# What is velocity ,momentum.

1. May 9, 2007

### guapig

As we know that bohr said we could not know anything without detecting.To talk about anything of a particle without measuring is meaningless,and we don't know what exactly happend during the process.

Someone think maybe it is just there ,you just don't know.But bohr said it is wrong.

And now I am thinking about something that confuse me.Considering bohr is right,according to heisenberg's uncertainty principle,we can't konw exactly the position and the momentum of a particle at the same time,and the process of motion is unkonw (or meaningless or something).then what is velocity (speed) at all.Without the defination of velocity (speed),what is momentum.

Thanks.
GUAPIG

2. May 9, 2007

3. May 9, 2007

### guapig

THanks Demystifier

I have checked the websites you gave,and I found that (maybe) it is still an unknow question,people are working on it.Right?

4. May 10, 2007

### Demystifier

Right! ....

5. May 10, 2007

### Fra

I'm not sure this is readable, but here are some non-mathematical elaborations. (I tried the forum-tex for the first time here, but notice it doesn't preview properly, and I found no test section either. so Idropped it - test $$\frac{testA}{testB}$$)

There are different philosophies to this, but I see no reason to think Bohr was relatively more wrong than anyone else at the time. That said QM still has some unclear logic, which I think even those who disagree on philosophy agree upon.

One way of seeing it is that in reality we can measure position exactly, for various reasons. Thus the more general starting point is to consider a probability distribution p(x), where p(x)dx is the probability to find the electron at [x,x+dx], given no further information, as infered from the past data flow.

Then one can introduce the QM wavefunction as a complex probability amplitude, whose square equals p(x).

At this initial point, the phase is arbitrary choice(*)

But then we take the fourier transform of this amplitude, which induces new variables, a complex field, which is the probability distribution of momentum and to each momentum-probability-field point there is an associated a complex phase.

_Relative to the original choice_(*) this phase is now determined.

However the observer will notice that in the general case as things change, the observed probability distribution is not constant. It changes. But in the tranformed picture, the momentum distribution (leaving out the complex phase) may be more predictable than is the original distribution (at least in the case of a freely moving particle), and we have constrained some information to the complex phase. This is why the tranformation is nice.

Thus in QM, momentum is defined as a kind of distribution in one of the event spaces induced by the fourier transform. So we can a connection between two eventspaces. Some information is still in unresolved in the phase. We simple find another event space, connected to the first one, where the uncertainty is lower.

Similiarly one can define energy relative to time, this further information is pulled out of the previous phase.

But that's of course assuming the concept of "time" makes sense to start with. In ordinary QM, space and time are considered as a background setting and is not questioned to it's fundaments. This is IMO unacceptable in the general case (GR and QG) but it's how the story goes in ordinary QM.

Anyway, this implies that since momentum is a propery of the position-probability distribution, changes is implicit in the construction, but ignored in the phase. So I would say you do not need a "clock" or "time-units" as such to define momentum. But you need change! So you need change, but you don't need to exactly "keep track" of the change with a clock.

You clock is in the phase at this point. But of course if you add a clock, you can pull energy distribution out that phase too.

But I consider this semiclassical reasoning, because space and time are not obvious to start with. But then again, ordinary QM is kind of semi-classical in a certain sense.

I think in order to appreciate this better, you need at least basic familiarity with fourier transforms, and the basic QM operator formalism. I doubt I'd make sense of my own plain words unless I had written it myself )

/Fredrik

6. May 11, 2007

### guapig

THANK U Fra .

now I have got some idea of what you said.It is really very wonderful.