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We know that position $\hat{r}$ and momentum $\hat{p}$ are both continuous spectrum operators, i.e. $$\hat{r}|r'\rangle=r'|r'\rangle, \quad \hat{p}|p'\rangle=p'|p'\rangle.$$ But the angular operator $\hat{L}=\hat{r}\times\hat{p}$ is not: $$\hat{L}^2|l\rangle=\hbar^2 l(l+1)|l\rangle.$$ Could anyone give some explanation?


marked as duplicate by Aaron Stevens, Qmechanic quantum-mechanics Oct 15 '18 at 4:06

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  • 4
    $\begingroup$ Perhaps it is related to the boundary conditions? $\endgroup$ – K_inverse Oct 15 '18 at 3:50
  • $\begingroup$ Are you asking for a mathematical reason from QM, or are you asking why the universe works this way? $\endgroup$ – Aaron Stevens Oct 15 '18 at 3:53
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/29655/2451 , physics.stackexchange.com/q/39208/2451 and links therein. $\endgroup$ – Qmechanic Oct 15 '18 at 4:01
  • $\begingroup$ Also note that the equations you give do not say anything about whether or not the eigenvalue Spectra are discrete or continuous. $\endgroup$ – Aaron Stevens Oct 15 '18 at 4:07

One possible answer: boundary conditions.

If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.

On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.

Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.

  • $\begingroup$ But even in bound states position is continuous. $\endgroup$ – Aaron Stevens Oct 15 '18 at 4:03
  • $\begingroup$ right, position is continuous, while momentum can be discrete, if the boundary conditions are such to force it to be. why that's true might be anthropomorphic - we experience the tangent space (position) more directly then the cotangent space (momentum). $\endgroup$ – levitopher Oct 16 '18 at 13:38

In general for a bounded state, its energy spectrum is discrete. Free particles on the other hand has continuous energy spectrum. This is true for a rotating motion as it is also bounded, i.e. the starting point is exactly equal to the end point upon rotation by $2\pi$, therefore its eigenvalues are also discrete.

  • $\begingroup$ But even in bound states position is continuous. $\endgroup$ – Aaron Stevens Oct 15 '18 at 4:08
  • $\begingroup$ Momentum is discrete though. This can be due to Heisenberg's uncertainty principle. More precise values of momentum (discrete selection of possible momenta) leads to less precise values of position (continuous selection). This can be mathematically seen by Fourier transform between the momentum and position eigenfunctions. $\endgroup$ – Gradient137 Oct 15 '18 at 4:15
  • $\begingroup$ Whether or not a spectrum is discrete has nothing to do with the HUP or the relation of Fourier transforms. Continuous does not mean less precise. By that logic then position and momentum cannot gave the same type of spectrum, which is not true. $\endgroup$ – Aaron Stevens Oct 15 '18 at 7:30

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