Bike and Integrals · [79 days ago]
My good friend tells me that it is hard for her to see why so many small bad things happen to me all the time while seemingly those around me seem to get all of their things fixed and made better all the time when they are around me? Perhaps I am a Karma machine of the second type?
Last week, I borrowed Lorenza’s bike [that’s Professor Viola] to use and get used to for going around in the incredibly small yet hilly town of Hanover. I had it aired and nice and took it for a couple of rides but realized that I needed a seat cushion. So I ordered a seat cushion and a bike lube from amazon and voila... when they arrived in mail my front bike tire is magically punctured! [I won’t mention the IRS, my house catching fire, etc.]
Now I am stuck with one problem to solve, for which I also offer a prize:
Suppose a function $\theta(t)$ has finite derivative and $\theta(0)=0$. I also know that \[\int_0^T e^{-i\theta(t)}dt=0\] and \[\int_0^T t e^{-i\theta(t)}dt=0.\] I need to show that somehow, $\theta(T)=0$ or something similar. Any hint will get you an acknowledgment in an upcoming paper. Direct solutions might land you a third author sweet spot on it. Or you could reverse engineer this puzzle to write the paper yourself!
Windows Vista and Wacom Drivers · [88 days ago]
I had no idea that getting a new laptop with Vista could mean so much trouble. I just solved one of them: When I used my Wacom tablet in Photoshop, sometimes in the middle of using the tip of my pen on the tablet, Photoshop would crash. To fix it go to “pen and input devices” in control panel [search for it in start menu to find it]. Select “press and hold” and click on settings and disable it. No more Photoshop crashes in the middle of photo editing!
Half a continent and back in 8 days · [107 days ago]
We drove from LA to Houston. I saw so much desert that the green hills of New Hampshire looked like ice cream to me. We went through the Joshua Tree national park and then all the way through I-10 and a couple of beautiful back country roads in Texas. It was fun.
New Hampshire · [116 days ago]
A certain business in New Hampshire is called “American Yankee”.
Benchmarking Simulations · [126 days ago]
What are good candidates for benchmarking, say a quantum error correction proposal or a specific qip proposal, for simulation or experimental tests?
Of course we all know that quantum computer is a device that does mainly quantum error correction and some little computation on the side [allegory due to Steane?]. So many suspect that the best benchmark might be a nice 5 qubit error correcting code… It has lots of entanglement, superpositions, mutual measurements, damn hard preparations, etc… Well, then maybe it is too hard then.
I don’t even want to think about factoring 35 or searching a database… Too hard to program, too hard to benchmark…
I have been thinking about preparing a cat state $|000\rangle+|111\rangle$, or Deutsch-Jozsa with 3 qubits… They need entangling operations, just one measurement step, and are simple enough.
Although, you can try simulating about 10 spin particles using exact Hamiltonian dynamics, I would still like to have most of those spins acting as some sort of bath… Or maybe I should go with some sort of bosonic bath with a small cut-off?
And then all of this is simply going to produce a simple graph…
Magnus Expansion · [130 days ago]
I have been occupied with the Magnus expansion for quite a while now, and I thought I might write something about it here and the way I think it describes a transition from classical to semi-classical to quantum.
The Magnus expansion is an example of a geometric integrator for differential equations on Lie groups. It means that using the Magnus expansion you get approximations that honor the geometry of the Lie group, while they are made of linear elements elevated later by the exponential map. In fact, they are somewhat natural extension of “integral” to Lie algebras but they are very useful.
That was just a bunch of big words. Take a time-driven differential equation: $$\frac{dy(t)}{dt}=-iA(t)y(t).$$ And let us think of $A$ as an operator or just a matrix. There billions of ways of approximating the solution, depending on the circumstances [of $A$]. Magnus expansion is about getting a solution in the form $$y(t)=e^{-i\Omega(t)}y(0)$$. If you think of $y$ as a quantum state and $A(t)$ as a time-dependent Hamiltonian, $\Omega(t)$ is like an effective Hamiltonian and $e^{-i\Omega(t)}$ is the propagator which will belong to some special unitary group. The operator $\Omega(t)$ satisfies a non-linear differential equation (due to Hausdorff) that I will not quote here but can be integrated approximately, in order of increasing powers of $A$ using the Magnus expansion:
$$\Omega(t)=-\int_{0}^t A(s)ds + i\frac{1}{2}\int_0^{s_1}\int_0^t [A(s_2),A(s_1)]ds_2 ds_1+\cdots$$
The stuff in [...] is very interesting and I like to call it third and higher order terms. Let’s just say that it is composed of all possible time-ordered commutators that there exists at that order.
What do I see here? The first order term is classical or semi-classical. The second term starts looking more like a quantum object as it will involve commutators and it gets more and more quantum as it progresses. The time-ordering also becomes more and more important, another quantum feature. Of course there is nothing specially quantum about the differential equation that we started with and it is our own choice to interpret it as the Schrodinger equation…
I will elaborate on a couple of applications of this expansion and how it might play with information correlations later.
Polish School of Negativism · [284 days ago]
It all has to do with reading this paper again:
Can one build a quantum hard drive? A no-go theorem for storing quantum information in equilibrium systems
quant-ph/0603260
The paper is about existence of a quantum memory in thermal equilibrium. More precisely it states that they don’t. Maybe it is besides the point that a quantum memory need not operate in thermal equilibrium, I was thinking about the probabilistic argument the authors give towards the end of the paper about the Kitaev model.
The deceptively pedagogical argument is something like: Suppose individual fluctuations need to form a band of length $L$ around the system to kill the quantum memory, where $L$ is the size of the lattice. The probability of doing so (like a random walk) is proportional to $1/L^2$ and the number of starting points for such a band is $L^2$. Then we take the product of the probabilities and come up with a probability for forming a band that is given by a constant independent of the size of the system, hence no improvement in the error probability as a function of the size of the system. My problems: (1) This argument applies to almost everything and (2) How are all these random walks independent of each other? Should these bands cross, I guess they will cancel each other.
I am not sure about (1)... maybe that is exactly what the authors have in mind. And they do talk about (2):
In our case, we do not have usual random walk, because coming back to the origin means annihilating the pair, and the walk is ended. If the path was short, then the process does not affect quantum memory. Thus we want to estimate the probability that a long path occur. Since the process is Markovian (i.e. memoryless) and the average number of anyons is constant, we can imagine, that the walk of the just annihilated pair is continued by a new pair that has been just created (see Fig. 2) In this way we have reduced the problem to the usual random walk.
The number of anyons will be constant on the average but not over the microscopic scale of the walk. Maybe this is exactly the point: You will need to be far enough from equilibrium (or from Markovian-ness) for this argument to break.
More sequences · [297 days ago]
Suppose we have a set of 4 vectors: $$\{ (1,-1,-1) , (-1,1,-1),(-1,-1,1),(1,1,1)\},$$ and we want to make a sequence $\{R_i\}_{i=1}^N$ of these vectors with a zero sum. You could easily convince yourself that you need to use all of them at least once with $N=4$.
Now suppose instead of the sum you want to cancel the following for all $\alpha\ne\beta$: $$\sum_{i<j} R_i^\alpha R_j^\beta $$ where $R_i^\alpha$ refers to the coordinate $\alpha$ of the $i$-th vector. Smallest such sequence is 8 steps long. It’s easy to show that the length of such sequence has to be a multiple of 4 and 4 itself doesn’t work but 8 does.
Now how could I do the same thing with a growing monster like this:
$$\sum_{i<j<k} R_i^\alpha R_j^\beta R_k^\gamma$$
In case... · [317 days ago]
I should really be setting up a mimetex plug-in for this blog but then. This caught my ear so much that I am going crazy:
SZ2 by Battle [Song profile from Pandora.com]
Comment [1]
Refresh · [326 days ago]
It was time to give my old photoblog page a nice redesign. I have to confess that I did not make the template but it sure looks good. And while I was at it, I started a new blog on the basics of photography. I would like to see how I can be a better writer dealing with things that are not my formal area of expertise.
On the science side, I finally found a reasonable way of dealing with $x$.