# How Many Basis Vectors in Hilbert Space?

Tags:
1. Sep 23, 2015

### RyanUSF

What is the dimensionality, N, of the Hilbert space (i.e., how many basis vectors does it need)?

To be honest I am entirely lost on this question. I've heard of Hilbert space being both finite and infinite so I'm not sure as to a solid answer for this question. Does the Hilbert space need 4 basis vectors to generate the 4x4 matrix of spin orientation (up up, up down, down up, down down). Could someone please clarify this for me?

2. Sep 23, 2015

### Strilanc

There is no "the" Hilbert space. Hilbert spaces are a type of mathematical object. For comparison, examples of other types of mathematical objects include groups, rings, and vector spaces.

To be a Hilbert space, you just have to follow a common set of rules like "have an inner product" and "the inner product of an element against itself is greater than or equal to 0". The Hilbert space rules don't limit the number of dimensions. Different Hilbert spaces have different numbers of dimensions, both finite and infinite. You could have any number of basis vectors.

3. Sep 23, 2015

### ddd123

There's this odd sentence on the Wiki for Hilbert Space though: <<Most spaces used in physics are separable, and since these are all isomorphic to each other, one often refers to any infinite-dimensional separable Hilbert space as "the Hilbert space" or just "Hilbert space">>. So it seems that there's a physics slang in which "The Hilbert space" means "isomorphic to l^2"?

4. Sep 23, 2015

### DEvens

5. Sep 23, 2015

### Strilanc

Oh, I see. Based on your quote, it sounds like "The" Hilbert space is infinite dimensional and so will have infinitely many basis vectors. It also sounds like a terribly misleading name (I guess saying "The Infinite Separable Hilbert Space" gets tiring).

Given that information, the space occupied by a finite number of qubits or spins is not "the" Hilbert space. The state space of $n$ qubits is $2^n$ dimensional and "a" Hilbert space, but it is not "the" Hilbert space because it is not infinite dimensional.

6. Sep 23, 2015

### ddd123

I suppose. Maybe it's because spin alone doesn't account for the whole physical configuration. Or Wikipedia is misinterpreting the reference.

7. Sep 23, 2015

### mathman

Finite dimensional "Hilbert spaces" are customarily called Euclidean spaces of given dimension.

8. Sep 23, 2015

### bhobba

I will be more succinct than some of the other posters.

A Hilbert space is an inner produc vector space that is complete.

I could spell out what each of those terms mean, but it is a good exercise to investigate them yourself:
https://www.math.washington.edu/~greenbau/Math_555/Course_Notes/555notes5.ps_pages.pdf

If you have any problems post here.

It is usually complex in QM but can be over the reals or quaternions. There is something in the back of my mind a theorem due to Soler shows that they are the only possible options, and since the real and complex number are subsets of the quaternions, a quaternion space would be the most general space. But don't hold me to it.

In QM it is usual to impose the condition of a countable basis. Again its a good exercise to investigate exactly what countable means.

If you want to understand at a deeper level what's going on in QM you should investigate Rigged Hilbert Spaces which are a generalisation of Hilbert spaces. Be warned though - that will require a fair amount of effort - its what mathematicians call non-trivial.

Thanks
Bill

Last edited: Sep 24, 2015
9. Sep 24, 2015

### tzimie

Talking about quaternions... Most of variables in QM are complex-valued. But as complex numbers are subset of quaternions, how do we know that they are not quaternions, "almost degenerated (but not exactly)" to complex numbers?

10. Sep 24, 2015

### bhobba

There are some theorems of Piron and Soler that almost, but not quite, proves it must be complex. If one adds the so called plane transitivity axiom then the proof is complete:
http://arxiv.org/pdf/quant-ph/0105107v1.pdf

Thanks
Bill

11. Sep 24, 2015

### ddd123

12. Sep 24, 2015

### bhobba

Ruled out in QM due to the theorems I mentioned.

Thanks
Bill

Last edited: Sep 24, 2015
13. Sep 25, 2015

### tzimie

Physical theorems are so full of "hidden assumptions" that I hardly believe them.

14. Sep 25, 2015

### bhobba

15. Sep 25, 2015

### gill1109

I believe that long, long, ago, l^2 was affectionately called "Hilbert's space". John von Neumann gave the first complete and axiomatic treatment of abstract Hilbert spaces, as well as naming this concept; and he did this specifically with a view to getting quantum theory onto firm mathematical foundations. As a side effect, Heisenberg's matrix mechanics and Schrödinger's wave mechanics could be shown to be mathematically equivalent.