Dimension of Rays in Hilbert Space

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SUMMARY

The dimension of a ray in a Hilbert space is defined as one complex dimension, which corresponds to two real dimensions. This is established by the fact that a ray represents all complex multiples of a particular vector, indicating that while the ray is one-dimensional in the complex sense, it encompasses two degrees of freedom when viewed through real numbers. When considering the Hilbert space as a space over the real numbers, the dimension of the ray becomes two, highlighting the distinction between complex and real dimensions.

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  • Understanding of Hilbert spaces
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  • Knowledge of vector spaces and their dimensions
  • Basic concepts of linear algebra
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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:

...the physical state corresponds not to a particular vector in the Hilbert space, but to the ray, or one-dimensional subspace, defined by the collection of all the complex multiples of a particular vector.

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

Thank you very much.

tom.stoer said:
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.

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.
 
mpv_plate said:
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.

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}##.
 
micromass said:
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}##.

Thank you, it's clear to me now.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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