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This is obviously a follow on question to the Phys.SE post Hilbert space of harmonic oscillator: Countable vs uncountable?

So I thought that the Hilbert space of a bound electron is countable, but the Hilbert space of a free electron is uncountable. But the arguments about smoothness and delta functions in the answers to the previous question convince me otherwise. Why is the Hilbert space of a free particle not also countable?

Community
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Jim Graber
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  • A free particle and a harmonic oscillator both have the same Hilbert space $L^2(\mathbb{R}^n)$. – tparker Jan 27 '18 at 15:36

1 Answers1

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The Hilbert dimension of the Hilbert space of a free particle is countable. To see this, note that

  1. The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$.

  2. An orthonormal basis of a Hilbert space $\mathcal H$ is any subset $B\subseteq \mathcal H$ whose span is dense in $\mathcal H$.

  3. All orthornormal bases of a given non-empty Hilbert space have the same cardinality, and the cardinality of any such basis is called the Hilbert dimension of the space.

  4. The Hilbert space $L^2(\mathbb R^3)$ is separable; it admits a countable, orthonormal basis. Therefore, by the definition of the Hilbert dimension of a Hilbert space, it has countable dimension.

Addendum. 2014-10-19

There is another notion of basis that is usually not being referred to when one discusses Hilbert spaces, namely a Hamel basis (aka algebraic basis). There is a corresponding theorem called the dimension theorem which says that all Hamel bases of a vector space have the same cardinality, and the dimension of the vector space is then defined as the cardinality of any Hamel basis.

One can show that every Hamel basis of an infinite-dimensional Hilbert space is uncountable.

As a result, the dimension (in the sense of Hamel bases) of the free particle Hilbert space is uncountable, but again, this is not usually the sense in which one is using the term dimension in this context, especially in physics.

joshphysics
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  • So the total universe, or at least all of quantum mechanics, is countable? I was once told so by a very famous physicist, but I didn't think that was the majority position. – Jim Graber May 22 '13 at 07:53
  • But then what are rigged Hilbert spaces good for? I thought their whole point was mixing countable and uncountable in a consistent way?

    A quick look at Wikipedia finds “They can bring together the 'bound state' (eigenvector) and 'continuous spectrum', in one place.”

     – Jim Graber May 22 '13 at 07:53
  • @Jim I (and no other person I know) can speak with any amount of certainty about the Hilbert space of the universe. What I can tell you is that most physicists model the state of a free particle as a vector in $L^2(\mathbb R^3)$ whose dimension is countable. I'm not sufficiently familiar with rigged Hilbert spaces to say anything intelligent about their utility unfortunately. – joshphysics May 22 '13 at 08:03
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    @Jim If you're using the rigged Hilbert space formalism, your state space is actually a subspace of a countable Hilbert space. See here: http://physics.stackexchange.com/questions/43515/rigged-hilbert-space-and-qm – user1504 May 22 '13 at 11:39
  • Eigenvectors and continuous spectrum are properties of operators, not of the Hilbert space. A closed subspace of a seperable Hilbert space is also a seperable Hilbert space, if its dimension is infinite. All seperable Hilbert spaces are unitary equivalent althought they might look very different. – jjcale May 24 '13 at 17:46
  • @jjcale: It's not totally clear who you are addressing. – user1504 May 24 '13 at 17:48
  • @user1504: I'm adressing you and Jim Graber. – jjcale May 24 '13 at 17:53
  • @jjcale The subspace I'm talking about isn't closed. – user1504 May 24 '13 at 17:55
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    @user1504 : ok, but then it's not a Hilbert space. – jjcale May 24 '13 at 18:02
  • @jjcale: No one said it was. It's the domain of the algebra of observables. – user1504 May 24 '13 at 18:22
  • @user1504 : But the question is about Hilbert spaces. – jjcale May 24 '13 at 19:13
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    @jjcale Jim asked a second question (in his comment) about rigged Hilbert spaces. In this formalism, the state space is not actually a Hilbert space. Instead, it is a vector subspace of some Hilbert space. This is sufficient to answer Jim's question. For more details about the formalism, go see the answer to the question I linked to above. – user1504 May 24 '13 at 19:19
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    @joshphysics- How can free particle Hilbert space be $L^2(\mathbb{R}^3)$? The free particle states are momentum eigenstates and hence of the form $\sim e^{i\vec k\cdot \vec r}$, which are not square-integrable. Is there any countable basis in the free particle case? – SRS Feb 21 '14 at 07:06
  • @Roopam The momentum eigenstates are not strictly speaking elements of the Hilbert space of the free particle. Every honest-to-goodness basis for the Hilbert space of the free particle is countable. One example of such a basis is given here http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions , as mentioned in the answer. – joshphysics Feb 22 '14 at 03:45
  • @joshphysics- Okay. But I don't understand why do we throw away this set ${e^{i\vec k\cdot\vec r}}$ out of the Hilbert space? Since these are not square integrable it implies that they does not belong to $L^2$. That is fine. But we could have extended the Hilbert space from $L^2$, to something that includes these functions as well. Isn't it? Does it imply this solutions are not physically acceptable? But during scattering problems and other cases we do use such functions? What is the Hilbert space there? Do we always use $L^2$ in quantum mechanics? To save Born's probability interpretation? – SRS Feb 22 '14 at 07:14
  • @Roopam I'd recommend that you investigate rigged Hilbert spaces which result precisely from attempts to include non-square-integrable functions into the state space. When we do scattering problems etc, it is useful to use plane waves to derive certain results, but ultimately, when we want to use probabilistic interpretations of quantum mechanics as you indicate, the physically allowable states need to be $L^2$. – joshphysics Feb 22 '14 at 08:12
  • @joshphysics-From Dirac's Principles of Quantum Mechanics "The space of bra or ket vectors when the vectors are restricted to be of finite length and to have finite scalar products is called by mathematicians a Hilbert space. The bra and ket vectors that we now use form a more general space than a Hilbert space." page-40. If we don't enlarge the hilbert space of quantum mechanics from $L^2$, scattering states of $H-$atom can't live there but don't think there is a reason to consider them unphysical. – SRS Feb 26 '14 at 08:05
  • @joshphysics- Rather, I will give up the absolute probability interpretation and look for relative probability as Landau and Lifshitz says. But indeed you were right, ${e^{ikx}}$-states are unphysical, so are ${\delta(x-x_0)}$. But all scattering states are not unphysical like those of H-atom. So those must be included in the hilbert space. Am I right? – SRS Feb 26 '14 at 08:08
  • @SRS description of scattering does not require us to step out of Hilbert space. The common usage of $e^{ikx}$ is due to fact that is makes calculations simpler. To do it more correctly, one should do the scattering calculations with wave packets described by normalized psi functions. So instead of representing the incoming particle as infinitely wide wave(which is obviously non-physical) one would represent it as a wave limited in its spatial extension (the width of the packet being determined by the experimental setup, e.g. wave width would be limited by apertures in the particle source). – Ján Lalinský Feb 02 '19 at 15:19