Hilbert space of harmonic oscillator: Countable vs uncountable?

Question

Hm, this just occurred to me while answering another question:

If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.

On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.

Clearly they can't both be right, so where is the flaw in my logic?

This post imported from StackExchange Physics at 2014-04-24 02:29 (UCT), posted by SE-user Lagerbaer

Comments

An important keyword here is separable.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Emilio Pisanty

Answer 1

The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^{2}(\mathbb{R})$ of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line.

The Dirac delta distribution $\delta(x-x_{0})$ is not a function. In particular, it is not square integrable, cf. this Phys.SE post.

One may prove that all infinite-dimensional separable complex Hilbert spaces are isomorphic to the set $${\ell}^{2}(\mathbb{N})~:=~\left\{(x_n)_{n\in\mathbb{N}}\mid\sum_{n\in\mathbb{N}} |x_n|^2 <\infty\right\}$$ of square integrable complex sequences.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Qmechanic

Answer 2

This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-)

The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a limiting case. This is essentially done by considering a UV regulator for your wavefunctions in space. Let's solve the simpler "particle in a box" problem, on a lattice. The answer for the harmonic oscillator will conceptually be the same. Also note that solving the problem on a lattice of size $a$ is akin to considering rectangular functions of width $a$ and unit area, as regulated versions of $\delta$-functions.

The UV-cutoff (smallest position resolution) becomes the maximum momentum possible for the particle's wavefunction and the IR-cutoff (roughly max width of wavefunction which will correspond to the size of the box) gives the minimum momentum quantum and hence the difference between levels. Now you can see that the number of states (finite) is the same in position basis and momentum basis. The subtlety is when you take the limit of small lattice spacing. Then the max momentum goes to "infinity" while the position resolution goes to zero -- but the position basis states are still countable!

In the harmonic oscillator case, the spread of the ground state (maximum spread) should correspond to the momentum quantum i.e. the lattice size in momentum space.

The physical intuition

When we consider the set of possible wavefunctions, we need them to be reasonably behaved i.e. only a countable number of discontinuities. In effect, such functions have only a countable number of degrees of freedom (unlike functions which can be very badly behaved). IIRC, this is one of the necessary conditions for a function to be fourier transformable.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Siva

Comments

Ah, that clears it up. I guess one problem is that while of course I can expand any wavefunction in terms of $\delta$-functions, those $\delta$-functions themselves are, among other things, not square-integrable.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Lagerbaer

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Right. And if I understand correctly, the discontinuity argument I gave should in essence be the same as a function being $\mathcal{L^2}(\mathbb{R})$, like @Qmechanic says. But I'm not able to think of a justification right away.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Siva

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-1: I don't mean to be mean, but I feel like all of this talk of regulators really obscures the mathematical crux of the issue. I feel like this response might lead people to feel that that introducing a regulator is necessary to make sense of this stuff, but, at least in my view, its really about getting your definition of dimension correct.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user joshphysics

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A regulator is just a tool to help you count the dimension. I used field theory inspired jargon, but I would think that the regulator notion would buttress the rigorous math of $\mathcal{L}^2(\mathbb{R})$ functions. After all, even to define something like a $\delta$-function but square integrable, you need a regulator. I'll rephrase my answer to make it a little clearer.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Siva

Answer 3

Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you mean by the "size" of a vector space.

The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension of a vector space as the cardinality of any basis.

Now let's examine the case of the harmonic oscillator. What is the dimension of its Hilbert space? As Qmechanic writes, the Hilbert space for the one-dimensional harmonic oscillator is $L^2(\mathbb R)$. We know that there exists at least one countable orthonormal basis for $L^2(\mathbb R)$. Therefore, the dimension of $L^2(\mathbb R)$ is $\aleph_0$.

Finally, as Qmechanic also points out, dirac deltas are not elements of $L^2(\mathbb R)$, so there is no contradiction.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user joshphysics

Comments

Can you briefly add why we know that there exists a countable orthonormal basis for $L^2(\mathbb{R})$?

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Lagerbaer

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Well it depends on what level of mathematical rigor you are looking for. Do you believe that the set of energy eigenvectors $|n\rangle$ is an orthonormal basis for $L^2(\mathbb R)$? If so, then you're done. Otherwise, one can prove this using standard theorems in functional analysis. This mathoverflow.net/questions/42310/when-is-l2x-separable also gives what you want if you note the fact that a Hilbert space is separable if and only if it admits a countable, orthonormal basis.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user joshphysics

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In general, it is usefull to remember that mathematically there is essentially one infinite-dimensional separable Hilbert space -- $l_2$, everything that you can imagine is isomorphic to it.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Peter Kravchuk

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@PeterKravchuk Tru dat.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user joshphysics

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@PeterKravchuk: Absolutely, but then another remark is in order. I find that many beginners tend to misuse this isomorphism, believing, for example, that the Schroedinger operator in $L^2$ can be mapped unitarily to the Schroedinger operator on $\ell^2$. This is not true, since (1) there are uncountably many possible isomorphisms to choose from (so equivalence of operators would not be canonically defined) and (2) if $A: L^2\rightarrow L^2$ and $B:\ell^2\rightarrow \ell^2$ are given operators, it does not follow that there exists an isomorphism $U: L^2\rightarrow\ell^2$ such that $UA = BU$.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user William

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@William, sure, this being possible is more or less the definition of (strong) unitary equivalence of $A$ and $B$.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Peter Kravchuk

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@PeterKravchuk: No, what I am saying is not quite the definition of strong equivalence. For every operator $A: L^2 \rightarrow L^2$ and a given isomorphism $U: L^2 \rightarrow \ell^2$ I can always find an operator $B: \ell^2\rightarrow \ell^2$ which is strongly equivalent to $A$ (simply define it as $B = UAU^{-1}$). On the other hand, it is not the case that given any operators $A: L^2\rightarrow L^2$ and $B:\ell^2\rightarrow \ell^2$, there exists an isomorphism $U: L^2\rightarrow\ell^2$ which makes $A$ unitarily equivalent (via $U$) to $B$.

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user William

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@William, sorry, probably my wording was not too clear. What I was trying to say is "yeah, not every two operators are unitarily equivalent", which is equivalent to your last sentence=)

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user Peter Kravchuk

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It might also be interesting to note that the harmonic oscillator Hamiltonian also has a continuous spectrum of unnormalisable functions that do not belong to $L^2(\mathbb{R})$ I can't recall what these are called (they're in the end of the harmonic oscillator chapter in that old Schiff QM book) and sometimes people talk about these, although I've never understood the need to invoke them and the Hilbert space concerned will not be separable (unseparable Hilbert spaces are a bit wild for me, so I tend to think of Hilbert spaces as separable anyhow).

This post imported from StackExchange Physics at 2014-04-24 02:30 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance