What is the Most Likely Position of a Particle in 1-Dimensional Wavefunction?

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Homework Help Overview

The discussion revolves around finding the most likely position of a particle described by a one-dimensional wavefunction. The wavefunction is defined within a specific interval, and participants are exploring the implications of this function on probability density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the wavefunction and the probability density, questioning how to determine the most likely position from the given wavefunction. There are attempts to clarify the difference between the most likely position and the expected value of the particle's position.

Discussion Status

The discussion is active, with participants providing insights into the mathematical relationships involved. Some have suggested taking derivatives to find maxima, while others are clarifying concepts related to probability density and wavefunctions.

Contextual Notes

There is an emphasis on understanding the definitions of terms such as "most likely position" versus "expected value," indicating that participants are navigating through foundational concepts in quantum mechanics.

nadeemo
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Homework Statement



Find the most likely position of the particle.

Homework Equations



\Psi = A[(x+1)^{2} - 1)]
between x = 0 and x = 1
\Psi = 0 anywhere else


The Attempt at a Solution



I found A to equal \sqrt{15 / 38}... but I am not sure how to do the rest of it
 
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You didn't even have to do that if you just want the most likely position. That's at the maximum of psi*conjugate(psi), the probability density.
 
that gives you expected value of the particles position <x>, which is the next question,
i need to find the most likely position of the particle...
is there another way to do this?
 
nadeemo said:
that gives you expected value of the particles position <x>, which is the next question,
i need to find the most likely position of the particle...
is there another way to do this?

No, <x> is the integral of x*psi*conjugate(psi) over the integral of psi*conjugate(psi). The 'most likely position' is the maximum of psi*conjugate(psi).
 
so take the derivative of |psi*cojugate(psi)| and set it to 0?

the conjugate would be A[x+1)^2 +1] ?
 
nadeemo said:
so take the derivative of |psi*cojugate(psi)| and set it to 0?

the conjugate would be A[x+1)^2 +1] ?

Yep. And since psi is real, conjugate(psi)=psi. You'll probably notice that the maximum of psi*psi is the same as the maximum of |psi|.
 
thanks a bunch =)
 

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