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roshan2004
Mar31-10, 12:54 PM
A particle moving on a straight line is described by \psi(x)=\frac{1+ix}{1+ix^2}.
Where is the particle likely to be found?
I took the derivative of probability density with respect to x and equated it to 0. I got my answer to be x=0.643,-0.643,1.554i and -1.554i.
Please tell me whether I am right or wrong or are there any other methods to solve this problem or not?

dx
Mar31-10, 02:20 PM
There is no reason for the expectation value to be at a stationary point of the probability density.

You have to evaluate the integral <x> = ∫ψ*(x)xψ(x)dx = ∫xP(x)dx.

roshan2004
Mar31-10, 02:25 PM
But the question is about maximum probability of finding the particle, isn't it?

dx
Mar31-10, 02:30 PM
"Where is the particle likely to be found" usually means that they want you to find the expectation value of x.

dx
Mar31-10, 02:32 PM
Unless the exact wording of the question was "where is the particle most likely to be found". Then you would find the x which maximises P(x).

roshan2004
Mar31-10, 02:38 PM
So if the question is where the particle is most likely to be found, is my answer correct.

dx
Mar31-10, 02:58 PM
x is a real number, how did you get imaginary values?

roshan2004
Mar31-10, 03:13 PM
By factorising

dx
Mar31-10, 03:29 PM
Ok. You have to ignore the imaginary solutions. Looking at this plot:

http://www.wolframalpha.com/input/?i=plot+%281+%2B+x^2%29%2F%281+%2B+x^4%29

it does look like the probability density has maxima at approximately x = 0.643 and x = -0.643

roshan2004
Mar31-10, 03:32 PM
Now I finally got it, thanks dx