Interpretation Schrödinger's formulae

AI Thread Summary
The discussion centers on the interpretation of Schrödinger's formulae, particularly the expression involving the wavefunction and the probability current. The probability current, represented as J, indicates the flow of probability density, analogous to electrical current. Participants explore the meaning of specific terms in the formula, such as ψ* ∇ψ, and its implications in quantum mechanics. There is an acknowledgment that while the formula is well-established in theory, direct measurement of certain components is not possible, as quantum mechanics relies on probabilistic outcomes. The conversation emphasizes the importance of empirical validation in understanding quantum phenomena.
Raparicio
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Dear Friends,

Does anybodi knows the meaning, or anything related to the term:

\Psi \nabla \Psi^*

or

\Psi \nabla \Psi^* - \Psi^* \nabla \Psi

Is the representation of something in the reality?

Best reggards.
 
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If you have a particle having a wavefunction \psi(\vec r, t), then:

\vec J(\vec r,t)=\frac{\hbar}{2mi}(\psi^* \nabla \psi-\psi\nabla \psi^*)

is the so-called probability current. It represents the flow of probability density, like electrical current is the flow of charge density.
 
Spinor?

Galileo said:
If you have a particle having a wavefunction \psi(\vec r, t), then:

\vec J(\vec r,t)=\frac{\hbar}{2mi}(\psi^* \nabla \psi-\psi\nabla \psi^*)

is the so-called probability current. It represents the flow of probability density, like electrical current is the flow of charge density.


Thanks, Galileo, but I'm trying to "imagine" what is, for example, one of the 2 terms:

\psi^* \nabla \psi

Has it any meaning? Is a rotor of the nabla operator?

:smile:
 
Nope,it's just the ~ to the integral nucleus of the momentum operator.

Daniel.
 
Galileo said:
If you have a particle having a wavefunction \psi(\vec r, t), then:

\vec J(\vec r,t)=\frac{\hbar}{2mi}(\psi^* \nabla \psi-\psi\nabla \psi^*)

is the so-called probability current. It represents the flow of probability density.

Sorry if this is really basic, I'm no quantum guru yet :-p, but could you say that the (classical) velocity is in the direction where \vec J has global max at a given t? Is it possible somehow to calculate \vec v from \vec J?
 
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"(Classical) velocity" has nothing to do with the probability current density...

Daniel.
 
empirical

empirical experiments verify this formulae is ok?
 
Which formulae...?We can't measure \vec{j}\left(\vec{r},t\right),but only probabilities.

Daniel.
 
Chr

dextercioby said:
Which formulae...?We can't measure \vec{j}\left(\vec{r},t\right),but only probabilities.

Daniel.

Daniel,

I mean that if exists any experiment or example in real word that confirms that formula or some of its components. For example, if the probability to find a particle in some place or time has this formula...
 
  • #10
I'm not an experimentalist and never will be,but i can assure that this simple part of QM has been fully checked and confirmed.We can't measure certain abstract things.Since QM is a probabilistic theory,all we can do is statistics.

Daniel.
 
  • #11
dextercioby said:
I'm not an experimentalist and never will be,but i can assure that this simple part of QM has been fully checked and confirmed.We can't measure certain abstract things.Since QM is a probabilistic theory,all we can do is statistics.

Daniel.

tks Daniel.
 
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