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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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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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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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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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