What is Geometry of Quantum Mechanics?

In summary: Hilbert spaces with a non-zero curvature)In summary, Projective Hilbert space is a space of rays in Hilbert space, which is non-linear and described by Riemann's metrics. It is a mathematical only, and has no physical significance.
  • #1
cartuz
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I read about Projective Hilbert space in quant/ph9906086. It is the space of rays in Hilbert space. Than Projective Hilbert space is non-linear with curves and described by Riemann’s metrics. But it is a mathematic only. Is here a physical sense? Is it the way of cooperation of Quantum Mechanic and General Relativity Theory?
 
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  • #2
Are you talking about removing parrallelism in (rigged) Hilbert spaces of QM and introducing a (pseudo-)Riemannian metric?

If we were to include some form of non-flat metric in Hilbert space then we'd have to modify the inner product as follows, I think,

Hilbert space inner product of QM: [tex]\langle \psi |\phi\rangle =\int^{\infty}_{-\infty}\psi^{*}\phi d^3x[/tex]

General Hilbert Space inner product: [tex]\langle \psi |\phi\rangle =\int^{\infty}_{-\infty}\psi^{*}\phi \rho (\mathbf{x})d^3x[/tex]

Where [itex]\rho[/itex] is a weighting function.

However, general relativity is based in space-time. Hilbert space is an infinite dimensional abstract vector space, not actual space-time itself. If we were to unify QFT and general relativity we'd need to change the geometry of the space-time we do QFT in, not Hilbert space.

I may be wrong...
 
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  • #3
Perturbation said:
Are you talking about removing parrallelism in (rigged) Hilbert spaces of QM and introducing a (pseudo-)Riemannian metric?

If we were to include some form of non-flat metric in Hilbert space then we'd have to modify the inner product as follows, I think,

Hilbert space inner product of QM: [tex]\langle \psi |\phi\rangle =\int^{\infty}_{-\infty}\psi^{*}\phi d^3x[/tex]

General Hilbert Space inner product: [tex]\langle \psi |\phi\rangle =\int^{\infty}_{-\infty}\psi^{*}\phi \rho (\mathbf{x})d^3x[/tex]

Where [itex]\rho[/itex] is a weighting function.

However, general relativity is based in space-time. Hilbert space is an infinite dimensional abstract vector space, not actual space-time itself. If we were to unify QFT and general relativity we'd need to change the geometry of the space-time we do QFT in, not Hilbert space.

I may be wrong...
May be you are right..
Here [itex]\rho[/itex] is sense of metric of Projective Hilbert space.
 
  • #4
Quantum mechanically, there is no physical difference between a state [itex]\psi[/itex], and any complex multiples [itex]k \psi[/itex]. (More precisely, multiplying the state by k doesn't change the expectation of any observable)

So, this gives a reason why one might want to take an entire ray and treat it as a single object -- if all of the states along that ray correspond to the same physical entity, why not make them all the same thing mathematically?


This construction is actually a common one -- for finite-dimensional spaces, anyways. It's the usual way of constructing, say, the projetive plane.


I doubt that this has anything to do with general relativity -- it looks like their thesis is just that homogenized QM is a really nice theory that one ought to learn. (Just like anyone serious about geometry ought to be comfortable with the projective plane)
 
  • #5
Could I have a link to that thesis?
 
  • #7
Hurkyl said:
Quantum mechanically, there is no physical difference between a state [itex]\psi[/itex], and any complex multiples [itex]k \psi[/itex]. (More precisely, multiplying the state by k doesn't change the expectation of any observable)
QUOTE]
Yes,
But in Projective Hilbert space it is different.
Of cause, it is more general case. Is it to complete the description of quantum states?
 
  • #8
Let me try this again.

All of the states [itex]k \psi[/itex] (for any complex k) all correspond to the same physical reality.

So, the ordinary Hilbert space contains a lot of irrelevant information.

So, what one might want to do is to get rid of that extra information -- all of those [itex]k \psi[/itex] from the original Hilbert space have been collapsed into a single point of the projective Hilbert space.
 
  • #9
Hurkyl said:
Let me try this again.

All of the states [itex]k \psi[/itex] (for any complex k) all correspond to the same physical reality.

So, the ordinary Hilbert space contains a lot of irrelevant information.

So, what one might want to do is to get rid of that extra information -- all of those [itex]k \psi[/itex] from the original Hilbert space have been collapsed into a single point of the projective Hilbert space.

Is it an analog to 2-dimentinal space and 3-dimentinal space?
Is 2-dimentinal space have extra information about position of the point in the space?
 
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  • #10
If you take 3-dimensional space and perform this construction, the result can be interpreted as a 2-dimensional space with some "points at infinity". And this is true in general.


For example, consider a photon. It has two states: spin up and spin down, which I'll write as |+> and |->.

The corresponding Hilbert space has two complex dimensions. However, this vector space is too big, because |+>, 2|+>, i|+>, (-3)|+>, et cetera, are all the same physical state.

So, we apply the construction described in the paper to produce a projective Hilbert space. Each point of this projective Hilbert space corresponds to an entire ray of the original Hilbert space; i.e. to a single physical state.

Since our original Hilbert space had two complex dimensions, the corresponding projective Hilbert space looks like one complex dimension, plus points at infinity. (Just one point, actually)

Topologically, the result is homeomorphic to an ordinary sphere. (It's the one-point compactification of the set of complex numbers)

This is called the Bloch sphere.
 
  • #11
Hurkyl said:
If you take 3-dimensional space and perform this construction, the result can be interpreted as a 2-dimensional space with some "points at infinity". And this is true in general.


For example, consider a photon. It has two states: spin up and spin down, which I'll write as |+> and |->.

The corresponding Hilbert space has two complex dimensions. However, this vector space is too big, because |+>, 2|+>, i|+>, (-3)|+>, et cetera, are all the same physical state.

So, we apply the construction described in the paper to produce a projective Hilbert space. Each point of this projective Hilbert space corresponds to an entire ray of the original Hilbert space; i.e. to a single physical state.

Since our original Hilbert space had two complex dimensions, the corresponding projective Hilbert space looks like one complex dimension, plus points at infinity. (Just one point, actually)

Topologically, the result is homeomorphic to an ordinary sphere. (It's the one-point compactification of the set of complex numbers)

This is called the Bloch sphere.

Thank you very much for your help.
 
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  • #12
I've never seen the merge of projective Hilbert space and rigged Hilbert space as to provide a unique & complete description of the mathematical foundation for quantum mechanics...

Daniel.
 

FAQ: What is Geometry of Quantum Mechanics?

What is the basic concept of geometry in Quantum Mechanics?

The basic concept of geometry in Quantum Mechanics is the mathematical framework used to describe the behavior of particles at the atomic and subatomic level. It involves the use of vector spaces, operators, and coordinates to represent the physical quantities and states of quantum systems.

How is geometry used in Quantum Mechanics?

Geometry is used in Quantum Mechanics to describe the spatial relationships between particles and their properties, such as position, momentum, and spin. It also helps to visualize and understand complex quantum phenomena, such as wave-particle duality and quantum entanglement.

What is the role of symmetry in the geometry of Quantum Mechanics?

Symmetry plays a crucial role in the geometry of Quantum Mechanics. It allows for the prediction and understanding of the behavior of particles, as well as the conservation of certain physical quantities, such as energy and momentum. Symmetry principles, such as rotational and translational symmetry, are fundamental to the mathematical foundations of Quantum Mechanics.

How does the geometry of Quantum Mechanics differ from classical geometry?

The geometry of Quantum Mechanics differs from classical geometry in several ways. Classical geometry deals with macroscopic objects and follows the rules of Euclidean geometry, while the geometry of Quantum Mechanics deals with microscopic particles and follows the rules of non-Euclidean geometry. Additionally, classical geometry is deterministic, while the geometry of Quantum Mechanics is probabilistic.

Can the geometry of Quantum Mechanics be visualized?

The geometry of Quantum Mechanics cannot be visualized in the traditional sense, as it deals with abstract mathematical concepts and states that do not have a physical representation. However, certain visualizations, such as Feynman diagrams and wave functions, can help to better understand and interpret the mathematical equations and principles of Quantum Mechanics.

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