Defining velocity distribution

[bob900]

Since a particle does not have a well defined position (before measurement), it also does not have a well defined velocity. But it seems like we can define a velocity distribution.

Suppose we observe a particle at position x=0 at time t=0. If at a later time t we observe it at x, the observed velocity is v=x/t. The probability of observing it at x at time t (and hence observing velocity v), is P(v,t) = |ψ(x,t)|^{2}.

The expected value of v, will then be <v> = ∫v |ψ(x,t)|^{2}dx = ∫(x/t) |ψ(x,t)|^{2}dx. Since momentum p = mv, <p> = m<v> = m ∫(x/t) |ψ(x,t)|^{2}dx.

Is this correct? Is this consistent with the known result that <p> = m d<x>/dt ?

 

[VortexLattice]

First of all, in your first sentence, I'm not sure if that's true. A particle can have no well defined position, but a well defined velocity.

Also, if you're talking about a free particle, that's not how it works, I believe... You have a particle at x = 0 at t = 0. There, its position is well defined and its probability distribution is a sharply peaked gaussian. Then, as time goes on, the gaussian widens, because you can find the particle at more places. But if you measure it, it can have done many things in between t = 0 and when you measured it. It may have gone to China and back, I dunno. So then the simple v = x/t doesn't quite hold.

 

[bob900]

First of all, in your first sentence, I'm not sure if that's true. A particle can have no well defined position, but a well defined velocity.

I was just quoting Griffiths, Introduction to QM 1ed, page 15, talking about a free particle :

"If the particle doesn't have a determinate position (prior to measurement), neither does it have well-defined velocity. All we could reasonably ask for is the probability of getting a particular value".

Also, if you're talking about a free particle, that's not how it works, I believe... You have a particle at x = 0 at t = 0. There, its position is well defined and its probability distribution is a sharply peaked gaussian. Then, as time goes on, the gaussian widens, because you can find the particle at more places. But if you measure it, it can have done many things in between t = 0 and when you measured it. It may have gone to China and back, I dunno. So then the simple v = x/t doesn't quite hold.

Prior to measurement it doesn't have a well defined position and so I don't see how you can meaningfully talk about it 'being' or having 'gone' anywhere at all, China or elsewhere. Post observation, however, you have a determinate position X and determinate time interval T that elapsed, and unless you redefine what "velocity" means, the (average) velocity = X/T.

 

[Khashishi]

A distribution is a statistical representation. It either represents the probability of a certain particle having a certain velocity, or the statistical breakdown of a bunch of particles. In either case, the measurement problem doesn't come into play.

 

[bob900]

A distribution is a statistical representation. It either represents the probability of a certain particle having a certain velocity

But how does one define a particle's 'velocity', when all one has is a wave function whose squared amplitude represents a probability of observing the particle at a certain position.

 

[tom.stoer]

<v> = ∫v |ψ(x,t)|^{2}dx

That's not correct; you have to use dx/dt instead of x/t; and dx/dt has to be replaced with p/m.

But how does one define a particle's 'velocity', when all one has is a wave function whose squared amplitude represents a probability of observing the particle at a certain position.

Velocity can be defined using the momentum-space wave function derived via Fourier transformation or via position-space wave function; these two approaches are strictle equivalent. Let's use momentum space, i.e.

\psi(p) = \int_{-\infty}^{+\infty}dx\,e^{-ipx}\,\psi(x)

Then let's define the expectation values for momentum

\langle p \rangle = \int_{-\infty}^{+\infty}\frac{dp}{2\pi}\,p\,|\psi(p)|^2

and velocity

\langle v \rangle = \frac{1}{m}\langle p \rangle