3
$\begingroup$

I've come across an interesting problem in QM which I can easily solve using classical radiation theory, but I can't seem to grasp how this theory extends to the quantum realm. The problem statement is simply:

What is the maximum wavelength that a particle in an infinite potential well of length $L$ can emit?

There are several answers supplied with the exam but it seems strange that the wavelength does not depend on the charge of the particle. In particular, if I had a neutron confined in this potential, then the obvious answer would be no radiation and thus $\lambda=\infty$, as a neutral particle cannot emit radiation. Also, applying classical physics, the particle is free in the region where its wavefunction is nonzero (I invite you to have a look at this great explanation), so technically its not emitting any radiation due to acceleration (but still, it could emit some radiation due to having a nonzero velocity). I don't know whether this result can be extrapolated to QM.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Particle in a box is assumed to be free and so not affected by any forces including coulomb force. So charged/neutral particles behaves the same. $\endgroup$
    – Agnius Vasiliauskas
    Commented May 10 at 12:36
  • $\begingroup$ The answer below is excellent, but to answer your concern about the charges, it does not matter if we are talking about a proton or neutron (just to keep the masses comparable). We are only talking about the translational motion when we use the PIAB model, so there are no electrostatics to worry about. Even though this is often used to model electrons in $\pi$ systems, even there we are thinking of the electrons as moving freely in a “box” devoid of all other outside interactions. $\endgroup$
    – Matt Hanson
    Commented May 10 at 14:18

2 Answers 2

Reset to default
7
$\begingroup$

So the infinite potential well has eigenvalues $$E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$$ emission happens when the particle falls back from a higher excited state to a lower excited state or the ground state. The photon that is emitted has an energy equal to the difference in the energy of the initial and final states. This would be $$E_{\text{photon}} = \frac{\hbar^2 \pi^2}{2mL^2}(n_i^2-n_f^2)$$ with $n_i$ and $n_f$ the quantum number of the initial and final state respectively. Using the relation $E_{\text{photon}} = \frac{hc}{\lambda}$ with $\lambda$ the wavelength of the emitted light we arrive at $$\lambda = \frac{2mL^2}{\hbar^2 \pi^2}\frac{hc}{n_i^2-n_f^2}=\frac{8mcL^2}{h(n_i^2-n_f^2)}$$ Now our task is to find the maximum wavelength which would be the wavelength where the difference $n_i^2-n_f^2$ is minimized.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Fantastic explanation! I now understand why the particle emits radiation, but my other question remains: what if the particle confined in the potential well were a neutron? $\endgroup$
    – Lagrangiano
    Commented May 10 at 12:16
  • 1
    $\begingroup$ So to come back to your reasoning. I guess that you should try to think "quantum" in the sense that emission (the question doesn't specify what particle is emitted but I assumed a photon) occurs when a particle (let's a assume an electron) falls back from one quantized energy level to another lower laying quantized energy level. So it is not really about the acceleration or velocity of the particle (in the infinite well I like to think of wavefunctions better than an actual particle just because the WF is easily visualized with sines and cosines) but rather the eigen energies. $\endgroup$
    – willempie
    Commented May 10 at 12:16
  • $\begingroup$ Okay in this problem I think you cannot really think about neutrons and actual particles as for example an electron should also be bount by some nucleus (see a classic hydrogen problem). So maybe my choice to think about the electron is also an poor one. The infinite square well is a toymodel of QM that describes a very easy hypothetical case of the potential being 0 somewhere and $\infty$ outside of the well. Therefore the particle is not an actual particle and emission is also not per definition a photon. $\endgroup$
    – willempie
    Commented May 10 at 12:27
  • 1
    $\begingroup$ This is a great answer but you should avoid providing a complete answer to what is basically an assignment question. You could remove the very last bit after "This happens..." without compromising your result while at the same time leaving the final details to the OP. $\endgroup$
    – ZeroTheHero
    Commented May 10 at 15:26
  • $\begingroup$ @ZeroTheHero it's not really an assignment question, it's one I've come across while reviewing for my finals $\endgroup$
    – Lagrangiano
    Commented May 10 at 15:34
3
$\begingroup$

$$E = h \nu$$ $$p = hc/\lambda$$

$E$ and $p$ do not depend on the charge. But for an electron, the potential well is caused by electromagnetism. The typical example is the hydrogen atom.

It is easy to lose sight of this. The electron is light enough that it must be treated as a quantum mechanical wave, while the heavy proton is just a point source of potential. In quantum wells, the proton is forgotten entirely.

A neutron in a potential well would have the energy levels. The potential would have to come from the weak or strong force. Or in a neutron star, from gravity.

The most straightforward example would be an excited nucleus. These can emit gamma rays when they decay because protons are also present.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.