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Which of these interpretations of the modulus squared of wavefunction is right? 
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#1
Nov2812, 02:07 PM

P: 283

Does [itex]\psi(\mathbf{x},t)^2d^3\mathbf{x}[/itex] or [itex]\psi(\mathbf{x},t)^2d^3\mathbf{x}dt[/itex] give the probability of a particle to collapse at the point [itex]\mathbf{x}[/itex] at time [itex]t[/itex]?
Griffiths sides with the former, but I'm having doubts. 


#2
Nov2812, 02:24 PM

P: 32




#3
Nov2812, 07:14 PM

P: 283




#4
Nov2812, 08:15 PM

P: 883

Which of these interpretations of the modulus squared of wavefunction is right?



#5
Nov2812, 09:38 PM

PF Gold
P: 1,168

Instead, if the wave function for a particle is normalized (the most clear approach), it gives us the probability that this particle is at some point of space (without the necessity to detect it there). You are right that the number of counts(clicks) of detector set in some definite distance from the piece of matter scattering charged particles will be proportional to time interval of the measurement, but this is because larger interval allows more [itex]\textit{distinct particles}[/itex] to come at the detector; however, each one can be ascribed by normalized wave function that gives density of probability in space by the rule [itex]\psi^2dV[/itex]. 


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