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In Quantum Mechanics the square of the wave function is compared to a probability density. Is there no similar relation to waves in the sense that something meaningful can be ascribed to the real part of the wave function, in the way that the real part of an EM wave is the actually measurable thing?

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  • $\begingroup$ Related: physics.stackexchange.com/q/53608, physics.stackexchange.com/q/17168 $\endgroup$
    – Kyle Kanos
    Commented Oct 17, 2013 at 19:51
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    $\begingroup$ Perhaps you would be interested in Madelung's formulation, cf. this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Oct 17, 2013 at 21:31
  • $\begingroup$ I have to say that I find the use of the word "enjoy" in that way to carry unwarranted implications that this is done on a lark. It is done because it works, which is always the final justification in all of science. $\endgroup$ Commented Oct 17, 2013 at 22:33

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The answer is negative. The real part of $\psi$ cannot have any direct physical intepretaton. This is because the wave function has physical meaning just up to a phase: $\psi$ and $e^{ia}\psi$ define the same quantum state, for every fixed real $a$. $|\psi|$ and things like $\bar{\psi} \nabla \psi$ are not affected by the choice of the phase and thus they may have physical meaning. Instead $Re(\psi)$ is affected.

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In Quantum Mechanics, The wave function has no direct physical meaning. It is just one way of "storing information". We interpret $\left | \psi \left ( x, t \right ) \right |^{2}$ as a probability density. This interpretation is possible because the product of a complex number with its complex conjugate is a real and positive number. For the probability interpretation to make sense, the wave function must satisfy certain conditions. We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one.

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    $\begingroup$ This is not entirely correct. The phase of the wavefunction (or at least its phase profile) does matter in experiments; for example, it has a huge influence on the momentum distribution. $\endgroup$ Commented Dec 6, 2013 at 13:41
  • $\begingroup$ Yes, you are right @EmilioPisanty. $\endgroup$
    – Fatima
    Commented Dec 6, 2013 at 14:10

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