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Physical intuition behind prequantization spaces
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Given a symplectic manifold $(M,omega)$ with integral symplectic form, that is $$omega in textIm(H_2(M,mathbbZ) to H_2(M,mathbbR)),$$ one can form a so-called prequantization space, that is a $S^1$ principal bundle $$ pi : (V, alpha) to (M,omega),$$ where $alpha$ is an $S^1$-invariant $1$-form on $V$ satisfying $$pi^* omega = d alpha.$$
This makes $(V,alpha)$ into a contact manifold.
What is the physical intuition behind this construction ? I know that it corresponds to the notion of geometric quantisation, but I have trouble seeing why $(V,alpha)$ could represent a "quantum" space associated with $(M,omega)$. For instance, what is the meaning of the fibres of $pi$ (indentified with the Reeb flow) ?
sg.symplectic-geometry contact-geometry quantization geometric-quantization
asked 4 hours ago
BrianT
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up vote
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Given a symplectic manifold $(M,omega)$ with integral symplectic form, that is $$omega in textIm(H_2(M,mathbbZ) to H_2(M,mathbbR)),$$ one can form a so-called prequantization space, that is a $S^1$ principal bundle $$ pi : (V, alpha) to (M,omega),$$ where $alpha$ is an $S^1$-invariant $1$-form on $V$ satisfying $$pi^* omega = d alpha.$$
This makes $(V,alpha)$ into a contact manifold.
What is the physical intuition behind this construction ? I know that it corresponds to the notion of geometric quantisation, but I have trouble seeing why $(V,alpha)$ could represent a "quantum" space associated with $(M,omega)$. For instance, what is the meaning of the fibres of $pi$ (indentified with the Reeb flow) ?
sg.symplectic-geometry contact-geometry quantization geometric-quantization
asked 4 hours ago
BrianT
3456
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given a symplectic manifold $(M,omega)$ with integral symplectic form, that is $$omega in textIm(H_2(M,mathbbZ) to H_2(M,mathbbR)),$$ one can form a so-called prequantization space, that is a $S^1$ principal bundle $$ pi : (V, alpha) to (M,omega),$$ where $alpha$ is an $S^1$-invariant $1$-form on $V$ satisfying $$pi^* omega = d alpha.$$
This makes $(V,alpha)$ into a contact manifold.
What is the physical intuition behind this construction ? I know that it corresponds to the notion of geometric quantisation, but I have trouble seeing why $(V,alpha)$ could represent a "quantum" space associated with $(M,omega)$. For instance, what is the meaning of the fibres of $pi$ (indentified with the Reeb flow) ?
sg.symplectic-geometry contact-geometry quantization geometric-quantization
asked 4 hours ago
BrianT
3456
Given a symplectic manifold $(M,omega)$ with integral symplectic form, that is $$omega in textIm(H_2(M,mathbbZ) to H_2(M,mathbbR)),$$ one can form a so-called prequantization space, that is a $S^1$ principal bundle $$ pi : (V, alpha) to (M,omega),$$ where $alpha$ is an $S^1$-invariant $1$-form on $V$ satisfying $$pi^* omega = d alpha.$$
This makes $(V,alpha)$ into a contact manifold.
What is the physical intuition behind this construction ? I know that it corresponds to the notion of geometric quantisation, but I have trouble seeing why $(V,alpha)$ could represent a "quantum" space associated with $(M,omega)$. For instance, what is the meaning of the fibres of $pi$ (indentified with the Reeb flow) ?
sg.symplectic-geometry contact-geometry quantization geometric-quantization
sg.symplectic-geometry contact-geometry quantization geometric-quantization
asked 4 hours ago
BrianT
3456
asked 4 hours ago
BrianT
3456
edited 4 hours ago
asked 4 hours ago
BrianT
3456
asked 4 hours ago
BrianT
3456
asked 4 hours ago
BrianT
3456
3456
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add a comment |Â
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If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $mathbbC$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions).
However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle.
If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $nabla_V_F+2pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick.
In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform).
I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.
answered 1 hour ago
Jonny Evans
3,38012333
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $mathbbC$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions).
However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle.
If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $nabla_V_F+2pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick.
In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform).
I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.
answered 1 hour ago
Jonny Evans
3,38012333
add a comment |Â
up vote
3
down vote
If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $mathbbC$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions).
However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle.
If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $nabla_V_F+2pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick.
In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform).
I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.
answered 1 hour ago
Jonny Evans
3,38012333
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $mathbbC$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions).
However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle.
If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $nabla_V_F+2pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick.
In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform).
I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.
answered 1 hour ago
Jonny Evans
3,38012333
If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $mathbbC$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions).
However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle.
If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $nabla_V_F+2pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick.
In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform).
I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.
answered 1 hour ago
Jonny Evans
3,38012333
answered 1 hour ago
Jonny Evans
3,38012333
answered 1 hour ago
Jonny Evans
3,38012333
answered 1 hour ago
Jonny Evans
3,38012333
3,38012333
add a comment |Â
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