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When we talk about current, we say electrons are "flowing" through a conductor. But if electrons are identical particles, how does it make sense to talk about them flowing?

To expand on that: imagine the simplest wire, just a 1-D chain of copper atoms, each with one conduction electron. If we apply a potetntial across the wire, what happens? Of course, we say there is a current, and the electrons "flow". But what does that really mean?

Suppose when the electrons "flow", each copper atom gives its electron to the next atom in the line. From a QM perspective, nothing has changed! The 'before' wave function is identical to the 'after' wave function, because all that we have done is exchange particles, and the wavefunction has to be symmetric upon particle exchange. The state of the system before and after the "flow" occured is exactly the same. So what does it really mean to say that there is a current flowing?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jun 7 '20 at 5:59
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    $\begingroup$ If I have a car, and I give it to a friend, and somebody else gives me a car of the same make, model, mileage and paint... can we say that I never gave away my original car in the first place, because my garage looks the same as it did before? No! If nothing else, fuel was consumed in moving the cars around. $\endgroup$ – Asteroids With Wings Jun 7 '20 at 16:49
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    $\begingroup$ a set of size N of indistinguishable electrons means that you cannot distinguish among any of them, you seem to be thinking that indistinguishability implies inability to distinguish the size of the set, which is wrong $\endgroup$ – lurscher Jun 7 '20 at 17:49
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Perhaps you're visualizing the electron flow as if it were a series of snapshots, timed so that the snapshots all look identical. But it's more than that. The wavefunction of a moving electron is different from that of a stationary electron: it includes a nonzero velocity-associated component. It's that added component (which is always there, even in the "snapshots" of electrons in a current-carrying wire) that equates to charge motion and thus to current.

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    $\begingroup$ The same is true for any classical system, too. It is not enough to know all positions at some time $t$. One also has to know the momenta. In a way, the spatial positions are only a projection of the "true" phase space positions onto a $n/2$-dimensional subspace. $\endgroup$ – scaphys Jun 5 '20 at 7:03
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    $\begingroup$ Given this is a question explicitly about indistinguishable particles in quantum mechanics, and this answer makes no reference to that, I don't see how it can be considered an answer to the question $\endgroup$ – isometry Jun 5 '20 at 15:52
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    $\begingroup$ Indistinguishability of the particles is irrelevant. $\endgroup$ – S. McGrew Jun 5 '20 at 16:13
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    $\begingroup$ .. then that is what you need to explain. $\endgroup$ – isometry Jun 5 '20 at 17:12
  • $\begingroup$ @BruceGreetham: Identical electrons are indistinguishable. But electrons in e.g. different atomic orbitals can absolutely be distinguished on the basis of their energy levels, and the same is true of momentum. The answer already says that ("which is always there, even in the 'snapshots' of electrons..."), so there's nothing to change. $\endgroup$ – Kevin Jun 6 '20 at 23:58
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In quantum mechanics there is what's called a probability current. It describes how probability density flows from one place to another. The electric current is just the probality current times the charge. See also this wiki page. If you impose that probability is conserved $$\frac d{dt}\int\psi(x,t)^*\psi(x,t)\,\text{d}x=0$$ then you can derive$^\dagger$ that $$\frac{\partial\rho}{\partial t}+\frac{\partial j}{\partial x}=0.$$ Here $\rho=|\psi|^2$ is the probability density and $j$ is the probability current. This last equation is a continuity equation, which tells you that if the density at a point increases it means it has moved there from neighbouring sites through a current. If you perform the calculation you get that $$j=\frac{\hbar}{2mi}\left(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\right)$$

Let's look for example at a plane wave $$\psi(x,t)=Ae^{i(kx-\omega t)}$$ The current becomes $$j=|A|^2\frac{\hbar k}{m}=\rho\frac p m=\rho v$$ Even though for a plane wave the density is the same everywhere there is still a current. The density is constant but the phase encodes the movement of the particle.

$^\dagger$To do this calculation first use the product rule, then $\frac{d}{dt}\psi=\frac{1}{i\hbar}\hat H\psi=\frac 1{i\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi\right)$ and finally partial integration.

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    $\begingroup$ But plane waves are not normalizable, so they are not valid wavefunctions... maybe a wave packet is a better example, although slightly more mathematically involved $\endgroup$ – giobrach Jun 5 '20 at 16:21
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    $\begingroup$ I think this nails the common (and OPs) misunderstanding: Momentum in QM - and thus current - is not the movement of probability (or amplitude), rather it's "all in the phase". The curious thing, however, happens at the ends of a wave-packet, where the probability rises or falls. The current gets imaginary! Would like to know more about that, have never seen it discussed in standard courses or books. The plane wave, obviously, hides these ends at infinity. $\endgroup$ – user257090 Jun 6 '20 at 13:34
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    $\begingroup$ Sure, if you force it to real, it is real. The derivative of $\psi$, however, includes a real part from the phase and an imaginary part from the change in absolute value of amplitude. I was not questioning your answer, btw., just extending on the understanding of phase vs amplitude. $\endgroup$ – user257090 Jun 6 '20 at 15:53
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    $\begingroup$ @non-user38741 I agree that the derivative of $\psi$ is generally complex but the current isn't. The imaginary part of a complex number can be defined by $\text{Im}(z)=\frac{1}{2i}(z-z^*))$ and is itself a real number. Since the current can be written in this form it is also real. It has to be real because it is related to the probability density by the continuity equation and the probability density is ofcourse real. $\endgroup$ – AccidentalTaylorExpansion Jun 6 '20 at 16:43
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    $\begingroup$ It isn't as involved as this answer implies. The simple truth is the o/p mis-stated the facts: in any example of QFT the logic of the existence of spacetime demands that the motion of a particle be modelled as a spacetime event, not as a series of snapshots designed to imply there is only space but not time, a presentation in which time is frozen, hence no phase. Further, there was a second mistatement by the o/p: he implied current can flow in a wire, which is impossible. Current can flow in a circuit: if you posit a wire that has open ends there can be no circuit, hence no flow of current. $\endgroup$ – Ed999 Jun 6 '20 at 23:30
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In the hopping model, in the absence of external field, the left hopping and the right hopping are equiprobable. But that symmetry is broken once the external field is applied. This way there’s a current.

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    $\begingroup$ +1 this is a very good answer, part of the confusion of the original question lies in the concept of "crystal momentum" and discrete translation symmetry $\endgroup$ – Kai Jun 5 '20 at 3:47
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The state of the system before and after the "flow" occured is exactly the same.

you forget that the wavefunction is complex. momentum is decoded in the phase of $\psi(x)$

e.g. an additional phase factor of $e^{ik_0x}$ giving momentum $\hbar k_0$.

so a wavefunction without current is distinguishable from a wavefuntion with current. E.g. the fourier transform of the latter would be shifted by $k_0$.

this doesn't change in the case of many particle wave functions

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Water molecules are exactly alike (excluding isotopes, for the nitpickers), so how can we speak of a flow of water?

In the case of electrons, think of say an old CRT display*. It works by sending a beam of electrons from an electrode through vacuum to strike phosphors on the face of the tube. Those electrons must flow through the vacuum, no? Those electrons must have gotten to the electrode somehow, and how else but by flowing through the wires?

*Or any other device that relies on sending electrons through vacuum: vacuum tubes, electron microscopes, electron beam machining, &c.

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    $\begingroup$ Wish I could up vote this answer by more than +1! The more interesting point it makes is that the CRT electron beam is scanning along the row of phosphors on the glass screen: the electrons not only flow thru the wires, they flow thru the "vacuum" too. How do they impart their energy to the phosphor dots on the CRT screen if they are not moving through the vacuum, i.e. even without wires, to strike the screen? $\endgroup$ – Ed999 Jun 6 '20 at 23:40
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I'd like to quickly tackle a fallacy I see lurking in the title to this question, i.e. "If electrons are identical and indistinguishable": while, from our perspective they may be "identical and indistinguishable", each electron is a distinct entity, and if our tools allowed it, we would be able to discriminate between them (hypothetically).

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    $\begingroup$ THIS . Distinguishable vs. indistinguishable particles just means what statistics apply, not that we can't see that there's a whole bunch of them moving. $\endgroup$ – Carl Witthoft Jun 5 '20 at 14:41
  • $\begingroup$ Indeed Mr. Witthoft! Also, further to the use of "identical" in the question title: logically, no two distinct electrons can be identical in all aspects, at the same time, since they must differ (at least) in their position. For if two electrons occupied the same exact position, at the same time etc., then there would only be one electron. $\endgroup$ – Latent Heat Jun 6 '20 at 16:52
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    $\begingroup$ I don't know if this answer really makes the point you want it to make. Isn't the point of indistinguishable particles that it's impossible to tell if two of them switch places (i.e. to discriminate between them), regardless of what tools you have? $\endgroup$ – David Z Jun 7 '20 at 6:10
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    $\begingroup$ This is wrong according to my understanding. The particles really are indistinguishable. It's just that you can have many indistinguishable particles at once. $\endgroup$ – Charles Hudgins Jun 7 '20 at 6:19
  • $\begingroup$ Hello Mr Hudgins! You are not totally incorrect your understanding of my comment: allow me to explain myself. Yes, in practically all aspects, from what we can tell, electrons are indistinguishable. My point was that, if we are referring to e.g. two electrons, then logically, these two electrons must differ in one key aspect: their position. As in, "Electron A at start of wire, and electron B at end of wire". Imagine perfectly identical human twins: they could never occupy the exact same space, at the exact same time, yet they are indistinguishable in all other aspects. $\endgroup$ – Latent Heat Jun 7 '20 at 16:17
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You can unravel your confusion about identical particles in quantum mechanics by considering the simplest case of a two electron system in a vacuum (the wire is not a necessary part of your confusion). We would typical describe this quantum state as consisting of one wavepacket centred around position $x_1$ with momentum centred around $p_1$ , and a second wavepacket centred around position $x_2$ with momentum centred around $p_2$ (the wavepackets need to be constructed so as not to violate Heisenberg's uncertainty principle). Then you are right that it is not meaningful to think of this two electron system as two distinct particles (we should form a symmetric combination of these two wavepackets to mathematically express that).

But then we can study the evolution of this system under Schrodinger's equation. Then we find it matches our intuition of electron movement : the larger the values of the momentums we choose for our quantum state, the quicker the positions of the wavepackets will change over time. In other words the multi-electron system of identical particles move depending on the momentums of the system and thus generates an electric current.

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Take a length of tube around 1.25 in (3 cm) diameter, maybe a plastic waste pipe. Fit a suitable funnel at the upper end and fill it with identical ping-pong or golf balls. Watch the balls roll down into the top end of the tube and out the bottom.

Does it make any sense to say that the balls are not moving through the tube?

Now try it in the bathtub with sloshes of water instead of balls. Watch the waves of water going in one end and out the other. Ask yourself the same question.

Another analogy is a lockdown queue at a supermarket: individuals leave at one end and join at the other, everybody stands on a designated spot, and periodically they all shuffle forward in waves of motion.

The equivalent with electrons would be to charge up a Van de Graaf generator and touch a wire to it. Place an oppositely-charged gold-leaf electroscope near the other end. The generator voltage falls steadily, a glow discharge appears at the other end of the wire and the elecroscope also discharges. Now try to argue that no electrons flowed along the wire in the direction from the generator to the electroscope.

The indistinguishability of one electron from another is a mathematical symmetry arising from conservation laws (the general correspondence between symmetries and conservation laws was first noted by mathematician Emmy Noethe around a century ago). It is not an ontological one. Similarly, identical twins are indistinguishable but they are not the same person. Wheeler and Feynman did once consider the idea that all electrons and all positrons are genuinely manifestations of one individual particle oscillating to and fro across Time, but the observed scarcity of positrons killed the idea.

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    $\begingroup$ This isn't the same thing, electrons in QM are fundamentally indistinguishable whilst ping pong balls are definitely not. $\endgroup$ – jacob1729 Jun 4 '20 at 19:19
  • $\begingroup$ OK I added a third experiment. basically, you have to consider where the electrons came from and where they are going: think of it in terms of states if you prefer more abstraction. $\endgroup$ – Guy Inchbald Jun 4 '20 at 19:37
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    $\begingroup$ @jacob1729: It doesn't matter whether the ping-pong balls are "fundamentally" indistinguishabe. It's sufficient for the purposes of the experiment that they are PRACTICALLY indistinguishable. $\endgroup$ – jamesqf Jun 5 '20 at 4:24
  • $\begingroup$ @Craig When charge moves from one place to another, the charge carriers do not throw photons at each other, they jump to it. $\endgroup$ – Guy Inchbald Jun 6 '20 at 12:11
  • $\begingroup$ @Craig The (negatively charged) Van de Graaf physically pumps out electrons into the wire, that is what discharging means. All current generators work in this way (or else they draw electrons in), that is what current is, the rate of change of charge: $I = dQ/dt$. It's just like a lockdown queue at a supermarket: individuals leave at one end and enter at the other, everybody stands on a designated spot, and periodically they all shuffle forward in waves of motion. I like that analogy, I'll add it to my answer. $\endgroup$ – Guy Inchbald Jun 9 '20 at 7:36
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Perhaps the problem is to take a QM notion (electrons are indistinguishable) and at the same time keep a classical intuition (electrons are small balls that move).

The QM model of a conductor is a band of available states for the valence electrons. That notion replaces the orbital for a single atom, because here they are "shared" by the atoms of the lattice.

They fill the states from the lowest energy upwards, and for each state corresponds a momentum.

Without an electric field, the distribution of momentum is balanced, but the effect of the field is to break that symmetry. There is now a net momentum in the direction of the E-field. The expected value of the electron velocity is defined as: $$\langle\mathbf v\rangle = \frac{\langle\mathbf p\rangle}{m}$$

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