New answers tagged

5

No, not really. The amount of entanglement and the amount of energy in a state are completely independent: entanglement (together with discord) is a property of the state itself and its relationship to a bipartite (or multipartite) structure of the Hilbert space in which it lives, whereas energy is a joint property of the state and the system's Hamiltonian....


6

The statement is either false or unproven, depending on how you take it. Scenario 1: "Quantum Computers Cannot Be Simulated, Even Given Infinite Time and Space" This statement would be one in which we wish to know the outcome of a quantum computation, but we lack a quantum computer. In this case, is there anyway we can make a classical computer do the ...


2

Quantum computations can be simulated in a quantum computer, at least theoretically. Quantum computers (if they end up existing) have a speed advantage over classical computers, such as being capable of speeding up a problem from, an exponential speed on the input length to a polynomial time in input length. This cannot seem like a lot, but it means that ...


0

The information paradox is 40-45 years old. AdS/CFT is not even 20 years old, modern string theory (after D-branes) in probably 25 years old. This is to say that the information paradox doesn't need any of these, even though of course you can try to solve it in the context of string theory. In a nutshell, the information paradox is a sharp theoretical ...


1

This has been open for a while so I will bite. The information paradox has two versions or iterations. The previous one is that information is demolished by black holes by the entropy of its event horizon and that Hawking radiation that is emitted is in a pure blackbody distribution. A blackbody distribution of radiation is maximally random. If you make a ...


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 ...


2

For context, any general quantum operation $\Phi$ on a bipartite system $AB$ with finite dimensional Hilbert space $H_{A} \otimes H_{B}$ has a Kraus representation $\Phi(\rho) = \sum_j K_j \rho K_j^{\dagger}$ where the "Kraus operators" $K_j$ are linear operators on $H_{A} \otimes H_{B}$ such that $\sum_j K_j^{\dagger} K_j = I_{AB}$. The Kraus ...


0

Erasure doesn't work the way you think it does. Basically it goes like this... The most obvious way to 'erase' entangled information is to just undo the operation that created it: Notice how the top qubit has been restored to a pure state by the end, despite that nasty entanglement in the middle. But you can get a bit fancier than just repeating the ...


1

Yes. We can do this for any number of qubits using N dimensional spherical coordinates. For two qubits we can write a general density matrix as a linear combination of direct products of Pauli matrices and identity , $$ \rho = \sum_{ij=0}^3 a_{ij}~ \sigma_{i} \otimes \sigma_{j}$$. Here $\sigma_{0} = I$, and the rest are the usual Pauli matrices. For a ...


1

A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix. Your qubit representation could be rewritten, more suggestively as: \begin{align} \rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \...


-1

A one qubit state may be written in general as \begin{equation} |\psi_1\rangle=\alpha|0\rangle+\beta|1\rangle \end{equation} where $\alpha,\beta \in \mathbb C$ and there is the further restriction that $\langle \psi_1|\psi_1\rangle=|\alpha|^2+|\beta|^2=1$. However, only the relative phase between $|0\rangle$ and $|1\rangle$ is physically meaningful (quantum ...


2

When don't you condition on the result, measurement of a qubit can only decrease its purity (you end up with less information than you started with). When you do condition on the result, measurement of a qubit will make it 100% pure but there are two possible results. One possible result is along the measurement axis you measured. The other is against the ...


2

This experiment has been done, first by Birgit Dopfer in 1998, then later by Dr. John Cramer of the University Of WA. In Dopfer's experiment, there was a "coincidence detector" which is basically an AND gate to filter out only the entangled pairs. By moving the detector in the beam of photons not going to the double-slit, the information about the photon's ...


0

It is not required to put the state in density matrix form prior to performing the unitary operation $U$. However, you want to learn properties about the reduced density matrix after tracing out some of the system post-unitary, so you need a density matrix eventually. So you could do the unitary, and then construct the density matrix: $$ \newcommand{\ket}[1]{...


1

Take for example the two particle wave function $$|\Psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle_1 \otimes |1\rangle_2 + |1\rangle_1 \otimes |0\rangle_2\right)$$ The probability to measure 0 for the first particle is given by $$ |\langle 0_1|\text{Tr}_2(|\Psi\rangle)|^2 = \frac{1}{2} $$ The wave function after the collapse if 0 was measured is $$ |\Psi_c\...


1

I don't think identical purity by itself suffices to make $\rho$ and $\rho'$ equivalent, which implies identical spectrum. Assuming this is what you are looking for and both states are known, one general approach is to diagonalize both $\rho$ and $\rho'$, then find the basis transformation ...


1

Your example matrix $A$ isn't unitary. The right-most column has a length of 2 instead of 1, and the left-most column is only perpendicular to the right-most column if $a=-1$. You also seem to have computed $\rho_{\text{new}} = A \rho$ instead of $\rho_{\text{new}} = A \rho A^\dagger$. You can tell your $\rho_{\text{new}}$ must be wrong at a glance because ...


2

\begin{equation} b_{j} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi pj/n}\beta_{p}\qquad b_{j+1} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi q(j+1)/n}\beta_{q} \end{equation} Then \begin{equation} b_{j}^{\dagger}b_{j+1}^{\dagger} = \frac{1}{n}\sum_{p,q}e^{\pi i(p+q)j/n}e^{\pi iq/n}\beta_{q}^{\dagger}\beta_{p}^{\dagger} = \sum_{p} e^{-\pi ip/n}\beta_{-p}^{\dagger}\beta_{p}^{...


0

Fidelity $F$ is a measure of how close two quantum states are, as mentioned in the comments. This measure is usually used to quantify how good an experimental scheme or experimental results are, by comparing a resulting state $|\psi\rangle$ (or $\rho$ for mixed state) with your targeted state $|\phi_0\rangle$, and can be defined by $F = |\langle \psi | \...


0

Basically you're asking if there's some kind of compact-yet-useful way of encoding a graph into the state of some qubits. I think the answer is no. Making a compact representation is easy. For example, suppose your graph is classically encoded by the bits $b_1, b_2, b_3, ..., b_{2^n}$. Then take $n$ qubits and output the superposition $\sum_{k=1}^{2^n} |i\...


-1

As stated in other answers, a single (non entangled) particle could be described by a mixed quantum state. But I would not say it can be in a mixed state. If there is any ontology to a quantum state, then it concerns pure states only. The density matrix merrily conflates classical and quantum probabilities because it is designed to represent our knowledge ...


8

A mixed state represents a lack of knowledge on the system (maybe caused by the observer, maybe more fundamental). It is a notion in my opinion closely related to the Bayesian interpretation of quantum theories (seen as non-commutative probability theories). Quantum states are (non-commutative) probabilities, i.e. they encode all the information on ...


1

Yes, one particle can be in a mixed state when there is classical uncertainty present. Or at least it is the cleanest visualization one can make of a quantum state with classical uncertainty to think of it as mixed. The classic example here is the state of the particle ("cat") in the Wigner Friend Thought Experiment as Wigner would calculate it if he had to ...


1

Yes, it can. "Ensemble" may mean many experiments with one particle, not obligatory many particles at the same experiment.


1

Yes, they're realizable. All unitary operations on qubits (or qudits) are physically realizable (with negligible error) (unless you're working with an extremely limited gate set). The 'shift' operator is just adding a constant number to the little-endian number represented by the qubits (as if they were bits). Here's a blog post about efficiently ...


2

I think the most interesting approach in this direction is Caticha's entropic dynamics, for example in his "Entropic Dynamics, Time and Quantum Theory", arxiv:1005.2357. "Quantum mechanics is derived as an application of the method of maximum entropy. No appeal is made to any underlying classical action principle whether deterministic or stochastic. ... Both ...


0

Bayesian probability is based on the classical logic of plausible reasoning, as described by Jaynes, Probability Theory: The Logic of Science. One can and should try to find a Bayesian interpretation of the wave function. Here I can recommend Caticha's "entropic dynamics" as one possible approach. But I don't think "quantum logic" will be helpful here. ...


1

To see that there are two sector, corresponding to the eigenvalues of $G$ note that $G^2=1$ since $$ G^2 = (\prod^n_{j=1}\sigma^{(j)}_x)^2 = \prod^n_{j=1}(\sigma^{(j)}_x)^2 = 1$$ Thus there are two eigenvalues to $G$ that are $\pm1$. These sectors need not be of the same size. Consider just 2 spins in theire singlet and triplet configurations. The singlet ...


2

1) The reduction they are referring to is explain in the middle of page 13: "We now write (4.5) in the invariant sector as a sum of $n/2$ commuting $2×2$ Hamiltonians that we can diagonalize." 2) They are "reducing" the difficult problem of diagonalizing a very large matrix to the much easier problem of separately diagonalizing $n/2$ different $2 \times 2$ ...


2

Let $[G,H]=0$, and consider an eigenstate $|\psi\rangle$ of $H$, $$ H|\psi\rangle = E\psi\rangle\ . $$ Then, $|\psi'\rangle := G|\psi\rangle$ is also an eigenstate of $H$ with the same eigenvalue, since $$ H|\psi'\rangle = HG|\psi\rangle = GH|\psi\rangle = EG|\psi\rangle = E|\psi'\rangle\ . $$ Thus, also $|\chi\rangle = \tfrac{1}{\sqrt{2}}(|\psi\rangle + |\...


2

The given ansatz includes two assumptions. (1) The approximate ground states is seeked on a subspace of the Hilbert space consisting of rotated versions of single constant vector. (This subspace is a $2$-sphere parametrized by a unit vector in $\mathbb{R}^3$ (2) The value of the spin projection in the direction of the unit vector is half of the number of ...


1

Both of the forms you propose, $$\sigma^{(1)}_x \otimes \sigma^{(2)}_x \ldots \otimes \sigma^{(n)}_x \tag1$$ and $$\left(\sigma_x \otimes \mathbb{I}_{n-1}\right)\cdot \left(\mathbb{I}\otimes \sigma_x\otimes \mathbb{I}_{n-2}\right) \cdot \left(\mathbb{I}_2\otimes \sigma_x\otimes \mathbb{I}_{n-3}\right)\cdot \ldots \cdot \left(\mathbb{I}_{n-1}\otimes \sigma_x\...


0

It's both - those two "long form" expressions you wrote are equivalent. Matrix multiplication distributes over tensor products, so when you multiply them together the $i$th factor in the tensor product just becomes $\sigma^{(i)}_x$ times a bunch of identity matrices.


1

First of all, let's clarify things: In Quantum Mechanics decomposing the wavefunction into a linear combination of eigenstates is not the same as decomposing the system into single particle basis. Say you have a unidimensional system of two particles, the wavefunction in terms of their position is $\psi(x_1,x_2)$ can be decomposed into a linear ...


0

Let $S$ be the stabilizer of some $n$-qubit code, with $r$ generators (what you call the size of $S$). The normalizer $N(S)$ is a group of logical operations on the code space. Given some input code state $|\psi\rangle$, the state $N|\psi\rangle, N\in N(S)$ is also in the code space. Consider that all the elements of the coset $nS, n\in N(S)$ act on the ...


2

By "Copenhagen interpretation", I assume that you mean the interpretation with instantaneous "collapse" one usually encounters in an introductory quantum theory course. Such collapse is a useful rule to do calculation but it is only a fiction. What typically happens is that the quantum system is correlated with the macroscopic measurement device and other ...


2

The first term (sum) in $\bar H$ obviously commutes with all $\sigma_x$ variables because it's a function of $\sigma_x$ only and they commute with each other. The second term (sum) in $\bar H$ also commutes with the product of all $\sigma_x$ because the first term in the summand is a $c$-number and the second term $\sigma_z^j \sigma_z^{j+1}$ anticommutes ...


1

A book that I highly recommend which covers the said topics is "Lectures on Quantum Theory: Mathematical and Structural Foundations" by Chris Isham. His treatment is more thorough than most textbooks and has some interesting insights. It is also well suited for someone without a deep quantum background. Two books which also covers these topics and are ...


0

If two particles are entangled and then separated, will affecting one of them affect the other? If the state space for particle $i$ is $H_i$ and $U$ is a unitary operator acting on $H_1$, then $U$ acts on $H_1\otimes H_2$ (that is, the state space of the entangled pair) as $U\otimes 1$. If affecting one particle affects the other, then how is it not ...



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