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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?

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  • $\begingroup$ Wonderful question. I'm thinking about its probably that only restriction we put on quantum theory is that it should be consistent with the observed spectrum of various physical quantities plus it should reproduce the classical result at plank's constant goes to zero. But whether it should match point wise i don't know. Waiting for a good explanation. $\endgroup$
    – Aftnix
    Commented Dec 24, 2013 at 12:52

2 Answers 2

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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.

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  • $\begingroup$ So you rely on lack of measuring instruments! What if they did converge pointwise, would that mean the lack of measuring device argument disappear from the QM scene? $\endgroup$
    – Rajesh D
    Commented Dec 25, 2013 at 6:55
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My question is, what can we do to make the pdf converge pointwise to the classical pdf

Answer : Nothing, if you accept Quantum principles.

You may see below an other example with the harmonic oscillator, with $N= 50$, and you see the same behaviour you are worrying about (classical behaviour in blue, quantum behaviour in red):

The fundamental reason is that Quantum Mechanics is about correlation amplitudes and probability amplitudes, so classical statistical concepts like correlations probabilities or probabilities is a "baby" view of the physic world. So, you cannot avoid " jiggling". The simplest approach is that probabilities amplitudes, or correlation amplitudes, for bound states, are "naturally" oscillating, this been traduced in probabilities as oscillations between zero and "maximum" values, while, for great quantum numbers, a "mean" looks like a classical probability.

enter image description here

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  • $\begingroup$ Merry Christmas!, and thanks for the answer!....What I find interesting from your answer is the phrase "probabilities is a "baby" view of the physic world. So, you cannot avoid " jiggling".... Well my big question is a bit hypothetical, Suppose, for suppose, If the QM is in tact with all its explanation of physical world, (spectra of hydrogen atom, Harmonic oscillator, blah..blah..blah.)...BUT its pdf converged pointwise to that of pdf of classical theory in limit n goes infinity, just like what I was expecting in this question...Physically and philosophically what would that mean? $\endgroup$
    – Rajesh D
    Commented Dec 25, 2013 at 4:13
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    $\begingroup$ Rajesh D: It would not mean anything physically or philosophically, because you would be proposing a model that cannot be reconciled with observable reality. Quantum mechanics converges with classical only because the wavelengths become fine enough so as to become negligible by classical standards. Spin is another example: You start out with up and down only, then start adding finer and finer angles as the spin number increases. The single greatest message of QM is that at the bottom, it's all waves - and waves insist on doing things like having null points that cannot be smoothly classical. $\endgroup$
    – Terry Bollinger
    Commented Dec 25, 2013 at 7:09
  • $\begingroup$ @TerryBollinger : Suppose, for suppose, If the QM (or a new version of it) is in tact with all its explanation of physical world, but still its pdf converged to classical one in pointwise.... $\endgroup$
    – Rajesh D
    Commented Dec 25, 2013 at 7:16
  • $\begingroup$ @Trimok : You have clculated the blue curve, what is the formal expression did you use for, in classical terms, I mean $x, V,t$ etc., $V$ being potential. $\endgroup$
    – Rajesh D
    Commented Dec 25, 2013 at 12:49
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    $\begingroup$ @RajeshD : In fact, you must write a particular solution of the equation of movement, so in the case of the harmonic oscillator : $x = A \cos \omega t$. Here the potential is $V(x)=\frac{1}{2} m \omega^2 x^2$. The total energy is $E=\frac{1}{2} m \omega^2 A^2$. The probability is proportionnal to the inverse of the speed, that is $p(x) \sim \dfrac{1}{|\dot x(x)|}$. So you have to express $\dot x$ as a function of $x$, here you have $|\dot x| = A\omega |\sin \omega t|=A\omega \sqrt{1-\dfrac{x^2}{A^2}}$, so finally $p(x) \sim \dfrac{1}{\sqrt{1-\dfrac{x^2}{A^2}}}$,... $\endgroup$
    – Trimok
    Commented Dec 25, 2013 at 13:02

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