If electrons are identical and indistinguishable, how can we say current is the movement of electrons?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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).

Answer 7

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.

Answer 8

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.

Answer 9

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: $$\langle\mathbf v\rangle = \frac{\langle\mathbf p\rangle}{m}$$