Consider probability densities for a particle in the lowest energy state of a simple harmonic oscillator. The quantum mechanical probability density peaks near the equilibrium point and extends beyond the sharp limits of motion predicted by classical physics. The classical probability density is inversely proportional to the classical velocity and is greatest at the endpoints of the motion, where the velocity vanishes. My question clearly is

Is there a inconsistency between Quantum and Classic in probability density of harmonic oscillator in it's ground state? Is the correspondence principle valid here?

Sec. 5.3 Eisberg Resnick Quantum Physics


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: Quantum SHO States
(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.

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  • $\begingroup$ Thanks Groy. The problem is ground state. When we increase quantum number n everything is Good, but ground state of QM contradict with ground state of classic. Can we compare these two with each other? $\endgroup$ – Abolfazl May 25 '14 at 19:22
  • $\begingroup$ I think the complication here is the notion of a "classical ground state". Having a "ground state" is really more of a quantum idea, in which everything is discrete (or "quantized", whereas there is no such property for classical mechanics. The correspondence principle is the statement that, at sufficiently high energies (which, here is related to the principle quantum number n) the quantum result will approach the classical one. $\endgroup$ – GJStein May 25 '14 at 19:34
  • $\begingroup$ So we shouldn't compare ground state of classic with QM. Am I right? $\endgroup$ – Abolfazl May 25 '14 at 19:42
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    $\begingroup$ Again, I would avoid using the term "ground state" to refer to anything classical, but you seem to have the general idea. The ground state of a quantum mechanical system may be quite different from a highly excited state, which, by the correspondence principle, should approach the classical probability density. $\endgroup$ – GJStein May 25 '14 at 19:45
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    $\begingroup$ The problem with ground state (both quantum and classical) is that it is the state with minimum energy. In the classical regime that means no oscillation and, therefore, position at the equillibrium. The quantum ground state then represents smearing of this equilibrium and one has to look at excited quantum states if one wants to compare that to classical oscillations. $\endgroup$ – Ondřej Černotík May 25 '14 at 20:17

A simple answer to your question is provided by the Wikipedia link you gave to the correspondence principle:

In physics, the correspondence principle states that the behavior of systems described by the theory of quantum mechanics (or by the old quantum theory) reproduces classical physics in the limit of large quantum numbers.

The quantum number of the ground state of the SHO is zero. Consequently, while, as the answer above showed, the correspondence principal does hold quite nicely, it is by definition not necessarily meaningful in the ground state.

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Consider that you are comparing the ground state wave function to the classical case of a moving particle. The classical case corresponding (as much as possible) to the ground state would be the harmonic oscillator at rest, a delta function probability distribution. Consider a higher energy state (limit of high quantum number) to make a better comparison. See for example:


The linked article explains how the higher energy state's probability distribution begin to look more and more like the corresponding classical probability distribution.

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  • $\begingroup$ Thank you for your answer, when we increase n for quantum numbers are we still in ground state in classic? $\endgroup$ – Abolfazl May 25 '14 at 19:13
  • $\begingroup$ Superb answer to point out that the ground state of the classical system is actually a bunch of probability at the origin, and so the blob of probability at the origin in the quantum state is really not that different (even though it is spread out a bit). If fact I suspect if you convolute both the classical and quantum systems at any chosen energy state, with a suitable 'uncertainty Gaussian', you'll get the same result in each case. $\endgroup$ – user183966 Apr 19 '18 at 22:18

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