Context: a particle of mass $m$ can move in 3D and is trapped inside of a sphere of radius $R$ and impenetrable walls (in a more mathematical sense, the potential energy is 0 inside of the sphere and $+\infty$ everywhere else).

What I want to derive is the particle's energy $E_0$ in the fundamental state using the de Broglie relationship $\lambda = \frac{h}{p}$, where $p$ is linear momentum and $h$ is Planck's constant.

If I visualize a standing wave inside of a sphere, it appears that its wavelength must be $\lambda = \frac{4R}{n} (n = 1, 2, 3...)$. Therefore:

$$\frac{4R}{n} = \frac{h}{p}$$

Since $p = \sqrt{2mE}$:

$$\frac{16R^2}{n^2} = \frac{h^2}{2mE}$$


$$E = \frac{n^2h^2}{32mR^2}$$

And in the fundamental state, $n = 1$, so:

$$E_0 = \frac{h^2}{32mR^2}$$

Should (?) be the answer. The actual answer to the problem however is

$$E_0 = \frac{3h^2}{32mR^2}$$

Which is 3 times the answer I found. According to the actual solution, the energy found using the de Broglie relationship is

[...]The energy in just one dimension. We must multiply by 3 to find the total energy, i.e., the energy in 3 dimensions.

Now, what bothers me about this is that de Broglie's relationship is $\lambda = \frac{h}{p}$, not $\lambda = \frac{h}{p_x}$ or $\frac{h}{p_y}$ or $\frac{h}{p_z}$. It's just $p$. Where does it say that the momentum in the relationship is just a component instead of the total momentum?

As far as I'm aware, we should use the total momentum, not just a component, so the energy found should be the total energy, not just a component. Where does my logic fail? Why does the de Broglie relationship give us the energy in just one of the dimensions instead of the total energy if nowhere in my attempt to solve the problem did I assume I was working in a single dimension? This has been bothering me and I haven't found an answer, but I have a guess.

Perhaps I am misunderstanding how the three-dimensional standing wave works and simply got the wavelength wrong. Either this or the de Broglie relationship is somehow meant to be one-dimensional, but I doubt it.

Or maybe it's something entirely different. Anyway, what is it?

(The same applies to the particle trapped inside of a bidimensional circle, being able to move in the $x$ and $y$ directions inside of the circle. But in this case the energy is multiplied by 2.)

EDIT: this is a high school physics problem. I have discussed with the professor and he has confirmed this is just an approximation, as I was meant to estimate the energy, and therefore the "multiply by $3$" thing is just for approximation by analogy with the 3D box.

  • $\begingroup$ For a spherical cavity, you should be solving the Schrodinger equation using spherical coordinates. It seems like you are just visualizing a one-dimensional wave stretching across the diameter. $\endgroup$
    – G. Smith
    Commented Feb 1, 2019 at 18:25
  • $\begingroup$ The point of the problem was not using the Schrodinger equation at all, only the de Broglie relationship. What I don't understand is why my logic doesn't work. After all, isn't the wavelength of a standing wave inside of a sphere just $\lambda = 4R/n$? $\endgroup$ Commented Feb 1, 2019 at 18:53
  • $\begingroup$ How can you know what the wavelength of a standing wave in a sphere is without solving the wave equation inside the sphere? I see no reason to assume it is 4R/n. I would have thought it had something to do with zeroes of Bessel functions, because I thought the radial dependence of a spherical wave was some Bessel function, not a sine wave. $\endgroup$
    – G. Smith
    Commented Feb 1, 2019 at 18:59
  • $\begingroup$ OK. en.wikipedia.org/wiki/Wave_equation#Spherical_waves explains that the spherically symmetric waves ($\ell=0$) turn out to obey a one-dimensional wave equation for $r\psi$. But this is not at all obvious until you solve the spherical wave equation. And I don’t see where a factor of 3 would apply. $\endgroup$
    – G. Smith
    Commented Feb 1, 2019 at 19:06
  • 3
    $\begingroup$ Where is this problem and “we must multiply by three” answer coming from? It works for a cube but it doesn’t work for a sphere and making you think it does is bad pedagogy. $\endgroup$
    – G. Smith
    Commented Feb 1, 2019 at 20:38

1 Answer 1


$\let\lam=\lambda \let\om=\omega$ You (or your teacher) shouldn't stress de Broglie's idea beyond its limits. Maybe a brief history sketch can help.

1913: Niels Bohr "explains" hydrogen spectrum introducing some new postulates:

  • The electron can't move in all orbits allowed by Newtonian mechanics. For circular orbits, only those radii are possibile which correspond to an angular momentum $L$ multiple of $\hbar=h/(2\pi)$.

  • The atom can absorb or emit radiation only by "jumping" from one allowed (stationary) state to another. Frequency (and wavelength) of emitted/absorbed radiation is determined by conservation of energy together with Planck's rule: $$E_2 - E_1 = h\,\nu = \hbar\,\om.$$

By combining these postulates with 2nd Newton's Law and Coulomb's Law for electrical force between electron and proton, Bohr got to the famous formula $$E_n = {m\,e^4 \over 2\,n^2 \hbar^2} \qquad \hbox{(Gauss' units)}$$ which accounted with extraordinary precision for the observed spectral lines.

However both postulates were incompatible with mechanics and electromagnetism as known at that time. So it was clear Bohr's idea was a first tentative step towards a new physics.

De Broglie's (1924) idea was to justify the first Bohr postulate as a consequence of a more general one:

  • to every particle there is "associated" a wave, whose wavelength is determined by particle's momentum, according to the relation $$\lam = {h \over p}.$$

He added another idea:

  • Bohr stationary states are those where the associated wave is a standing wave along electron's orbit.

It's easy to see how Bohr follows from de Broglie. In a standing wave ve must have $$2\,\pi\,r = n\,\lam = n\,{h \over p}.$$ Then $$L = r\,p = {n\,h \over 2\,\pi} = n\,\hbar.$$

Wonderful, but... It sounds peculiar to have a wave oscillating along a circle in the form of a standing wave, more or less like a guitar string. What keeps the wave bound to that orbit? How to explain a unidimensional wave in a 3D space? What constrains it that way?

The answer came just two years later and was given by Schrödinger, who wrote a wave equation for de Broglie waves. He could also solve it for several cases, included the hydrogen atom. It wasn't an easy task, but the way towards solution had been traced in the 19th century, thanks to the work of great mathematical physicists on wave equations.

I believe you haven't still heard of Schrödinger equation, so I'll stop here my historical sketch. My aim was only at showing you that it's incorrect to use de Broglie waves to solve 3D physical problems. Its proper place and its full merit is in the history of physics, in having traced part of the way that brought a few years later to the fully developed quantum mechanics.

To be fair de Broglie's waves are still used sometimes. There are cases where they can give reasonable results, and are often an easy way to get an order-of-magnitude value for significant quantities (e.g. density of states in statistical calculations). Even in your problem using de Broglie is ok if you content yourself of the order of magnitude, not the exact value of energy. This I'm going to show.

Now for your problem. You found $$E_0 = {h^2 \over 32\,m\,R^2}$$ and ask why the "right" solution brings a factor of 3. My answer is simple: your solution is an order of magnitude, for the reasons I explained above, and within these limits it's correct. The "actual" answer (where did you find it?) is plainly wrong.

It would be right for a cubical box (I mean as a solution of Schrödinger equation) but yours is spherical and can't be treated viewing the three space dimensions as if they were independent and simply additive as to energy. I wish to give you the exact solution for a spherical box. The wave function corresponding to the lowest energy is $$\psi(r) = {1 \over r}\>\sin{\pi\,r \over R}$$ (an irrelevant normalization factor apart). If you try to plot this function and to compare it with the one you had found, you'll see they are rather alike.

The exact energy eigenvalue is $$E_0 = {h^2 \over 8\,m\,R^2}$$ i.e. 4 times the one you had found. Note that it's still larger than the one was sold you as exact. You may wonder why.

I'll give you a handwaving answer. I already said that the factor 3 is right for a cubical box (of side $2R$). Incidentally, the right wavefunction for a cubical box is $$\psi(x,y,z) = \cos {\pi\,x \over 2\,R}\,\cos {\pi\,y \over 2\,R}\, \cos {\pi\,z \over 2\,R}.$$ The spherical box is enclosed in the cubical one and has a smaller volume. Then the wavefunction for the cubical box is more "constrained" and the uncertainty principle says that we must expect a greater value for $\langle p^2\rangle$. This entails a greater energy.


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