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In the von Neumann axioms for quantum mechanics, the first postulate states that a quantum state is a vector in a separable Hilbert space. It means it is assumed the Hilbert space has a basis with at most infinite countable elements (cardinality). In other words, it states that the energy of an arbitrary system is discrete. Is it always true? If not, can you make an specific example in nature that its eigen-energies are uncountable?

ACuriousMind
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heaven-of-intensity
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  • This might help: http://physics.stackexchange.com/q/65457/ –  Dec 25 '15 at 22:16
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    The spectrum of linear operators contains both discrete and continuous sets, so energy eigenvalues can be discrete and continuous (see the hydrogen spectrum). Once you couple these systems to the vacuum fields, all spectra, even line spectra, become continuous. – CuriousOne Dec 25 '15 at 22:33
  • @CuriousOne that is actually a sleight of hand –  Dec 25 '15 at 22:34
  • @brucesmitherson: Not at all. One is an important result of linear operator theory, the other one is physical reality: there are no and there can be no systems with infinite Q. – CuriousOne Dec 25 '15 at 22:35
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    I am talking about "classical" quantum mechanics here, I might have misunderstood your comment, but if so will be too cryptic for the OP –  Dec 25 '15 at 22:38
  • @brucesmitherson: Even in non-relativistic quantum mechanics the atomic state couples to the electromagnetic field and because of the third law of thermodynamics that field has a finite temperature. You can't escape the finite line width of atomic transitions in either theory and certainly not in real phenomenology. – CuriousOne Dec 25 '15 at 22:41
  • @CuriousOne see, your comment was too cryptic, I totally missed it. I believe then that we agree that a continuous spectrum is unphysical? –  Dec 25 '15 at 22:45
  • or I drank too much in xsmass –  Dec 25 '15 at 22:45
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    @brucesmitherson: It's the other way round: there are no discrete spectra. That's a good approximation, though, for transitions that have very narrow line width. Like with anything else in physics, you pick your approximation that is appropriate for the precision of your measurements. If your spectrograph has low resolution, you see line spectra, with high resolution you see continuous lines. In some cases the line width is so small that it becomes the reference for all other time/frequency measurements, that's the systems we use in atomic clocks. Those clocks still have finite Q resonators. – CuriousOne Dec 25 '15 at 22:50
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    @CuriousOne Then I drank too much, I'll check tomorrow :) –  Dec 25 '15 at 22:51
  • @CuriousOne yes. The specturum of linear operators can be continuous. Can you make an specific example in "nature" that its energy is continuous? – heaven-of-intensity Dec 26 '15 at 06:37
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    The energy of the free particle is continuous, so is the spectrum of any atom, molecule etc. past the ionization energy. The electrons in metals are occupying an essentially continuous spectrum etc.. – CuriousOne Dec 26 '15 at 06:46
  • @CuriousOne I see more clearly today! nice point I have not thought about it that way, however, it seems to be against any other answers (see previous link). Perhaps you should post it as an answer so it can be voted or commented by a wider audience. –  Dec 26 '15 at 15:12

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One counterexample is so simple as to be trivial:

The free particle has a completely continuous energy spectrum, since the Hamiltonian $H = \frac{p^2}{2m}$ has $[0,\infty)$ as its spectrum (this follows directly from $p$ having completely continuous spectrum $(-\infty,\infty)$. The reason this does not violate the spectral theorem/the countability of the basis of the Hilbert space is that this $H$ is an unbounded operator, and the "eigenstates" $\lvert p \rangle$ (which are $\psi_p(x) = \mathrm{e}^{\mathrm{i}px}$ in the position wavefunction representation) are not inside the Hilbert space (just note that $\psi_p(x)$ is not square-integrable on $\mathbb{R}$ to see that it is not in the canonical space of position wavefunctions $L^2(\mathbb{R},\mathrm{d}x)$). Only wavepackets, i.e. square-integrable superpositions of the plane waves $\psi_p(x)$, lie inside the Hilbert space of states.

In fact, the "eigenstates" belonging to continuous eigenvalues are never "normalizable", cf. this phys.SE question, and thus never are vectors in the actual Hilbert space, but only in the larger space of the so-called rigged Hilbert space, cf. this phys.SE question

Another counterexample would be the hydrogen atom, where the energies above a certain threshhold are continuous, and are again essentially free states.

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ACuriousMind
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  • Thanks for your clear answer. I didn't understand your latter counter-example. The hydrogen atom energies are discrete, isn't it? – heaven-of-intensity Dec 28 '15 at 18:28
  • @K.N.O.: The bound state energies are discrete. But there is a continuum of free (or "scattering") states (or rather eigenvalues, since the associated things are not, strictly speaking, prober states) above the "ionization energy", see this phys.SE question for an introduction and further references. – ACuriousMind Dec 28 '15 at 18:34