Charge carrier density: Hall experiment vs Fermi-Dirac statistics

[fluidistic]

Many times, the charge carrier density of a material is determined from a Hall effect experiment, from $R_H=1/(ne)$ (SI units). Where $R_H$ is determined from a measured voltage and other controllable parameters. As far as I know, this simple formula comes from the obsolete Drude's model (see the very beginning of Ashcroft and Mermin textbook). I wonder what $n$ represents when it comes from that formula.

In reality, in metals at least, only electrons that have an energy near the Fermi energy can conduct electricity. That's because they are the only ones that can get accelerated by an applied electric field since the lower energy electrons cannot increase their energy due to the already occupied states and they have to follow the Pauli exclusion principle, being fermions. This can be seen with the Fermi-Dirac distribution. As as a result, the $n$ value that can be computed this way is much smaller than the one that comes from Drude's model (around 3 orders of magnitude smaller for Cu at 300K according to some calculations on the Internet).

On hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/Fermi2.html) they show a formula to compute $n$ which is valid at 0K. The moral of the story is that it indeed involves dealing with the Fermi-Dirac statistics, unlike what the Drude's model does.

Therefore, I can understand what the $n$ coming from Fermi-Dirac statistics is, but not the one coming from Drude's model. They seem to differ by several orders of magnitude. Can someone explain what the $n$ coming from the Drude model physically represents?

 

[Lord Jestocost]

Whether one uses the classical free electron gas model or the quantum (Fermi-Dirac) free electron gas model, one gets for the Hall coefficient in both cases R_H = 1/(nq).

[PDF]
The Hall Effect

 

[phyzguy]

Many times, the charge carrier density of a material is determined from a Hall effect experiment, from $R_H=1/(ne)$ (SI units). Where $R_H$ is determined from a measured voltage and other controllable parameters. As far as I know, this simple formula comes from the obsolete Drude's model (see the very beginning of Ashcroft and Mermin textbook). I wonder what $n$ represents when it comes from that formula.

In reality, in metals at least, only electrons that have an energy near the Fermi energy can conduct electricity. That's because they are the only ones that can get accelerated by an applied electric field since the lower energy electrons cannot increase their energy due to the already occupied states and they have to follow the Pauli exclusion principle, being fermions. This can be seen with the Fermi-Dirac distribution. As as a result, the $n$ value that can be computed this way is much smaller than the one that comes from Drude's model (around 3 orders of magnitude smaller for Cu at 300K according to some calculations on the Internet).

On hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/Fermi2.html) they show a formula to compute $n$ which is valid at 0K. The moral of the story is that it indeed involves dealing with the Fermi-Dirac statistics, unlike what the Drude's model does.

Therefore, I can understand what the $n$ coming from Fermi-Dirac statistics is, but not the one coming from Drude's model. They seem to differ by several orders of magnitude. Can someone explain what the $n$ coming from the Drude model physically represents?

Drude's model is a phenomenological model, so you don't calculate the density of electrons directly. You can measure them directly with a measurement like the Hall Effect. The n in Drude's model is the density of "free" electrons. To calculate the appropriate n to use in Drude's model, you start with the Fermi-Dirac statistics. As you said, only electrons near the Fermi energy are able to move, so only a small fraction of the total number of electrons contribute to conduction. You can calculate this fraction by integrating the Fermi-Dirac distribution and using the Sommefeld approximation. I don't remember the exact details, but I think the fraction of electrons which can contribute to conduction is proportional to T/TF, where TF is the Fermi temperature. Since TF for a metal like Cu is ~ 10^5 K, at room temperature this fraction is small.

 

[fluidistic]

Drude's model is a phenomenological model, so you don't calculate the density of electrons directly. You can measure them directly with a measurement like the Hall Effect. The n in Drude's model is the density of "free" electrons. To calculate the appropriate n to use in Drude's model, you start with the Fermi-Dirac statistics. As you said, only electrons near the Fermi energy are able to move, so only a small fraction of the total number of electrons contribute to conduction. You can calculate this fraction by integrating the Fermi-Dirac distribution and using the Sommefeld approximation. I don't remember the exact details, but I think the fraction of electrons which can contribute to conduction is proportional to T/TF, where TF is the Fermi temperature. Since TF for a metal like Cu is ~ 10^5 K, at room temperature this fraction is small.

Thanks. I would have worded the bolded sentence slightly differently: "only electrons near the Fermi energy are able to accelerate (or feel the electric field)". Thanks for the pointer on how to calculate $n$.
So, I am a bit confused. If I perform the Hall experiment and get $R_H$ from which I can get $n$. The $n$ corresponds to the density of "free" electrons as the Drude's model predicts, or does it correspond to the density of the electrons that are really able to conduct, as described by a QM (Fermi-Dirac statistics) treatment? The two are definitely not equal, I believe. You mention that to calculate the appropriate n to use in Drude's model I should start with the F-D statistics, but people usually assume that X atom has Y free or valence electrons and that each such atom contributes to the total number of electrons by Y amount (which is obviously way too high compared to reality).

 

[phyzguy]

So, I am a bit confused. If I perform the Hall experiment and get $R_H$ from which I can get $n$. The $n$ corresponds to the density of "free" electrons as the Drude's model predicts, or does it correspond to the density of the electrons that are really able to conduct, as described by a QM (Fermi-Dirac statistics) treatment? The two are definitely not equal, I believe.

Well, I am not certain, but I think the are the same. You would take X atoms which have Y valence electrons, and this would give you the total number N of atoms in the Fermi-Dirac gas. Then you would calculate the fraction of those which are near the Fermi energy EF, which would be the small number ~T/TF that I discussed above. This would give you n ~ N * T/TF << N which should be the same number you measure with a Hall effect measurement. Why do you think this would not agree?

 

[fluidistic]

Sorry, I am extremely confused myself.
A related question that bothers me is that if current is due to a few electrons going at high speeds rather than many electrons going at a low drift velocity, then I would expect some order of magnitude to match. More precisely, if we take copper at 300 K, if $n_\text{Sommerfeld} \approx n_\text{Drude} \times 10^{-3}$, and $v_\text{Sommerfeld} \approx v_\text{F} \approx v_\text{Drude} \times 10^{10}$. Then there is no way that the current can be the same in both models, because $J \sim nv$, there's a $10^7$ order of magnitude discrepancy. I have to reread some books.

I just do not get what $n$ physically represents. Even if the n matched between the two models, the predicted current would differ by several orders of magnitude. There is some huge hole in my understanding.

 

[Lord Jestocost]

The Drude model and the Sommerfeld model result in the same expression for the electron conductivity, so n is in both models the free electron density (see table 1.1 in Ashcroft's and Mermin's "Solid State Physics"). Despite the expressions are the same, the physics behind both models are very different. As P. Ravindran remarks in his lecture notes ( http://folk.uio.no/ravi/cutn/cmp/ ) on the Sommerfeld model: "So the two pictures, one of which is conceptually incorrect, give the same answer, because of a cancellation between a large number of particles acquiring a small extra velocity and a small number of particles acquiring a large extra velocity.”

Maybe, these lecture notes might be of help, too: [PDF]
6.730 Physics for Solid State Applications - MIT

 

[fluidistic]

The Drude model and the Sommerfeld model result in the same expression for the electron conductivity, so n is in both models the free electron density (see table 1.1 in Ashcroft's and Mermin's "Solid State Physics"). Despite the expressions are the same, the physics behind both models are very different. As P. Ravindran remarks in his lecture notes ( http://folk.uio.no/ravi/cutn/cmp/ ) on the Sommerfeld model: "So the two pictures, one of which is conceptually incorrect, give the same answer, because of a cancellation between a large number of particles acquiring a small extra velocity and a small number of particles acquiring a large extra velocity.”

Maybe, these lecture notes might be of help, too: [PDF]
6.730 Physics for Solid State Applications - MIT

Thanks a lot Lord Jestocost. I think you highlight an interesting thing. When we apply an E field to a metal, it is not really that the Fermi sphere that gets shifted along the E field direction, it is that the electrons that had originally a momentum in the opposite direction than the E field (or same direction actually, but it doesn't matter) and with magnitude near $p_F$, now will have a momentum in the opposite direction, with a magnitude near $p_F$. Thus the change in momentum of the few electrons affected by the E field is roughly equal to $2 p_F$. Your link claims $2 p_F/3$, I am not sure where this factor of 1/3 comes from, but it doesn't matter, the order of magnitude is the same.

So again, this somewhat confirms what I thought, and my point raised in post #6 still remains.

 

[fluidistic]

To be more explicit: Quantum model predicts $\sim 10^{-3}$ times less conducting electrons than Drude's model. That could come from phyzguy's answers. Thus, one would expect the velocity of the conducting electrons to be about $10^3$ times greater than in Drude's model, if both model account correctly for the current (they do). However, I do not get this at all mathematically. The drift velocity of copper at room temperature for a reasonable current is below $10^{-4}$ m/s. So one would expect the fermi velocity to be near $10^{-1}$ m/s, but it is of the order of $10^6$ m/s, insanely higher. There is an enormous discrepancy.

Edit: Even more confusing is that it seems like in a QM treatment, the drift velocity doesn't seem to depend on the applied E field (or current) in an appreciable way. I cannot grasp what's going on.

 

[Henryk]

In reality, in metals at least, only electrons that have an energy near the Fermi energy can conduct electricity. That's because they are the only ones that can get accelerated by an applied electric field since the lower energy electrons cannot increase their energy due to the already occupied states and they have to follow the Pauli exclusion principle, being fermions

This statement is actually incorrect !. In fact, all the electrons in a metal contribute to electrical conduction.
Let us, for simplicity, consider a simple metal with a spherical Fermi surface. The electrons occupy (at 0 K) what is called the 'Fermi sphere' .
If you apply an electric field, the potential energy of an electron is modified by $V = e\vec E \vec r$.
Going back to Heisenberg representation, the rate of change of electron momentum is given by $\frac d{dt} p = \frac d{dt} (\hbar \vec k) = - \nabla V = -e\vec E$.
The meaning of above is that when an electric field is applied to a metal, the momentum (or k-vector) of ALL electron state starts changing in time in a coherent way. Therefore, the Fermi sphere will, in reciprocal space, move with the (reciprocal space) velocity proportional to and in the direction of the applied field (times electron charge). And that means all the electrons do it, not only those near the Fermi energy!.
Now, the counteracting this movement is the scattering processes. The scattering process take place from a filled to empty ones. Application of en external field moves the Fermi sphere along the direction of the applied force (field times charge). That mean in the direction of the applied field states above the 'zero field' Fermi energy are become occupied while at the other direction of the momentum space, the opposite is true (states of lower energy become empty). Scattering processes are therefore possible that dissipate both, momentum and energy of electron acquired due to the presence of the electric field.

 

[fluidistic]

This statement is actually incorrect !. In fact, all the electrons in a metal contribute to electrical conduction.
Let us, for simplicity, consider a simple metal with a spherical Fermi surface. The electrons occupy (at 0 K) what is called the 'Fermi sphere' .
If you apply an electric field, the potential energy of an electron is modified by $V = e\vec E \vec r$.
Going back to Heisenberg representation, the rate of change of electron momentum is given by $\frac d{dt} p = \frac d{dt} (\hbar \vec k) = - \nabla V = -e\vec E$.
The meaning of above is that when an electric field is applied to a metal, the momentum (or k-vector) of ALL electron state starts changing in time in a coherent way. Therefore, the Fermi sphere will, in reciprocal space, move with the (reciprocal space) velocity proportional to and in the direction of the applied field (times electron charge). And that means all the electrons do it, not only those near the Fermi energy!.
Now, the counteracting this movement is the scattering processes. The scattering process take place from a filled to empty ones. Application of en external field moves the Fermi sphere along the direction of the applied force (field times charge). That mean in the direction of the applied field states above the 'zero field' Fermi energy are become occupied while at the other direction of the momentum space, the opposite is true (states of lower energy become empty). Scattering processes are therefore possible that dissipate both, momentum and energy of electron acquired due to the presence of the electric field.

I still believe your explanation does not hold. While it is true that if your explanation was correct, the end result would be exactly the same than in the case where only a few electrons would be affected (with respect to the Fermi sphere), in your case the transient period in which the electrons respond to the electric field would make no sense. In your case, as soon as the E field is applied, the electrons that have a momentum near $p_F$ get an increase in momentum, followed by the electrons with momentum $p_F-\epsilon$, etc., until the very last electrons with momentum near $-p_F$. According to Ashcroft and Mermin's textbook near page 222, this is incompatible with Liouville's theorem. In the case I describe, in the absence of E field, there is virtually no scattering (because we're at 0K, or near that temperature, i.e. we have a well defined boundary of the Fermi sphere, we also neglect boundary effects.). When an E field is applied, it is assumed that it's a small perturbation to the system (i.e. the energy involved is very small compared to the Fermi energy). That small perturbation can only affect, at least at first, electrons that have an energy near the boundary of the Fermi sphere. It turns out that the few electrons having a Fermi momentum near $-p_F$ are the ones that will interact with the E field and end up with a momentum near $+p_F$ (actually slightly greater than $+p_F$). And that is it, when looking at the Fermi sphere, the end result looks like the whole sphere has shifted against the E field direction, but physically this is not what happened. You can see a correct explanation in Ziman's "Electrons in metals - A short guide to the Fermi surface" where it is implied that your derivation is correct because mathematically it is equivalent to the end result of what really happens when a quantum mechanical treatment is applied, but that the QM treatment (i.e. the closest to reality) only involves the scattering of charges near the surface of the Fermi sphere.

Nevertheless, thanks in part to you, things become clearer to me now. Indeed, there is something fishy in my post 6 and 9 in that in the QM treatment, the number of electrons that scatter should depend on E (and thus on the current) while it doesn't in what I wrote. In fact, for small electric fields, the "displacement" of the Fermi sphere should be proportional to |E|. I'll have to redo the math with this in mind, i.e. calculate a drift velocity in the QM model for instance.

I also now think that $n$ in Drude's model and Sommerfeld or a QM treatment is indeed the same, as phyzguy wrote. That $n$ corresponds to the total number of free electrons (not only the few that contribute to the electrical current) divided by the volume. So it will be roughly equal to about 1 electron per atom, per volume element. It therefore has to do with the total number of "valance" electrons that constitute the whole gas (and not the core electrons), even though most of these free electrons have no impact on the electrical current nor any other transport process.

So, doubts resolved. I still need to do the math, but I finally understand.Edit:
So, at 0K or near it (even room temperature is near it for common metals), the electrical current is due to a small fraction of electrons that were moving in the field direction at near $v_F$ and which got their direction reversed due to the E field, at slightly higher speeds.

If I really want to calculate the number of these affected electrons, an estimate on the electrons that can react to the E field based on temperature is not really good. Instead, I must calculate the area of the crescent between the original Fermi sphere and the "shifted" one, and compute the ratio with the total Fermi surface area. This ratio would be equal to the ratio of the number of electrons that reacted to the E field, to the total number of free electrons. In reality, I should do it in 3d, so it would be a volume.
In 2D, if $h =2p_F - e\tau |E|\hbar$, then the area of the crescent is worth $(\pi -2) p_F^2 \arccos\left ( \frac{1-h}{2p_F} \right )+2\sqrt{p_Fh-h^2/4}(p_F-h/2)$.

 

[fluidistic]

This statement is actually incorrect !. In fact, all the electrons in a metal contribute to electrical conduction.

Another argument in favor of the truth of the statement is when a temperature difference is maintained in a single material. This time the sphere does not merely move out. Its surface "sharpens" where the coldest electrons are, and broaden/blur where the hotter electrons are. It is consistent with the fact that the only electrons that are affected by the temperature gradient are the one near the Fermi surface, and not all of them. The same holds for any small (i.e. with energy much less than the Fermi energy) perturbation. Be it an electric field, magnetic field or temperature gradient.

By the way, I am advancing a bit regarding what $n$ really is. I now think it isn't anything simple at all. It is completely unrelated to both the total number of free electrons and the ones that participate in conduction. Brief reason: the latter number should depend on the current I, or the electric field E, while n doesn't. The former only works for alkali metals where the Fermi surface is almost spherical and no holes are present in the Fermi "sphere". As soon as the Fermi surface gets much more complex, the number of free electrons becomes vastly greater than the number "n" that one gets from the Hall experiment.

Therefore, so far I can conclude that whatever $n$ is, it is neither related to the number of free electrons nor the number of "charge carriers", i.e. the electrons that establish a current. It looks like some number related to the topology of the Fermi surface. Clearly Drude model cannot give any clue on it. So it is much more complicated than I thought. I am not sure why it isn't well known that "charge carrier density" is a misnomer, because the number that's associated to this name has nothing to do with the number of charge carriers.Edit: Nevermind. n is probably the number of electrons that do take part in electrical conduction. This number should not be proportional to the electric field strength (so I was wrong on this). On the other hand it cannot be derived with a Sommerfeld expansion as phyzguy said (that would be valid for the electrons participating in the specific heat). So it does not give information on the topology of the Fermi surface by itself. But if the Fermi surface is "big" while n is small, then it means there are pockets or other strange things. Great. Mystery solved... at last!