10
$\begingroup$

The energy level of a simple harmonic oscillator is $E_n=(n+\frac{1}{2})\hbar\omega$. Is there any physical explanation why these levels are equally spaced ($= \hbar\omega$)?

Maybe this link can be helpful.

$\endgroup$
12
$\begingroup$

Yes. There is a simple explanation.

The classical harmonic oscillator has a well defined frequency $\omega$, independent of initial conditions. This can only happen if the quantum system has precisely equally spaced energies with gap $\hbar \omega$.

The reasons for this is that motion in a quantum system can only happen if more than one energy level is occupied. The frequency of the motion is then set by the energy difference of the different occupied energy levels. Thus if every classical trajectory has the same frequency $\omega$, then any combination of quantum states must also have frequency differences that are multiples of $\hbar\omega$.

The argument can be made even more refined by thinking about the existence of coherent states and semi-classical superpositions in the quantum system, but the basic argument is as above.

The reverse argument can also be used to conclude that the hydrogen atom will not have equally spaced energy levels, since the frequency for motion in a Coloumb potential depends on the initial conditions.

$\endgroup$
  • 1
    $\begingroup$ I was going to bring up the hydrogen atom as a counterexample but you brought it up first: classical motion in an inverse distance potential seems to me to have a well defined frequency - isn't the earth-sun system an example of such? $\endgroup$ – NickD Jul 18 '17 at 16:23
  • 1
    $\begingroup$ @MikaelFremling Isn't the frequency always independent of initial condition? What is an example where the frequency depends on initial condition? $\endgroup$ – SRS Jul 18 '17 at 17:24
  • 4
    $\begingroup$ @Nick the length of a year in the Earth-sun system is a function of the semi-major axis (that is it is a function of energy). The frequency of states in the hydrogen atom is also a function of energy. The frequency of the harmonic oscillator (both classical and quantum) is independent of energy. $\endgroup$ – dmckee Jul 18 '17 at 17:26
  • 1
    $\begingroup$ In a sense that is a re-statement of your question, but as there are potnetial curves in which frequency falls with energy (like the hydrogen atom) and those in which it increases with energy (like the square well) then there must be a(t least one) potential where it is a constant function of energy. $\endgroup$ – dmckee Jul 18 '17 at 17:28
  • 1
    $\begingroup$ Now, finding the special case from a standing start is an interesting problem. See the brachistochrone problem for instance. I think the easy way to show that the harmonic oscillator is that model is probably best approached int he Lagrangian formalism. $\endgroup$ – dmckee Jul 18 '17 at 17:32
2
$\begingroup$

I might as well condense the neat phase-space argument of Royer's 1996 pedagogical paper, since it is so short and standard for the phase-space behavior of the QHO, and the other answers did not utilize it.

The essential point is that, both classically and quantum mechanically, the time evolution of the oscillator is rigid rotation in phase space, as discovered by Groenewold in 1946.

That is to say that, just as in classical evolution, any quantum configuration is uniformly rotated in phase space by the hamiltonian. Absorbing all constants into the variables to rescale elliptic trajectories to circles, the hamiltonian (classical and quantum) presents as $$ H=\tfrac{1}{2} (p^2+x^2). $$ Consequently, remarkably, the action of the hamiltonian on any phase space configuration, classical Liouville density or Wigner function, amounts to $$ \{\{ H,f\}\}= \{ H,f\}= (x\partial_p- p\partial_x)~ f(x,p), $$ readily recognized as rigid rotation in phase space, since, exceptionally here, the quantum Moyal Brackets $$\{\{H,f\}\}=\left (H\left (x+{i\hbar\over 2} \overrightarrow{\partial }_{p} ~,~ p-{i\hbar\over 2} \overrightarrow{\partial }_{x}\right ) ~ f(x,p) -H\left (x-{i\hbar\over 2} \overrightarrow{\partial }_{p} ~,~ p+{i\hbar\over 2} \overrightarrow{\partial }_{x}\right ) ~ f(x,p)\right )/i\hbar $$ collapse to classical Poisson Brackets (!).

Rigid rotation around a circle, of course, has integral eigenvalues, as its angular eigenfunctions are single-valued exponentials. This is a feature of both the Liouville operator and the QM Hamiltonian.

The Fourier modes of the classical propagator also involve integrally spaced energy levels, but in classical mechanics this type of separation of variables is not necessary, or useful. Standard phase-space QM reviews further solve the radial $\star$-genvalue stationary equation (a Laguerre equation, not a Hermite one!) to confirm the eigenvalues are necessarily integrally spaced, as demonstrated here.

Note the integral eigenvalue spectrum is not a mere consequence of phase space U(1) symmetry: it is a feature of the Hamiltonian being the very generator of rotations itself! A Hamiltonian such as $(p^2+x^2)^a_\star$ corresponding to the a-th power of the Hilbert space oscillator Hamiltonian would still be symmetric under x-p rotations, but would have a spectrum of $(n+1/2)^a$, of course.

The argument, sadly, does not prove uniqueness of this hamiltonian for that spectrum, but a reasonable presumption given such a spectrum is the existence of a map proving equivalence of the fortunate hamiltonian to that of the oscillator, above.

$\endgroup$
  • $\begingroup$ Rotational symmetry. One of the more powerful tools we have. $\endgroup$ – user93146 Jul 1 '18 at 1:44
0
$\begingroup$

This is probably not a physical explanation but it's worth to track the speciality of the Harmonic oscillator that leads to equispaced energy levels in quantum mechanics.

$\bullet$ A Hamiltonian of the form $H=\hbar(a^\dagger a+1/2)$ leads to equally space energy levels.

$\bullet$ The Hamiltonian of a simple harmonic oscillator is given by $$H=\frac{1}{2}(p^2+x^2)$$ in units where $m=\omega=1$. Both the kinetic and the potential energies are quadratic. Whenever the potential is quadratic, the Hamiltonian can be reduced to $$H=\hbar(a^\dagger a+1/2)$$ which is not possible if the potential $V(x)$ is not quadratic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.