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Simple harmonic oscillator- the probability density function

  1. Mar 1, 2009 #1
    How to find the probability density function of a simple harmonic oscillator? I know that for one normal node is should be a parabola but what is the formula and how do we derive it?
  2. jcsd
  3. Mar 1, 2009 #2
    Probability is equal to the time spent in interval (x,x+dx) normed to total time it takes mass to run through entire available x space ( from -A to A, A is amplitude ). So you have

    dP = 2*dt/T .

    Take your dt from energy conservation law ( v = dx/dt ! ) and integrate from -A to A. IIRC your distribution should have two poles at amplitudes, maybe some pi somewhere ... don't know, do your math =) Let us know what have you found!

    Your integration limits should not be from -A to A but over the range where you want to calculate probability of finding your mass! Upper integral will give you 1, of course.
    Last edited: Mar 1, 2009
  4. Mar 1, 2009 #3
    You will find that although the particle average position is zero, the most of the time it spends at the ends where its velocity is small.

  5. Mar 1, 2009 #4
    I know that, I was more curious about the function p(x)....
  6. Oct 12, 2009 #5
    I'm struggling with the same problem here. Trying to find the probability density function for a simple harmonic oscillator.

    I'm having trouble deriving P from the information suggested here.

    It was suggested that we should use:

    dP = 2*dt/T

    and the fact that

    v = dx/dt

    Now we have:

    x(t) = Acos(ωt)


    v(t) = -Aωsin(ωt)

    Now if we integrate dP we get

    P = int( 2/(T*v) dx)

    But since I only have v(t), but not v(x), I'm not sure how to go about this.

    I know that a(x) = -ω2x

    but I'm not sure if that can be of any aid?

    Anyone have a clue?

  7. Oct 12, 2009 #6
    dx=vdt so P =2 int(dt)/T = 1.
  8. Oct 12, 2009 #7
    Which is what we started with ( dp = 2 dt/T ), with normalization added: P(-infinity < x < +infinity) = 1

    But I'm supposed to find P as a function of x, which probably looks something like a parabola. Now isn't it neccesarry for me to substitute that dt with dx/v ?

    Then I would need to define v as a function of x, right?

    Frímann Kjerúlf
  9. Oct 12, 2009 #8
    Maybe Asin(ωt) = A[1-cos2(ωt)]1/2 = A[1-x2/A2]1/2, so you obtain v(x)?
  10. Oct 12, 2009 #9
    Thats seems to be about right, although I don't understand how you got there :)

    I got the same results doing:

    Isolate x:

    x(t) = Acos(ωt) => t(x) = arccos(x/A)/ω

    differentiate t(x):

    dt/dx = 1 / (A^2 - x^2)^1/2

    which is equal to 1/v :

    v(x)= dx/dt = (A^2 - x^2)^1/2

    which is the same result you got.

    Thanks :)
  11. Oct 12, 2009 #10
    I used the trigonometry, but your way is also right.
  12. Aug 24, 2011 #11
    What is about a Probability density to find particle with amplitude A or -A?

    Near point A and -A.
    Last edited: Aug 24, 2011
  13. Oct 24, 2012 #12
    I tried to solve for the functional form of the probability.
    I set A = w = 1 and ignored 2/T for simplicity.

    So, probability of finding oscillator at position x is proportional to the amount of time it spends at every dx---

    Prob. prop to ...
    int(dt) = int(dx/v)

    v(x) = 1/(1-x^2)^1/2 (*There's two ways to get this--see post #8 and #9*)

    Integration of 1/(1-x^2)^1/2 dx

    yields Arcsin(x), which is negative when x is negative--->doesn't make sense as a probability distribution.

    Any thoughts?
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