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Suppose I have a massive particle in non-relativistic quantum mechanics. Its wavefunction can be written in the position basis as

$$\vert \Psi \rangle = \Psi_x(x,t)$$

or in the momentum basis as

$$\vert \Psi \rangle = \Psi_p(p,t)$$.

$\Psi_x$ and $\Psi_p$ are related to each other via a Fourier transform.

However, if I write $\vert \Psi \rangle$ as an integral over infinitely-many "position basis vectors"

$$\vert \Psi \rangle = \int_{-\infty}^\infty \Psi_x(x)\vert x \rangle$$

then the position basis vectors $\mid x \rangle$ are Dirac delta functions - they aren't really functions. If we try to represent them in the momentum basis, we get non-normalizable plane waves. These basis vectors are not members of the physical Hilbert space.

My undergraduate quantum text explains that the Dirac deltas and plane waves are calculational tools and demonstrates their use. The Dirac deltas do not represent true wavefunctions. A real particle with low position uncertainty would simply have a wavefunction with a high but finite peak.

I'm fine with this; I think I understand how to do the calculations and what they mean. However, I am still unsure of how to find a basis for the physical Hilbert space that consists of vectors actually in the space.

In a bound state with no degeneracy, the energy eigenfunctions form a basis. The physical Hilbert space then consists of all linear combinations of the energy eigenfunctions. However, when we move to a scattering state, the spectrum of energy eigenvalues becomes continuous and the energy eigenfunctions are not normalizable because they are essentially the same as the plane-wave momentum eigenfunctions.

Because the scattering state has a physical Hilbert space of normalizable wavefunctions, shouldn't I be able to find a basis that consists of elements of the physical Hilbert space itself, even if this basis is not convenient for calculations?

Is there an example of such a basis for a free particle?

Mark Eichenlaub
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  • You say "However, I am still unsure of how to find a basis for the physical Hilbert space." -> So am I because there is an infinite number of bases (even reasonable ones), so I have no idea which one you are looking for. – Marek Aug 01 '11 at 06:12
  • If you are looking for some "smoothing out" of "eigenstates" of continuous spectrum then it might be good to first ponder the idea that working with vectors might not be a best idea from the conceptual point of view (you are obviously not satisfied with calculational point of view because for that you don't need anything beyond Dirac deltas). For this, you might instead consider forgetting about vectors altogether and just consider the algebra of operators (the famous von Neumann's C*-algebra) that already contains all the information and no Dirac delta stuff. – Marek Aug 01 '11 at 06:15
  • @Marek By "a basis" I meant that any basis would do to satisfy my curiosity about the matter. – Mark Eichenlaub Aug 01 '11 at 06:44
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    I would like to be pedantic and point out that wavefunctions should be written: $\left\langle x \middle| \Psi \right\rangle = \ldots$. – genneth Aug 01 '11 at 09:24
  • Please fix the question: you most certainly can write a wavefunction in the form you say you can't write it in. – Ron Maimon Sep 05 '11 at 18:47
  • @Ron Please read more carefully. I know you can write the wavefunction in the position basis. My question correctly says that the "position basis" is not a basis for the physical Hilbert space because its elements are not in the Hilbert space. Thus, I cannot write the wavefunction as a sum over "position basis vectors" because there are no such vectors in the Hilbert space. – Mark Eichenlaub Sep 05 '11 at 19:17
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    But you can still expand in a pseudobasis using integrals. The notion of "basis" in quantum mechanics includes distributional bases, and so the formula you write is a correct expansion. It's exactly the same as the Fourier transform formula, where the expansion in p states is just as much a limit process, because the p states are not in the Hilbert space but in its distributional completion. – Ron Maimon Sep 05 '11 at 20:47
  • @Ron I don't disagree. I'll change the wording slightly. – Mark Eichenlaub Sep 05 '11 at 21:03
  • Also, https://physics.stackexchange.com/questions/95594/non-separable-solution-for-the-schr%C3%B6dinger-equation – Cheng Dec 02 '22 at 07:24

3 Answers3

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If I understand correctly, your question basically comes down to identifying a basis for the space of square-integrable functions, $L^2(\mathbb{R})$, since any physical state $|\Psi\rangle$ can be constructed by performing the integral you listed in your question with a function $\Psi_x(x)\in L^2(\mathbb{R})$. $L^2$ is known to be a vector space, so a basis has to exist. Off the top of my head, I think an example would be

$$f_k(x) = e^{-x^2/a^2 - ikx}$$

which is just the normal plane wave basis $e^{-ikx}$ multiplied by a Gaussian envelope $e^{-x^2/a^2}$ where $a$ is some constant. Multiplying by this Gaussian envelope ensures that the functions will be square-integrable, but since you use the same envelope for every element of the basis, you can factor it out of the Fourier transform, so it doesn't change any of the essential properties of momentum-space decomposition.

P.S. I found a question on math.SE that seems related, and which motivated this answer.

Community
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David Z
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  • Right. Similarly, we know dozens of other bases that correspond to eigenstates of some Hamiltonian or other. Of course none of these is in any way natural (or even useful) to an arbitrary general problem. – Marek Aug 01 '11 at 06:32
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The eigenfunctions of a self adjoint operator lie outside the Hilbert space of square integrable functions on the line. One solution is to work with a basis of eigenfunctions of a non-self adjoint operator such as $x+ip$. Of course these are the coherent states. For the coherent states, one has an ovecomplete basis and a partition of unity, thus it is not difficult to decompose any vector in $L^2(\mathbb{R})$.

An other option is to work with a rigged Hilbert space, which is a formalization of Dirac's bra and ket method. A very good exposition of rigged Hilbert spaces is given in Francois Gieres' article. According to this option, you work with the eigenstates of the position operator but you remember that they do not belong to the Hilbert space, but rather to a Banach space which contains also distributions.

David Bar Moshe
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    I would point out that coherent states are not true eigenfunctions but only left or right (but not both) eigenfunctions. Also, overcomplete basis is kind of oxymoron. I'd just use the term spanning set. – Marek Aug 01 '11 at 08:41
  • Overcompleteness is a well defined mathematical concept. http://en.wikipedia.org/wiki/Overcompleteness I don't think that I have seen use of the terminology of a spanning set in conjunction with resolution of the unity. May be you would prefer the term tight frame defined in the Wikipedia page (which is true of course for the coherent states) – David Bar Moshe Aug 01 '11 at 09:22
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    @Marek: however, the term "overcomplete basis" is common in textbooks and literature --- it'll be hard to overcome the network effect by now. – genneth Aug 01 '11 at 09:23
  • @David: I have no problem with overcompleteness. I just don't like overcomplete basis (which the wikipedia article avoids for a good reason). Though I agree with genneth that it'll be hard to change it now... – Marek Aug 01 '11 at 09:46
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    In the future, please link to the arXiv abstract page rather than the pdf file, e.g. http://arxiv.org/abs/quant-ph/9907069 – Qmechanic Mar 26 '12 at 20:53
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You can expand any function in $L^2(R)$ in the (not necessarily normalizable) eigenbasis of an arbitrary Hamiltonian. So you just need to pick one with a discrete spectrum. Then the eigenvectors are normalizable, and your expansion is a sum. (The Hilbert space is the same for all nonsingular Hamiltonians with one dof, just expanded in different ways to get different physics.)

Arnold Neumaier
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