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Probability of finding a particle at X?

  1. Jan 1, 2016 #1
    Hey all,

    Here's my question : Given X is the position of a particle at time T, how would I go about finding the probability of finding said particle at any given position if i randomly pick a time out of the interval T1 to T2?

    Let's say that my X(t) = cos(t). How can I find the probability of observing the value of X(t) to be equal to some value X if i randomly select a value for time from my interval T1<T<T2?

    This is kind of inspired by quantum physics; in particular the time independent solutions to the shrodinger wave equation for a harmonic oscillator. Thanks :)
     
  2. jcsd
  3. Jan 1, 2016 #2

    mfb

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    Staff: Mentor

    Integrate from T1 to T2?
    For a classical motion with a non-zero velocity, this leads to a sum over ##\displaystyle \frac{1}{v_i}## where vi are the velocities at times i where the particle crosses the position you are looking at.
     
  4. Jan 1, 2016 #3
    Integrate what from T1 - T2? X(t)? 1/V(t)? What I'd ultimately want would be a function of X that gives me the probability.

    Kinda confused, sorry :/
     
  5. Jan 1, 2016 #4

    mfb

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    Staff: Mentor

    Integrate the probability to find the particle at position x at time t from T1 to T2 (and divide by (T2-T1) to normalize it properly). With an exact position that needs distributions to do it properly, but with the result I posted above.

    It is probably easier with the example:
    X(t) = cos(t)
    X'(t) = -sin(t)
    Let's say T1=0 and T2=2 pi and we are interested in the probability to find the particle at x=0. The particle crosses this point twice, at 1/2 pi and 3/2 pi. The derivative there is -1 and 1 respectively. The sum I suggested (forgot to take the absolute value) gives ##\frac{1}{|-1|} + \frac{1}{|1|}##, dividing it by 2 pi gives 1/pi.
    The probability to find ##x=\sqrt{3/4}##, following the same steps, is ##\frac{1}{2\pi} \left( \frac{1}{|-1/2|} + \frac{1}{|1/2|} \right) = \frac{2}{\pi}##.
     
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