Does it make sense to speak about the speed of electrons in a current-carrying wire (non perfect conductor)? If so, what is their speed?

Here are my thoughts:

On the Internet (Wikipedia, physicsforums, here on PSE, etc., and many other websites), one can read that the electrons move randomly at high velocities but that their average velocity, called drift velocity, is $\vec 0$ when no current is applied and very small (a few cm/s at most) when a current is applied. However, as Ron Maimon wrote, this assumption is based on Drude's model of a conductor, which is known to be incorrect in many ways. In that model, electrons are like particles of a classical ideal gas, with a well defined position and speed at all times. However, it has been many decades since that model was been supplanted by QM models that invoke a wave function to describe the electrons in the material. I don't know which models exactly (tight binding for instance?).

Ron Maimon wrote:

the electronic wave functions are spread out in a metal. The correct notion of electron velocity is the Fermi velocity, which is enormous typically, because the wavelength is about 1 atomic radius. While it isn't the same as the speed of electricity going down the wire (which is the speed of the field perturbations, some significant fraction of the speed of light), it is enormously high.

So he speaks about a Fermi velocity, as if a velocity made sense. I've also read (from him and I think Ashcroft and Mermin's book "Solid State Physics") that only electrons near the Fermi energy contribute to the electric conductivity. If that's correct, then I can understand why the Fermi velocity makes some sense, because that's the speed an electron in vacuum would have if it had an energy equal to the Fermi energy. I involved the vacuum because I think that the electron can have a well definite momentum (and hence velocity), unlike in a solid conductor metal. Am I wrong?

So the correct answer would be that in reality, it doesn't even make sense to speak about the speed of electrons in a current carrying metallic wire. If the conductive electrons (the ones responsible for the electric conductivity) were somehow instantly put into a vacuum without changing their energy, they could have a well definite (assuming a measurement is done, I suppose? I.e. a wave function collapse into an eigenstate of the momentum operator.) speed/velocity equal to the Fermi velocity. There is therefore no such thing as a drift velocity, and the common assertion that an AC almost impedes the motion of the electrons is also false. Indeed, I've seen the claim that the electrons move at the drift velocity back and forth and so that the electrons are almost static (see here for instance). This view is completely erroneous. Am I wrong here, too?


6 Answers 6


Like everything else in physics it makes sense to talk about some quantity like that in the context of a model. And we bother because the model is (at least occasionally) useful. Consider the question

"Why doesn't the current in a household circuit with a bend in the wire radiate at high power?"

Clearly charge is accelerated by going around the bend, and acceleration implies radiation. But how much charge and subject to how much acceleration makes a difference.

Some workable notion of the speed of electrons is one way to approach the problem (and one that is accessible to students in the introductory class).

  • $\begingroup$ I agree with you that it sometimes makes sense to talk about some quantity that turns out to be useful in a model in that it gives "correct values" of some other quantities, but when the model is outdated, wrong and has been supplanted by others, it should be mentioned it is so. I still don't know, whether or not, there is a well defined speed/velocity for the electrons in a metallic conductor, and it's not Drude's model that will give us the answer. $\endgroup$ Commented Feb 22, 2018 at 21:02
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    $\begingroup$ Also, using the naive Drude model, each time electrons bump into the nuclei, thereby randomizing "magically" their velocity, they should radiate. There would be no need to invoke a bent wire (to justify radiation). $\endgroup$ Commented Feb 22, 2018 at 21:08
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    $\begingroup$ @no_choice99 naively, all the electrons are radiating, but since all their accelerations are also random, this radiation, on average, (mostly) cancels out. The remainder that doesn't cancel out should be on the level of black body radiation. $\endgroup$
    – LLlAMnYP
    Commented Feb 23, 2018 at 11:40
  • $\begingroup$ Regarding radiation, typical AC currents oscillate at 50 or 60 Hz, depending on where you live. Therefore, the radiation emitted by these currents is strongest around 50 or 60 Hz, which is far lower in frequency than even most useful radio waves (which are usually at least in the tens of kHz). Since radiated power is typically proportional to frequency to the fourth power, radiation at 50 to 60 Hz isn't going to emit much power at all. That said, it can still be detected, and has to be accounted for in some precision experiments. $\endgroup$ Commented Feb 23, 2018 at 17:47
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    $\begingroup$ @no_choice99 dmckee beat me to the explanation, but... the electrons are responsible for something like half of the heat capacity in a metal, so I wouldn't be surprised that they radiate just as much. $\endgroup$
    – LLlAMnYP
    Commented Feb 24, 2018 at 19:04

I’m going to second dmckee’s notion that concepts like velocity need context. Every variable in every theory needs interpretation to connect it to our everyday intuition. For example, here’s a theory: $F=ma$. This theory has no meaning beyond the math of a differential equation if we don’t interpret $F$ to be something that makes sense in light of our intuition of what is a “force”.

So what do we mean by velocity of an electron? It depends on the theoretical context. If you have a classical, e.g. Drude, context, then yes, electrons are just balls of charge that bounce around like billiard balls. And the velocity you would compute in this context is the real velocity just as much as a classical velocity you would compute via $F=ma$. Just because your theory isn’t the “deepest” doesn’t mean it isn’t real (as long as it is consistent with experiment). If that were the case, then none of physics would be “real” because no one’s discovered the ultimate theory of everything.

As a side note, why all the Drude model hate? Considering how simple it is, it is remarkably accurate, especially for the simple metal conduction you’re describing. Check out, for instance, this paper. The gold conductivity is extraordinarily well-described by the Drude model from DC all the way to optical frequencies. The Drude model is no more outdated and incorrect than $F=ma$. When it’s relevant, it works (and the same can be said for any established theory). It’s actually a remarkable feature of solid state physics that media as complex as crystals of atoms with jillions of interacting charged particles usually just act according to Drude.

So getting back to your question, I would infer that you are looking for an answer at the level of abstraction of Bloch electrons and band theory. So, say that electrons move at the Fermi velocity until they bump into something, which happens on average as often as is consistent with the phenomenological Drude scattering time (10’s of femtoseconds for metals). But remember that when we’re talking about Bloch electrons, we aren’t really talking about little balls of stuff that fly around any more; electrons are now waves, with phase and group velocities. The peculiarities of wave physics are now all on the table. For example, at the top of a semiconductor’s valence band, electron effective mass is negative. So what does that mean for our classical intuition of observables like velocity?

What it means is that every measured and computed quantity needs to be interpreted in the context of the relevant model.

  • $\begingroup$ I feel this answer misses the point of the question and doesn't answer it. Furthermore I disagree with several things:1) It is less wrong than $\vec F = m\vec a$, 2) Complex crystals usually just act according to Drude model. 3) Electrons bump into something in average at a rate predicted by phenomenological Drude's scattering time. I address 1):You're comparing a physical law (call it a theory, ok) to a theoretical model, which is based on assumptions and laws. In particular Drude model is based on Newton's laws with some magic (like electrons don't interact with each other despite having $\endgroup$ Commented Feb 23, 2018 at 17:20
  • $\begingroup$ a charge) amongst others. A law can have a range of validity, while if a model's assumptions are wrong (which is the case with Drude model), at best it can be helpful to compute quantities that happen to match experiment values. I do not deny that it is indeed the case with Drude model. But it sometimes fails, and it always fails everywhere for many properties. I address 2): Drude model fails to account (sometimes by yielding values off by several orders of magnitude and with the wrong sign) for at least 10 physical properties (see chapter 3 of Ashcroft and Mermin). it's also incomplete in $\endgroup$ Commented Feb 23, 2018 at 17:21
  • $\begingroup$ that it can't predict the behavior of some properties but that's beside the point. It is simply not a correct description of the electrons in a solid. I address 3): The relaxation time depends on the band, position and wave vector of the electrons. $\endgroup$ Commented Feb 23, 2018 at 17:22
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    $\begingroup$ @no_choice99 I love A&M chapter 3! But you have to read it like a recitation of the failings of classical physics that drove the quantum revolution. It’s all true...and yet we still teach and use classical mechanics and E&M because they’re the most relevant theories describing everyday life. Similarly, I agree with your accounting of the failings of Drude. But you asked about simple conduction and velocity, which is a classical concept. These are Drude’s strengths! Don’t use a complex model when a simple one will do. They’re all just models of a possibly-unknowable underlying reality anyway! $\endgroup$
    – Gilbert
    Commented Feb 23, 2018 at 18:17

The Hall effect measures the drift velocity, essentially by balancing the Lorentz force $q\ \vec{v}_{drift} \times \vec{B}$. That speed is often as large as one would expect from measuring the current and knowledge of the density and sign of charge carriers, for example in doped semiconductors.

In simple metals, it is consistent with the number of valence electrons per atom. It is consistent with the interpretation that all valence electrons participate in conduction.

  • $\begingroup$ But you're already assuming that the electrons are "classical particles" with a well definite position and velocity, to begin with. No surprise there's a drift velocity involved in that case. You're not starting with the Hamiltonian of the conductor and the corresponding wave function $\Psi (\vec r,t)$. Would there be a drift velocity if you had done so? $\endgroup$ Commented Feb 22, 2018 at 21:10
  • $\begingroup$ @no_choice99 Is it even possible to calculate the Hall voltage from $\Psi$? Or to calculate the Drude peak in the optical conductivity? The quantum free-electron gas gives a Fermi velocity. Together with the Drude relaxation time $\tau$ this gives a mean free path $\lambda = v_F \tau$ that makes sense with measurements. It may be mixing quantum with classical images, but it works. $\endgroup$
    – user137289
    Commented Feb 22, 2018 at 21:33
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    $\begingroup$ I do not know the answer to your first 2 questions and I agree with you on all the rest. That's an example why outdated and supplanted models are sometimes useful. I do not claim the contrary. $\endgroup$ Commented Feb 22, 2018 at 21:40

In a solid, the electronic structure is described by energy bands of energy $E_{n\mathbf{k}}$, where n is a quantum number labeling the band, and $\mathbf{k}$ is the momentum eigenvalue. For an electron residing on that energy band, the velocity would be $v_\mathbf{k}=\frac{1}{\hbar}\frac{\partial E_{n\mathbf{k}}}{\partial\mathbf{k}}$. Depending on the kind of band structure, the functional form of this velocity may tell us how the solid would behave. For example, a conductor with a quadratic band dispersion $E_\mathbf{k}\sim k^2$ would be described quite well by the Drude model. On the other hand, if the conductor is like graphene, and has a Dirac dispersion, $E_\mathbf{k}\sim k$, then a different model has to be applied to understand its behavior under applied potential.

The drift velocity makes sense as a statistical quantity. To compute it, one needs to sum the velocities $v_\mathbf{k}$ for all the momentum states occupied up to the Fermi level. As it turns out, only the states between $E_F$ and $E_F+V$ where $V$ is the applied potential, carry current. The contributions from the other states in the Fermi sea cancel out.

The Fermi velocity is different for different bands intersecting the Fermi level. In the Drude model, where it is implied that you have a single band with a dispersion $E_\mathbf{k}=\frac{\hbar^2 k^2}{2m^*}$, where $m^*$ is the effective mass, the Fermi velocity is just $v_F=\sqrt{2m^*E_F}$, where $E_F$ is the Fermi energy. The Fermi energy is usually obtained knowing the electron density, as explained in Mermin and Ashcroft's book. But, in solids, there may be more occupied bands.

The simplest example is when magnetic field is present and induces spin splitting. Assume this Zeeman splitting is $-B\sigma_z$, where B is proportional to the magnetic field, and $\sigma_z$ is the Pauli matrix. In that case, $E_\mathbf{k}=\frac{\hbar^2 k^2}{2m^*}\pm B$. The Fermi velocity is $v_F=\sqrt{2m^*(E_F\pm B)}$, where the "+" sign is for spin "up" and "-" sign for the spin "down" band.

Edit in response to the comment. To calculate the velocity of any particle in quantum mechanics, one starts from $\mathbf{v}=\frac{1}{i\hbar}\left[\mathbf{x},H\right]$, where $H$ is the hamiltonian, and the square brackets denote a commutator. If the particle is in a certain state, maybe $\left| \Psi \right>=\left| n\mathbf{k} \right>$, you need to take the expectation value of that operator and then you're all set.

Truth is your electrons are in a many body state $\left| \mathbf{k}_1 \right> \left|\mathbf{k}_2 \right> \dots \left| \mathbf{k}_N \right>$, where $\mathbf{k}_i$ are all the possible momenta up to the Fermi level, and $N$ is the number of electrons in the solid. The velocity operator is a many body operator. If your solid is modeled as an electron gas of noninteracting quasiparticles, this operator is diagonal in particle space, so each particle sitting in a state $\left| \mathbf{k}_i \right>$ will have a corresponding velocity $\mathbf{v}_i=\frac{\hbar\mathbf{k}_i}{2m}$.

Saying that electrons in the solid should have these velocities is not entirely accurate. If one injects in the solid an electron, this electron will not belong to the Fermi sea. One can inject a particle that looks like a gaussian wave packet $\left| \Psi \right>$ and its energy will be $E=\left<\Psi |H|\Psi\right>$ and velocity will be calculated as above $\mathbf{v}=\frac{1}{ih}\left< \Psi|\left[\mathbf{x},H\right]|\Psi\right>$. If we are in the linear response, i.e. we apply a small electric field on our crystal that induces a current, this electron will be injected in one of the states above the Fermi level, so its velocity will be close to $v_F$.

  • $\begingroup$ This is the answer that I like the most so far. There is, however, a gigantic step in going from $E_{n, \vec k}$ and claiming that $v_{\vec k} = \frac{1}{\hbar} \nabla _{\vec k}E_{n, \vec k}$ would be the speed of an electron on the nth band. It is so because $\hbar \vec k$ is not the electron's momentum, but the crystal momentum. Thus, I do not understand why would electrons have such (mean) speed. I've checked A&M's book p.141 as well as the Appendix E and I still don't understand the claim. (to be continued in the next comment) $\endgroup$ Commented Mar 12, 2018 at 17:47
  • $\begingroup$ Could you elaborate on why the claim holds? $\endgroup$ Commented Mar 12, 2018 at 17:47
  • $\begingroup$ Response to your edit: Ok so the definition for the velocity is just equivalent to using $\hat{\vec p}/m$. But for Bloch electrons (our case), $\hat{\vec p} \neq \text{constant}\psi$ while $\hat H \psi _{n\vec k} = \varepsilon _\{n, \vec k}$. In other words, Bloch electrons have a well definite energy but not a well definite momentum. So I cannot just assume that electrons are in some $|n\vec k \rangle$ state. The eigenstates of the Hamiltonian are not eigenstates of the momentum, both operators are not diagonalizable using the same basis. (to be continued) $\endgroup$ Commented Mar 13, 2018 at 17:47
  • $\begingroup$ This leaves me the impression that the electrons have no well definite momentum (and hence velocity). What am I missing? $\endgroup$ Commented Mar 13, 2018 at 17:47
  • $\begingroup$ In a solid, you have translation invariance. That means that the translation operator $T=e^{i\mathbf{p}\mathbf{R}}$, where $\mathbf{R}$ is a direct lattice vector, commutes with the Hamiltonian. The state $\left|n\mathbf{k}\right>$ is an eigenstate of that operator. It is also an eigenstate of the momentum operator. $\endgroup$ Commented Mar 14, 2018 at 7:38

I see the mention of drift velocity most often not in physics, but in an engineering context where beginners often think about electrons shooting around wires at the speed of light like magic electrical energy bullets. This leads to a lot of false intuition about how electrical devices work, again from a practical engineering perspective.

As an example of such, beginners often get confused about the negative charge of electrons, and wonder why schematics and equations are all "backwards", as if the direction the electrons are moving is important. In reality, for most cases an engineer will encounter it's not relevant at all.

In this context, drift velocity makes a lot of sense. Like many phenomena, advanced physics can be invoked to say "that's not really how it works". To give an example, see this question on diffraction works. The QED explanation isn't wrong, but it's also overcomplicated and not of much use for an optical engineer. But the classical explanation is still a valid theory which correctly predicts how optics work in many cases.

Likewise, drift velocity gets beginners thinking in the right direction, that electrical circuits aren't powered by energy bullets, but rather forces transmitted through wires by a slowing moving "fluid" of electrons.

It may not have an elegant relation to advanced physics as we understand them, but it is a model that fits observations and has a use. Ohm's law is in a similar boat. Would you say Ohm's law "doesn't make sense" just because the definition of it has no direct relationship to the underlying physics?

So is drift velocity the most advanced physical theory which explains all the phenomena we've observed? No. But does it make sense to talk about it for some applications? I think so.


As a non-quantum-expert I may be completely wrong (it happens), but I don't see any reason to think that the concept of electron velocity doesn't make any sense. Momentum is a quantum vector quantity, and sure, the electrons in a metal are delocalized, but it still remains true that the average momentum of the electrons should be subject to acceleration by an electric field. And if there is an average momentum vector, then there is an average velocity vector.

Besides which, consider something like cathode emission in a vacuum tube. Once they're emitted, we have your independent electrons in a vacuum, pretty much. But we have to obey conservation (of momentum, of mass, of charge) at the point of the emission, don't we? So if average momentum, velocity, and mass flow rate exist on one side of the equation, they must correspond to something on the other.

  • $\begingroup$ In a vacuum tube, the momentum at emission is negligible, I think. You can probably model it (for electrical purposes) as if the electrons come out at rest, and are then accelerated by the electric field. $\endgroup$ Commented Feb 23, 2018 at 7:24
  • $\begingroup$ Also, electrons don't just "flow" out into the vacuum; I think they're emitted with more momentum than average drift, and in doing so exert an impulse on the atom they were emitted from. But yes, the average drift velocity has some momentum. $\endgroup$ Commented Feb 23, 2018 at 7:26

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