# Validity of naively computing the de Broglie wavelength of a macroscopic object

Many introductory quantum mechanics textbooks include simple exercises on computing the de Broglie wavelength of macroscopic objects, often contrasting the results with that of a proton, etc.

For instance, this example, taken from a textbook:

Calculate the de Broglie wavelength for

(a) a proton of kinetic energy 70 MeV

(b) a 100 g bullet moving at 900 m/s

The pedagogical motivation behind these questions is obvious: when the de Broglie wavelength is small compared to the size of an object, the wave behaviour of the object is undetectable.

But what is the validity of the actual approach of applying the formula naively? Of course, a 100g bullet is not a fundamental object of 100g but a lattice of some $10^{23}$ atoms bound by the electromagnetic force. But is the naive answer close to the actual one (i.e. within an order of magnitude or two)? How does one even calculate the de Broglie wavelength for a many body system accurately?

The de Broglie wavelength formula is valid to a non-fundamental (many body) object. The reason is that for a translation invariant system of interacting particles, the center of mass dynamics can be separated from the internal dynamics. Consequently, the solution of the Schrödinger equation can be obtained by separation of variables and the center of mass component of the wave funtion just satisfies a free Schrödinger equation (with the total mass parameter). Here are some details:

Consider a many body nonrelativistic system whose dynamics is governed by the Hamiltonian:

$\hat{H} = \hat{K} + \hat{V} = \sum_i \frac{\hat{p}_i^2}{2m_i} + V(x_i)$

($K$ is the kinetic and $V$ the potential energies respectively). In the translation invariant case, the potential $V$ is a function of relative displacements of the individual bodies and not on their absolute values. In this case, the center of mass dynamics can be separated from the relative motion since the kinetic term can be written as:

$\hat{K} = \frac{\hat{P}^2}{2M} + \hat{K'}$

Where $P$ is the total momentum and $M$ is the total mass. $K'$ is the reduced kinetic term. In the case of a two-body problem, for example the hydrogen atom $K'$ has a nice formula in terms of the reduced mass, for larger number of particles, the $K'$ formula is less nice, but the essential point is that it depends on the relative momenta only.

For this type of Hamiltonian (with no external forces), the Schrödinger equation can be solved by separation of variables:

$\psi(x_i) = \Psi(X) \psi'(\rho_i)$

Where $X$ is the center of mass coordinate, and $\rho_i$ is a collection of the relative coordinates.

After the separation of variables, the center of mass wave function satisfies the free Schrödinger equation:

$-\frac{\hbar^2}{2M}\nabla_X^2\Psi(X) = E \Psi(X)$

Whose solution (corresponding to the energy $E = \frac{p^2}{2M}$) has the form:

$\Psi(X) \sim exp(i \frac{p X}{\hbar})$

from which the de Broglie wave length can be read

$\lambda = \frac{2 \pi \hbar}{p}$

• This is practically contradictory to John's answer except if one interprets it as "a wavelength" for an extended body. Won't the wavelengths of the constituents extend spatially by far over this wavelength? Where does decoherence come in? – anna v Mar 20 '13 at 11:41
• @Anna What I tried to emphasize is that the de Broglie wavelength is a property of a free degree of freedom. The internal state of the composite system is not free and will not be characterized with its constituents de Broglie wavelengths but rather with its bound state energies for example its vibrational modes. – David Bar Moshe Mar 20 '13 at 13:51
• @Anna cont. When conditions are provided such that the system stays in a single (ground state) internal state, it is possible to approximate it by its center of mass quantum dynamics governed by its de Broglie wavelength, for example in the case of the buckyball and other large molecule interferometry. – David Bar Moshe Mar 20 '13 at 13:52
• Anna cont. It is true that in the composite case the decoherence conditions are harder to achieve, because not only can this system loose coherence by being "kicked" by interacting particles but also it can loose coherence by random excitations of its internal degrees of freedom for example when the thermal excitations exceed its vibrational energies. – David Bar Moshe Mar 20 '13 at 13:52
• @John Rennie You are correct, but I don't know if it is due to the present limitations in knowledge or technology or there is a fundamental theoretical limitation of decoherence. In the first case it might be possible in the future to increase the experimental bounds where interference is observable. – David Bar Moshe Mar 20 '13 at 15:50

If you've read about optical diffraction experiments like the Young's slits, you may have noticed they all refer to coherent light. This is the requirement that all the light in the experiment is in phase. If you aren't using coherent light you won't observe any diffraction because different bits of the light will diffract differently and the diffraction pattern is washed out.

Exactly the same applies to observing the wavelike behaviour of quantum objects. If you're diffracting electrons this isn't a problem, but if you're trying to diffract a bullet you require all parts of the bullet to be coherent. In principle you could prepare a bullet in a coherent state, but even if you could manage this the bullet would immediately decohere due to interactions with it's environment. This process is known as quantum decoherence. I've linked a Wikipedia article but be warned that the article isn't well written for non-nerds. If you want to know more you'd be better off Googling for popular science articles on decoherence.

Anyhow, as you obviously suspected from the way you've worded your question, because of decoherence it doesn't make sense to talk about a single de Broglie wavelength for macroscopic objects like bullets. As far as I know, the largest object ever to show quantum behaviour is an oscillator built by Andrew Cleland's group at Santa Barbara. This was around 50 - 100 microns in size, which is actually pretty big. However this is something of a special case and took enormous care to build. A more realistic upper limit is a buckyball, which is around a nanometre in size.

• It's also interesting to note that when you can isolate a large-ish object like a buckyball sufficiently from its environment to prevent environmental decoherence, then interference can still be measured even if the object's size is larger than its de Broglie wavelength--the page here on buckyball interference experiments says "The de Broglie wave length is thus ~ 400 times smaller than the size of the particle". – Hypnosifl Dec 22 '14 at 23:40
• I think the reason this works is because, as noted on this page, the separation between fringes is equal to the de Broglie wavelength times D/d, where D is the distance from the slits to the screen, and d is the spacing between slits--typically D is much larger than d, so the spacing between fringes will be larger than the de Broglie wavelength by the same factor. – Hypnosifl Dec 22 '14 at 23:41
• sorry to come into this so late - but is not the issue to have all bullets coherent with each other rather than to have a bullet coherent with itself? – tom Mar 23 '15 at 10:19
• @tom: well, a single photon will diffract therefore so should a single bullet. – John Rennie Mar 23 '15 at 16:09
• Coherent light isn't required (though it helps). Here's a double-slit image from sunlight (via). – rob Jul 20 '17 at 12:07

"But is the naive answer close to the actual one"

There is no "actual one" - the de Broglie formula defines the de Broglie wavelength for body of mass $m$. It is useful for description of microscopic particles that are observed to exhibit diffraction phenomena and the values for electrons seem to be in agreement with experimentally measured diffraction pattern.

100 g bodies were not observed to exhibit diffraction phenomena. According to the de Broglie formula, their wavelength is so short that diffraction of the wave would be unobservable, so the formula was not disproven. But it is not useful either.

I just spent a few hours researching this question, and it seems to me that:

It's important to understand what is meant by a particle having a wavelength. This great link has more information (http://electron6.phys.utk.edu/phys250/modules/module%202/matter_waves.htm) It emphatically does not mean that if we were to somehow isolate the particle and set it moving with some velocity v, it would move sinusoidally in space as a function of time. The key idea is that although deBroglie's relation about wavelength applies both to photons and other particles, it means something different in both cases! For the photon, it makes pretty intuitive sense - it refers to the EM wave that is the photon. What about other particles?

The wavelength refers to the particle's wavefunction, which, under the statistical interpretation, tells you the probability you would find the particle at a certain position x (http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c5). The previous link also makes sense of what it means to talk about the wavelength of a particle (if the wavelength isn't sinusoidal - basically, you take some sort of average).

Does a macroscopic object, like the aforementioned bullet, have a wavelength? I believe so, since we can take all the individual particles and consider the interactions between all the particles in the bullet (giving us entanglements and whatnot). We can imagine the bullet having some complicated wavefunction that describes the probability of it being found somewhere. This is the point on which I'm unsure, but the bigger point is that it doesn't really matter, since the object does not live in isolation to the environment. The idea is that it decoheres by interaction with the environment (a great description here: www.ipod.org.uk/reality/reality_decoherence.asp), which basically means that the environemtn acts differently on each part of the bullet, so even if the bullet might have originally acted like a quantum system, it no longer does after a measurement (i.e. when we see it). And so it doesn't make sense now to talk about the wavelength of the bullet since we really need to consider it as part of a bigger system, the bullet plus the environment.

• But your final argument apply to small and elementary particles as well. I am not a downvoter, tough. – Alchimista Apr 12 '18 at 12:18