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Fermi-Dirac statistics at the Fermi level

by Tipi
Tags: fermi, fermidirac, statistics
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Tipi
#1
Jul26-10, 08:47 AM
P: 46
Hi all,

I've search for my question and found no answer. I think it should be pretty simple...

Fermi energy corresponds to the last occupied energy, as I understand it. So, energy levels in the Fermi gas are all filled with two electron of opposite spins, up to the Fermi energy. Saying it that way, one could conclude that at zero Kelvin, there is a probability of 1 of finding all the electrons in energy levels equal or below the Fermi energy.

But, and this puzzle me, the probability of finding an electron at the Fermi level is one half. To see this, take the Fermi-Dirac statistics and put E = Ef which gives one half. So, at zero Kelvin (or other temperatures...), the probability of finding all the electrons in energy levels lower or equal to Fermi energy is one half. This is in contradiction with what i've said in the first paragraph.

Could someone explain me what is the matter about this one half? I don't understand its physical origin, and I don't see at what energy(ies) the other one half of the time the electron are supposed to be...

Thanks in advance,

TP
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Tipi
#2
Jul26-10, 10:13 AM
P: 46
Maybe my question is more clear this way :

The average number of electrons in an energy level is given by the density of states multiplied by the distribution function (here, Fermi-Dirac statistics). In the Fermi gas, the density of state is 2 states per energy level.

So, the average number of electron in the state E is

[tex]n_\text{av}(E) = 2 f(E)[/tex],

where [tex]f(E)[/tex] is the F.-D. distribution. This mean that, at absolute zero, there is in average 2 electron per level, except for the Fermi level, where there is only one electron on average.

Why it is so? Why it is not 2 electrons on average per level, including the Fermi level?

Thanks,

TP
weejee
#3
Jul26-10, 11:29 AM
P: 199
Quote Quote by Tipi View Post
To see this, take the Fermi-Dirac statistics and put E = Ef which gives one half.
This is indeed true, but at zero Kelvin, if you move infinitesimally below Ef, the Fermi-Dirac distribution gives 1.

Also, when we talk about average occupation number per energy level in the context of Fermi-Dirac statistics, different spin states are considered as different levels.

(I'm sorry for your confusion if you've seen my previous, now deleted, reply. I misunderstood your question before.)

Tipi
#4
Jul26-10, 11:44 AM
P: 46
Fermi-Dirac statistics at the Fermi level

Dear weejee,

First of all, thanks for your answer.

(I'm sorry for your confusion if you've seen my previous, now deleted, reply. I misunderstood your question.)
I was answering when a saw this new post...

Also, when we talk about average occupation number per energy level in the context of Fermi-Dirac statistics, different spin states are considered as different levels.
Yes. This explain the factor 2 in [tex]n_\text{av}[/tex] of my last post.

Quote Quote by weejee View Post
This is indeed true, but at zero Kelvin, if you move infinitesimally below Ef, the Fermi-Dirac distribution gives 1.
So, at absolute zero, the fermi level is at the chemical potential minus an infinitesimal? If yes, this would mean that there is a probability of one-half of finding an electron in an unoccupied level (E = mu) and my question would be : why it is so?

If no, that is, if the chemical potential is exactly the last occupied state at absolute zero, I still don't understand why there is, on average, only one lectron in the last occupied level, even if there is an even number of electrons.

I'm sorry if I look confused (but I am!),

Thanks again for your time,

TP
Tipi
#5
Jul26-10, 01:55 PM
P: 46
It seems that the occupied states are the ones with energies below the Fermi energy, and the unoccupied with energies over this energy. So, the fermi energy is somewhat undefined.

When we do an integral till the Fermi energy, we consider [tex]f(\epsilon<\epsilon_F)[/tex] to be one everywhere and we don't care of the value of [tex]f(\epsilon_F)[/tex] juste between occupied and unoccupied levels.

Every solid state book ask the reader to note that [tex]f(\epsilon_F)=1/2[/tex]. My question is then : What can we do with this information?

And still, I would like very much if I could understand what the Fermi energy means (it is not the energy of an occupied level nor of an unoccupied one), and why there is a probability of one half of finding an electron at that energy.

Thanks in advance,

TP
Pete99
#6
Jul26-10, 03:08 PM
P: 43
I might be wrong, but I always though that the fermi energy is the energy at which the probability of occupancy is 1/2, by definition.
Tipi
#7
Jul26-10, 03:26 PM
P: 46
Hi Pete,

Quote Quote by Pete99 View Post
I might be wrong, but I always though that the fermi energy is the energy at which the probability of occupancy is 1/2, by definition.
And do you have any hint on why one would like to define it as such? I've seen this definition somewhere, but there was no a priori (nor a posteriori) justification. Thanks in advance.

A colleague told me that the one-half at E=mu is only a thermal average. Maybe this is the key. I would appreciate more information on that. I've look at the derivation of the statistic but couldn't trace back the justification of the one-half.

Why, at absolute zero, the F.-D. distribution is not one everywhere and then, at E>Ef, zero? I think the answer to this question maybe linked to the thermal average discussed by my colleague. We need a specialist of thermal physics...

TP
Tipi
#8
Jul26-10, 03:36 PM
P: 46
Kittel, Thermal Physics 7th edition, p. 156 :

If there is an orbital of energy equal to the chemical potential (E=mu), the orbital is exactly half-filled, in the sense of a thermal average.
Pete99
#9
Jul26-10, 04:52 PM
P: 43
As far as I remember, when you derive the equation of the fermi distribution, the fermi level appears, and it has to be calculated such as the integral of the fermi distribution times the density of states equals the total number of carriers. For me, the meaning of the fermi level is the energy level that satisfies this requirement.

You asked, why at zero kelvin there is a probability 1/2 of finding an electron at the fermi level, and not a probability of 1 or 0. I think that the point here is that the density of states is actually discrete, but it is usually approximated as a continuum. I would say that at 0 K, the fermi level that will satisfy the condition mentioned above, will be in between two states, so that all states below it are filled, all states above it are empty, and there are no states exactly at the fermi level.
Tipi
#10
Jul26-10, 06:49 PM
P: 46
Hi Pete,

Quote Quote by Pete99 View Post
As far as I remember, when you derive the equation of the fermi distribution, the fermi level appears, and it has to be calculated such as the integral of the fermi distribution times the density of states equals the total number of carriers. For me, the meaning of the fermi level is the energy level that satisfies this requirement.
I agree. I will look in this direction.

You asked, why at zero kelvin there is a probability 1/2 of finding an electron at the fermi level, and not a probability of 1 or 0. I think that the point here is that the density of states is actually discrete, but it is usually approximated as a continuum. I would say that at 0 K, the fermi level that will satisfy the condition mentioned above, will be in between two states, so that all states below it are filled, all states above it are empty, and there are no states exactly at the fermi level.
I think this make sense and throw away the need for an interpretation of the Fermi level. But in fact, you can derive de Fermi-Dirac distribution function for discrete spectrum. The normalization stated above is than made by a sum instead of an integral (the sum over all occupied states of the distribution must be equal to the number of particles). The fermi level found is one of the discrete ones and my question persists. I think the first part of your post give the path to a better understanding.

Thanks a lot for your contribution,

TP
Pete99
#11
Jul26-10, 07:08 PM
P: 43
Quote Quote by Tipi View Post
Dear weejee,
If no, that is, if the chemical potential is exactly the last occupied state at absolute zero, I still don't understand why there is, on average, only one lectron in the last occupied level, even if there is an even number of electrons.
I know you can derive the fermi distribution for a discrete density of states. What I meant with the second part of my previous post was that the probability of occupation at the fermi level is 1/2, but this does not mean that at 0 K, there will be a state at this energy. Hence, the last occupied level can be completely filled, and no half-filled.
weejee
#12
Jul26-10, 08:05 PM
P: 199
Let's consider a system in contact with a reservoir, exchanging energy and particle. In such case, I think the occupation number of the energy level right at the chemical potential should fluctuate between being occupied and being empty. The average time spent in each case should be the same, accounting for the average occupation 1/2.

This agrees with the fact that an equilibrium is a dynamical state, in which the system moves around the allowed phase space in a pretty random fashion.
Tipi
#13
Jul26-10, 08:07 PM
P: 46
Quote Quote by weejee View Post
I think the occupation number of the energy level right at the chemical potential should fluctuate between being occupied and being empty.
Even in the ground state?
weejee
#14
Jul26-10, 08:12 PM
P: 199
Quote Quote by Tipi View Post
Even in the ground state?
I assumed that our system is in contact with a reservoir. If we assume an isolated system, and if the levels are non-degenerate, I think the chemical potential cannot coincide with any energy level, by construction.

If we have a two-fold degeneracy, such as due to spin, the chemical potential can coincide with that degenerate level, and only one out of two will be occupied.
Tipi
#15
Jul26-10, 09:12 PM
P: 46
Hi weejee,
Quote Quote by weejee View Post
If we have a two-fold degeneracy, such as due to spin, the chemical potential can coincide with that degenerate level, and only one out of two will be occupied.
I am looking for a physical argument that could help me to understand that fact.

Suppose a simple problem : two (free) electrons in a box of length L at absolute zero. Then, the possible k are
[tex]k_n=\frac{n\pi}{L},[/tex]
where L is finite. Since we can put two electrons per state, there is only one occupied state. I think this occupied state is the Fermi level. Without any equations, I can state that the probability of finding an electron in the Fermi level is one. If you think it is not one, than you understand what I don't.

If we add electrons in the box, we rise the Fermi level, but if the number of electrons is even, there will always by two electron by level. Without any equations, I can state that the probability of finding an electron in the Fermi level is one, no matter how many electron there is.

Then, come back to the two electrons case. If I enlarge gradually the box, the energy levels will become closer and closer. In this continuous spectrum limit, there is still two electrons to sit in one level, which I think is still the Fermi level. If I add electrons in even number, then I continue to think that there will be two electrons sitting in each state, the last occupied one still being the Fermi Level.

In each case, I always conclude that there is a probability of one of finding an electron in the fermi level. From [tex]f(\epsilon)[/tex] I conclude, in contradiction, that this probability is one-half.

This is how my mind pictures the situation. I really hope there is a problem in it, so I'll finally understand the fact that [tex]f(\epsilon_F)=1/2[/tex].

Thanks for your help.

TP
weejee
#16
Jul26-10, 10:04 PM
P: 199
Hi Tipi,

Quote Quote by Tipi View Post
If we add electrons in the box, we rise the Fermi level, but if the number of electrons is even, there will always by two electron by level. Without any equations, I can state that the probability of finding an electron in the Fermi level is one, no matter how many electron there is.
To comment on your scenario, it means the chemical potential is actually (at least slightly) higher than the highest occupied level, by construction.

'Chemical potential' is not generally equal to 'the energy of the highest occupied level' in zero temperature. I'd rather say that,

1. If the highest occupied level is fully occupied, the chemical potential lies higher than that.
2. If the highest occupied level is partially occupied (inevitably means it is degenerate), the chemical potential lies at that level.
3. If the spectrum is continuous, we don't really need to distinguish between the energy of the highest occupied level and the chemical potential.

WJ
Gokul43201
#17
Jul26-10, 11:29 PM
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Haven't read all the posts here but the essence of the apparent problem raised in the OP is an artifact of using a continuous function (the Fermi function) to describe occupancies over a discrete set.
Tipi
#18
Jul27-10, 12:00 PM
P: 46
First of all, I would like to thank you all for your help in understanding the interpretation of the Fermi-Dirac distribution at the Fermi-Level. I think everything have been said and the issue is well outlined in the last posts.

To conclude this thread, I would like to quote Kiréev* who give a good summary of what have been said here :
Quote Quote by Kiréev
When E=F the function become indeterminate and discontinuous. [...] We can see that when T = 0, all states with E < F are occupied by electrons, while those where E > F are unoccupied. The state E = F can be considered as occupied with a probability of 0.5 (see below).
In the case considered the physical interpretation of the Fermi energy is evident - it's the maximal energy that electrons of a metal can own at absolute zero. We can also say that the Fermi level is the edge between occupied and unoccupied levels. In general, the Fermi level represents the Gibbs thermodynamical potential reported to a particle alone ; also the Fermi energy is often called the chemical potential. This means that the Fermi energy is numerically equal to the work to be done to rise by one the number of particles in the system.
TP

* P. Kiréev, La physique des semiconducteurs, MIR (1975). The quote is my translation of the french text.


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