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I've heard, that Schrodinger initially tried to interpret the wavefunction that he obtained for an electron as a charge density, but it wasn't correct. I know, that nowadays the modulus squared of his wavefunction is interpreted as a probability density of obtaining the particle in a given space interval, but what were the arguments against the charge density interpretation?

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  • $\begingroup$ Where did you hear that? $\endgroup$ Jul 18 '18 at 15:38
  • $\begingroup$ It was mentioned in a video in the Khan Academy tutorial about the Bohr's model of an atom. $\endgroup$ Jul 18 '18 at 15:48
  • $\begingroup$ Since the time-dependent wavefunction is complex, what interpretation would you give to the imaginary part? In addition, it is the modulus squared of the wavefunction that is a probability density, not the wavefunction itself. $\endgroup$ Jul 18 '18 at 15:55
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    $\begingroup$ how could it be charge density in 3D space when $|\psi(r_1,r_2)|^2$ has 6 space variables for a tw0-electron system? $\endgroup$
    – hyportnex
    Jul 18 '18 at 16:18
  • $\begingroup$ @ZeroTheHero Is there an interpretation of the complex part of the wavefunction in terms of probability? The modulus squared gives us the probability density, but i think there is no interpretation of this complex part in the wavefunction itself. $\endgroup$ Jul 19 '18 at 9:00
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As long as you don’t collapse the wave function, its behavior would be similar to a charge density indeed. The density would have a linear relation to $|Ψ|^2$, so that the total charge gives you $-e$. However, when you measure its position, the wave function collapses, and that would not happen with a charge density. That would mean that all of the charge density would suddenly gather around one point in space, which would violate energy conservation in all kinds of ways.

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I assume your question is, since the modulus square of a wave function is the probability density, then if a particle is charged, its probability density is also the charge density, with up to a constant in front.

Now if you would like to interpret wave function $\psi$ as probability density, not only its imaginary part would cause problem in interpretation, you would also have to reinvent the Schr$\mathrm{\ddot{o}}$dinger equation as the relation between $\left| \psi \right|^2$ and $\psi$ is not linear.

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A wave function in quantum description of an isolated system is a probability amplitude having real and complex parts and the probabilities for the possible results of measurements made on the system can be derived from it.

The wave function is a function of the number of degrees of freedom to a maximal set observables. For a given system, the choice of which commuting degrees of freedom one may use is not unique, and correspondingly the wave function is also not unique.

For instance, it may be taken to be a function of all the position coordinates of the particles over position space, or the momenta of all the particles .

Some particles, like electrons/photons have non zero spins , and the wave function for such particles includes spin as an intrinsic, discrete degree of freedom.

When a system has internal degrees of freedom, the wave function at each point in space assigns a complex number for each possible value of the discrete degrees of freedom (e.g., z-component of spin).

According to superposition principle , wave functions can be added together and multiplied by complex numbers to form new wave functions. Or a state of the system can always be expanded in linear combination of all possible states with finite densities.

The inner product between two wave functions is a measure of the overlap between the corresponding physical states, and is used in the probabilistic interpretation of quantum mechanics, relating transition probabilities to inner products.

The Schrödinger equation determines how wave functions evolve over time.

In statistical interpretation the squared modulus of the wave function [psi]^2 is a number interpreted as probability density of measurement for

a particle's being detected at a given place – or having a given momentum – at a given time, and possibly having definite values for discrete degrees of freedom.

The integral of probability density , over all the system's degrees of freedom, must be unity in accordance with the probability interpretation.

The above requirement that a wave function must satisfy is called the normalization condition.

Since the wave function is complex valued, only its relative phase and relative magnitude can be measured—its value does not, in isolation, tell anything about the magnitudes or directions of measurable observables;

One has to operate the physical quantum operators , whose eigenvalues correspond to sets of possible results of measurements, to the wave function psi and calculate the statistical distributions for measurable quantities.

Therefore I just wonder how one is going to construct a wave function as solution of schrodinger equation which can represent charge density. Perhaps it is not possible.

reference-https://en.wikipedia.org/wiki/Wave_function

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Yes, while it can be argued $e\psi^2$ can be interpreted as a charge density in an operational way based on the charge distribution's effect on other quantum objects, as is done in this paper, Schrodinger wanted to really view the electron cloud as a spread out charge density in a semi-classical way (i.e. that charge was really smeared out in space). This would lead to all kinds of problems like self interaction of electron clouds with themselves.

For instance, if the charge of an electron were really spread out (or separable), then all electron clouds, being negative at all points, due to self interaction should self repel all parts of the cloud and push itself out to infinity. There is no evidence for this type of self interaction of the electron cloud. Also, upon wavefunction collapse after measuring the location of the electron, a great current of charge would be rushing from all parts of the cloud toward the location of the measured electron, potentially causing currents and anomalous magnetic fields that again, are not measured.

So even though you might be able to identify $e\psi^2$ as a charge density in an operational way based on its effect on other objects, the fact that the "charge distribution" does not have any self interactions within non-relativistic QM, means that at least in non-relativistic QM, it can't be viewed in the semi-classical charge distribution way that Schrodinger appears to have desired. Which is partly why people settled on the Born rule as interpreting $\psi^2$ as probability density instead of charge density.

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  • $\begingroup$ I don't disagree, but wave function collapse is more of a postulate than an actual phenomenon. Extrapolating effects from collapse like induced currents is a stretch. The very idea of "parts of a single quanta of charge" producing anything observable seems like an oxymoron to me. A single electron could have a fully distributed charge in an orbital, and still only be detected at a particular x,y,z locus; the uncertainty principle should fully describe both conditions without needing to invoke collapse. $\endgroup$
    – JPattarini
    Dec 3 '18 at 15:36
  • $\begingroup$ That's my point essentially. If you were to try and view the wavefunction (and its spread out nature) as a physical object, then you would have to identify wavefunction collapse as a physical process. Since we don't see evidence for this we don't think of the wavefunction (or its collapse) as a physical object (be it charge, mass, or otherwise). $\endgroup$ Dec 3 '18 at 22:52

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