Timeline for Coherent states vs quantization of Lagrangian submanifold
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29 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 17, 2012 at 15:06 | comment | added | Vanessa | Umm, yes actually all of these are invariant under the Cartan, in my notation they are $z^k w^{d-k}$ | |
| Feb 17, 2012 at 14:00 | comment | added | Alexander Chervov | Well, you want to look at C^n as S^n(C^2)... Okay, but I would prefer to work with C^n - it has a natural basis - eigenvectors for Cartan, which I denote as z^k . My point of view that z^0 and z^maximal - are two coherent states. Morever I remember what corresponds to z^k for all k - this are just paralles to equator on sphere, with the conditions that they are Bohr-Sommerfeld. And this is easy to generalize for any toric variety. | |
| Feb 17, 2012 at 13:51 | comment | added | Vanessa | The full symmetry group is $SU(n+1)$ so the stabilizer is more or less $SU(n)$. Actually it's a bit bigger but the additional elements amount to phase rotations on the invariant vector so they're not interesting | |
| Feb 17, 2012 at 13:48 | comment | added | Vanessa | @Alexander our Hilbert space consists of complex homogeneous polynomials of degree $d$ on the 2-dimensional Hilbert space from which we formed the projective space $mathbb{CP}^1$. Since it's 2-dimensional, any vector in it has a unique-up-to-scalar orthogonal vector. | |
| Feb 17, 2012 at 12:56 | comment | added | Alexander Chervov | I do not quite understand: "z^d and w^d where w is orthogonal to z." What is w ? Any vector is Hilbert space is polynom of degree "d" in z. So I might ask what polynom ? Any way I think the correct answer is that poles correspond to z^0 and z^2d | |
| Feb 17, 2012 at 12:52 | comment | added | Alexander Chervov | Full stabilizer ? su(n-1) ? | |
| Feb 17, 2012 at 12:50 | comment | added | Alexander Chervov | @Squark CP^(n-1) fixed points from down-to-earth - also "n". Projective space is n points up-to equivalence: (a1, ... a_n). Consider the points of the form (0,..,0,,0,...0). They are actually fixed points for the torus action. And we have "n" of them. The Cartan-torus acts like (a1,a2,...an)-> (a1, ..a_kt,a_{k+1}*t^{-1},...) | |
| Feb 17, 2012 at 12:48 | comment | added | Vanessa | @Alexander, for the Cartan subgroup yes but I can take the full stabilizer of a single point in $\mathbb{CP}^n$ | |
| Feb 17, 2012 at 12:46 | comment | added | Vanessa | Of course all of this doesn't contradict my proposed form for the Lagrangian submanifold. Also note that in the complexified picture it is holomorphic Lagrangian submanifolds which correspond to pure states. | |
| Feb 17, 2012 at 12:45 | comment | added | Alexander Chervov | @Squark From toric geometry vision: for CP^n the polytope is simplex and we have "n" invariant points the vertexes of simplex. | |
| Feb 17, 2012 at 12:40 | comment | added | Vanessa | Oops... @Alexander your states have to be invariant up to a scalar. Indeed we have two invariant states corresponding to the two poles: $z^d$ and $w^d$ where w is orthogonal to z. Note that the degree $d$ of the polynomials is fixed (each degree corresponds to a different superselection sector from p.o.v. of deformation quantization and different normalization of symplectic form -> choice of line bundle from p.o.v. of geometric quantization). Btw for $\mathcal{CP}^n$ with $n > 1$ this "problem" doesn't happen: there is only 1 invariant point | |
| Feb 17, 2012 at 10:41 | comment | added | Alexander Chervov | @Squark I begin to remember something... I think that z^0 and z^{2d} - these are two coherent states which correspond to zero and to the North and South pole points ! If this correct it will correspond to my conjecture about the correspondence with Lagrangian submanifolds quantization, because we know how to quantize some Lagrangians submanifolds in toric manifolds: the holomorphic sections of the line bundles correspond to integer points in the polytope, which in turn corresponds to Lagrangian submanifolds which are tori which are preimages of the moment map project M->polytope. Not clear ? | |
| Feb 17, 2012 at 8:22 | comment | added | Alexander Chervov | @Squark Great ideas, I really like what you write! However I am missing something... If we consider invariance with respect to Cartan subgroup - it has TWO invariant points - North pole and South pole... under quantization we should get coherent states which are invariant up to scalar(?) Then all "z^d" are Okay. Or do you think that we should consider invariance not up to scalar but exact invariance ? Then yes I understand you idea we should choose z^d where "d" is half dimension of the representation space - it will be invariant with respect to Cartan... But if dimension is even - problem!!! | |
| Feb 17, 2012 at 8:19 | comment | added | Vanessa | Btw a small ad: I think this question would be great for theoreticalphysics.stackexchange.com too :) | |
| Feb 17, 2012 at 8:12 | comment | added | Vanessa | I think this argument also generalizes to $CP^n$. I never saw the quantization of $CP^n$ but I think geometric quantization yields holomorphic sections of the $d$-th power of the tautological line bundle i.e. the $d$-th sector is complex homogeneous polynomials of degree $d$ in $n+1$ variables | |
| Feb 17, 2012 at 8:07 | comment | added | Vanessa | For $\mathbb{CP}^1$ the correctness of my suggestion follows from symmetry considerations alone. There is exact 1 quantum state invariant under the subgroup of $SU(2)$ fixing $p$: in the $d$-th superselection sector (for spin $2d$) it corresponds to the polyonmial $z^d$ where $z$ is the coordinate in the direction of $p$. The coherent state is invariant under this group by functoriality of construction so it must be it. The Lagrangian submanifold I constructed is also invariant under this group so it corresponds to the same state. | |
| Feb 15, 2012 at 9:41 | comment | added | Alexander Chervov | @Squark Wow ! Cool idea. Let think. For S^2=CP^1 coherent states - just Veronese embedding CP^1 -> CP^n (a:b)->(a^n, a^(n-1)b, a^(n-2)b^2 ,..., b^n), so we can try to check.... | |
| Feb 15, 2012 at 8:32 | comment | added | Vanessa | $M \times \bar{M}$ | |
| Feb 15, 2012 at 8:32 | comment | added | Vanessa | I meant $M \times \hat{M}$ and $M \times p$ | |
| Feb 15, 2012 at 8:30 | comment | added | Vanessa | I think there is a simple answer. Given a Kahler phase space $M$, its complexification is the complex symplectic manifold $M \cross \bar{M}$. This manifold contains Lagrangian submanifolds of the form $M \cross p$ each of which intersects the real (diagonal) cycle at one point corresponding to $p$ | |
| Feb 8, 2012 at 19:49 | comment | added | Alexander Chervov | @Squark I tried to correct... | |
| Feb 8, 2012 at 19:49 | history | edited | Alexander Chervov | CC BY-SA 3.0 |
**Vague attempt for correction**
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| Feb 8, 2012 at 19:04 | comment | added | Alexander Chervov | @Squark thank you for the comment, Oops... you right ... somehow I missed $\omega=2$ ... | |
| Feb 8, 2012 at 18:56 | comment | added | Vanessa | It might be worthwhile to look for a generalization of x+ip to arbitrary Kahler manifolds | |
| Feb 8, 2012 at 18:53 | comment | added | Vanessa | In case of a linear simple tic manifold your idea is simple to formalize. For example for the plane coherent states are eigenstates of $x+ip$ which makes clear the corresponding complex Lagrangian sub manifold and indeed it intersects the real cycle at one point | |
| Feb 8, 2012 at 18:48 | comment | added | Vanessa | The example is not quite correct since the ground state energy of the quantum harmonic oscillator is positive, namely it's $\hbar \omega / 2$ (in your normalization $\omega = 2$) | |
| Feb 8, 2012 at 16:30 | answer | added | David Bar Moshe | timeline score: 5 | |
| Feb 6, 2012 at 20:52 | history | asked | Alexander Chervov | CC BY-SA 3.0 |