Simple harmonic oscillator- the probability density function

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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?
 
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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?
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!

edit:
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.
 
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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.

Bob.
 
I know that, I was more curious about the function p(x)....
 
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)

and

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?

regards
Frímann
 
...

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.
dx=vdt so P =2 int(dt)/T = 1.
 
dx=vdt so P =2 int(dt)/T = 1.
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?

Regards
Frímann Kjerúlf
 
Maybe Asin(ωt) = A[1-cos2(ωt)]1/2 = A[1-x2/A2]1/2, so you obtain v(x)?
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 :)
Frímann
 
I used the trigonometry, but your way is also right.
 
What is about a Probability density to find particle with amplitude A or -A?

Near point A and -A.
 
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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---

i.e
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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