1
$\begingroup$

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 $t$?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ What I am saying is just that $\psi$ cannot be assumed to be a scalar field. $\endgroup$ – Valter Moretti Apr 22 at 11:09
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.