Quantum Number of a Tennis Ball A tennis player has a tennis ball container with a single ball in it (it normally holds three). He shakes the tennis ball horizontally back and forth, so that the ball bounces between the two ends. We model the tennis ball as a quantum particle in a box.
The questions: what is the quantum number n for this ball? If the ball were to absorb a photon and jump to the next energy level, what should the energy (in eV) of that photon be?
For both of these questions, I am confused about how to apply quantum mechanical principles to the tennis ball. For the former, I suppose that a quantum number of n would make sense...if we model it as a particle in the box, the probability distribution across the container would match that of a particle at the n=2 energy level (remember--we are moving back and forth, and therefore the particle would be most likely to be at one of the ends and not in the middle). Would this be the correct reasoning? Is there something else I'm missing? For the latter question, I would use
$$E_n=\frac{n^2\hbar^2\pi^2}{2ma^2}$$
where m is the mass of the tennis ball and a is the length of our container. Say that we are now in the $n=2$ energy level. To go to $n=3$, we have to apply an energy of $\Delta E=E_3-E_2$ to make that jump. Is this the correct procedure for both of these questions? I'm just having a hard time applying quantum mechanical thinking to these macroscopic objects.
Thank you in advance.  
 A: You are right that the question is making a very crude model of a tennis ball. However, it does capture some important qualitative features of the classical limit that are worth understanding.
The intuition that you are supposed to get is that the classical limit corresponds to large $n$. The key thing to look at is $\Delta E_n/E_n$, the fractional difference in energy between two adjacent energy levels:
\begin{equation}
\frac{\Delta E_n}{E_n} = \frac{(n+1)^2-n^2}{n^2} = \frac{2n +1}{n^2}
\end{equation}
When $n$ is small, the jump to the next energy level is relatively large, and you notice that the tennis ball is quantum mechanical. The tennis ball is not free to move freely because the energy levels of this bound system are discrete, and it can't bounce around however it likes. 
When $n$ is huge, the jump to the next energy level is very small (it scales like $2/n$). In that case you can approximate  the energy levels as continuous, and you don't notice quantum behavior at all.
Based on all of this, you should notice that $n=2$ is probably not a good guess for the energy level of a tennis ball.
Indeed, taking $a=10 {\rm cm}$ and $m=10 {\rm g}$, I find
\begin{equation}
E_1 = \frac{\hbar^2 \pi^2}{2 m a^2} \approx 10^{-64} {\rm J}
\end{equation}
which is muuuuuch less than the average kinetic energy of a tennis ball! And $E_2$ is only a factor of 4 bigger than this, which is really no better
Some parting thoughts:
(1) If it bothers you that a tennis ball is a system with many internal degrees of freedom, but here we are modeling it as a single quantum particle with no structure, well congrats because that is definitely a very crude model. However we can say we are only looking at the center of mass of the tennis ball: quantum mechanics allows us to separate out the center of mass for special treatment in a similar way as occurs in classical mechanics. 
(2) I said the limit $n\rightarrow \infty$ is the classical limit. We can also phrase it as $\hbar\rightarrow 0$. To see this, note that $E_1\rightarrow 0$ as $\hbar\rightarrow 0$, so any particle with finite energy must have $n\rightarrow \infty$ to compensate.
A: The first question is what is the quantum number $n$ for the ball. This question makes the assumption that the wavefunction for the ball is in an energy eigenstate. I see no reason why this would be the case. I would think that shaking the ball back and forth would excite many modes in the container, but I am not sure about that. What I am sure of, though, is that for any energy eigenstate, $\langle X \rangle$ is always at the center of the tube. This can easily be verified by looking at the symmetry of the wavefunction. Thus there is no bouncing back an forth if the ball is in an energy eigenstate. However if the ball is, say, in a superposition of the ground state and the first excited state, then you can verify that $\langle X \rangle$ does oscillate back and forth with time. (disclaimer: I didn't actually do the computation myself, I just imagined the sum of the first excited mode with the ground state being nonsymmetrical, so $\langle X \rangle$ will be off to one side. After some time the first excited state will accumulate and extra $\pi$ of phase and so you will essentially subtract it from the ground state wavefunction and you see $\langle X \rangle$ is now the mirror image of what it was).
You have the correct procedure for the second part. (Assuming you can find an $n$ for the first part.)
