# Interpretation of the wave function in newtonian spacetime

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 $$t$$?

• of finding the particle, but by whom? The observer. However, the observer could just be the environment – Wolphram jonny Apr 21 at 23:35
• i am not understanding what you mean – amilton moreira Apr 21 at 23:41
• 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. – 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.
• 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 – amilton moreira Apr 22 at 10:02
• 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 $\psi$. – Valter Moretti Apr 22 at 11:06
• What I am saying is just that $\psi$ cannot be assumed to be a scalar field. – Valter Moretti Apr 22 at 11:09
• 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. – Valter Moretti Apr 22 at 11:19