# Manifold of manifolds

1. Feb 24, 2005

### Mike2

So the equations of QM give eigenfunctions and eigenvalues. The eigenfunctions form a complete set with which any state is a combination of such. When measuring, the superposition of states collapse to one of the eigenfunctions. And the probability that some state with be measured in a particular eigenfunction is formed like an inner product of two states, etc.

All this to ask the question: the eigenfunction is a function that maps one manifold to another. And each eigenfunction is a different manifold from the others. There is an inner product between these manifolds to form the probability of going from one to the other. I wonder if the inner product tells us that all the separate eigenfunctions rest within a larger manifold. This would be a manifold of manifolds. Is this a valid way of looking at things? If so, then is there some more general equation that specifies this manifold of eigenfunctions, perhaps some symmetry principle?

Thanks.

2. Feb 24, 2005

### dextercioby

What's your acception/definition of a manifold...? I don't think it matches mine...

Daniel.

3. Feb 25, 2005

### Mike2

As I understand it, the eigenfunctions are a manifold in the usual since. But then there seems to be an inner product for these eigenfunctions (which are manifolds), and this inner product gives you the probability of going from one state to the next. I have to wonder if this inner product is the same as a metric which is normally defined on a manifold. In this case, it would be a manifold whose points are eigenfunctions.

If that is the case (and I could very well be wrong), then I wonder what would be specifying this overall manifold of eigenfunctions. Could it be some sort of symmetry principle?

Thanks.

4. Feb 25, 2005

### Fredrik

Staff Emeritus
If the set of all eigenfunctions is a manifold (I'll take your word for it), let's call it M. The inner product is then just a function from M x M into C (the set of complex numbers). How could just one function be a manifold?

5. Feb 25, 2005

### dextercioby

The set of eigenfunctions (this mean only discrete spectrum) span a HILBERT SPACE...A manifold is something else...

Daniel.

6. Feb 25, 2005

### masudr

A n-manifold is a Hausdorff space such that any point has a neighbourhood corresponding to points in a Euclidean n-space. I don't think all sets of eigenfunctions will necessarily form an n-manifold; though this is not the issue of real importance here.

Secondly, an inner product does not define a new manifold.

Thirdly, the word manifold has been mercilessly thrown about in the first post, without even stopping to see what it actually means. I think you are clutching for straws in looking for some new abstract symmetry that will reveal all the secrets of the world!

Masud.

7. Feb 26, 2005

### Mike2

OK, so a Hilbert space is not a manifold. My mistake, sorry. What about an underlying symmetry principle that distinguishes the Hilbert space. Any clues on that?

8. Feb 26, 2005

### dextercioby

What do you mean by "distinguishes the Hilbert space"...?And yes,the QM version of Emmy Noether's theorem is pretty handy.

Daniel.

9. Jun 23, 2005

### Weizensnake

A Hilbert space is indeed a manifold, namely a Hilbert manifold. Like ordinary manifolds are locally modelled after a Euclidean space, a Hilbert manifold is locally modelled after a Hilbert space. So in particular in can be overcountably infinite dimensional. Ordinary Euclidean space, e.g. R^3 is of course also a manifold, but a special (flat) one, because of course Euclidean space is locally modelled after a Euclidean space. Same goes for Hilbert manifolds.

In the case of the state space of a physical system, the formalism of manifolds is not needed, though. The decisive feature of the Hilbert space is that it is a vector space with an inner product. As all vector spaces, it is spanned by a base, which however can comprise overcountably many vectors for a Hilbert space. As a base you can use the eigenVECTORS of an arbitrary hermitian operator. Viewed this way, they are just elements of this space, not functions. You label this set of vectors by the eigenvalues to which they belong (assume they are non-degenerate for now). Then any vector can be represented in this basis by calculating the appropriate coefficient for the linear combination. Now an eigenFUNCTION is an eigenvector of some operator, e.g. the Hamilton operator (i.e. an energy eigenstate), which has been represented in the basis of the eigenvectors of the position operator. Like this it becomes a function Psi(x), where x is the eigenvalue of the position operator and the function value of Psi at x is the coefficient of the linear combination.

Hope I could help a little.