Can a Measurable Function V Exist for Given Quantum Energy Eigenvalues?

[jostpuur]

This is about quantum mechanics, but it is sufficiently difficult existence question dealing with DE and operator theory, that I think it fits the DE subforum the best:

Let E:\mathbb{N}\to\mathbb{R} be an arbitrary map, but so that \textrm{Im}(E) is bounded from below. Does there exist a measurable function V:\mathbb{R}\to\mathbb{R} such that the eigenvalues of the operator

<br /> H = -\frac{1}{2}\partial_x^2 + M_V<br />

are the given E(n)?

Here M_V:\mathbb{C}^{\mathbb{R}}\to \mathbb{C}^{\mathbb{R}} is the multiplication operator (M_V\psi)(x) = V(x)\psi(x).

For the sake of rigor we can give the following definition for the domain of H,

<br /> D(H) = \{\psi\in L^2(\mathbb{R},\mathbb{C})\;|\; \psi\;\textrm{is piece wisely}\;C^2\quad\textrm{and}\quad \int dx\;\Big|-\frac{1}{2}\partial_x^2\psi(x) + V(x)\psi(x)\Big|^2 &lt; \infty\}<br />

So H is a mapping H:D(H)\to L^2(\mathbb{R},\mathbb{C}).

 

[atyy]

I'm not a math person, so just a wild question:is your question equivalent to asking whether a Hamiltonian exists whose eigenvalues are the zeros of the Riemann zeta function?

 

[jostpuur]

I don't know enough of Riemann zeta function to answer that. I'll say that I don't know how precisely my question would be related to it now.

If T:\mathcal{H}\to\mathcal{H} is some linear operator on some inner product space, we can define a new non-linear mapping

<br /> f:\mathcal{H}\to\mathcal{H},\quad f(\psi) = T\psi - \frac{(\psi|T\psi)}{\|\psi\|^2}\psi<br />

which has the property

<br /> f(\psi)=0\quad\Leftrightarrow\quad T\psi\propto\psi,<br />

so in this sense every eigenvalue problem is related to a problem of finding zeros of some function. So... who knows?

 

[jostpuur]

Now I read your post more carefully. You were speaking about eigenvalues being zeros of some function, and I was about eigenvectors being zeros of some function. So this seems to be different thing.

 

[atyy]

Now I read your post more carefully. You were speaking about eigenvalues being zeros of some function, and I was about eigenvectors being zeros of some function. So this seems to be different thing.

Let's see, I understood your question to be:
Given a countable set of real numbers, does a Hamiltonian exist such that its eigenvalues are the elements of the given set?

(A little more strictly, you asked if a V(x) exists such that the eigenvalues of the Hamiltonian are the elements of the given set.)

 

[atyy]

Here is an example of what I was thinking about. My understanding is that you ask whether an explicitly given countable set of reals are the eigenvalues of a Hamiltonian. The link to the Riemann zeros is to ask whether an implicitly given countable set of reals are the eigenvalues of a Hamiltonian.

http://mathworld.wolfram.com/BerryConjecture.html
http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.4223v1.pdf

 

[jostpuur]

I see. So the problem I described, seems to be similar to the Berry Conjecture.

 

[atyy]

I see. So the problem I described, seems to be similar to the Berry Conjecture.

Hmm, reading the Berry conjecture, it seems they need a more general form of the Hamiltonian than in your problem. I wonder whether they've already ruled out "nice" Hamiltonians, and if so, how they did that.

 

[atyy]

Just out in PRL:

Landau levels and Riemann zeros
German Sierra, Paul K. Townsend
http://arxiv.org/abs/0805.4079