Why Schrodinger's wave function cannot be interpreted as charge density? 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?
 A: 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.
A: 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.
A: 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

A: 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.
