left( \begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array} \right);$

The first condition is
$M=\left( \begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array} \right)\cdot M \cdot\left( \begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array} \right)^T$
(^T is transpose) which gives
$M=\left( \begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array} \right);$

The second conditions is
$M=\left( \begin{array}{ccc}-1&0&0\\-1&0&0\\0&0&1\end{array} \right)\cdot M \cdot\left( \begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&1\end{array} \right)^T$
which gives
$M=\left( \begin{array}{ccc}a&0&0\\0&a&0\\0&0&a\end{array} \right);$

So the tensor is isotropic and can in fact be replaced by a simple scalar.

This holds for each and any rank-2 tensor in each and any cubic crystal because the two symmetries used are part of all 5 cubic crystal classes.

If you want to learn how to generalize this approach and maybe use it for lower symmetry systems, then you should look into group theory.

For the diamond structure there are in fact many more symmetries, but as you can see it is not necessary to use them.

For the conductivity to be asymmetric you need non-linear terms, e.g. 3rd order.

The other possibility are effects of spatial dispersion, i.e. dependence of the linear conductivity on the wavevector. In Si, only even powers of k (i.e. second in practice) can contribute, hence the anisotropy would stem from a fourth rank tensor.

scivet
So, if we consider only linear conductivity of Si, then it must be isotropic.
But in reality, the conductivity of Si is not linear and it would have higher order (greater than 2) rank tensor properties.
The origion of anisotropic conductivity of Si comes from non-linear conductivity, not from its crystal structure.

Is this what you mean?

The other possibility are effects of spatial dispersion, i.e. dependence of the linear conductivity on the wavevector. In Si, only even powers of k (i.e. second in practice) can contribute, hence the anisotropy would stem from a fourth rank tensor.

So, if we consider only linear conductivity of Si, then it must be isotropic.
But in reality, the conductivity of Si is not linear and it would have higher order (greater than 2) rank tensor properties.
The origion of anisotropic conductivity of Si comes from non-linear conductivity, not from its crystal structure.

Is this what you mean?

No, as I said, I am talking about linear conductivity. I am talking about non-local corrections beyond the local $j(x)=\sigma E(x)$. In Fourier space you still have $j(k)=\sigma(k)E(k)$ but with $\sigma$ now not being a constant tensor (or function of ω, only) but dependent on the vector k. If you expand the k-dependence you get $\sigma_{ij}(k)\approx \sigma^0_{ij}+\sum_{kl}\alpha_{ijkl} k_k k_l$. You can see that now the fourth order tensor α appears which is no longer isotropic but gives rise to birefringence, see the section "spatial-dispersion induced birefringence" in the article I cited. There is also explained how this effect can be traced back to the band structure.

M Quack
A macroscopic tensor like the DC conductivity σ you measure in the lab has to have the macroscopic symmetry of the crystal.

If you go to microscopic properties like σ(k) with k inside the first Brillouin zone, then it must have the microscopic symmetry of that particular k-vector. This is lower than the macroscopic symmetry (= the symmetry at k=0, also called the Gamma point), hence less constraints and the possibility of anisotropy.

For optical wave lengths k is very small compared to the Brillouin zone, so that you can use a Taylor series. Linear terms drop out because of inversion symmetry, σ(k)=σ(-k), therefore the leading order terms are quadratic.

scivet
Thank you so much guys. All the replies were really helpful.

If you go to microscopic properties like σ(k) with k inside the first Brillouin zone, then it must have the microscopic symmetry of that particular k-vector. This is lower than the macroscopic symmetry (= the symmetry at k=0, also called the Gamma point), hence less constraints and the possibility of anisotropy.

I don't understand this, why is sigma (k) with k inside the Brillouin zone a microscopic property?
I only considered the transformation of the fourth order tensor alpha under the "macroscopic" point group symmetry.

M Quack
My understanding is that macroscopic properties concern large length scales, much much larger than the lattice parameter and hence k~= 0, certainly k very very small.

By considering σ(k) you are looking at the BZ, i.e. Fourier transforms taken with resonably large k. I would consider this a microscopic property. I guess I should have written k a non-negligible percentage of the BZ boundary.

I agree that there must be a crossover as k->0 that should be relevant to (visible) optical properties.

Effective mass=band curvature for small k etc probably also fall into this crossover regime.

Yes, but the small k limit is exactly what you are looking at when you consider a Taylor expansion of conductivity.
A similar example are classical sound waves which are just low k phonons.

M Quack
Yes.

So how do you "feed" k into the macroscopic DC conductivity? Or is this a k=0 property and thus isotropic?

My point is that to observe anisotropy you have to measure at a finite value of k.

For DC conductivity this is in deed not very interesting. At omega=0 spatial dispersion can be described in terms of magnetic permeability. Longitudinal variations of E will be screened completely, anyhow.
I was refering mainly conductivity in the optical range.

M Quack