Probability of finding a particle in 1D box from classical view

The probability of finding a particle in a 1D-Box (from the classical view, the particle behaves as a particle but not a wave) is the same everywhere in the box. This seems to me quite counterintuitive since i expect that the probability of finding the particle would be 1 at one specific value of $$x$$ and 0 at all other $$x$$. I think like so because even when the probability is the same everywhere, you still cant be sure where the particle is. But from the classical view, the particle should be localized at a point, shouldnt it?

• Classically, the particle is bouncing back and forth between the walls. – Pieter Nov 24 '18 at 12:16
• Based on your question you don't appear to understand probability and probability distributions. And when you say "classical" you also need to specify a velocity/position profile, because classically you might have anything. Your question is under-specified. – Bill N Dec 4 '18 at 23:14

What the probability distribution looks like depends on how much knowledge you are assuming that you have about the initial conditions and the time at which you're looking at the particle.

If you know the particle's initial position and velocity to infinite precision, and you know exactly how long you want to wait before making a measurement, then yes, the probability distribution will be a delta function.

But if there's any uncertainty in any of these quantities, then the probability distribution will be wider. If there's an uncertainty in the initial position $$\sigma_x$$, but the initial velocity and time interval are known exactly, then the probability distribution at any time will have width* $$2\sigma_x$$. If there's an uncertainty in the initial velocity $$\sigma_v$$, but the initial position and time interval are known exactly, then the probability distribution will start as a delta function and widen as a function of time, with a width of $$2\sigma_v \Delta t$$ after a time interval $$\Delta t$$. If there's an uncertainty in the time of interest $$\sigma_t$$, but the initial position and velocity are known exactly, then the probability distribution will have a width $$2v\sigma_t$$ for a particle moving at velocity $$v$$. For uncertainties in multiple quantities, the expressions are more complicated, but will in general be wider than the sum of the two widths assuming exact knowledge.

In the particular situation you're talking about, typically the assumption we make is that we don't know the particle's initial conditions, nor do we know how much time has passed since the particle started moving (so the uncertainty in every quantity is large). As such, there's no reason to expect that it's more likely for the particle to be at one point in the box more than another point, which gives you a uniform probability distribution.

*The definitions of uncertainty and width I am using here are as follows: if you know that your particle has position $$x=1\pm 0.05$$, then the uncertainty in position $$\sigma_x=.05$$ and the width of the probability distribution is $$0.1$$. This also refers to the maximum width of the probability distribution as a function of position of the center; if the center is close enough to the boundary of the box, then the width will be smaller.

You can certainly think about a classical particle which is stationary at a point $$x_0$$ in the potential well, with a probability distribution resembling $$\delta(x-x_0)$$ (note that $$\int^\infty_{-\infty}\delta(x-x_0)\mathrm{d}x=1)$$. There's no need to think of a wavefunction and think about the counterpart of $$\left|\psi\right|^2$$ or anything. I'm just talking about the position probability distribution, and a simple delta function is a perfectly valid one, classically speaking.

But the energy of this particle will be zero, and it will have no quantum mechanical counterpart (because we know that a zero-energy quantum particle would have a stupid wavefunction like $$\psi(x)=0$$). If you want to compare a classical particle in a box to a quantum model, you'll need to look at two systems with the same total energy, otherwise you're comparing apples and oranges. And if you consider a classical particle with nonzero kinetic energy, the probability of it being in any particular $$\mathrm{d}x$$ is the same, because it's undergoing elastic collisions with the walls of the well and will hence be moving with a constant speed. Of course, we'll need some trivial assumptions: like the time spent in contact with the walls must be negligible.

• And the comparison with classical physics may be more enlightening when one regards the harmonic oscillator and the classical turning points. – Pieter Nov 24 '18 at 12:44

Others answered already, but I would offer an analogy that may help.

If someone showed you a black box and said that there's a classical particle in there, then asked you to guess where the particle is, you would probably answer that it could be anywhere, even though it is actually somewhere. In probability language you are assigning a uniform prior probability to the position of the particle. "Prior" as in "before you have a chance to look for yourself."

If that someone asked you to guess if that particle is in the left half of the box or the right half, and you picked at random, you would be correct half of the time.

To summarize, in statistical mechanics the particle always knows where it is; we don't, and so we guess.