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Can anyone give an example of when infinite-dimensional Hilbert spaces are required to describe a physical system? The standard answer to this question is yes, and I'm sure some of you will be quick to point out several clear examples:

Ex. 1: Photon number states $\left\{ |0\rangle,|1\rangle,...,|n\rangle \right\}$

Ex. 2: Harmonic oscillator number states (same as above).

Ex. 3: Continuous-variable basis of a single free particle $\left\{ |x\rangle \right\} \forall x \in \mathbb{R}$

From the infinite number of basis vectors in these examples, it is generally concluded that the dimension of the Hilbert space is also infinite.

However, it is clear that a photon number state such as $|\psi\rangle=|100\rangle$ is unphysical (to be clear, this is the Fock number state with $n=100$, not a tripartite state with one photon in one channel). If you dispute this, I would suggest you try preparing such a state in a single-mode fiber in an optics lab. For this reason, we can apply a (somewhat arbitrary) cutoff value of $n$ and recover a finite-dimensional Hilbert space.

Similarly, the cardinality of the basis set for a free particle is actually $\mathbb{N}$, not $\mathbb{R}$ (due to the properties of the function space), and although this set of orthogonal functions is infinite, we can also apply a similar cutoff for any reasonably behaved wavefunction.

Can anyone give an example of a system described by an infinite-dimensional vector space for which such a cutoff cannot be applied?

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    $\begingroup$ I'm confused about what you're asking exactly. Infinite dimensional complex Hilbert spaces certainly exist as mathematical objects. Moreover, they can be used to model certain physical systems accurately. It seems to me that you are asking if you can somehow always get away with modeling these same systems with finite-dimensional Hilbert spaces. Is my impression correct? If so, I'd humbly suggest a rewording of the question along these lines. If not, then please clarify. $\endgroup$ – joshphysics Jun 27 '14 at 3:28
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    $\begingroup$ $$[x,p]=i$$ does not have a finite dimensional representation. $\endgroup$ – user26143 Jun 27 '14 at 6:46
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Isn't your Ex 3 such an example?

In any event, you might have trouble producing a pure photon state $\left|\,001\right>$, but you should have no trouble creating a coherent state. Just turn on your laser pointer. The coherent state is a superposition of single photon states, summing over all single photon number states. You can't impose a cutoff.

See this Wikipedia article, in particular search for the phrase "This can be easily seen" and look at the equation that follows.

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  • $\begingroup$ Well, the occupation probability for every particle number is a Poissonian distribution, so it is peaked at a finite occupation number and then decreases rapidly for higher occupations. It seems to me you could impose a cutoff at an occupation number much higher than the expected number, and the resulting finite-dimensional wavefunction would only differ imperceptibly from an exact coherent state. $\endgroup$ – tparker Jul 25 '18 at 16:29
  • $\begingroup$ @tparker Yes, that's a good refinement to my answer. $\endgroup$ – garyp Jul 26 '18 at 1:28

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