User austen - MathOverflow

Eigenvalues of random Hamiltonian matrices

2013-01-31T10:34:09Z

A real $2n\times 2n$ Hamiltonian matrix has the general form

$$H=\begin{pmatrix} A & B \cr C & -A^T \end{pmatrix} $$

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

$$ J=\begin{pmatrix} 0 & 1 \cr -1 & 0 \end{pmatrix}, $$

so it would be natural to consider the Gaussian orthogonal ensemble for $JH$. Generically, the eigenvalues come in quadruples: $\lambda, -\lambda, \bar\lambda, -\bar\lambda$

Answer by Austen for Solvable models in quantum mechanics

2013-03-22T10:20:23Z

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 $\hbar=1$)

$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} - \int d^3\mathbf{r}'\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|} V(\mathbf{r}')\psi_{\mathbf{k}}(\mathbf{r}'),$$

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

$$f_i^{(1)}=-\frac{a_i}{1+ia_ik},$$

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 $r\to\infty$

$$\psi_{\mathbf{k}}(\mathbf{r}) \to e^{i\mathbf{k}\cdot\mathbf{r}} + f(\theta,\phi)\frac{e^{ikr}}{r}$$

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

$$\psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}+4\pi\sum_i \tilde f(\mathbf{r}-\mathbf{r}_i) f_i^{(1)}\psi_{\mathbf{k}}(\mathbf{r}_i),$$

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 $\alpha_i=-(4\pi a_i)^{-1}$).

I believe this is called the Foldy--Lax method in scattering theory.

Answer by Austen for Similarity between Cauchy-Riemann eqs and Hamilton equations.

2012-12-07T08:54:51Z

Doesn't this only work if $H(p,q)$ is a harmonic function in the plane?

If that's the case, we have

$\{u,H\}=\frac{\partial H}{\partial p}\frac{\partial u}{\partial q}-\frac{\partial H}{\partial q}\frac{\partial u}{\partial p}=|\nabla H|^2=-|\nabla u|^2$

where $\{\cdot,\cdot\}$ is the Poisson bracket. Since $\frac{du}{dt}=\{u,H\}$ this tells us how fast $u$ is changing.

Answer by Austen for Hamiltonians which commute both as operators and as connections

2012-12-06T16:01:53Z

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

$H = i\left[\frac{\partial P}{\partial t},P\right]$

where $P$ is a projector onto your chosen instantaneous eigenstate. This comes from T, Kato, J. Phys. Soc. J. Jpn. 5, 435 (1950).

Answer by Austen for Can a sphere be a phase space?

2012-11-27T09:57:07Z

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.

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

Answer by Austen for A mysterious Heisenberg algebra identity from Sylvester, 1867

2012-07-16T11:29:18Z

The following instance of your identity is well known in physics (and is sometimes called an "operator disentangling" identity)

$:\exp\left[\left(e^W-1\right)_{ij}a_i^\dagger a_j\right]:\;=\exp\left(W_{ij}a^\dagger_i a_j\right)$

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).

For general (rather than just quadratic) $\phi$ the formula is completely new (indeed remarkable) to me.

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):

Combinatorics and Boson normal ordering: A gentle introduction American Journal of Physics, 75 (7), pp. 639 (2007)

The authors' comments after Eq. 30 seem to imply that the formula doesn't generalize simply.

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.

Comment by Austen

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.

Comment by Austen

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?

Comment by Austen

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.

Comment by Austen

2013-01-31T19:56:00Z

Thanks. Edited.