# 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.

21. Oct 20, 2004

### humanino

I very much agree. Of course it is not physical, but for all practical purposes it works fine. If $$\Psi_{\vec{p}}(\vec{x},t)=N_{\vec{p}} \,e^{\frac{\imath}{\hbar}(\vec{p}\cdot\vec{x}-E_{\vec{p}}\,t)}$$ and we impose $$\langle\Psi_{\vec{p}},\Psi_{\vec{q}}\rangle=\delta(\vec{p}-\vec{q})$$
then we find : $$\langle\Psi_{\vec{p}},\Psi_{\vec{q}}\rangle=N^*_{\vec{p}}N_{\vec{q}}\int_{-\infty}^{\infty}dxe^{\frac{\imath}{\hbar}(\vec{p}-\vec{q})\vec{x}} = N^*_{\vec{p}}N_{\vec{q}}2\pi \delta(\frac{\vec{p}-\vec{q}}{\hbar})=|N_{\vec{p}}|^22\pi\hbar\delta(\vec{p}-\vec{q})$$ whence $$\Psi_{\vec{p}}(\vec{x},t)=\frac{1}{\sqrt{2\pi\hbar}} e^{\frac{\imath}{\hbar}(\vec{p}\cdot\vec{x}-E_{\vec{p}}\,t)}$$

Last edited: Oct 20, 2004
22. Oct 20, 2004

### vanesch

Staff Emeritus
I agree with that, and in fact it made me worry , in that it would be strange that we can assign, after the fact, as well a precise momentum and a precise position to a quantum particle. I tried to argue that the measurement of the momentum necessarily has to correspond to _another_ quantum state than the measurement of position corresponded to (one that came later in time).

Ok, let's analyze what you propose. You propose that we prepare a state in such a way that it will be essentially confined to a small position range, say, by setting up a converging beam or something like it (but without using a slit), so that there cannot be any momentum transfer at that point, and that a later momentum measurement has to make us conclude that the particle already had that precise momentum when it went through the bottleneck in the wave, and so that at that precise position, it had also a precise momentum. Interesting

I guess a way to prepare such a state would be to have an ellipsoid mirror, where you have a point-like source of particles in one focal point, and we consider the bottleneck at the other focal point, and measure the momentum far away. So did the particle have a well-defined position (the second focal point) and a well-defined momentum (the one that is measured) in the transverse (to the focal axis) direction ?

I wonder if this doesn't displace the problem towards the source. Imagine we make, at the first focal point, a slit so small that it is a fraction of a wavelength of the radiation. Then this slit works essentially as a dipole radiator (see Jackson, chapter 3) and I think that the size of the slit is "forgotten" by the outgoing wave. So I wonder if you really can make a monochromatic beam focalise much below a few wavelengths (in order to be sure that the particle was in that narrow area), without introducing boundary conditions that impose interaction (such as with a conducting slit).
But I agree you're pushing me in dark waters

cheers,
Patrick.

23. Oct 20, 2004

### ZapperZ

Staff Emeritus
I think I may see the problem here...

Let me clarify that by imposing the slit, you are automatically ALREADY making a position measurement. I think there is some confusion that the technique I described can somehow make it irrelevant which one is measured first or last.

We can look at it a different way... Confine a particle in an infinite square well. Make the width of the well smaller and smaller. THEN measure the momentum of that particle. I claim that this is no different than the photon passing through a single slit. The momentum that it has due to the confinment of the slit is what we measure. I suppose I was over-extending the definition of "simultaneous" measurement because it has now become the issue of how quickly the "collapse" occur as soon as you impose a measurement. My over-extended simultaneous idea simply reflects the effect that as soon as you impose a position constraint, then "immediately" the momentum properties assumes its QM behavior.

In ANY case, I hope it is very clear that I am not invoking any hidden variables here. That's a nasty can of worms to be opening! :)

Zz.

24. Oct 20, 2004

### vanesch

Staff Emeritus
Yes, but now it is open, and that's your fault :tongue2:

The problem I'm having now, is inspired by what you wrote, and now I'm confused myself. Imagine the scenario I already discribed a bit:
We have an almost point-like source (of the size of a few wavelengths) of monochromatic light (or electrons, or neutrons, doesn't matter), in the focal point F1 of a huge (say 0.1 lightyear across), perfect, half-elliptical mirror. The waves going out from this pointlike source and hitting the mirror will converge in a very narrow region around F2. Now, as I tried to weasel out in my previous post, this "pointlike" convergence will spread over a volume a few wavelengths across at least. As there is nothing to interact with in the neighbourhood of this focal point, we can assume that whatever momentum is measured later, it must also be the momentum the particle had right at that point.
But now we put a detector one lightyear away in a specific direction from this focal point. It will of course take a year, but when we get a click in that detector, we know the direction of the particle with mindboggling accuracy. So even if we knew that the particle was confined only in a zone a few wavelengths across (and hence delta-x is a few wavelengths) we know the angle with such a precision (and hence, because we know the wavelength) so that we know p with a delta-p which is as small as we like. So we can track back that the particle, when it was in the region of the focal point, had a delta-x and had a delta-px which didn't satisfy Heisenberg. :surprised:

But aaaahhhhh

I think I see where the problem comes from. In order to be able to calculate px from the angle, we need to know the wavelength with an amazing precision. This means that we have to consider a signal with a very narrow bandwidth, and as such, a LONG WAVETRAIN. So we cannot say that the particle was at focal point F2, it was on LINE going through focal point F2, but at any instance, we didn't know where it was on that line (so it had a quite large delta-x, due to the projection of the wavetrain length on the x-direction). If we wanted to make sure it was near F2, we'd have to chop the signal in time, but that would introduce a spread in frequency in such a way, that we wouldn't be able to deduce p_x from the angle alone.

cheers,
Patrick.

25. Oct 20, 2004

### seratend

I agree too. Your ellipsoid mirror is a must have! I don't even think in such an apparatus. (I thought you were not very good with quantum optics :tongue:).
In fact, I think we accept the complementarity’s of position, momentum. So, we really need to explain the position-momentum measurement of zapper like a time-frequency mesurement of an electrical signal with the slit and a position measurement apparatuses. We can also view the position measurement apparatus, the detector, as another slit apparatus .

So here is my small contribution (up to you to say yes or no):

The second measurement is a position measurement that selects ("filter") only a spatial part of the wave state after the first slit (the preparation).
Thus like any filter, we can say, yes, we have after the second slit/detector (spatial "after", not time "after" because time "after" is a consequence of the setup), detected with a given uncertainty the "p_x frequency" of the incoming wave (after the first slit) that depends mainly on the geometry of the detector.

This p_x value (the click of the detector) may have a very small uncertainty, orders of magnitude lower than the momentum extension of the state of the particle after the slit: it depends only on the type of detector/slit we have.

But, now notice that this measurement does not say that the quantum state after the slit is a particle with a defined position and a defined momentum. It is like saying that an electrical signal is a point (a dirac pulse) in time and it has a single frequency).

Instead, we may build a momentum filter that may select the p_x part of the particle state.
See picture below (if it works).

Text in the picture:
the State |Ψ1> of the particle after the 1st slit
|Ψ1>= Σ_x f(x)|x>
= Σ_xk f(x)exp(-ikx)|k>

where f(x) is the spatial extension of the state.

Detection of the spatial location of particle in state |Ψ1> by the second slit (or a detector) given by the State |Ψ2> of the particle after the 2nd slit.
|Ψ2>= <xo|Ψ1> |xo> (projection)
where |xo> has a small spatial extension (dimension of the detector).

|Ψ2>= <xo|Ψ1> |xo>
= Σ_x f(x)<xo|x> |xo>
= Σ_x_ko [f(x)<xo|x> <ko|xo>]|ko>
= Σ_x_k_ko [f(x)exp(-ikx)<xo|k> <ko|xo>]|ko>
= Σ_k_ko [F(k) <xo|k> <ko|xo>]|ko>
= Σ_ko [Σ_k[F(k) <xo|k>] <ko|xo>|ko>

[Σ_kF(k) <xo|k>] is the part of the |Ψ1> selected by the second detector/slit. And <xo|k> (or <xo|ko>) is the momentum extension of the 2nd detector.

We thus see, that in order to get a momentum p_x with a low uncertainty the spatial size of the detector should be very large ([<xo|r>].[<xo|k>] > ~hbar) comparatively to the | Ψ1> (F(k) ~ constant when <xo|k> <>0). We are in a classic approximation (the particle is a point). There are no surprises. (Fourier bless the QM! )

Therefore, at the other side, if we say that the localisation of the detector is very small, we collect all the momentum parts of the particle quantum state | Ψ1>. Thus, when we interpret the location of the slit/detector as giving the momentum of the particle respectively to the first slit, we only measures the “group momentum” of the particle (may be the mean value, I’m not sure). There are no surprises: we have a position with a “group momentum” all defined with the precision we want.

Seratend

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