# Rays in Hilbert space

1. Apr 28, 2013

### mpv_plate

I am wondering, what is the dimension of a ray in a Hilbert space? For example here (page 2, bottom of page) I have read:

I understand why a state is represented by all multiples of a vector, not just the vector. But is the ray really one-dimensional? It would be one-dimensional if we multiply the vector by real numbers. But we are multiplying it by complex numbers. Along one-dimensional ray we could fully determine the vector multiple by a single real number. However with complex numbers there is one additional degree of freedom. Thus for each position along one-dimensional ray there are infinitely many choices for one additional parameter.

Or "one-dimension" here means a complex dimension, i.e. 2 real dimensions?

2. Apr 28, 2013

### Bill_K

Yes, one complex dimension. The state space is a complex Hilbert space, so the scalars are understood to be complex numbers.

3. Apr 28, 2013

### tom.stoer

Introducing a second dimension would mean to introduce a direction perpendicular to a given ray. This can't be done by multiplication with a scalar.

4. Apr 29, 2013

### mpv_plate

Thank you very much.

Thank you for clarifying that, I did not realize it is a complex dimension. However if I "split" each dimension of the Hilbert space into 2 dimensions (the Real and Imaginary part of each particular complex dimension), then multiplication of a vector by a complex scalar does introduce movement in 2 dimensions: the Real and Imaginary part of each complex dimension. Is that correct? Because a complex number has 2 components, so it can carry information about 2 dimensional shifts.

5. Apr 29, 2013

### micromass

Staff Emeritus
That's right. But when you do that, you're looking at the Hilbert space as a space over $\mathbb{R}$. Nothing wrong with that. But the rays of the Hilbert space as a space over $\mathbb{R}$ will have dimension $2$.
So if you encounter the word one-dimensional, then you should know it is the complex dimension (the space over $\mathbb{C}$), and not the dimension of the space over $\mathbb{R}$.

6. Apr 29, 2013

### mpv_plate

Thank you, it's clear to me now.