quantum mechanics - 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?

Answer

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}$