Estimating our de Broglie wavelength

Question

This could be a very naive question, but I feel always disturbed by it since I couldn't ever figure out a perfect answer.

Why is our de Broglie wavelength (dBw) so tiny and how could we roughly estimate it?

1) Say I am running at average speed of 10 m/s. So $v$=10 m/s. My body weight is $m$ = 66.3 kg. Then my dBw is

$\lambda=\frac{h}{mv}=\frac{6.63\times10^{-34}\,\rm{kg}.\rm{m}^2.\rm{s}^{-1}}{66.3\,\text{kg}\,\times\, 10\, \rm{m}.\rm{s}^{-1} }=10^{-36}\,\rm{m}$.

2) I got the speed of earth's rotation 460 m/s. So my dBw is

$\lambda=\frac{h}{mv}=\frac{6.63\times10^{-34}\,\rm{kg}.\rm{m}^2.\rm{s}^{-1}}{66.3\,\text{kg}\,\times\, 460\, \rm{m}.\rm{s}^{-1} }=0.02 \times 10^{-36} \rm{m}.$

3) My body temperature is close to the room temperature. Now room temperature is ~ 300 K. And from kinetic theory, we know average kinetic energy=1/2 $m <v>^2$ = 3/2 $k_B T$ ($k_B$ is the Boltzmann constant). Considering myself made out of protons only ($m=1.67\times10^{-27}\,\rm{kg}$), $<v>=\sqrt{1.5\,k_B T/m}=\sqrt{\frac{1.38\times 10^{-23}\,\rm{kg}.\rm{m}^2.\rm{s}^{-2}.\rm{K}^{-1}\times 300\,\rm{K}}{1.67\times 10^{-23}\times 10^{-4}\, \rm{kg}}}=1.57\times 10^{3}\, \rm{m/s}.$ Then the dBW of the constituent protons will be $\lambda=\frac{6.63\times 10^{-34}\,\rm{kg}.\rm{m}^2.\rm{s}^{-1}}{1.67\times 10^{-23}\,\rm{kg}\,\times\, 1.57\times 10^3 \rm{m}.\rm{s}^{-1}}=2.53\times 10^{-14}\,\rm{m}.$

So which one could be the most appropriate? I don't trust much the first one, since I doubt: What will happen if I decide not to move at all ($v=0$)? Then $\lambda$ will become infinity which is ridiculous.

Last thing, if the 3rd one seems more reasonable, can one compare to experiments on fullerenes and ultra-cold atoms?

PS. If there is a mistake in the calculations, please bring into my notice. Thanks.

You could factor in the Earth's rotation but why bother? De Broglie wavelengths are relative. — Jimmy360, May 20 '15 at 00:47
I'm afraid to buy the 'rigid body' argument. You can replace me by any other rigid object of the same mass. There's a common exercise in high school physics, which asks one to find the dBw of a moving bullet. I hope someone talks about fullerene here. — hbaromega, May 20 '15 at 09:59
And why is our de Broglie wavelength so tiny? Because we are so damn heavy, and thus should not show quantum effects (which show, when the de Broigle wavelength is on the range of your spatial extension). — Sebastian Riese, May 24 '15 at 13:08
It appears unethical to attempt an answer to this question now, since a large part of that would be along the lines of what @CuriousOne mentioned in the comment. Maybe you could develop this a little more (emphasizing on the large m part), and post as an answer instead of a comment. — 299792458, May 25 '15 at 07:07
The answers at this link (http://physics.stackexchange.com/questions/111850/de-broglie-wavelength-of-slow-moving-macroscopic-objects) seem to indicate that you are trying to apply the de Broglie wavelength to an unsuitable object. — Ernie, May 26 '15 at 17:42
I disagree with the comments saying that you cannot define a DeBroglie wavelength for a macroscopic object, see the accepted answer here: http://physics.stackexchange.com/questions/57390/validity-of-naively-computing-the-de-broglie-wavelength-of-a-macroscopic-object — Rococo, May 29 '15 at 04:32

score 1 · Answer 1 · answered May 29 '15 at 12:37

I would say that neither of your alternatives are reasonable. The main argument to dismiss them is that the de Broglie wavelength is a semi-classical concept. It makes sense when the dBw is larger or on the same order of magnitude as the object (or length scales) under consideration.

Addressing your alternatives:

2) is just a choice of coordinate system, and as such has no physical relevance on the dBw.

1) can be used as a back of the envelope calculation to show that if the dBw can be defined it has to be ridiculously small (might as well be zero). For comparison the radius of a proton is ~$10^{-15}$ m. You could never really get away with saying that you average speed is zero. Just by breathing you are definitely moving.

3) is interesting albeit dodgy. I admit that if you would cool the human body to very very close to zero K then you would probably see quantum effects, and then maybe you could talk about the dBw again.

That aside I doubt that it is fair to say that you are made up of protons. I find it more likely that you are made up of atoms (primarily carbon and oxygen?) of even of molecules (lots of H$_2$O in the body), as these are the objects that will carry the majority your kinetic energy.

Secondly, I doubt the equipartition theorem holds for something as complicated as the human body. What I mean is that you are presumably not free to access all of the degrees of freedom all at once. Some degrees of freedom will likely be frozen out by their high energy thresholds.

To conclude: Calculating the dBw of a human is nonsense, and 1) if proof of this.

score 0 · Answer 2 · answered May 20 '15 at 01:39

An infinite de Broglie wavelength isn't completely ridiculous. Indeed, what the de Broglie relation does is connect a wavelength to a particle of definite momentum-- and if the momentum is definitely zero, then the wavelength is infinite. In other words, the probability wave does not have any perceptible change with position, hence the location of the particle is completely undetermined. Of course, the uncertainty principle prevents this from occurring in the first place.

Regarding your calculations, I see no problem with 1). As Jimmy360 has eluded to, there's not much physical value in factoring in the Earth's rotation in 2), and 3) seems too roundabout to be useful (if such a calculation can be considered useful in the first place).

Stack Tracer · Answer 3 · 2015-05-28T14:33:50.350

First, I just want to say that this question has no reasonable answer without taking general relativity into account.

First, if we assume that you can apply De Brogile wavelengths to an entire human at once, both 1) and 2) are true simultaneously. If I were standing on the sidewalk as you ran past, your wavelength would be what you found in 1). If I were on a satellite which was co-orbiting the sun alongside the earth, I would see your wavelength as 2). Both of these are true, but only from different reference frames. (Many people would consider calculation #2 to be useless, just due to the fact that we are not likely to be in that location)

However, De Brogile wavelengths only apply to particles. Although introductory physics courses have students calculate the wavelength of something like a baseball, this calculation is nonsense. The baseball is made up of lots of smaller particles, for which the concept of wavelength makes sense.

It's like asking what the temperature of a car is. The engine is hot, the air inside is cold, and the rest of the car is somewhere in between. Or like measuring the diffraction pattern of people passing through a door. It's not real.

Therefore, 3) is the only answer that makes any sense.

score 0 · Answer 4 · answered May 29 '15 at 04:11

De broglie wavelength makes sense only if you are in one of the momentum eigenstates,therefore it should only be used for particles.And secondly you never have particles with 0 speed as it will lead to infinite uncertainty in conjugate position. Therefore i feel that only the third calculation seems reasonable.

score 0 · Answer 5 · edited Apr 13 '17 at 12:39

I feel sorry to say that even after putting a bounty none of the replies could reach my expectation. However, a reasonable amount of support goes for my answer 3) and I thank for that and I am grateful to the people providing additional thoughts. Hence I'll attempt to put my own understanding by collecting some of the arguments from the responses.

I think, in the beginning scientists more-or-less accepted the notion of wave having particle-like discrete behaviors. However, after the wave-particle duality hypothesis by de Broglie in 1924, people felt forced to look for wave-nature coming out of a particle. So de Broglie's hypothesis got its validity when electron diffraction was observed by Davisson, Germer, and Thomson. So diffraction and later interference by Jönsson became common tool to verify the existence wave nature of a quantum object.

Now one should remember that any object is quantum as quantum theory is valid in all time-space domains, but to see it's effect we may need to look at smaller systems. Since diffraction and interference are the ways to test the wave nature, on such experiments one needs slits of the size of the de Broglie wavelength. Let me pick up some of the issues below.

A) Wavelength of a large molecule (Fullerene) : The quantum interference on fullerene, whose radius is about 700 pico meter (7 Å) , has been done by Arnst group in Austria (https://vcq.quantum.at/fileadmin/Publications/2003-17.pdf). Now the temperature of the experiment was 900 K, which gives rise to most probable speed: $\bar{v}=\sqrt{2k_B T/m}=144\,\rm{m/s}$. However, the experiment uses a velocity filter and find most probable speed: $\bar{v}=200\,\rm{m/s}$. Using this one may expect: $\lambda_{\rm{dB}}=\frac{h}{m\bar{v}}=2.8\times 10^{-12}\,\rm{m}$.

From the paper we may find: $w=35\,\mu\rm{m}$, $D=1.25\,\rm{m}$, and $d=100\,\rm{nm}$ where they are fringe width, detector-to-gratings distance, and grating spacing respectively. Then using $\lambda=\frac{wd}{D}$ we find $\lambda_{\rm{dB}}=\frac{35\times10^{-6}\,\rm{m}\times 10^{-7}\rm{m}}{1.25\,\rm{m}}=2.8\times10^{-12}\,\rm{m}$.

B) Infinite de Broglie wavelength paradox : As I asked whether $v=0$ permitted or not, one can recall that the wavelenth is the measurement of the periodicity in length dimension (say, distance between two repated peaks). So when it becomes infinity, it means that the peak gets its next repetition at very large distance. That means no periodicity and hence wave nature is not detectable using our physical equipments. But getting bigger wavelenth by slowing down is the key idea of the ultracold atomic physics. Here average speed of atoms are discreased significantly by going up to a very low-temperature (see this: https://physics.aps.org/assets/e694a42b-2bf3-4a3c-98fe-66f55811896a/e10_1.png).

C) Uncertainty/wave packet argument against $v=0$ case: First of all, the argument doesn't hold for very large objects. Why it doesn't hold that could be a subject for debate on macro-realism (http://arxiv.org/abs/1412.6139). Secondly, one can always talk of expectation value even when quantum mechanics holds. Thirdly, when we use wave packet argument to localize a quantum object, then we associate a group of waves, that implies that there is no single de Broglie wavelength, i.e. uncertainty remains in $\lambda$.

D) Earth's rotation effect: For an experimenter on earth this velocity contribution won't be detectable. Considering this only helps to avoid zero velocity situation.

E) My 3rd answer: Actually my 3rd answer not only works for human beings, it is the same any macroscopic stuff such as egg, baseball, bullet, etc for a given temperature. One crucial thing is that it is very difficult send such big objects through very narrow slits. Another important aspect one may consider: To see successful interference or diffraction pattern, the waves need to be coherent. A discussion in Stackexchange has been made by John Rennie already.

So in one line (though it's not a conclusion), one can say, existence of de Broglie wavelength doesn't make much sense for human or other large objects since it cannot be tested within our present limitation.

Estimating our de Broglie wavelength

5 Answers5