[Physics] Particle in a 1-D box and the correspondence principle

classical-mechanicsobservablesprobabilityquantum mechanicswavefunction

Consider the particle in a 1-d box, we know very well the solutions of it. I'd like to see how the correspondence principle will work out in this case, if we consider position probability density function (pdf) of the particle.

In the classical case, under the absence of any potential, its fair enough to assume the position pdf $P_{classic}$ of the particle to be constant inside the box and zero outside of it.

In the quantum case we know it is $$P_{quantum}(x) = \frac{2}{L} \sin^2(\frac{n\pi x}{L})$$ inside the box and zero outside.

A picture's worth a thousand word so here I grab a picture from a university website (actually google) for which I am thankful to the authors.

particle in a 1-D box

From the picture we can see that when we consider large enough quantum number $n$, we see that the $P_{Quantum}$ is still jiggling and no matter how large $n$ we consider, it keeps jiggling and doesn't really bother to converge to $P_{classic}$, but the consolation is, they match in an average sense.

Ideally, for such a fundamental problem like this, I would expect the function $P_{quantum}(x)$ to converge to $P_{classic}(x)$, pointwise. Am I asking for too much? There are infinite number of types of $P_{quantum}$ matching $P_{classic}$ in an average sense, so that would mean there can be infinite number of quantum mechanical theories which obey correspondence principle in average matching, which I think is not ery appealing property of a theory.

My question is, what can we do to make the pdf converge pointwise to the classical pdf?

Best Answer

It is worth noting that while the quantum PDF still exhibits rapid fluxuations, they get to be very high frequency, and when you go to measure the distribution with a physical instrument you introduce a resolution.

At some point that resolution gets to be much larger then the wavelength of the fluxuations and you in fact detect the classical distribution.

Related Solutions

Quantum vs Classical – Probability Density in Harmonic Oscillator Ground State

The correspondence principle is valid here, though it may not be immediately apparent from observation of the lowest order mode of the quantum mechanical oscillator. The higher order modes, by contrast, begin to redly display this feature as you increase the principle quantum number n. The following line of Mathematica code (taken from the Neat Examples section from the Plot function documentation) represents the probability amplitudes for the quantum simple harmonic oscillator:

f[n_, x_] := Abs[((1/Pi)^(1/4) HermiteH[n, x])/(E^(x^2/2) Sqrt[2^n n!])]^2

Plotting a series of these (which is also given in the documentation), gives the following:
_{(source: wolfram.com)}

As you can see, as you increase the quantum number (shown as an increase in height), the probability density actually starts to favor the region farther from the origin, as in the classical case. In fact, I've plotted the probability amplitude for n=50, which gives the following: n=50 State As you can see, this is much closer to the classical probability density. Naturally, this trend increases as we move to even higher n values.

[Physics] Particle in a box: simultaneously bounded momentum and position

The eigenstates for a particle in a box are neither states of definite momentum nor superpositions of just two momentum eigenstates of equal magnitude and opposite sign. For a box of length $L$ centered at the origin, the energy eigenstates are

$$ \psi_n(x) = \begin{cases} \sqrt{\frac{2}{L}} \cos \left( \frac{n \pi x}{L} \right) \qquad &\text{for $n$ odd} \\ \sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) \qquad &\text{for $n$ even} \end{cases}$$

inside the box and zero elsewhere. Fourier transforming via $\psi_n(k) := \int_{-L/2}^{L/2} e^{-i k x} \psi_n(x)\, dx$ gives

$$ \psi_n(k) = \begin{cases} (-1)^{(n-1)/2} 2 \pi n \sqrt{2 L} \frac{\cos(k L/2)}{(n \pi)^2 - (k L)^2} \qquad &\text{for $n$ odd} \\ (-1)^{n/2} 2 \pi i n \sqrt{2 L} \frac{\sin(k L/2)}{(n \pi)^2 - (k L)^2} \qquad &\text{for $n$ even} \end{cases}$$ and as we'd expect from the Paley-Weiner theorem, the wavefunctions have unbounded support in momentum space.

The reason that the eigenstates don't have definite momentum magnitude (as we'd expect) is that momentum is only a good quantum number in translationally invariant systems, and the hard walls in the Hamiltonian break translational invariance. If we were to impose periodic rather than hard-wall boundary conditions, then translational invariance would be restored and the energy eigenstates $\psi_n(x) = \frac{1}{\sqrt{L}} e^{i n \pi x/L} $ would indeed be states of definite momentum $k_n = n \pi / L$. But in this case, the spatial manifold is compact and periodic and so is topologically equivalent to the circle $S^1$ rather than the real line $\mathbb{R}$, so we have Fourier series that are only defined at discrete values in Fourier space instead of Fourier transforms, and the Paley-Weiner theorem doesn't apply.

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