Difference between expectation value and probabilty

MGWorden
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Homework Statement



Psi(x) = Ax -a<x<a

I am trying to find the probability that my measured momentum is between h/a and 2h/a


Homework Equations



I have normalized A= sqrt(3/(2a^3))

I know that if I was finding the expected momentum I would use
\int\Psi * p \Psi dx



The Attempt at a Solution




so far I have done
\int \Psi * p \Psi dx
with bounds from from h/a and 2h/a


but I know this is the expected value of momentum and I don't think that is the same thing as the probability of momentum.

Could someone please explain the difference in finding the probability of momentum and the expected value of momentum. Or if they are the same thing let me know there is no difference.

Thanks for your time.
 
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The probability that the momentum lies between p1 and p2 is not related to the expected value of momentum. The expected value of momentum is the most likely momentum value we'd find in the state \Psi(x). The probability to find the particle with a given momentum p is | \psi(p) |^2, where \psi(p) is the normalized wavefunction in momentum space. To find the probability of finding the momentum in a range of values, we have to integrate this.
 
Thanks for your help. I converted over to momentum space using a Fourier transform then integrated.

Your explanation saved me much time searching.

Cheers!
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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