In solid state physics the electron density is often equated to $e|\psi|^2$. However, the Sakurai says (Chapter 2.4, Interpretation of the Wave Function, p. 101) that adopting such a view leads "to some bizarre consequences", and that Born's statistical interpretation of $|\psi|^2$ as a probability density is more satisfactory.

I am aware that a position measurement of an electron leads to (in the Copenhagen interpretation) a collapse of the wave function into a position eigenstate $x$ with the probability given by $|\psi(x)|^2$, and that it is known from some scattering experiments that the electron behaves as a point-like particle.

However, the electron density is experimentally observable, e.g. by X-ray scattering. One could argue that X-ray scattering is done with a large ensemble of atoms, so that what we actually observe is the average over many position measurements.

I am wondering if this the only "bizarre consequence", since this distinction seems to be a minor difference to me. My questions aims at clarifying the difference between the probabilistic interpretation of $|\psi(x)|^2$ and the interpretation as an electron density.


The statement that the right interpretation of $|\psi(x)|^2$ is probabilistic means that the value of this expression may only be "measured" in the same sense as other probabilistic distributions – by a repeated experiment starting with the same initial conditions. Every probability distribution may be approximately reconstructed by "throwing the dice" many times and recording the frequencies.

However, $|\psi(x)|^2$ cannot be measured in one repetition of the situation. There can't exist a gadget that would show $|\psi(x)|^2$ on its display. That's why we say that the wave function (or its squared absolute value) isn't observable: it's not observable in one copy of the situation. This word "observable" may be interpreted colloquially as well as technically. Technically, an "observable" is a Hermitian operator. The squared absolute value of the wave function isn't an operator, so it's not an observable, so it can't be measured.

The charge density is a classical quantity in classical physics and may be an observable in quantum field theory – like for other observables, only the probabilities of different values may be predicted. However, in one-particle quantum mechanics, a particle is either located in the volume $dV$ or it's not. Nothing in between is possible and a measurement may answer whether the question is Yes or No. The probability of Yes is given by $|\psi(x)|^2 dV$.

If you understand the things above and your wording indicates that you do, then you understand why the right interpretation of the wave function is probabilistic. This assertion means nothing else than the fact than the fact that to "measure" $|\psi(x)|^2$ at a given point, one needs to repeat the same situation many times and use the laws of probability and statistics. The function $|\psi(x)|^2$ doesn't correspond to a property of the region of space that may be measured instantly, by one measurement in one repetition of the experiment.

  • $\begingroup$ Thanks for the clarification. I'm still wondering about the "bizarre consequences" mentioned in the Sakurai. Is there any experiment which would give a fundamentally different result if $|\psi|^2$ actually was an electron density instead instead of a probability density? $\endgroup$
    – m4r73n
    Dec 17 '12 at 13:33
  • $\begingroup$ Hi, if it were an electron (and electric charge) density, the results would be completely different. In a double-slit-like interference experiment, each electron would actually create a whole cloud on the photographic plate. Atoms would lose their sharp identity - the nuclei's charge would be neutralized by thousands of "piece of electrons" from various other places. And so on. When we see that the electron may always be seen at particular locations, it's clear that assuming that it's actually spread would lead to a totally different world. $\endgroup$ Dec 20 '12 at 7:04
  • $\begingroup$ @LubošMotl: I don't get that: "Technically, an "observable" is a Hermitian operator. " Isn't it that technically the observable corresponds to a hermitian operator? Such that the possible values of the observable are the eigenvalues of that operator (or similar with expectation values). So the task, rather would be to come up with an operator $\mathcal{O}$such that we would get a (non-sensical) $$ \mathcal{O} \psi(x) = |\psi(x)|^2 \psi(x) $$? $\endgroup$ Jan 23 '19 at 9:25
  • $\begingroup$ Dear Rudi, feel free to say "correspond" but there is really no difference. "Observable" is meant to be a physical-philosophical term, Hermitian operator is a mathematical object, but within quantum mechanics, the first is always represented by the latter. Otherwise the right hand side of your equation is not linear in psi, it doesn't show an action of a linear operator on psi (it is cubic), so it if your equation is meant to be a definition of an operator O, then it is a nonlinear operator. Equivalently, using physics language, O isn't an observable because it's cubic, not linear. $\endgroup$ Jan 25 '19 at 17:23
  • $\begingroup$ Because O isn't a linear operator i.e. because O isn't an observable, the basic postulate of quantum mechanics says that O isn't observable without "a", either. There is no way to create any gadget whose measurements would have anything to do with the cubic expression of yours. $\endgroup$ Jan 25 '19 at 17:25

Obviously, the interpretation of $e |\psi|^2$ as an electron density will hold as soon as we consider an ensemble of electrons, since the probability density then results in a statistical distribution. In the case of just one electron we can still interpret $e |\psi|^2$ as an electron density as long as we think about longer time scales. Please note, that this only works as all observables (that I can think of now) depend linearly on the electron charge.

In any other case the electron still has to be considered point-like and is not smeared out in space, only that its position in the phase space is not determined. This means e.g. that the detection of an electron will always (after the aforementioned collapse of the wave function) result in one single point of detection.

  • $\begingroup$ Sorry, many electrons must be described by a wave function $\psi(\vec x_1, \vec x_2, \vec x_3,\dots)$ of many vectors of the individual positions, not just by a 3D wave function. There can't really be a sensible realization of your electron ensemble described by one psi because the electrons are fermions that can't be in the same state, so you can't make a condensate out of them. On top of them, they repel each other. Just think about psi equal to the 1s ground level of the hydrogen atom. Clearly, you can't imagine "ensemble of many electrons" in that state, right? $\endgroup$ Feb 11 '19 at 15:43

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