User austen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:53:09Z http://mathoverflow.net/feeds/user/25145 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120397/eigenvalues-of-random-hamiltonian-matrices Eigenvalues of random Hamiltonian matrices Austen 2013-01-31T10:34:09Z 2013-05-15T08:04:58Z <p>A real $2n\times 2n$ Hamiltonian matrix has the general form</p> <p>$$H=\begin{pmatrix} A &amp; B \cr C &amp; -A^T \end{pmatrix}$$</p> <p>where $A$, $B$ and $C$ are $n\times n$ matrices, and $B$ and $C$ are symmetric. Are there any results regarding the eigenvalue distribution of an ensemble of such matrices? For example, the above condition is equivalent to the symmetry of $JH$ with</p> <p>$$J=\begin{pmatrix} 0 &amp; 1 \cr -1 &amp; 0 \end{pmatrix},$$</p> <p>so it would be natural to consider the Gaussian orthogonal ensemble for $JH$. Generically, the eigenvalues come in quadruples: $\lambda, -\lambda, \bar\lambda, -\bar\lambda$</p> http://mathoverflow.net/questions/123672/solvable-models-in-quantum-mechanics/125261#125261 Answer by Austen for Solvable models in quantum mechanics Austen 2013-03-22T10:20:23Z 2013-03-22T10:20:23Z <p>I don't know if this will satisfy, but here is a physics proof' of the above formula. The wavefunction satisfies the Lippmann--Schwinger equation (with units mass $m=1/2$ and $\hbar=1$)</p> <p>$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} - \int d^3\mathbf{r}'\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|} V(\mathbf{r}')\psi_{\mathbf{k}}(\mathbf{r}'),$$</p> <p>where $V(\mathbf{r})=\sum_i V_i(\mathbf{r})$ is the potential due to the scatterers. Now a single point scatterer has the (angle independent) scattering amplitude</p> <p>$$f_i^{(1)}=-\frac{a_i}{1+ia_ik},$$</p> <p>where $a_i$ is the scattering length (i.e. the effective range' of the potential is zero). Recall that the scattering amplitude is defined by the asymptotic behavior as $r\to\infty$</p> <p>$$\psi_{\mathbf{k}}(\mathbf{r}) \to e^{i\mathbf{k}\cdot\mathbf{r}} + f(\theta,\phi)\frac{e^{ikr}}{r}$$</p> <p>Imagine iterating the Lippmann--Schwinger equation. Since we know the answer for a single scatterer, we can sum up repeated scattering off the same scatterer in the resulting (Born) series to give the scattering amplitude above, only keeping track of when we switch to a different scatterer</p> <p>$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}+4\pi\sum_i \tilde f(\mathbf{r}-\mathbf{r}_i) f_i^{(1)}\psi_{\mathbf{k}}(\mathbf{r}_i),$$</p> <p>where $\tilde f$ is the function you defined above. $\psi_{\mathbf{k}}(\mathbf{r}_i)$ on the right hand side should be understood in the asymptotic sense i.e. as one approaches the $i^{\text{th}}$ scatterer. This is for the scattered wave. If you want a bound state, it should be a solution of the homogenous equation for the wavefunction at each of the $N$ scatterers i.e. without any incoming plane wave. Then the appropriate determinant must vanish, which is just the condition stated in your question (with the identification $\alpha_i=-(4\pi a_i)^{-1}$).</p> <p>I believe this is called the Foldy--Lax method in scattering theory.</p> http://mathoverflow.net/questions/114761/similarity-between-cauchy-riemann-eqs-and-hamilton-equations/115693#115693 Answer by Austen for Similarity between Cauchy-Riemann eqs and Hamilton equations. Austen 2012-12-07T08:54:51Z 2012-12-07T08:54:51Z <p>Doesn't this only work if $H(p,q)$ is a harmonic function in the plane?</p> <p>If that's the case, we have </p> <p>$\{u,H\}=\frac{\partial H}{\partial p}\frac{\partial u}{\partial q}-\frac{\partial H}{\partial q}\frac{\partial u}{\partial p}=|\nabla H|^2=-|\nabla u|^2$</p> <p>where $\{\cdot,\cdot\}$ is the Poisson bracket. Since $\frac{du}{dt}=\{u,H\}$ this tells us how fast $u$ is changing.</p> http://mathoverflow.net/questions/95116/hamiltonians-which-commute-both-as-operators-and-as-connections/115621#115621 Answer by Austen for Hamiltonians which commute both as operators and as connections Austen 2012-12-06T16:01:53Z 2012-12-06T16:01:53Z <p>To expand on Greg's answer regarding the adiabatic theorem. You are looking for situations where the adiabatic evolution is exact. This is the case for a Hamiltonian of the form </p> <p>$H = i\left[\frac{\partial P}{\partial t},P\right]$</p> <p>where $P$ is a projector onto your chosen instantaneous eigenstate. This comes from T, Kato, J. Phys. Soc. J. Jpn. 5, 435 (1950).</p> http://mathoverflow.net/questions/114640/can-a-sphere-be-a-phase-space/114642#114642 Answer by Austen for Can a sphere be a phase space? Austen 2012-11-27T09:57:07Z 2012-11-27T09:57:07Z <p>To answer the question in the title: if by phase space we mean a symplectic manifold, then only for $k=1$ is there a symplectic structure. This is the phase space of a classical spin.</p> <p>It is not necessary for a manifold to be identified with $T^*M$ for some $M$ to qualify as a phase space. This is the first place we encounter the idea, with $M$ being the configuration space of a system, but the concept is more general</p> http://mathoverflow.net/questions/102281/a-mysterious-heisenberg-algebra-identity-from-sylvester-1867/102344#102344 Answer by Austen for A mysterious Heisenberg algebra identity from Sylvester, 1867 Austen 2012-07-16T11:29:18Z 2012-07-16T15:36:30Z <p>The following instance of your identity is well known in physics (and is sometimes called an "operator disentangling" identity)</p> <p>$:\exp\left[\left(e^W-1\right)_{ij}a_i^\dagger a_j\right]:\;=\exp\left(W_{ij}a^\dagger_i a_j\right)$</p> <p>where $::$ denotes normal ordering, $a_i$ and $a^\dagger_j$ are canonical Bose annhilation and creation operators satisfying $[a_i,a^\dagger_j]=\delta_{ij}$, and W is an arbitrary matrix (summation implied).</p> <p>For general (rather than just quadratic) $\phi$ the formula is completely new (indeed remarkable) to me.</p> <p>Frustratingly, it's hard to track down the origins of the above formula. Here's a recent discussion that includes the above version for a single boson mode (Eq. 30): </p> <p>Combinatorics and Boson normal ordering: A gentle introduction American Journal of Physics, 75 (7), pp. 639 (2007)</p> <p>The authors' comments after Eq. 30 seem to imply that the formula doesn't generalize simply.</p> <p>EDIT: I realized that Sylvester initially states the quadratic form above, and then limits his generalization to functions $\phi$ "linear quantic in $\delta_x$, $\delta_y$, $\delta_z$,...". Still, this generalization appears to contradict Eq. 31 of the above article.</p> http://mathoverflow.net/questions/130554/a-heat-kernel-for-schrodinger-operator-with-low-order-terms Comment by Austen Austen 2013-05-14T09:41:25Z 2013-05-14T09:41:25Z Shouldn't I be able to simplify the exponentials in terms of the shifted variables $\tilde x=x+b/2a$, $\tilde y=y+b/2a$? That is the origin of the $\exp(b^2t/4a)$, which reflects the shift of the quadratic potential. The terms you have don't seem to simplify nicely, however. http://mathoverflow.net/questions/120397/eigenvalues-of-random-hamiltonian-matrices/130458#130458 Comment by Austen Austen 2013-05-13T14:56:05Z 2013-05-13T14:56:05Z Do I understand correctly that both real and imaginary axes have a $\sqrt{n}$ density of eigenvalues, but there is only repulsion around the imaginary axis? http://mathoverflow.net/questions/120397/eigenvalues-of-random-hamiltonian-matrices/120501#120501 Comment by Austen Austen 2013-02-01T15:36:23Z 2013-02-01T15:36:23Z Thanks Carlo! I'm not sure that the answer lies in the literature on the transfer matrix, however, as that mostly involves the (real) transmission eigenvalues, rather than the (generically complex) eigenvalues of the matrix H itself. http://mathoverflow.net/questions/120397/eigenvalues-of-random-hamiltonian-matrices Comment by Austen Austen 2013-01-31T19:56:00Z 2013-01-31T19:56:00Z Thanks. Edited.