I am just wondering what does the SHO system from quantum mechanics actually physically represent? Is it just a SHO of a quantum particle, seems a little too obvious for quantum theory?

I'm from a straight maths background so I don't usually pry into the physical aspect as long as the maths is solid but I've recently found that examining the meaning of the maths often makes it a lot easier to understand. Better late than never I guess.

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    $\begingroup$ My first theoretical physics professor motivated the importance of the classic harmonic oscillator with the following joke: "When theoreticians have to describe a cow, they will automatically assume that she is spherical. And if that's not enough to solve the problem, they will additionally assume that she is homogeneously covered in milk". Ergo, WYSIWYG, a harmonic quantum oscillator. It's one of the few systems that we can solve completely, the solution space is easy to understand and there is hope that it's a useful toy system for everything from phonons to quantum gravity. $\endgroup$
    – CuriousOne
    Dec 20, 2014 at 18:30
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    $\begingroup$ @CuriousOne: "The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction." - Sidney Coleman. $\endgroup$
    – JamalS
    Dec 20, 2014 at 18:41
  • $\begingroup$ @JamalS: LOL! That's a really good one! $\endgroup$
    – CuriousOne
    Dec 20, 2014 at 18:45
  • $\begingroup$ @PeterBrown: I am always suspicious on math results s.t. I always check if they make sense. $\endgroup$
    – Sofia
    Dec 20, 2014 at 20:00
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    $\begingroup$ @PeterShor: Indeed, it is extremely close to being harmonic. The deviations from harmonic behavior come only from the fact that the electromagnetic field can interact with charged particle fields. At high enough energies (really ridiculously high) the electromagnetic wave can create (virtual) charged particles which then interact back with the electromagnetic wave and spoil the harmonic behavior. $\endgroup$
    – DanielSank
    Dec 21, 2014 at 2:03

3 Answers 3


I had missed this question.

Most potentials physicists use to model the behavior of atoms and molecules, are symmetric in space. When the potential is unknown it is usual to guess at a first term in a Taylor expansion . For symmetric potentials the first term in a Taylor series expansion is the x**2, and that is why the use of the harmonic oscillator is ubiquitous.

The current prominent theory of everything for particle physics is string theory , where particles are vibrational modes on the basic one dimensional string. It is amusing to think that maybe this also is a first approximation to a much more complicated formalism.


It turns out, as a general fact, that theories which contain only quadratic terms in position and momentum can be solved as a linear combinations of SHO (just look at the SHO Hamiltonian and you will find why). The quantum treatment of the harmonic oscillator enriches the interpretation adding the creation and annihilation operators which (as you can guess) can create or destroy the excitations of the SHO. Now: imagine you have a Lagrangian containing only quadratic terms in both position and momentum, and you resolve it finding a combination of SHO. The $\hbar\omega$ found for this SHO brings the physical meaning: if $\hbar\omega = \sqrt{m^{2} + p^{2}}$ there you have a particle, if $\hbar\omega = 2\omega|sin(\frac{ka}{2})|$ there you have phonons (sound).


If you have two particles with an interparticle potential of $$V(\vec{r}_1,\vec{r_2})=\frac{1}{2}k(\vec{r}_1-\vec{r_2})\cdot(\vec{r}_1-\vec{r_2})$$ then you can find solutions by separation of variables for the center of mass (as a free particle) as for the relative separation. The relative separation solution will be a SHO solution. If one of the particles is much more massive, then you can interpret the position of the relative position as the location of the lighter particle.

In practice a uniform potential just shifts the energy which doesn't affect much and if a potential has a minimum, then that minimum might locally look similar to a upward opening quadratic, so the lowest energy solutions to that potential might look somewhat similar to a SHO.

But really physicists study it because: 1) they can, it's a solvable problem an d2) it's a model problem in that techniques to solve it can be used in other situations.


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