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?
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asked Dec 25, 2015 at 21:40
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2$\begingroup$ 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. $\endgroup$– CuriousOneDec 25, 2015 at 22:33
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1$\begingroup$ I am talking about "classical" quantum mechanics here, I might have misunderstood your comment, but if so will be too cryptic for the OP $\endgroup$– user83548Dec 25, 2015 at 22:38
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2$\begingroup$ @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. $\endgroup$– CuriousOneDec 25, 2015 at 22:50
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1$\begingroup$ @CuriousOne Then I drank too much, I'll check tomorrow :) $\endgroup$– user83548Dec 25, 2015 at 22:51
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1$\begingroup$ 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.. $\endgroup$– CuriousOneDec 26, 2015 at 6:46
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1 Answer
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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.
answered Dec 28, 2015 at 1:58
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$\begingroup$ Thanks for your clear answer. I didn't understand your latter counter-example. The hydrogen atom energies are discrete, isn't it? $\endgroup$ Dec 28, 2015 at 18:28
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$\begingroup$ @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. $\endgroup$– ACuriousMind ♦Dec 28, 2015 at 18:34