# A Does electron energy contribute to current

#### fluidistic

Gold Member
It is often said that in metals only the electrons having an energy within $k_BT$ of $E_F$ can contribute to a current. I do not understand where this $k_BT$ comes from. I know it enters the conductivity tensor because such a tensor can be written in terms of a surface integral over the Fermi surface that contains a term with $\frac{\partial f}{\partial \varepsilon}$ where $f$ is the Fermi-Dirac function, so this introduces the $k_BT$ factor. But I fail to see exactly how it translates to the current being created by those electrons.

In the textbook "Electronic transport in mesoscopic systems" by Datta, page 37 reads
Datta said:
It is easy to see why the current flows entirely within a few $k_BT$ of the quasi-Fermi energy.
but then completely diverges and never actually show why this statement holds. In fact, 1 or 2 pages below, it is shown that the difference in energy of the conduction electrons (those which produce the current) is worth $2e|\vec E|L_m$ where $L_m$ is the mean free path. In other words, the energy width around the Fermi level is a quantity that does not have any $k_BT$ term, and is proportional to the electric field's strength (which makes entirely sense to me). So what's going on here?! "It is very easy to see X" and then he went to prove "Y"? Am I missing something?

Now using intuition and the free electron model, I can understand why in the case of a thermal perturbation of order $k_BT$, only electrons that aren't below $E_F-k_BT$ can interact and modify their energy with the field. It is due to Pauli exclusion principle which prevents electrons much below the surface to gain an energy of $k_BT$ because they would gain an energy leaving them into already occupied states, which is forbidden.

When I apply the exact same reasoning but instead of using a perturbation of order $k_BT$ (usually between $10^{-5}$ eV and $10^{-3}$ eV), I use a voltage of 1 V across a 1 cm long sample, so $2e|\vec E|L_m$ is about $10^{-6}$ eV, which is extremely small compared to $E_F$ (order of 1 eV), I get a width around $E_F$ that has a magnitude like Datta's calculations, namely $e|\vec E|L_m$ which has nothing to do with $k_BT$ and is, in this case, about 40 times smaller than raising the temperature from absolute 0 to 1 K.

I do not see where I go wrong. How does one make $k_BT$ appear regarding the electrons that create a current? And where do I go wrong?

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#### mfb

Mentor
All states much lower than EF will be fully occupied all the time, all states much higher will be empty all the time. Much higher/much lower here means large multiples of kT. Current comes from a slight imbalance in the occupation for states with different momenta: States where electrons move along the current flow are more likely to be occupied than the states in the opposite direction. That asymmetry is only possible if there are both occupied and empty states around.

#### fluidistic

Gold Member
All states much lower than EF will be fully occupied all the time, all states much higher will be empty all the time. Much higher/much lower here means large multiples of kT. Current comes from a slight imbalance in the occupation for states with different momenta: States where electrons move along the current flow are more likely to be occupied than the states in the opposite direction. That asymmetry is only possible if there are both occupied and empty states around.
I know this, and as I wrote in my post, the momentum imbalance leads to an energy difference around $E_F$ (that Datta calculates) which is proportional to $|\vec E|$, not $k_BT$. So I still don't know where $k_BT$ comes from.

#### mfb

Mentor
... but that's what I wrote? At EF-5kT basically all states will be occupied, applying an electric field doesn't change that. At EF+5kT basically all states will be free, applying an electric field doesn't change that. This is just how the Fermi distribution works.

There is no sharp cut-off for the states that contribute. States at EF contribute most, states at EF +- kT a bit less, states at EF +- 2 kT even less, and so on.

#### nasu

I think you are confusing thermal phenomena with electrical conduction. Only electrons withing a few $k_B T$ contribute to the electron heat capacity. There is no such a requirement for electrical conduction. After all the expression for electrical conductivity contains the total free electron number density.

#### fluidistic

Gold Member
I think you are confusing thermal phenomena with electrical conduction. Only electrons withing a few $k_B T$ contribute to the electron heat capacity. There is no such a requirement for electrical conduction. After all the expression for electrical conductivity contains the total free electron number density.
Good point, but no, I am not confusing the two. The calculations for the specific heat is exactly what I had in mind when I wrote my 3rd paragraph "Now using intuition and the free electron model, I can understand why in the case of a thermal perturbation of order kBT (...)" where I make clear that, to me, it is intuitive than only the electron within a few $k_BT$ around $E_F$ can be excited by the thermal perturbation. I also state that when I apply the same logic to a current, I see no way to involve $k_BT$ at all, which agrees with your thoughts, but goes against what I have read in textbooks (such as Datta's textbook as I mentioned).

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#### fluidistic

Gold Member
... but that's what I wrote? At EF-5kT basically all states will be occupied, applying an electric field doesn't change that. At EF+5kT basically all states will be free, applying an electric field doesn't change that. This is just how the Fermi distribution works.

There is no sharp cut-off for the states that contribute. States at EF contribute most, states at EF +- kT a bit less, states at EF +- 2 kT even less, and so on.
To make things simple, take T= absolute zero. According to your view (and the ones I found in most textbooks, including Datta's), only the electrons exactly with energy $E_F$ can contribute to a current. However, according to Datta's math (that I could follow, it's really no deep calculations), the difference in energy between the most and least energetic electrons that contributes to the current is worth $2e|\vec E|L_m$. To me, this means that states that are between $E_F - e|\vec E|L_m$ and $E_F$ can be brought to higher energy states (above $E_F$), when T= absolute zero.

But anyway, mfb, you made me understand what's going on, I think. I'll post an answer probably later. It is not as simple as saying that only states within a few kbT will contribute to the current (while ignoring the electric field strength), nor is the view that ignores kbT (and only focuses on the electric field strength) accurate. It's a mix of both. I'll write up an answer.

Edit: For those who can't wait, here's the answer I posted on stack exchange: https://physics.stackexchange.com/questions/483318/how-to-show-that-the-electrons-responsible-for-a-current-have-an-energy-within/483601#483601. Might post here too.

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#### mfb

Mentor
If 2e|E|Lm is of the order of (or larger than) kT then what I said is not true any more but that doesn't seem to be the case OP is looking at.

• fluidistic

#### fluidistic

Gold Member
If 2e|E|Lm is of the order of (or larger than) kT then what I said is not true any more but that doesn't seem to be the case OP is looking at.
I am very interested in what would happen in such a case. Feel free to post!

"Does electron energy contribute to current"

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