# Why, when you eject an electron from a metal, do you get exactly one electron and not part of an electron?

Let's assume you have a metal with an electron sea so the electrons are very delocalized. The electrons are described by some wave function $$\Psi(\vec x_1,\dots,\vec x_N).$$ Now an electron is ejected for example by a high energy photon. This will result (if I understand correctly) in a wave packet moving away from the metal. Let's say this wavepacket is described by $$\psi(\vec x)$$. My intuition tells me the metal can only eject 'whole' electrons. To define 'whole' let's take a small volume $$V$$ that surrounds the wavepacket. If it's a whole electron then $$\int_V|\psi(x)|^2dx\approx 1$$ In other words the probability of the wavepacket containing an electron is 1. But what stops us from emitting half an electron? This would mean that $$\int_V|\psi(x)|^2dx\approx 1/2$$ and there would still be a 50% chance of finding the electron in the metal. Are these half electron states possible in free space? If not what mechanism prevents it? And since the electron was very localized before it got emitted how did the probability density somehow collect to form a wavepacket? Is the answer the same in quantum mechanics and in QFT?

Electrons are indivisible
Within the context of condensed matter physics, electrons are indivisible particles, i.e. their particle number operator has integer eigenvalues. Therefore one cannot meaningfully speak about half of an electron.

Probabilities can be less than 1
However, the setup described can be thought of as a scattering problem: a photon is incident on a metal and generates an outgoing electron wave. In such a formulation the electron wave will resemble a free electron only far away from the metal, but one cannot really say that electron is strictly outside of the metal, without destroying the solution, i.e., there is always a non-zero probability that electron is still inside. In this sense, the integral over the wave function outside of the metal is less than one. This is not a fraction of the electron, but the probability that the electron did leave the metal.

Decoherence in the real world
If the problem is described more completely, one has to take into account the interaction with the environment, which leads to the collapse of the joint wave function photon-metal-electron, so that the electron eventually gets localized somewhere outside of the metal. This happens, for example, when the ejected electron lands on an electrode measuring the photocurrent.

Why all this complexity?
If all this complexity seems exaggerated, it is because it is unusual when viewed from the point of view of the classical point like particles, where the elementary parts of complex world can be often described without relation to the rest.

Catastrophe
Finally, just to further complicate the things, ejecting an electron from a metal results in X-ray singularity, also known as the Anderson orthogonality catastrophe, according to which the matrix element for ejecting an electron tends to zero, i.e., the process is impossible.

• So can I say that the decoherence can happen long after the electron has left the metal? Does that mean the probability density will have to be transported over long distances? Nov 25, 2020 at 20:23
• It depends on coherence length and coherence time, which usually requires very low temperatures. Josephson junctions are one example. Nov 25, 2020 at 20:38

Simple scattering theory would predict exactly the sort of superposition you are describing. Your factor of $$1/2$$ would essentially be the modulus square of some scattering amplitude (possibly integrated over a range of angles).

An electron in a metal, however, is part of a (typically strongly interacting) many-body system, and so the fact you have removed a particle is not going to go unnoticed. To put it another way you are not simply ejecting an electron from the metal but creating an electron-hole pair, and while the electron may escape, the hole will remain and continue to interact with the other electrons in the metal. This interaction with a macroscopicly large system will cause the 2 branches of the scattering wavefunction (the one where scattering occurred and the one where the electron remained trapped) to decohere, resulting in a mixed state with a more or less classical probability that scattering occurred.

• The two branches of the eject/no-eject outcomes will also contain corresponding outcomes of photon vanished/not-vanished—so that the total energy remains conserved in both cases. Nov 25, 2020 at 20:17

The electron is an elementary particle described by a quantum mechanical wave function, normalize to 1, because the wavefunction is connected with the probability of finding the electron at (x,y,z,t), and probabilities have to be normalized to 1 by construction.

All of our data validate the statement that the electron is elementary,and is always whole.