A Newtonian spacetime is a quintuple $(M, \mathcal{O}, \mathcal{A}, \nabla, t)$ where $(M, \mathcal{O}, \mathcal{A}, \nabla)$ is a 4 dimensional differentiable manifold with a torsion free connection, and $t$$\in C^{\infty}(M)$ is such that $dt\neq 0$ and $\nabla dt=0$.

Since time $t$ is absolute there exist 3 dimensional plane of simultaneity. Now my question is, what is the interpretation of probability density $|\psi(x)|^2$ ?

Is it the probability that the particle is at position $x$ in relation to an observer or, is it the probability that the particle is at a point $x$ in the plane of simultaneity defined at time $t$?

  • $\begingroup$ of finding the particle, but by whom? The observer. However, the observer could just be the environment $\endgroup$ – Wolphram jonny Apr 21 at 23:35
  • $\begingroup$ i am not understanding what you mean $\endgroup$ – amilton moreira Apr 21 at 23:41
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    $\begingroup$ I do not understand why someone wants to close this question. Maybe it is not written into an explicit way, but it deals with a physically relevant and subtle issue regarding the nature of the non-relativistic quantum wavefunction: whether or not it is represented by a classical scalar field. The answer is negative for $\psi$ but positive for the associated probabiliy distribution. $\endgroup$ – Valter Moretti Apr 22 at 11:28

Even restricting to the class of observers where the connection coefficients vanish (inertial observers), the function $\psi$ depends on the observer, since the action of Galileian group is not trivial on $\psi$: it cannot be considered a scalar field over the 3-surfaces at constant absolute time in view of the appearance of a phase depending on $x$ and $t$ (and on the element of the group). Conversely, its squared absolute value you consider (the probability) is independent from the observer. Therefore $x$ can be viewed as a point in the absolute space at time $t$ if dealing with $|\psi(t, x)|^2$, since this object is a scalar field over that manifold (at fixed time) as the phase disappears when computing the absolute value, whereas this intepretation is impossible when directly dealing with $\psi(t,x)$, because every chart associated to every different observer assigns a different value even if $x$ defines the same point in the absolute space.

  • $\begingroup$ Say we are in 2 dimension 1 for spatial and 1 for time. and we have $X \phi=x \phi$. what am trying to understand if this $x$ is a point in the manifold or the position of the particle in relation to the observer $\endgroup$ – amilton moreira Apr 22 at 10:02
  • $\begingroup$ In your book (A5 (b)) you defined like this . why can we not defined independently of an observer ? $\endgroup$ – amilton moreira Apr 22 at 10:21
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    $\begingroup$ Because when changing the observer you sould also change the value of $\psi$ at the same point of absolute space, since a phase appears in front of $\psi$. $\endgroup$ – Valter Moretti Apr 22 at 11:06
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    $\begingroup$ What I am saying is just that $\psi$ cannot be assumed to be a scalar field. $\endgroup$ – Valter Moretti Apr 22 at 11:09
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    $\begingroup$ However, yes when writing $\psi(x)$, $x$ is referred to the observer, it is a point in the coordinate system of a given observer and not a point in the common manifold shared by the observers. $\endgroup$ – Valter Moretti Apr 22 at 11:19

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