Can Physical Motion be Transformed into a Probability Distribution?

[Crosson]

I am interested in the following topic:

I would like to transform a physical motion X(t) into a probability distribution P(X). Specifically, if I know the position of a particle at all times I want to know the probability that I will find it in a specific location (integrate P(x) over a small volume to get probability ).

I am thinking of macroscopic systems for which we can write down a Hamiltonian. Of course, in my fantasy, the probability transform will take the form of an operator which is applied to the dynamic equation for X(t) to systematically produce a static diff eq for P(x).

I have tried to define various limiting processes that would intuitively construct P(x), but none of them have led to results which I can inteperet as probability.

It seems to me that this sort of analysis has obvious applications in Chaos theory, and potential applications towards moving beyond Quantum Mechanics.

My question is: Where has this already been done? If it hasn't been done, I presume it can't be done.

 

[HallsofIvy]

If you KNOW the position of the particle at all times, probability has nothing to do with it!

 

[Crosson]

Perhaps you are totally right, but perhaps you are misunderstanding me.

Suppose you have a classical pendulum in a sealed box. Every couple of seconds a camera takes a snapshot of the pendulum and its postion (like a poincare map). At the end of the experiment I look at all the photographs and say something like:

"Okay, it was in the middle 2 times, just to the left or right 3 times, and to the extreme left or right once".

So I build a graph, N(x) where for each position x I record the number of snapshots at that position. If I divide N(x) by N, the total number of snapshots, I get P(x), the probability of observing the pendulum at point x.

The question is, if I know the equation of motion for the harmonic oscillator, how do I predict the probability distribution that I would observe?

 

[Berislav]

Interesting. This ties to what I suggested in my thread.
For the simple (non-damped) pendulum case, a probability distribution makes sense only if we do not know the initial conditions; then we could talk about the probability associated with the first measurement, in which case every position would be equally probable. After that the harmonic equation would tell us with 100% probability where we would find the pendulum when we measure it's position after a certain time interval.

If one knows the equation of motion it would be redundant to ask for a probability distribution.

But, I think that I do see where you are going with quantum mechanical case. It seems to me that you think if the probability distribution and the equations of position are on one-to-one corespodence then we could find the equation describing the motion of a quantum state from it's probability distribution.

This is what I suggested (in a way), but sadly that is not the case.

Consider a simple quantum mechanical system (e.g, an electron in a infinite potential well.) The probability distribution through space is defined as the square of the wave equation. If there were some equations describing the motion of the electron in the well (even extremely chaotic ones with high deegres of randomness), when we make the first measurement we get the position of the electron, then in the next measurement the probability of finding the particle in a certain volume would depended on our first measurement (i.e. the more closer the space to the first observed position of particle the more it is more probable to observe the particle in that volume). This would affect the wave equation and therefore can not be.

 

[PBRMEASAP]

The question is, if I know the equation of motion for the harmonic oscillator, how do I predict the probability distribution that I would observe?

The amount of time the mass spends in a small space dx is given by dt = dx/v(x), where v(x) is the velocity expressed as a function of position. That would be proportional to the probability of finding it there.

 

[Crosson]

It seems to me that you think if the probability distribution and the equations of position are on one-to-one corespodence

This is almost certainly not the case.

then we could find the equation describing the motion of a quantum state from it's probability distribution.

Sort of, but I know enough QM that it won't be so simple. I would like to think more generally: the differential equation of motion is transformed into a differential equation of probability. So we only need to show that our new laws of motion lead to the schrodinger equation when applicable. Again, it won't be that simple for a page long list of reasons.

If one knows the equation of motion it would be redundant to ask for a probability distribution.

Wrong. What about in cases where the probability distribution is simpler to solve than the original motion? I am talking about chaos here. If we had a systematic way to reduce non-linear dynamical equations to dynamic probability equations, that would be "order in chaos".

The amount of time the mass spends in a small space dx is given by dt = dx/v(x), where v(x) is the velocity expressed as a function of position. That would be proportional to the probability of finding it there.

This is along the line I was thinking, but of course dx and dt tend toward zero and you are telling me that the probability is propotional to zero. The trick is to ratio this with another vanishingly small quantity, so that the ratio is finite. I tried measuring the time it takes to get from one point to another relative to "standard time" like this:

$$P(x) = lim_{\tau -> 0 } \frac {t(x + v(x)\tau) - t(x)}{\tau}$$

But this always gives me a result of 1

 

[PBRMEASAP]

This is along the line I was thinking, but of course dx and dt tend toward zero and you are telling me that the probability is propotional to zero. The trick is to ratio this with another vanishingly small quantity, so that the ratio is finite. I tried measuring the time it takes to get from one point to another relative to "standard time" like this:

$$P(x) = lim_{\tau -> 0 } \frac {t(x + v(x)\tau) - t(x)}{\tau}$$

But this always gives me a result of 1

It gives you a continuous probability distribution, just like the square of a wavefunction of a particle gives you the probability distribution in position (or momentum). You are correct that the probability of finding the mass is located at a specific point is zero, but the distribution will tell you the probability that it is in a certain interval.

edit: by the way, it looks like you were taking the derivative of time with respect to time, so 1 is a very reasonable answer.

 

[Crosson]

t looks like you were taking the derivative of time with respect to time, so 1 is a very reasonable answer.

Look again. This looks most similar to the derivative of time with respect to distance, but the term v(x) changes things (if v(x)*tau was just tau, then this would be the derivative of time wrt position).

You are correct that the probability of finding the mass is located at a specific point is zero, but the distribution will tell you the probability that it is in a certain interval.

This is correct, P(x) is not the probability of finding it being at point x. It is much more sensible to say that:

$$\int_{Vol} P(x) dV$$

Is the probability of being in a small volume V.

None of the responses so far have indicated they have seen this elsewhere, which is surprising.

 

[PBRMEASAP]

None of the responses so far have indicated they have seen this elsewhere, which is surprising.

"Introduction to Quantum Mechanics", by Richard Liboff. He does this near the beginning of the book. It is a very neat result.

 

[reilly]

Try a delta function, essentially an extremely narrow Gaussian distribution. As in if a particle obeys x=f(t), then P(x,t)= delta(x-f(t)) -- or a Gaussian with the mean of
x=f(t) and the standard deviation say = 1/100000000000.
Regards,
Reilly Atkinson

 

[Crosson]

x=f(t), then P(x,t)= delta(x-f(t))...standard deviation say = 1/100000000000.

This is not a proper use of the delta function. As far as you say, the argument of the delta function is 0 for all x and t. But the value of the function is undefined at this point. Undefined in a sense that 1000000000... will never be enough.

"Introduction to Quantum Mechanics", by Richard Liboff. He does this near the beginning of the book. It is a very neat result.

Does richard addresses the problem of converting a dynamic equation of motion into a dynamic probability distribution (if so how)?

Or are you being facetious? I am aware that I am borrowing notation from simple QM, but the origin of P(x) is completely different than (and much more physically grounded than) that of the wave function.

 

[reilly]

Crosson -- What I suggest is commonly done. Justification can be found in Lighthill's classic book, Generalized Funtions and Fourier Analysis. If you don't like the delta function, use, as I suggest, a very narrow Gaussian. (Among the formal definitions of a delta function, is the idea of the delta as a limit of increasingly narrow Gaussians.) One place where this delta funtion approach is used is in describing charges, moving or otherwise, and currents in classic E&M, so that Poisson's eq. can be used for point charges -- you need a current density, so that magnetic fields can be properly described. And, people have been doing this sort of thing for a long, without controversy, at least in the trade.

Regards,
Reilly Atkinson

 

[PBRMEASAP]

This is not a proper use of the delta function. As far as you say, the argument of the delta function is 0 for all x and t. But the value of the function is undefined at this point. Undefined in a sense that 1000000000... will never be enough.



Does richard addresses the problem of converting a dynamic equation of motion into a dynamic probability distribution (if so how)?

Or are you being facetious? I am aware that I am borrowing notation from simple QM, but the origin of P(x) is completely different than (and much more physically grounded than) that of the wave function.

Man you just can't be helped can you? Reilly and I both gave you honest, reasonable answers. In both cases, your immediate reaction was "you're wrong!" or "I already thought of that and it doesn't work". And then you accuse me of misrepresenting Liboff's fine book. You deserve to remain ignorant.

 

[Crosson]

Man you just can't be helped can you? Reilly and I both gave you honest, reasonable answers.

I appreciate your responses very much.

your immediate reaction was "you're wrong!"

My problem with Reilly's post was just misunderstanding the notation (after stating x = f(t), he switched over to a different meaning of x), I am really sorry I came out swinging like I did. Everything I have done in EM with delta functions has involved integration, and the sifting property (the delta does not appear in the final equation).

And then you accuse me of misrepresenting Liboff's fine book.

I was skeptical because I have never read a QM book that transforms an ODE of motion into an ODE for probability. Can you describe at all how this is done?

 

[Berislav]

PBRMEASAP, correct me if I'm wrong, but is this what you are basically saying:

$$P(\Delta x)= C\int_{x_1}^{x_2} \frac{1}{v} dx$$
where C is some constant, which can be found from the fact that,

$$C\int_{x_{min}}^{x_{max}} \frac{1}{v} dx=1$$
where for the pendulum case it would be 1/T.

 

[PBRMEASAP]

I appreciate your responses very much.



My problem with Reilly's post was just misunderstanding the notation (after stating x = f(t), he switched over to a different meaning of x), I am really sorry I came out swinging like I did. Everything I have done in EM with delta functions has involved integration, and the sifting property (the delta does not appear in the final equation).



I was skeptical because I have never read a QM book that transforms an ODE of motion into an ODE for probability. Can you describe at all how this is done?

Okay then. Forget what I said in my last post.

The equation of motion is not directly transformed into a probability or ODE for probability. I'm not sure how that would be done. Following Liboff (I think it is around page 200 or so), you use the total energy of the harmonic oscillator

$$E = \frac{1}{2}m \omega^2 x^2 + \frac{1}{2}m v^2$$

to write v = v(x), velocity as a function of position. Since we are assuming the EOM are known exactly, we know the total energy. Then the probability distribution function for the position p(x) is proportional to 1/v(x). You can normalize it by requiring that the probability of finding the mass in the interval [-A, A] is 1, where A is the amplitude of the motion. Again, we know A because we have the EOM.

It comes out to something like $$p(x) = \frac{const.}{\sqrt(A^2-x^2)}$$

Liboff then shows that in the limit of high energy/quantum number, the local average of the quantum mechanical probability distribution (in terms of Gaussians and Hermite polynomials) converges to this classical distribution. I think that is very hip. The actual wavefunction^2 distribution oscillates very rapidly, so local average just means taking the average over a small interval about each point. Anyway, this shows the correspondence principle between quantum energy eigenstates and a classical "energy state".



Berislav: yes, that is what I meant.

 

[Berislav]

The most general treatment of the subject I could find on the net is this article on Wikipedia about Liouville's theorem (scroll down to the part labeled Mathematical expression):

Liouville's theorem (Hamiltonian)

 

[Crosson]

Thanks Berislav, this is what I was looking for. I have not done very much statistical mechanics, but now I see that I have some interest in it.