Probability distribution of classical momentum

[bouken]

Suppose I solve the Schrodinger wave equation described in terms of position (as opposed to momentum based description), it gives me the wave function from which I can determine the probability distribution function (pdf) for position with a parameter as time.

I view it as the following, the probability distribution for a random Variable X with parameter t. Now what I would like to do is find the probability distribution for the variable m dX/dt - the classical definition of momentum. Could someone guide me as to how to go about this?

 

[bouken]

Maybe I should elaborate on my purpose. I would like to see whether there is any connection at all between the classical notion of momentum and the quantity derived using the momentum operator

 

[jtbell]

You use the momentum-space wave function:

http://en.wikipedia.org/wiki/Wave_function#Momentum-space_wavefunction

 

[LostConjugate]

The expectation value of any quantum operator is the result of a classical measurement.

$$m\frac{d}{dt}<x> = m<\frac{p}{m}>$$

 

[bouken]

@Jitbell

I looked at that, but what I want is the distribution of m dX/dt the classical momentum instead of the quantum mechanical description of it.


@LostConjugate

I am looking for the entire probability distribution? Is there a way to find that?

 

[LostConjugate]

@Jitbell

I looked at that, but what I want is the distribution of m dX/dt the classical momentum instead of the quantum mechanical description of it.


@LostConjugate

I am looking for the entire probability distribution? Is there a way to find that?

The probability distribution is a quantum mechanical expression. I am not sure what your looking for.

 

[bouken]

@LostConjugate

I am looking for the probability dstribution of m dX/dt which is of course a quantum mechanical object.

Edit: though it represents a classical definition of momentum

 

[LostConjugate]

@LostConjugate

I am looking for the probability dstribution of m dX/dt which is of course a quantum mechanical object.

Edit: though it represents a classical definition of momentum

Well the probability distribution of x is just

$${f(x)}^*f(x)dx.$$

The probability distribution for momentum is

$$\frac{{f(p)}^*f(p)}{2\pi\hbar}dp$$

 

[Avodyne]

What you wish to do cannot be done.

Consider measuring the velocity by doing two position measurements at nearby times. That is, we measure the position at time t₁, and get result x₁. Then we measure the position at time t₂ and get result x₂. We infer that the velocity was approximately v(x₂-x₁)/(t₂-t₁).

The probability that the first measurement would yield a result between x₁ and x₁+dx was P(x₁,t₁)dx, where P(x,t)=|ψ(x,t)|².

The problem is, we do not know the probability that the second measurement would yield x₂. This is because the first position measurement completely changed the wave function. We have to know the wave function after the first position measurement (which requires a model of just how the measurement was made, as well as the wave function just before it was made), and then we have to time evolve the modified wave function to get to time t₂.

Speaking more generally, the probability distribution P(x,t) does not have enough information in it. What we really need is the joint probability distribution that tells us the probability of the result of a position measurement at one time, given that a measurement at an earlier time yielded a particular result.

 

[bouken]

What you wish to do cannot be done.

Consider measuring the velocity by doing two position measurements at nearby times. That is, we measure the position at time t₁, and get result x₁. Then we measure the position at time t₂ and get result x₂. We infer that the velocity was approximately v(x₂-x₁)/(t₂-t₁).

The probability that the first measurement would yield a result between x₁ and x₁+dx was P(x₁,t₁)dx, where P(x,t)=|ψ(x,t)|².

The problem is, we do not know the probability that the second measurement would yield x₂. This is because the first position measurement completely changed the wave function. We have to know the wave function after the first position measurement (which requires a model of just how the measurement was made, as well as the wave function just before it was made), and then we have to time evolve the modified wave function to get to time t₂.

Speaking more generally, the probability distribution P(x,t) does not have enough information in it. What we really need is the joint probability distribution that tells us the probability of the result of a position measurement at one time, given that a measurement at an earlier time yielded a particular result.

Avodyne you made my day. I understand and love this answer

Ok, now here is the next question. How do I find this correlation between successive measurements that you are talking about?

 

[bouken]

@ avodyne.. You could just give me pointers to the analysis I seek. Any reference which does a treatment that I am looking for would be fine :)

 

[Avodyne]

How do I find this correlation between successive measurements that you are talking about?

In quantum mechanics, as I explained, the first measurement changes the state, and we have to specify the new wave function to get the probability of the result of a second position measurement. The new wave function will depend on the mathematical model of the first measurement (such as, set the wave function to zero outside some interval, and keep it unchanged, except for overall normalization, inside that interval).