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Register a new userPolyas conjecture in the presence of a constant magnetic field
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We consider the Dirichlet Laplacian with a constant magnetic field in a two-dimensional domain of finite measure. We determine the sharp constants in semi-classical eigenvalue estimates and show, in particular, that Polyas conjecture is not true in the presence of a magnetic field.
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The different types of orbits in the classical problem of two particles with equal masses and opposite charges on a plane under the influence of a constant orthogonal magnetic field are classified. The equations of the system are reduced to the problem of a Coulomb center plus a harmonic oscillator. The associated bifurcation diagram is fully explained. Using this information the dynamics of the two particles is described.
We consider a non-relativistic electron interacting with a classical magnetic field pointing along the $x_3$-axis and with a quantized electromagnetic field. The system is translation invariant in the $x_3$-direction and we consider the reduced Hamiltonian $H(P_3)$ associated with the total momentum $P_3$ along the $x_3$-axis. For a fixed momentum $P_3$ sufficiently small, we prove that $H(P_3)$ has a ground state in the Fock representation if and only if $E(P_3)=0$, where $P_3 mapsto E(P_3)$ is the derivative of the map $P_3 mapsto E(P_3) = inf sigma (H(P_3))$. If $E(P_3) eq 0$, we obtain the existence of a ground state in a non-Fock representation. This result holds for sufficiently small values of the coupling constant.
We prove that the Nazarov-Sodin constant, which up to a natural scaling gives the leading order growth for the expected number of nodal components of a random Gaussian field, genuinely depends on the field. We then infer the same for arithmetic random waves, i.e. random toral Laplace eigenfunctions.
In this note we consider a quantum mechanical particle moving inside an infinitesimally thin layer constrained by a parabolic well in the $x$-direction and, moreover, in the presence of an impurity modelled by an attractive Gaussian potential. We investigate the Birman-Schwinger operator associated to a model assuming the presence of a Gaussian impurity inside the layer and prove that such an integral operator is Hilbert-Schmidt, which allows the use of the modified Fredholm determinant in order to compute the bound states created by the impurity. Furthermore, we consider the case where the Gaussian potential degenerates to a $delta$-potential in the $x$-direction and a Gaussian potential in the $y$-direction. We construct the corresponding self-adjoint Hamiltonian and prove that it is the limit in the norm resolvent sense of a sequence of corresponding Hamiltonians with suitably scaled Gaussian potentials. Satisfactory bounds on the ground state energies of all Hamiltonians involved are exhibited.
Let $V$ be a finite dimensional inner product space over $mathbb{R}$ with dimension $n$, where $nin mathbb{N}$, $wedge^{r}V$ be the exterior algebra of $V$, the problem is to find $max_{| xi | = 1, | eta | = 1}| xi wedge eta |$ where $k,l$ $in mathbb{N},$ $forall xi in wedge^{k}V, eta in wedge^{l}V.$ This is a problem suggested by the famous Nobel Prize Winner C.N. Yang. He solved this problem for $kleq 2$ in [1], and made the following textbf{conjecture} in [2] : If $n=2m$, $k=2r$, $l=2s$, then the maximum is achieved when $xi_{max} = frac{omega^{k}}{| omega^{k}|}, eta_{max} = frac{omega^{l}}{| omega^{l}|}$, where $ omega = Sigma_{i=1}^m e_{2i-1}wedge e_{2i}, $ and ${e_{k}}_{k=1}^{2m}$ is an orthonormal basis of V. From a physicists point of view, this problem is just the dual version of the easier part of the well-known Beauzamy-Bombieri inequality for product of polynomials in many variables, which is discussed in [4]. Here the duality is referred as the well known Bose-Fermi correspondence, where we consider the skew-symmetric algebra(alternative forms) instead of the familiar symmetric algebra(polynomials in many variables) In this paper, for two cases we give estimations of the maximum of exterior products, and the Yangs conjecture is answered partially under some special cases.
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