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Coherent states http://en.wikipedia.org/wiki/Coherent_states are vectors in the Hilbert space which in certain sense are strongly localized and "corresponds" to points in classical phase space (see below).

On the other hand vectors in Hilbert space in classical limit corresponds to Lagrangian submanifolds (see below).

How these two ideas correspond to each other ?

Is the following reasonable/known: coherent states corresponds to Lagrangian submanifolds in the COMPLEXIFICATION of the phase space, which intersect with real points of the phase space by the just one point ?

Motivating example(NOT quite correct thanks to "Squark"): consider R^2 and H=p^2+q^2. Consider p^2+q^2=0 - the real slice is just one point. On the other hand the in complexification we have 1-dimensional Lagrangian submanifold. The wave function $\psi$ corresponding to Lagrangian submanifold H=0, is constructed in a simple way $\hat H \psi=0$. $\hat H$ is hamiltonian of the harmonic oscillator and its eigenfuction is well-known to be coherent state.

So in this example idea seems to work.

Vague attempt for correction It seems we need to take into account Sommerfeld's 1/2 correction (it is related to Maslov's index). I mean we should consider H=r, and choose such "r" which satisfies Bohr-Sommerfeld quantization condition i.e. symplectic form integrated over interior of H=r should be integer +1/2. But we need somehow subtract this 1/2 from quantum eigenvalue to fit into the desired picture...

Background.

Quantization of Lagrangian submanifolds

Down-to-earth idea of the construction of the vector in Hilbert space from the Lagrangian submanifold.

Consider sumanifold defined by the equations $H_i=0$. Consider "corresponding" quantum hamiltonians $\hat H_i $, consider vector $\psi$ in the Hilber space such that $\hat H_i \psi = 0$. This $\psi$ we are talking about. Why it is important "Lagrangian" ? It is easy. If $A \psi =0$ and $B\psi = 0$ then it is true for commutator $[A,B]\psi = 0$. In classical limit commutator correspond to Poisson bracket so we see that even if we start from $H_i$ which is not close with respect to Poisson bracket we must close it - so we get coisotropic submanifold. Lagrangian - just restiction on the dimension - that it should be of minimal possible dimension - so after quantization we may expect finite dimensional subspace (in the best case 1-dimensional).

There are related by slightly different points of view. Lagrangian Submanifolds in Deformation Quantization

Coherent states

Sometimes one can realize Hilbert space corresponding to classical symplectic manifold $M$, as holomorphic functions on $M$. Each point $p\in M$ defines a functional $ev_p: Fun(M) \to C$ just evaluation of the function at point $p$. On the other hand in Hilbert space any functional corresponds to a vector $v$, such that $<v|f>= ev_p(f)$. So $v$ is coherent state corresponding to $p$. This approach is probably due to J. Rawnsley.

See Mauro Spera "On Kahlerian coherent states" http://www.emis.de/proceedings/Varna/vol1/GEOM19.pdf

The idea that complexification of the phase space is important plays a role in "Brane Quantization" approach by Witten and Gukov. See e.g. http://arxiv.org/abs/1011.2218 Quantization via Mirror Symmetry Sergei Gukov

There are actually different generalizations of the coherent states. In particular group theoretical approach was developed by Soviet physicists Askold Perelomov who worked in ITEP:

A. M. Perelomov, Coherent states for arbitrary Lie groups, Commun. Math. Phys. 26 (1972) 222-236; arXiv: http://arxiv.org/abs/math-ph/0203002

A. Perelomov, Generalized coherent states and their applications, Springer, Berlin 1986.

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Alexander Chervov
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  • 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 $\omega = 2$) – Vanessa Feb 08 '12 at 18:48
  • 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 – Vanessa Feb 08 '12 at 18:53
  • It might be worthwhile to look for a generalization of x+ip to arbitrary Kahler manifolds – Vanessa Feb 08 '12 at 18:56
  • @Squark thank you for the comment, Oops... you right ... somehow I missed $\omega=2$ ... – Alexander Chervov Feb 08 '12 at 19:04
  • @Squark I tried to correct... – Alexander Chervov Feb 08 '12 at 19:49
  • 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 $p$ – Vanessa Feb 15 '12 at 08:30
  • I meant $M \times \hat{M}$ and $M \times p$ – Vanessa Feb 15 '12 at 08:32
  • $M \times \bar{M}$ – Vanessa Feb 15 '12 at 08:32
  • @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.... – Alexander Chervov Feb 15 '12 at 09:41
  • 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. – Vanessa Feb 17 '12 at 08:07
  • 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 – Vanessa Feb 17 '12 at 08:12
  • Btw a small ad: I think this question would be great for http://theoreticalphysics.stackexchange.com/ too :) – Vanessa Feb 17 '12 at 08:19
  • @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!!! – Alexander Chervov Feb 17 '12 at 08:22
  • @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 ? – Alexander Chervov Feb 17 '12 at 10:41
  • 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 – Vanessa Feb 17 '12 at 12:40
  • @Squark From toric geometry vision: for CP^n the polytope is simplex and we have "n" invariant points the vertexes of simplex. – Alexander Chervov Feb 17 '12 at 12:45
  • 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. – Vanessa Feb 17 '12 at 12:46
  • @Alexander, for the Cartan subgroup yes but I can take the full stabilizer of a single point in $\mathbb{CP}^n$ – Vanessa Feb 17 '12 at 12:48
  • @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},...) – Alexander Chervov Feb 17 '12 at 12:50
  • Full stabilizer ? su(n-1) ? – Alexander Chervov Feb 17 '12 at 12:52
  • 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 – Alexander Chervov Feb 17 '12 at 12:56
  • @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. – Vanessa Feb 17 '12 at 13:48
  • 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 – Vanessa Feb 17 '12 at 13:51
  • 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. – Alexander Chervov Feb 17 '12 at 14:00
  • Umm, yes actually all of these are invariant under the Cartan, in my notation they are $z^k w^{d-k}$ – Vanessa Feb 17 '12 at 15:06

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This is a partial answer. I can understand the correspondence in the special case of coadjoint orbits $\mathcal{O}$ of compact Lie groups and Hamiltonians consisting of symbols of magnetic Laplacians on their cotangent bundlles. On one hand a solution of the Hamilton-Jacobi equation corresponding to these Hamiltonians define to a Lagrangian submanifold in $T^{*} \mathcal{O}$, and on the other hand the energy eigenspaces of the magnetic Laplacians are finite dimensional, thus the symbols of their orthogonal projectors in $L_2(\Gamma(L))$ (where, $L$ is the line bundle over $\mathcal{O}$ corresponding to the magnetic part of the symplectic form on $T^{*} \mathcal{O}$) are reproducing kernels which can be viewed as coherent states parametrized by points of the coadjoint orbit.

Now, for real coadjoint orbits, their complexifications are equivariently homeomorphic to their cotangent bundles, thus the correspondence can be passed to the complexification.

David Bar Moshe
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