# Localization vs De Broglie wavelenght

1. Jul 15, 2006

### coccoinomane

Localization vs De Broglie wavelength

First of all, I would like to say hello to everybody since this is my first post, even if it's been some time since I read Physics Forums.

Second, sorry for my bad English, I'm Italian :)

Third, the issue:

I read from my teacher's notes, that a particle (a proton in that case, but I don't think it matters) cannot be localized with precision superior to its De Broglie wavelength.
In my mind, if $$\Delta x$$ is the uncertainty on the particle x coordinate, and $$\lambda$$ is the particle's associated De Broglie wavelength, then my teacher's statement can be expressed as

$$\Delta x\geq\lambda$$

at least in one dimension.

But all I get from the Uncertainty principle, assuming $$\Delta p=\frac{h}{\Delta\lambda}$$, is

$$\Delta x\Delta p=\Delta x\frac{h}{\Delta\lambda}\geq\frac{\hbar}{2}\Rightarrow\Delta x\geq\frac{\Delta\lambda}{4\pi}$$

Where am I wrong? Maybe in considering $$\Delta p=\frac{h}{\Delta\lambda}$$? Or am I missing some reasoning as a result of which $$\Delta\lambda\simeq\lambda$$?

Guido

Last edited: Jul 15, 2006
2. Jul 15, 2006

### wavemaster

Yup.

$$p = \frac{h}{\lambda}$$
then, by differentiating both sides
$$\Delta p = -\Delta \lambda \frac{h}{\lambda^2}$$
But I see no point in messing with $$\Delta \lambda$$. Your teacher is talking about $$\lambda$$.

Also note that these deltas can be quite misleading. What Heisenberg means with "uncertainty of something", represented by delta, is the standard deviation of that quantity's outcomes.

3. Jul 15, 2006

### coccoinomane

I understand my mistake on differentiating, but how do you justify the fact that we cannot localize the particle with precision greater than $$\lambda$$?

In the free particle case

$$\Delta x\geq\lambda$$ since $$\Delta x=\infty$$;

but what about a bound particle?

Guido

Last edited: Jul 16, 2006
4. Jul 16, 2006

### wavemaster

I think your teacher's statement was rather "intutive". (Note that he seems to be talking about a precise wavelength, no $$\Delta \lambda$$ is involved. That'd mean a precise momentum).

More likeky, his pointly was this: if you have a minimal thing that would represent a wave, maybe a pulse or a fairly localized wavepacket, it would have a spread about a wavelength. Just like you cannot talk about the position of a cloud better than it's width, you cannot define position of that fairly localized thing better than it's spread.

5. Jul 16, 2006

### coccoinomane

Am I corrent in assuming that, with the statement

you mean that, for a fairly localized wavepacket, $$\Delta x \simeq \lambda$$ ?

If the answer is "Yes", wouldn't that imply that the momentum has to be similar to his incertainty?
In fact, $$\Delta x \Delta p \geq \frac{\hbar}{2} \quad \textrm{and} \quad \Delta x \simeq \lambda = \frac{h}{p} \quad \Rightarrow \quad \Delta p \gtrapprox p$$

Am I taking my teacher assertion too seriously? :)

Guido

6. Jul 16, 2006

### wavemaster

What i mean is this: if you have a "piece of wave" that is too short -such as a sudden small bump-, would you call it a wave? To talk about a wave, you'll need a wavelength, roughly. (Of course, if you're pedantic, you actually need something that is periodic, but one-wavelength piece is fairly okay.)

7. Jul 17, 2006

### Eye_in_the_Sky

Hello, coccoinomane.

I am wondering:

Did your teacher have a specific context in mind to get the relation ∆x≥λ?

For example, is the teacher talking about shooting a particle through a single slit after which the particle arrives at distant screen?

If so, then we are talking about "diffraction". In that case, if we take ∆x to be roughly the width of the central bright maximum associated with the overall diffraction pattern, then we can write

∆x ≈ λ(2L/w) .

Here, w is the width of the slit, L is the distance from the slit to the screen, and λ is the wavelength of the incoming particle.

This is the only (simple) way I can come up with to get something 'close' to what you have mentioned.

8. Jul 17, 2006

### coccoinomane

Hi Eye_in_the_Sky,

Thanks for the answer! I thought about the diffraction pattern, but I don't think it's the case.

In fact, my teacher was talking about a gas of protons at very high temperature. Precisely, the context was an exercise that required calculating the temperature at which protons get close to a distance similar to their De Broglie wavelength.
After the question mark, there was a sentence that sounded like this:

I solved the problem (I hope correctly), the result is

$$T = 5.22 \cdot 10^6 K$$

but I never used the "Note"!

What do you think?

Guido

edit: my browser can't "read" the latex formula I wrote; just in case somebody else has the same problem, the formula is
T = 5.22 * 10^6 K degrees

Last edited: Jul 17, 2006
9. Jul 18, 2006

### Eye_in_the_Sky

I think the note is not given to help you solve the problem. Rather, I think it is given in order to help you with interpreting the meaning of your solution.

Let's go back to the example I gave in terms of the single slit. This time, I'll write the relation like this:

Δ = (λ/w)∙(2L) .

When does the quantum mechanical behavior of the system become significant? ... Well, quantum effects become significant when the slit width w is comparable to or smaller than the de Broglie wavelength λ. On the other hand, for w>>λ, quantum effects are negligible.

It is the same thing with the proton gas. When does the quantum mechanical behavior of the proton gas become significant? ... Quantum effects become significant when the mean interparticle spacing d≡(V/N)1/3 is comparable to or smaller than the de Broglie wavelength λ. On the other hand, for d>>λ, quantum effects are negligible.

In terms of the temperature which you calculated (let's call it "Tc"), the above tells us that for temperatures T>>Tc , quantum effects are negligible: the proton gas can be treated as a classical ideal gas (note: I am ignoring the charge). But for T~Tc quantum effects begin to become important, and as T becomes smaller and smaller, those effects become more and more significant. (Note: I am keeping d fixed in value.)

... But, still, the wording used in that note does seem rather strange to me. "Localization" (of the proton) sounds like the idea of a "wavepacket". Vaguely, this is almost making some sense – but it is 'vague' and only 'almost'. ... I think I'll have to think about it some more.

Last edited: Jul 18, 2006
10. Jul 18, 2006

### coccoinomane

an eye opener, so I tried to deduce it using the Heisenberg Uncertainty Principle.

Correct me if I'm wrong!

Let's think of the particle-particle collision in terms of the collision parameter b (the minimum distance the two protons would be in a collision if the interaction could be "turned off")

Classically b could assume every value without consequences on the momentum, but not in QM: making b --> 0 means restricting the particle to an arbitrary small volume and hence increasing its moment's uncertainty.
If we want the classical limit to be respected, b and p (the related momentum) should respect

$$\Delta b \ll b$$
$$\Delta p \ll p$$

using the Uncertainty Principle, the two conditions become

$$p \cdot b \gg \frac{\hbar}{2}$$

but $$p = \frac{h}{\lambda}$$; hence

$$b \gg \lambda$$

If the two particles are in an ideal gas, the collision parameter b, averaged over all the collisions, is the interparticle spacing d, so the classical limit apply when

$$d \gg \lambda$$

Do you think this derivation is correct?

Concerning my teacher's note, I found out that it can be easily derivated in the hypothesis of molecular chaos (no preferential direction in the particles' motion). In fact, in this case we can take $$\Delta p \simeq p$$, with $$\Delta p$$ intended as the standard deviation of the momentum distribution.

$$\Delta x \Delta p \simeq \Delta x \cdot p = \Delta x \cdot \frac{h}{\lambda} \geq \frac{\hbar}{2} \Rightarrow \Delta x \gtrsim \lambda$$

which is my professor "note".

I'm sure I have made a lot of mistakes (both in Physics and in English), don't worry in underlining them :)

Guido

11. Jul 19, 2006

### Eye_in_the_Sky

You have made a very nice derivation.

I do have some difficulty, though, with the expression

∆b << b .

I would rather have seen

∆b << d .

In other words, I think it is better to go right away to the average and not leave that for the end.

Also, I would like to 'sharpen up' the language a little bit.

So, in slightly different notation, and with the benefit of the insight you have provided, I would like to rewrite your derivation as follows:
_________

In the classical limit, we expect to be able to think of the system of particles in terms of sufficiently localized wavepackets. That is, we require

∆x << d ,
∆p << paverage ,

for each particle.

From the above two conditions, it follows that

d∙(h/λ) = d∙paverage >> ∆x∙∆p ≥ h/(4π) ,

in which case d >> λ .

Therefore, a necessary condition for the classical limit to apply is d >> λ .
_________
_________

Note: Your insight has completely solved my problem of how to use the idea of "wavepackets" in arriving at d >> λ .
____________________

It gives us the desired conclusion immediately: d >> ∆x ≥ λ .

12. Jul 20, 2006

### coccoinomane

Now everything is clear (al long as QM can be clear )!

Thanks a lot for the explanations,

Guido