Let's say we have a quantum particle with mass $m$ in a 1-Dimensional box. The potential outside the box is infinite. Say that $n=2$, so that $|\psi|^2$ will have two maxima. How would the gravitational attraction between this particle (particle1) and another particle (particle2) work? Toward which maxima the particle2 is going to be pulled by the particle1? Or what will be the point toward which the particle2 is going to be pulled?


closed as off-topic by StephenG, Kyle Kanos, Aaron Stevens, FGSUZ, ZeroTheHero Nov 4 '18 at 2:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ I'm voting to close this question as off-topic because this requires a working theory of quantum gravity to answer properly and we don't have one. $\endgroup$ – StephenG Oct 25 '18 at 21:10
  • 2
    $\begingroup$ Assuming Newtonian gravity, there's an interesting answer from perturbation theory that I'm currently preparing. $\endgroup$ – probably_someone Oct 25 '18 at 21:14

For this answer, we will assume Newtonian gravity, and calculate the effects on particle 1 from the gravitation of particle 2 (which we will assume is a fixed point mass).

Suppose we have a 1-dimensional "particle in a box" of mass $m$ confined to the interval $[0,L]$. The energy eigenstates are labeled by $n=1,2,...$ and have energies


and wavefunctions

$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

Suppose now we introduce a small perturbation to that system, namely, the gravitational attraction from another particle of mass $M$ at position $x_G$, which we will restrict to lie outside the box so as to keep the perturbation small and avoid singularities. The potential introduced by this attraction at position $x$ is


We treat this as a perturbation of the original wavefunction, and proceed using first-order perturbation theory. The change in the energy of the $n=2$ state, to first order in $GmM$, is

$$\Delta E_2 = \langle\psi_2|V|\psi_2\rangle=\int_0^L -\frac{2GmM}{L}\frac{\sin^2\left(\frac{2\pi x}{L}\right)}{|x-x_G|}dx$$

This integral has no elementary expression for its solution, but it can be plotted. For example, here is the change in energy as a function of distance, assuming the attracting particle is to the left of the box:

enter image description here

In the above, the $y$-axis is in units of $E_0=\frac{2GmM}{L}$. As you can see, the closer the attracting particle is to the box, the more the energy of the particle in the box is lowered.

The first-order correction to the wavefunction can also be calculated using the well-known formula

\begin{align*} \Delta\psi_2(x) &= \sum_{k\neq 2}\frac{\langle\psi_k|V|\psi_2\rangle}{E_2-E_k}|\psi_k\rangle \\ &= \sum_{k\neq 2}\frac{\int_0^L -\frac{2GmM}{L}\frac{\sin\left(\frac{2\pi x}{L}\right)\sin\left(\frac{k\pi x}{L}\right)}{|x-x_G|}}{\frac{(2^2-k^2)\pi^2\hbar^2}{2mL^2}}\sin\left(\frac{k\pi x}{L}\right) \end{align*}

Once again, there is no closed-form solution, but here is an approximation of this state's probability $|\psi_2+\Delta\psi_2|^2$ for a particle with $x_G=-1$, $L=1$, and $GMm=.01$, with the $y$-axis in arbitrary units (where the corrections have been taken out to $k=6$):

enter image description here

As you can see, it's basically identical to the original wavefunction, which is a sign that we did things right, because first-order perturbation theory is only valid for very small perturbations to the wavefunction. To get a closer look on what has changed, let's take the ratio of the perturbed probability density to the original probability density $\frac{|\psi_2+\Delta\psi_2|^2}{|\psi_2|^2}$:

enter image description here

As you can see, the particle is now slightly more probable to be on the left side of the box than before. (Curiously, it also appears to be more probable to be on the edges of the box either way, but looking at the asmmetry in this ratio, it clearly prefers to be on the left side). Given that our attracting particle is on the left side of the box, this makes sense. So, overall, it seems that our probability distribution has shifted slightly leftward. Calculating the average position of the particle, in the above units:

$$\langle x\rangle = \frac{ \langle \psi_2+\Delta\psi_2|x|\psi_2+\Delta\psi_2\rangle}{\langle \psi_2+\Delta\psi_2|\psi_2+\Delta\psi_2\rangle} = \frac{\int_0^L (\psi_2+\Delta\psi_2)^2 x dx}{\int_0^L (\psi_2+\Delta\psi_2)^2 dx}\approx 0.499996$$

which is slightly left of the average position of the unperturbed particle, $0.5$.

  • 1
    $\begingroup$ the gravitational attraction from another particle of mass M at position xG, which we will restrict to lie outside the box The question was about the gravitational attraction between two particles in the box. A particle outside the box makes no sense at all given the potential outside is infinite. $\endgroup$ – StephenG Oct 25 '18 at 22:42
  • 2
    $\begingroup$ @StephenG The question does not specify whether the other particle is inside or outside the box. The only situation that is likely amenable to perturbation theory is when the other particle is outside the box, and that is the situation that this answer deals with. If you would like to cover another situation, you are more than welcome to. $\endgroup$ – probably_someone Oct 25 '18 at 23:04
  • 2
    $\begingroup$ @StephenG Also, the question only seems to specify that the potential outside the box is infinite for one of the particles (chosen WLOG to be particle 1). That does not mean that it is also infinite in the same region for particle 2. For this answer, I assume that particle 2 is also confined by a sufficiently narrow potential well to have an essentially fixed position. $\endgroup$ – probably_someone Oct 25 '18 at 23:05

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