# Tag Info

81

The blurring is not randomised, it is predictable. See Can someone please explain what happens on microscopic scale when an image becomes unfocused on a screen from a projector lens? for a basic explanation. Each point of the in-focus image is spread out into a diffraction pattern of rings called a point spread function (PSF), and these ring patterns ...

47

No, it is not. Quantum computers can factor large numbers efficiently, which would allow to break many of the commonly used public key cryptosystems such as RSA, which are based on the hardness of factoring. However, there are other cryptosystems such as lattice-based cryptography which are not based on the hardness of factoring, and which (to our current ...

36

Yes, it's called deconvolution. Here are some examples of deconvolved images from microscopes: I found these by Googling for "example of deconvolved image from a microscope". It's possible by shooting lasers through the lens and generating a map of ray data that can be used in the deconvolution algorithm. I have also seen this done with an in focus image ...

32

Yes, it can be (partially) undone, because the process is not random and only part of the information is lost. Physics You comment that you are interested in the physics aspect of the question, so let's first clear that up: an image is focused when every point of the object corresponds to a point of the image - all the rays emanating from a given point of ...

18

There is actually an entire complexity class devoted to the answer, which is "no, it cannot break any code." The class is known as BQP, or "bounded error quantum polynomial time." It is the class of decision problems which can be solved by a quantum computer in polynomial time, with no more than a 1/3 error margin (this error term is accounted for in a ...

17

The goal with a camera lens system (whether a microscope or not) is to deliver light from one point on the object to one point on the sensor. This however cannot be perfectly achieved for several reasons. Light will diffract off the apeture, the smaller the apeture the more the diffraction. The system will only be perfectly in focus for an infinitely thin ...

17

I've work on exactly this problem and the answer is extremely controversial! At the heart of the controversy is a disconnect between tomography (3D imaging) and the surface maps (2.5D, "we all live in a hologram") approaches to image formation. The "we all live in a hologram" approach has recently seen a lot of commercial/experimental success much to the ...

16

In this article the limits of the details that can be recovered using deconvolution are derived. It's explained that noise leads to limits on how effective deconvolution can be to recover details. In the ideal case there will only be Poisson noise due to the finite number of detected photons. The smallest recovarable details scale as $N^{-\dfrac{1}{8}}$. So, ...

9

In the factoring algorithm, there are three kinds of qubits. In the OP's notation, there are "input qubits", which start in a superposition of all possible values, and which you eventually take the Fourier transform of. There are "value qubits", in which you compute the function $y^a \pmod{N}$, where $a$ is the value in the input qubits. And there are "...

8

Not really. Quantum computations are reversible in a very specific sense, which is not amenable to what you are thinking of. Suppose you want to compute a function $f$ of some bitstring $x$. Then you encode the bitstring into the state $\newcommand{\ket}{|#1\rangle}\ket{x}$ of your registry and append a "blank" qubit to get $\ket{x}\ket{0}$. Applying $f$ ...

7

The XOR gate is almost always called the NOT gate in quantum information and computation. To implement the XOR gate unitarily one must do it in a reversible way, since unitary gates must be invertible; conversely, any reversible logic gate defines a physical unitary operation. The XOR gate is itself not invertible, since e.g. both inputs "00" and "11" yield ...

5

Has anyone ever tried to formulate physics base on computer science? No, what do you mean by computer science? Data structures, algorithm, cryptography, artificial intelligence? No. Programming, computer architecture, networking, virus, brain computer interface? No. Computer graphics, visualization, database, linux kernel, Windows 7, ... Noooo. I can't ...

5

MPS are ansatz wavefunctions that need to be optimized to describe the ground-state of a given Hamiltonian. DMRG is one of the best method we have to optimize the MPS. Therefore you can think of MPS as a framework, and DMRG as an algorithm. Of course, this is not how things where developed historically, but that is the current reinterpretation.

5

Matrix product state (MPS) is a way to write down many-body quantum states. It's a natural representation for infinite 1D states that bipartite entanglement entropy obeys area-law ($S = constant$). This doesn't mean that it can't represent finite systems which are not 1D and $S = F(L)$, where L is some dimension of the system. Depending on the dimensionality ...

5

Update: I originally thought the question was referring to the "value" qubits when the asker said "auxilliary". This answer explains why you don't need to measure the value qubits. For the actual auxilliary qubits, which are used as workspace while computing the value qubits, it should also be okay to measure them afterwards but only because a proper circuit ...

5

Implementing such a method is not a trivial task. Just FYI, there are multiple programs purpose-built to perform Hartree-Fock calculations. In terms of a good introduction to the theory of Hartree-Fock calculations, I found this pdf extremely helpful. First of all, you have given the expression for the psuedo-exact (psuedo, because we have made the Born-...

4

No, you cannot run it in reverse if you are going to get an answer using a quantum computer. Even though the intermediate step with quantum gates are reversible. It can be easily understood by the following general quantum algorithm: Preparing an initial state $\left|\psi\right\rangle$ Get the final state $\left|\phi\right\rangle = U\left|\psi\right\rangle$ ...

4

Usually, a XOR quantum gate is implemented by the function : $XOR(Q) = a_1|00 \rangle + a_2|01 \rangle + a_4|10\rangle + a_3|11 \rangle$ The first bit is conserved, while the second bit is the result of an XOR operation between the first and second bit. For instance, if we have the combination $|11\rangle$, this means, after the transformation : $1$ ...

4

It is only exactly at the critical temperature that this CFT result works. You haven't mentioned if you have used the critical temperature when you did the monte-carlo. At/near critical point, autocorrelation time becomes huge. (If I am not mistaken, autocorrelation time must blow up exactly at critical temperature, however it is cut-off due to finiteness of ...

4

No, you can't do that. First of all because it doesn't make much sense to talk about the wave function corresponding to a spatial dimension. A wave function $\psi(x_1,\dots,x_n,t)$ gives you the probability amplitude of the system being at the position $(x_1,\dots,x_n)$ at the time $t$. It must by definition be dependent on all the coordinates necessary to ...

4

The best way to understand this is to work through an example. Here is how you factor the number 15 using Shor's algorithm. (If you prefer, view the problem not as factoring 15 but as solving $2^r=1 \hbox(mod 15)$, and skip Step Seven). As you work through this, imagine replacing 15 with a much larger composite number, and you'll see the advantage of ...

4

A trivial implementation of a Grover's-algorithm oracle is a controlled Z gate involving every wire: The highlighted gate flips the amplitude of the $|010001111\rangle$ state, and nothing else. (The Hadamards are just surrounding context. Part of Grover's algorithm, but not part of the oracle.) Of course no one would use Grover's algorithm to figure out ...

4

The standard way of checking for stability in such simulations is to plot the total energy over a long simulation time. A small amount of drift is acceptable (in fact, inevitable) but if it's too large then you either need a smaller time-step or a more stable integration scheme. What counts as 'too large' though depends on your aims. Another method is to ...

3

Simple climate models use the fact that the total amount of water is conserved, and assume that there are three processes that affect it: evaporation, precipitation, and moisture flux (i.e. how the moisture already in the atmosphere moves around). We can write the moisture flux vector simply as the product of the velocity vector and specific humidity: $\... 3 I have tried to reproduce your problem using my own code, but I couldn't: I got the correct value for the exponent. I can't tell you what went wrong in your calculation, but I can tell you exactly what I did! You can inspect my code on GitHub. It was the first thing I ever wrote in C++ and today I would do many things differently, so please don't judge the ... 3 Modern theoretical computer science is a good bit deeper than digitalization and computer graphics, etc. Modern computer science is effectively the study of formal systems called type theories. Superficially this are nothing but formalizations of collections of (data) types and functions between their terms (instantiations). But the study of these systems ... 3 Finding a radioactive item is conceptually the same as finding a light source. You detect what it emits with a sensor that measures the angle the radiation comes from and project it back. Do this with a few detectors and find the common point. The problem comes if you can only absorb the radiation without measuring the direction. What radiation is it ... 3 The underlying problem of your trap design is that two dipoles can't make a quadrupole! You are probably aware of the multipole expansion of electric (or magnetic) fields. This is where the terms "dipole" and "quadrupole" come from. As a physics student, one often encounters the exterior multipole expansion. This is what you use when there are charges ... 3 Notice that we want the system to remain in the lowest eigenstate of the Hamiltonian. Both Hamiltonians are two dimensional and their eigenstates span a two dimensional Hilbert space. The Hamiltonians contain a projector to$|\Psi_0\rangle$($|\Psi_1\rangle$). Define$|\Psi_0^O\rangle$($|\Psi_1^O\rangle$) as being normalized and orthogonal to$|\Psi_0\...

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