# De Broglie wavelength of composite systems

Can the De Broglie wavelength of a composite system (like a molecule) be derived as opposed to being calculated from the composite mass?

EDIT: @Dr jh, interesting relation you have derived. However, that is a rewrite of the original DeBroglie equation albeit using the λ variables. De Broglie guessed his solution by setting relativity constraints and came up with the frequency of the particle in its rest frame as f=m0c2h. I guess we can reformulate my question as to why does a particle acquire such a frequency as it seems to be a property of the bound system so we can for a moment ignore it is composite. What underlying physics gives the particle this frequency. Put another way, if we had a large ball over water and we see it bobbing up and down, we would assume something is pushing it, something like a water wave. Can we do something similar here and assume that the bound system acquired it's proper time frequency somehow by resonating to an underlying wave in the vacuum? What's interesting in the way De Broglie derived his solution, and I have seen this only in rare QM books, is that the particle has an associated spatially flat (constant phase) wave which then when observed from the point of a moving frame looks like a plane wave along the direction of motion with the known λ. No other wave shape in the particle's frame except the constant phase wave would produce such a relationship.

Can the De Broglie wavelength of a composite system (like a molecule) be derived as opposed to being calculated from the composite mass?

Yes, it can. This is standard material for the hydrogen atom in sufficiently solid quantum-mechanics textbooks, and the extension to larger systems is (largely, though not completely) straightforward. However, you do have to start from fully-grown QM, including the canonical commutation relations and the Schrödinger equation.

The way it works is that you start with the Schrödinger equation in the form $$\left[ \frac{\mathbf p_p^2}{2m_p} +\frac{\mathbf p_e^2}{2m_e} -\frac{e^2}{|\mathbf r_e-\mathbf r_p|} \right]\Psi(\mathbf r_p, \mathbf r_e,t) = i\hbar\frac{\partial}{\partial t}\Psi(\mathbf r_p, \mathbf r_e,t)$$ (where $$\mathbf p_p$$ and $$\mathbf p_e$$ are the operators for the proton and electron momenta), and you do a change-of-variables transformation to center-of-mass and relative coordinates, \begin{align} \mathbf R & = \frac{m_p \mathbf r_p + m_e \mathbf r_e}{m_p+m_e} \\ \mathbf r & = \mathbf r_e - \mathbf r_p, \end{align} with corresponding momenta $$\mathbf P$$ and $$\mathbf p$$, and you can show that this results in $$\left[ \frac{\mathbf P^2}{2M} +\frac{\mathbf p^2}{2\mu} -\frac{e^2}{|\mathbf r|} \right]\Psi(\mathbf R, \mathbf r,t) = i\hbar\frac{\partial}{\partial t}\Psi(\mathbf R, \mathbf r,t) ,$$ where $$M=m_p+m_e$$ is the total mass and $$\mu = \frac{m_pm_e}{m_p+m_e}$$ is the reduced mass.

What that means is that the dynamics factorizes completely, with the center-of-mass dynamics obeying the simpler Schrödinger equation of a free particle: $$\frac{\mathbf P^2}{2M} \Psi(\mathbf R, \mathbf r,t) = i\hbar\frac{\partial}{\partial t}\Psi(\mathbf R,t) .$$ Now, the de Broglie relation (itself encoded in the canonical commutation relations) tells you that $$P = h/\lambda_\mathrm{COM}$$, where $$\lambda_\mathrm{COM}$$ is the de Broglie wavelength of the center of mass, but you also know that (in the limit where it makes sense to talk about velocities) $$P = M v_\mathrm{COM}$$.

The relationship you want comes from putting those two together.

That's an interesting question. The de Broglie wavelength of any object is given by

$$λ_c = \frac{h}{mv}$$

where m is the mass of the object, v is it's velocity and h is Planck's constant. For composite objects, like molecules, we can simply add the masses together, but a relation for combining the wavelengths of the constituent atoms might be a little bit more complicated. Let's try to do this by induction. The two-atom (two-body) case (e.g., the $$NaCl$$ molecule) we combine the two masses of each atom so that

$$m_T=m_1 + m_2$$

where $$m_T$$ is the total mass. The de Broglie relation for the mass of each of the atoms are:

$$m_1= \large \frac{h}{λ_1v}$$
$$m_2= \large \frac{h}{λ_2v}$$

(and obviously the velocity for each component is identical to the velocity of the whole) whereas for the composite de Broglie wavelengths we have

$$λ_T = \large \frac{h}{(m1+m2)v}$$ = $$\frac{h}{[(h/λ_1v)+(h/λ_2v)]v}$$ = $$\frac{1}{1/λ_1 + 1/λ_2}$$

We can then apply this to the 3-body case, and assuming that the two components do indeed combine as before, then we can inductively combine a third wavelength, i.e.,

$$λ_T = \large \frac{1}{1/λ_1 + 1/λ_2 + 1/λ_3}$$

And again by induction we can extend this to find a general relation for the composition of de Broglie wavelengths for any number of component wavelengths (or component atoms) N:

$$λ_T = \frac{1}{1/λ_1+1/λ_2+...+1/λ_N}$$

This relation shows that the resulting de Broglie wave of the molecule, is the reciprocal of the sum of the reciprocals, of the constituent (atoms) wavelengths. This is an interesting result.

• This... has nothing to do with actual quantum mechanics. Aug 19, 2020 at 8:49
• Interesting shortcut, thanks. De Broglie guessed his solution by setting relativity constraints and came up with the frequency of the particle in its rest frame as $f=\frac{m_0 c^2}{h}$. I guess we can reformulate my question as to why does a particle acquire such a frequency as it seems to be a property of the bound system so we can for a moment ignore it is composite. What underlying physics gives the particle this frequency Aug 19, 2020 at 15:18