Zero points of particle wave function?

[tuomas22]

Hi. I'm trying to study the very basics of quantum physics and I ran into a problem.

Does a free particle which is at zero potential wavefunction have some points where it's zero? So the probability of finding it would be zero? I know there is if the region has boundaries like in infinite square well example, but if there isn't any boundaries?
Does the uncertainty principle say something about this (because the momentum is definite??) and I just don't see it?

 

[victorphy]

A free particle is in a momentum eigenstate,i.e. it has a definite momentum which doesn't change with time.According to uncertainty principle, the uncertainty of position must be infinite.That is you can find the particle everywhere with the same probability.

 

[Bob_for_short]

For a free particle you have a plane wave which is complex. Its absolute value is constant, nowhere zero.

 

[tuomas22]

Thanks for answers!

The wavefunction is this?
$$\Psi(x,t) = Ae^i^(^k^x^-^\omega^t^)$$
and with my math the absolute value of this is $$|A|$$
which is constant. How is this constant determined then?

 

[Bob_for_short]

If you know that your particle is for sure in some volume V, then A=V^-1/2. It is called also a normalization factor. The probability to find the particle is some sub-volume δV is equal to δV/V = |ψ|²δV.

 

[sokrates]

But of course "knowing that your particle is for sure in some volume V" invalidates the "free particle" assumption, and the "pure" momentum eigenstate postulate, because that knowledge means that the particle is not completely "delocalized".

So OP is right, in the case of a truly free particle the integral lies between -inf and +inf and A cannot be determined.

I have yet to see a satisfactory workaround for this difficulty, but apparently replacing the infinities with a very large volume solves the problem for most practical purposes...

 

[nnnm4]

There is no difficulty in the position representation of the momentum eigenstate being not normalizable. It just means that it's a mathematical construct that can be used to make physical wavefunctions.

 

[jtbell]

in the case of a truly free particle the integral lies between -inf and +inf and A cannot be determined.

In order to get a normalizable free-particle wave function, you have to construct a wave packet: an integral superposition of plane-wave functions. Its amplitude approaches zero for points far enough from the center of the packet. Such packets invariably must include a range of momenta $\Delta p$ in inverse proportion to the packet's spatial size $\Delta x$, which leads directly to the Heisenberg Uncertainty Principle.

These packets are just as much "truly free" as the individual plane-wave solutions.

 

[zonde]

We have particle source. After activating source if particle have definite velocity we definitely know the volume where it is. If we say that say that volume is less definite than we think than we invalidate statement about definite velocity.
So "truly free particle" is emitted infinite time ago.
Maybe we can look at possibility that uncertainty of position is about direction only?

 

[Bob_for_short]

I have to remind you that the wave function describes an ensemble of measurements whereas you think in terms of one particle.

To understand the difference consider an interference pattern. It cannot be obtained in one-particle measurement (one point). Only a sufficiently big number of measurements reveals the true quantum state. The same is valid in a free particle case.

 

[zonde]

Isn't it so that wave function describes distribution of measurements in case of ensemble and probability in case of one measurement?

To me it looks like definite momentum already imply ensemble. So I don't quite see how this changes what I have said?

 

[Bob_for_short]

We have particle source. After activating source if particle have definite velocity we definitely know the volume where it is. ... Maybe we can look at possibility that uncertainty of position is about direction only?

It is clear that you speak of one particle, classical in addition. The particle source does not emit one particle but many. We do not know when each of them is emited. This creates a huge uncertainty in position measurements. If the source is far enough and the particle energy is constant (constant ω, for example), then the wave vector (momentum) is quite definite but not the position. The plane wave solution describes an ensemble of measurements in this situation.

 

[Bob_for_short]

But of course "knowing that your particle is for sure in some volume V" invalidates the "free particle" assumption, and the "pure" momentum eigenstate postulate, because that knowledge means that the particle is not completely "delocalized".

If the volume V is sufficiently large, the position uncertainty is rather small. For example, if your volume is larger than and includes the interference picture you study, then there is no effect of the volume size. All you have is a uniform particle flux j which is sufficient to calculate/measure the cross sections, interference pattern, etc.

 

[zonde]

If the source is far enough and the particle energy is constant (constant ω, for example), then the wave vector (momentum) is quite definite but not the position.

If you say that momentum is quite definite and position is quite uncertain then of course there are no objections.
I was more arguing against infinities like exact momentum and infinite uncertainty in position.

 

[sokrates]

In order to get a normalizable free-particle wave function, you have to construct a wave packet: an integral superposition of plane-wave functions. Its amplitude approaches zero for points far enough from the center of the packet. Such packets invariably must include a range of momenta $\Delta p$ in inverse proportion to the packet's spatial size $\Delta x$, which leads directly to the Heisenberg Uncertainty Principle.

These packets are just as much "truly free" as the individual plane-wave solutions.

I see your point. What I was saying then, is not really related to the particle being free or not. Thanks for the correction. Then I'd conclude that the issue of not being normalizable makes sense only when we talk about a "pure" momentum eigenstate ( a plane wave solution) - and in that case position is not really defined (due to uncertainty) and this poses no problems at all..
Is this view more accurate?