Is it possible to formulate quantum mechanics with some Schroedinger equation that describe only the absolute value of the wave function (without the phase)?
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1$\begingroup$ Related: physics.stackexchange.com/q/32422/2451 , physics.stackexchange.com/q/8062/2451 , physics.stackexchange.com/q/116595/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 16, 2021 at 23:54
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$\begingroup$ If you don't like wave functions and/or complex numbers, check out the formulation of QM in terms of the Wigner probability distribution. In QM, the probability distribution on the phase space is allowed to take negative values on the scales of order $\hbar$. Nevertheless, as one evaluates expectation values of physical observables, those obey all the usual properties. $\endgroup$– mavzolejCommented Nov 17, 2021 at 0:39
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2$\begingroup$ Does this answer your question? Form of Schrödinger equation for the probability density $\endgroup$– AccidentalFourierTransformCommented Nov 17, 2021 at 1:10
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$\begingroup$ @mavzolej That's an extremely misleading comment. As the title of the Wiki article you linked says, it is quasiprobabilities -- not probabilities. You simply can't have probabilities over the phase space in QM. That's the whole point. Also, obviously, probabilities are never allowed to take negative values, as made clear by Kolmogorov many years ago. QM doesn't change rules of consistency or logic. That's another way to see that it's quasiprobabilities that Wigner is talking about and not probabilities, if they were probabilities, they wouldn't be allowed to be negative. $\endgroup$– user87745Commented Nov 17, 2021 at 9:23
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In short, no. Of course, you can find out the time-evolution law for the absolute value of the wavefunction using the standard Schrodinger equation, but, you can not describe everything that quantum mechanics describes using just the absolute value of the wavefunction.
The phases in the full wavefunction $\psi(x)\equiv\langle x\vert\psi\rangle$ have a whole bunch of physical "information" that is lost when you take the absolute value. In particular, for example, you can't even predict the expectation value of momentum using only the absolute value of the wavefunction in position space. Proof: all eigenstates of momentum have the same absolute value of the wavefunction in position space.
The basic point is that the phases of a wavefunction in a given basis encode the information about the probability distribution over eigenvalues of observables that don't commute with the observable that is being diagonalized in the given basis. And since there are non-commuting observables in quantum mechanics, you can't throw away the phases.
However, notice that the probability distribution over the eigenvalues of an observable that is being diagonalized by the given basis does not depend on the phases of the wavefunction in that basis -- trivially so. So, what we can do is describe the full physical content of a system using the absolute value of the wavefunction in several bases at once. If we describe the absolute value of the wavefunction over the bases of "a maximal set of non-commuting observables" then we ought to have encoded all the information contained in the phases when we describe the system using the wavefunction in one basis. So, what is this "maximal set of non-commuting observables"? Well, it is the tomographically complete set of observables for the given system -- or, in other words, the set of observables that form a complete basis in the vector space of observables defined on the given Hilbert space of the given system.