New answers tagged

0

Quantum computers don't work with exponentially big inputs, they work with exponentially big intermediate values. For example, people often frame Grover's algorithm as a database search. That would be a dumb way to use it, since you'd spend as much effort getting all the values into superposition as you would just doing the search classically. The correct ...


0

As far as I understand, all quantum computing purpose is to accelerate exponentially (upon input length) computation time of given task. It's not what it is about, even wiki is better start then that. Although I'm not expert. Better approximation, which allows to test if task is suitable for QC or not, probably will be parallel data processing or signal ...


1

The director (living on $\mathbb{R}P^1$ )is given only only modulo the interval$ [-\frac{\pi}{2}, \frac{\pi}{2} ]$ , however the winding number has to be computed on the universal covering space $\mathbb{R}$. The same situation happens for $S^1$ parametrized by $[-\pi, \pi]$ but the winding angles can be arbitrary integer multiples of $2 \pi$. Thus in ...


1

0) Note that in general, there is no need for the lattice spacings in different direction to be the same, as long as they all go to zero in the continuum limit. We can, for example, take the lattice spacing in the $x$ direction to be $a$, and the spacing in the $y$ direction to be $2a$. This is frequently done to achieve better resolution without the ...


2

Necessary and sufficient condition for conjugacy: see Sylvester's law of inertia. In your notation, the two symmetric matrices $K$ and $K'$ are conjugate by a transformation $W^TKW = K'$ if and only if they have the same number of positive and negative eigenvalues (no null eigenvalues since both are nonsingular). The eigenvalues themselves need not be ...


0

The short answer: the unitary time-evolution operator in quantum mechanics is $$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$ where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer... The exponential of a matrix is defined by $$\mathbf{M} = \sum\...


1

While looking up a ref. for this answer google search produced this thesis on "Numerical methods for solving the TDSE". It is right on the topic, contains all of the following, and more. I am still posting this because it may prove useful as a fast guide. Stable numerical methods for the TDSE are usually implicit methods. That is, each time-iteration ...


4

The Euler integration $$ \psi(t+\Delta t) = (1-\mathrm{i}H\Delta t)\psi(t)\implies \psi(t) = (1-\mathrm{i}H\Delta t)^{t/\Delta t}\psi(0)$$ is not numerically stable even for time-independent or even constant $H$ (that is, an eigenstate) because the normalization of the wavefunction is not preserved - the operator $1-\mathrm{i}H\Delta t$ is not unitary: $$ \...


0

The linearized formula for the "evolution by $\Delta t$" is exactly as accurate for a time-dependent $H(t)$ as it is for a time-independent one: the error is of order $O(\Delta t)^2$ in both cases. You must just remember to substitute the right relevant $H(t)$ every time you use the formula at a new time $t$. It's not hard do see why the accurate doesn't ...



Top 50 recent answers are included