# Free particle

1. Oct 19, 2004

### broegger

One solution to the time-independent Schrödinger equation for a free particle (moving in 1 dimension) is:

$$\psi(x) = Ae^{ikx}$$

This has a definite momentum p = h-bar*k, but it can't be normalized since:

$$\int_{-\infty}^{\infty}\lvert\psi(x)\rvert^2dx = \int_{-\infty}^{\infty}|A|^2dx = \infty$$

Does this mean that we cannot have a free particle like this?

2. Oct 19, 2004

### Galileo

Yes, it means the free particle cannot be in a stationary state or any finite linear combination of stationary states.

3. Oct 19, 2004

### broegger

How can a free particle exist then??

4. Oct 19, 2004

### ZapperZ

Staff Emeritus
No, you CAN have a free particle "like this". It just means that its allowed position is everywhere. This, if you notice, is a direct manifestation of the uncertainty principle, since you have already determined that a free particle has a definite momentum.

Free particle approximations are used to describe (as a zeroth'th order approximation) the conduction electrons in conventional metals. This results in the semiclassical Drude model. So yes, they can "exist".

Zz.

5. Oct 19, 2004

### vanesch

Staff Emeritus
Well, it is mathematical nitpicking. Because in order to "exist" it has to be, say, in our visible universe. So although it can be "almost anywhere", its spatial spread is not infinite. So its momentum spread is not 0. But so tiny that for all practical purposes we'll have a function that has the form h(x) exp(-ikx) with h(x) such a slowly turning off function for x-> +/- infinity, that we can forget about it.

This is the same as saying that you have an electrical signal which is a sine wave. You can't have such a signal, because you have to turn on the oscillator at a certain moment, and you'll have to switch it off a certain day. But if the oscillator is already running for half an hour, for all practical purposes you can say it is a sine function.

cheers,l
Patrick.

6. Oct 19, 2004

### broegger

But how do you go about the problem of normalization? I can't see how a valid wave function can yield an infinite probability of finding the particle somewhere in the universe...

7. Oct 19, 2004

### ZapperZ

Staff Emeritus
It isn't normalizable, at least not if you assume infinite space!

Note that it doesn't have infinite probability. The infinity comes in because your space is unbounded. This is seldom true, but only an approximation if one ignore the edges and boundaries of the system. That is why I said this is valid as the zeroth order approximation!

Zz.

8. Oct 19, 2004

### broegger

But then it is localized and the momentum spread can't be zero according to the uncertainty principle? I don't understand this...

Also, I have read that a free particle is represented as a wave packet, which is an infinite linear combination of the functions I mentioned.

9. Oct 19, 2004

### ZapperZ

Staff Emeritus
Let's tackle ONE thing at a time and not open all the can of worms that you have.

If you have a very localized particle, then PREDICTING what its momentum going to be will have a large uncertainty. So if you prepare 1000 system, each with a particle that has a very small Delta(x), then if you make a momentum measurement of all 1000 systems, you will obtain a large statistical spread in the value of the momentum that you measure. This means that if you have such a system, predicting what it's momentum is will be very difficult.

Not that it doesn't mean you cannot measure it's momentum. It DOES have a definite momentum, depending on how well you measure it using your instrument. The uncertainty principle does NOT say that you can't measure both position and momentum simultaneously. This is a common error most people make in trying to understand this principle. All it says that, unlike classical mechanics, having an identical system and knowing the same value of a position, does not necessarily mean you can make a reliable prediction of its corresponding momentum each time.

I have previously, in a couple of other threads, given the example of the diffraction from a single slit as the clearest example of the manifestation of the uncertainty principle. I have a feeling most people either didn't understand it, or didn't see the relevance.

As for wave packet, this is a whole different matter. A "wave packet" is a superposition of a number of waves. You can take many plane-waves (which are your A*exp(ikx)) with different wavelengths and add them to make these wave-packets. Whether these corresponds to a "particle" depends on the nature of the situation. You cannot simply assume that a "wave-packet" = "particle". This would be wrong.

Zz.

10. Oct 19, 2004

### vanesch

Staff Emeritus
This sounds very hidden variable, no ?

cheers,
Patrick.

11. Oct 19, 2004

### ZapperZ

Staff Emeritus
No, it is more mundane than that. There's nothing to prevent you from making as an accurate of a measurement of the position and momentum simultaneously. If I have a very small slit along the x axis, and light passes through it, I know that the uncertainty of the x position of a photon is the width of the slit. Now, I can put a photodetector behind the slit, such as a CCD, and depending on the number of pixel on the CCD, I can accurately detect where the photon hits. The accuracy of this position depends entirely on my instrument. This image position then translates directly into the x-momentum. This means that my single-shot measurement of p_x does NOT depend on the size of the slit (Delta(x)) - it depends entirely on the accuracy of my instruments.

However, if I have the same exact setup, and I send another photon through the same slit, where it hits my detector behind the slit now depends on the width of the slit. If the width is small enough that it is comparable to the wavelength of the photon, then if I do this experiment many times, the spread in the statistics of all the momentum values that I measure will be HUGE. The smaller I make the slit (i.e. the smaller Delta(x) is), the larger the spread. THIS, is the "uncertainty" that is meant in the HUP, and NOT the uncertainty in a single-shot measurement.

Zz.

12. Oct 19, 2004

### vanesch

Staff Emeritus
Ok, but that's not what is usually called "simultaneously". It is an x-measurement, followed by a p_x measurement. While the x-measurement measured the x of the original state, the p_x measurement only measures the p_x of the x-projection of the original state (and if the slit is very narrow, there is no correlation left with the original state).

cheers,
Patrick.

13. Oct 19, 2004

### ZapperZ

Staff Emeritus
You're correct. I shouldn't have used the word "simultaneously". What I meant was that the very same photon of which we had restricted its position can also be measured of its momentum up to arbitrary precision, limited only by our instrument. The precision of this measurement isn't what is referred to in the HUP.

However, my reply to you was to make sure we're NOT talking about "hidden variables" here.

Zz.

14. Oct 19, 2004

### vanesch

Staff Emeritus
Ok, thanks. I had the impression that you meant somehow that there WAS a given momentum and a given position, but when we tried to measure it, we got out a statistically spread result. Now I *know* you don't believe in hidden variables (nor do I), so I was a bit surprised

cheers,
Patrick.

15. Oct 19, 2004

### ZapperZ

Staff Emeritus
Actually, I take that back....

While in the example I gave it appears that you do make a position measurement and THEN a momentum measurement, this isn't so.

Unless the photon that left the slit scattered off another photon, the x-component of its momentum directly corresponds to the x-component of its momentum that it had right at the slit! There's no mechanism for it to change its trajectory from the slit to the detector. So it's momentum is what it had at the slit. I only made the measurement LATER in time, but this info corresponds to what it had earlier.

I can have the same slit, but fill the slit with particles. A photon comes in, and we already know it is now restricted in its x-position. If it scatters off one of the particles, via analyzing the momentum of that particle and the scattered photon, I can immediately tell you its x-component of the photon momentum while it was in the slit, even though I measure the scattered photon and the particle later in time. This info tells me what was going on right at the slit when the scattering occured.

Zz.

16. Oct 19, 2004

### vanesch

Staff Emeritus
You got me thinking about that one. However, I don't give up
It exchanged momentum with the slit until it was a few wavelengths away from it, so the momentum you measured only had a meaning after the photon got through the slit and travelled through the "near-field" zone. I didn't do the calculation, but I'm pretty sure that this "near-field" zone has the right spatial extension to avoid you saying that when the photon was at the slit (well-known position) it had momentum px.

cheers,
Patrick.

17. Oct 19, 2004

### ZapperZ

Staff Emeritus
But where I put the screen or detector is arbitrary. I can put it as close to the slit as I want. I put it further from the slit simply because I want to be able to get a measurable divergence of the photon beam path to make a good measurement of its momentum. But there's nothing that says I can't make it as close as I want to.

Furthermore, the particles that I had scattered off the photons are directly in the slit. They are sampling exactly what the transverse momentum of the photon is. I also don't think the walls of the slit comes into play. It is there only to restrict the x-position of the photons that are allowed to pass through.

Zz.

18. Oct 19, 2004

### reilly

At the risk of being simple minded, I'll say that plane waves can indeed be normalized in infinite space-- formally the normalization is taken to be a delta function. Virtually all work in QFT uses the delta function normalization, in fact, it's used in virtually all scattering theory. With the advent of generalized functions (see Lighthill on Generalized Functions and Fourier Integrals, I think that's the right title) and Distribution Theory. delta funtions, derivatives of delta function, and so on can be made as rigorous as anybody wants. Normalization of eigenstates of continuous eigenvalues was a big deal in the 1930s, but not really anymore.

19. Oct 19, 2004

### vanesch

Staff Emeritus
But there is! You need to be in the far-field region in order to make sure that the photon is locally describable by a plane wave (so that the position measurement IS a momentum measurement). If you are in the near-field region, the interaction between the photon and the slit (or the particles) where they exchange x-momentum is not over yet, so it is still "bending" (the lines perpendicular to the wavefronts are not straight lines, if you apply geometrical optics).

Yes, but the interaction isn't over as long as you aren't a few wavelengths away: momentum can still be exchanged.

cheers,
Patrick.

20. Oct 20, 2004

### seratend

I will try to help zapper as I think its way of thinking is interesting :tongue2: .
So, now think that the photon or the particle, before the slit is in a state such that the spatial extension is null outside the slit. Therefore, we can say that the photon or particle who passes the slit does not interact with it and we still may apply the zapper measurements (we have a free wave packet with detla_p and a delta_x).

Seratend.