Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Schwartz kernel of spectral projection of Laplacian and integrated density of states

+ 3 like - 0 dislike
147 views

I'm reposting here a question I asked on MSE which did not receive an answer.

I am considering the Dirichlet Laplacian $\Delta$ on some smooth domain $U$. For now assume that $U$ is bounded, and later we will change that.

For $E\in\mathbb R$, I can use the functional calculus (or spectral theorem) to define the operator $\chi_{(-\infty, E]}(\Delta)$, which basically corresponds to projecting onto all eigenspaces up to energy $E$. This operator has an associated Schwartz kernel $S(x,y)$, i.e., an integral kernel: $$\chi_{(-\infty, E]}(\Delta)f=\int_{y\in U} S(x,y)f(y)dy.$$

I am interested in a relation between $S(x,y)$ and the integrated density of states of $\Delta$. Since the spectrum of $\Delta$ is discrete, the integrated density of states is just the normalized counting function: $$N(E)=\frac{\#\{\lambda\in \text{spec}(\Delta):\lambda \leq E\}}{|U|}.$$

Using some elementary arguments, I was able to show that $$S(x,y)=\sum_{\lambda\leq E}g_{\lambda}(x)\overline{g_{\lambda}(y)},$$ where the summation is over all eigenvalues up to $E$ and $g_{\lambda}$ is the noramlized eigenfunction associated to $\lambda$. Using this identity, straightforward integration shows that $$N(E)=\frac{\int_{x\in U} S(x,x)dx}{|U|}.$$

So far so good. But my real interest is actually to study $U$ which is unbounded, and so the spectrum of $\Delta$ is no longer discrete. In the unbounded setting, we define the integrated density of states as $$N(E)=\lim_{n\rightarrow \infty} \frac{\#\{\lambda\in \text{spec}(\Delta|_{U_n}):\lambda \leq E\}}{|U_n|},$$ where the increasing sequence of subdomains $U_n\subset U$ converges to $U$ "nicely" (don't worry about technicalities, just assume it makes sense in our setting).

In this case, the relation above between $S(x,y),N(E)$ clearly no longer holds, as there are (usually) no eigenfunctions and $|U|$ is infinite, so the expression no longer makes sense.

I believe that one can replace the summation over $g_\lambda$ with integration over spectral projections (with respect to certain spectral measures), but I am not very well versed in these sorts of things. I am also not sure what would replace the normalization factor, as it is infinite now.

Does anyone have any idea on how to prove a similar relation in the infinite setting? Or if it already exists somewhere, I'd also appreciate a reference.

Thanks in advance.


This post imported from StackExchange MathOverflow at 2024-11-02 20:47 (UTC), posted by SE-user GSofer

asked Oct 6 in Mathematics by GSofer (15 points) [ revision history ]
recategorized Nov 2 by Dilaton

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...