Particle in 1D Infinite potential well = a ball

Question

Here are the two equations I'm concerned with: $$\Psi = \sqrt{\frac2a}\sin\left(n\frac{\pi x}a\right)$$ $$E = n^{2}\frac{\hbar^2π^2}{2ma^2}.$$ If we have a ball with mass 1 kg, confined in a 1 m infinite potential well, and its kinetic energy is 1 J, then the corresponding $n$ will be very high. As a result we get a very frequently oscillating probability density function. Which essentially says that the probability of finding the ball is pretty much the same everywhere for human eyes/senses.

But why is the probability of finding the ball the same everywhere?

In reality, we would find (by seeing) the ball at one particular place at a particular time.

So what is the probability function telling me? Which probability is equal everywhere (because the probability of finding ball changes with time if it has some energy)?

Answer 1

Which essentially says that the probability of finding the ball is pretty much same everywhere for human eyes/senses.

But why is the probability of finding the ball same everywhere?

Firstly, because we are talking about probability, and then because we are talking about quantum mechanics:

• The probabilistic aspect means that we have the same probability to find a ball in any part of the well. That is, every time we perform our experiment, we find the ball in a different place, but the frequency with which it is found in different regions of equal size (e.g., in the right and left halves of the well) is the same
• Quantum mechanics means that our measurement is not only probabilistic, but that we cannot perform the measurement twice on the same system, since a measurement collapses the wave function and changes the system state (at least in the Copenhagen interpretation.) Thus, we necessarily work with an ensemble - a large set of wells with balls, all prepared according to the same protocol, and each measured only once. The probability then predicts the averages over the ensemble.

Finally, what might be still confusing here is thinking of a macroscopic object as a quantum one. That is, why a real ball does not behave as a quantum particle? The question has been asked many times in different versions in this community. Here are a few links:
Do double slit experiment with big balls
Will tennis ball produce same interference pattern in double slit experiment if everything was scaled up?
Do physicists distinguish things that can be explained through the movement of matter as opposed to some unseen or non-material forces?

Answer 2

The equation you gave for $\Psi$ is an eigenstate of the system. This means that is has one single energy and also that the probability of finding a particle at a certain position doesn't change in time. The probability density is the square of the wavefunction $|\Psi|^2$ and for eigenstates it doesn't oscillate in time, in fact it is constant. Realistic states are often a sum of multiple of these eigenstates. The probability functions of these states is not constant. The wavefunction can move from left to right and change shape and do all sorts of stuff. Below I showed an eigenstate, $\sin x$, a sum of two eigenstates and finally a cleverly constructed state that is localized in a single spot. This last state is how the wavefunction of your ball would look like: the bigger an object, the more localized your object will become. If your ball has velocity, this peak will move at constant velocity from left to right and bounce when it reaches the edge of the potential. If you want I could make a gif of this.

Note: the scale and normalization of these functions is completely wrong, you should just look at the shape.

Answer 3

It would indeed have a homogeneous probability distribution, if one could somehow manage to put the ball into a pure quantum state with a fixed $n$. However, under the conditions you gave, the energy spacing between neighboring levels is about $\delta E \sim 10^{-10}$eV. Given that for an optical photon $\hbar \omega \sim 1$eV, you cannot just "find (by seeing)" anything without seriously messing everything up/entangling billions of different eigenstates together.

Once you do mess/entangle everything up "by seeing", you end up with a random gaussian probability distribution (very narrow) for the ball position corresponding to classical physics.

Answer 4

The probability of finding the ball is not the same everywhere. The frequency of oscillation is very high, but the amplitude is exceedingly tiny. It wiggles a tiny amount but overall moves at the classically expected speed.

This is the "classical limit" of quantum physics: quantum physics gives the same results as Newtonian physics at large (i.e. non-microscopic) scales.

Answer 5

A clear explanation is provided by AccidentalTaylorExpansion's answer. The wave packet( not an energy eigenstate wave function, but a linear combination of some energy eigenstates) moves as if it is a classical particle. This is a time-dependent problem, you cannot get the answer by considering the energy eigenstates by the time-independent Schrodinger equation $H\psi = E\psi$ only, you have to solve the time-dependent Schrodinger equation $i\hbar\frac{d}{dt}\psi=H\psi$. Although you solve TDSE by expanding it with energy eigenstates, they are two different equations. You can consult quantum mechanics textbook by Cohen-Tannoudji about the movement of a wave packet and the information it provides.

Besides, one point can be added. You can find that the possibility that the particle is not located at its classical position is never zero. You will never get a moving delta function. But that concerns nobody because nothing will happen if a ball in a 1D infinite potential well has no specific position. This ball is constrained in a well! It can harm nothing!

A wave function $|\psi \rangle$(, a "state" equivalently,) is an abstract object with a lot of information. You can represent it in the position basis, get a wave function in the position basis $\psi(x)\equiv\langle x|\psi\rangle$, and calculate the possibility of finding it at the position range $[a,b]$ using Born's rule $P([a,b])=\int_{a}^{b}|\psi(x)|^{2}dx$. But you can do these just because the position basis is complete in this problem, not because the observable position $X$ is "physical" in this problem. In classical mechanics, the position $X$ is always a physical observable. You use the pair $(x,p)$ to label a state in Hamiltonian mechanics. However, quantum mechanics tells that this experience is wrong, the position is not necessary when labeling a state.