7

Classical Hamiltonians By the spectral theorem, for every Hamiltonian there exists a basis in which it is diagonal. Thus, it is not correct to say that diagonal Hamiltonians are classical since this would apply to all Hamiltonians. A Hamiltonian $H$ which is diagonal in a product basis $\mathcal{B}$ is sometimes described as classical, because the evolution ...


6

Quantum computers can run classical computations using exactly the same algorithms, and hence have the same running time in terms of scaling. For example, if you look at shor’s algorithm, a major component of that is modular exponentiation, but nobody ever draws the circuit because they just say “use the classical algorithm”. In terms of absolute running ...


6

Yes, since the trace norm is the sum of the absolute value of the singular values, and the singular values can be found for each of the $a$ blocks independently.


5

Can implementing error correction in this case be any easier than in the case of a general quantum circuit? Yes, for example you could use a classical error correcting code such as a repetition code. Or, and I really want to emphasize how much more efficient this would be as a strategy for running the computation, you could throw the quantum computer into a ...


5

That quantity appears to be identical to Holevo information, which turns out to be the upper bound on how much classical information you can transmit using a quantum channel [1]. More generally the Holevo information is an upper bound for a quantity called "accessible information" which is (roughly speaking) the maximum information you can learn ...


5

Your mistake is that you assume that $\rho$ and $\sigma$ are classical-quantum in the same classical basis on $X$. However, there is no need to do so -- all which is necessary is that there exists such a basis, which can however depend on the state. As soon as you choose a different classical basis for the two states, your argument breaks down.


5

There aren't many examples! The main reason for advantages in quantum computers is the ability to constructively combine amplitudes - if you've only got 1 qubit, there aren't any amplitudes to combine! The best use case I can think of is randomness. A quantum computer (implemented with arbitrary error) could theoretically be a near perfect source of entropy,...


4

Classical computers are inherently deterministic, so they either generate pseudorandom numbers, or use an external physical process with statistically random noise to generate random numbers. Quantum computers are inherently probabilistic, so generating true random numbers is very natural for them. Quantum random number generators are already on the market ...


4

Intuition The expression $\|\mathcal{A} - \mathcal{I}\|_\diamond$ quantifies how close the channel $\mathcal{A}$ is to the identity channel $\mathcal{I}$ which is the channel that preserves quantum information perfectly. In order for a channel to transfer quantum information well, it must preserve both diagonal and off-diagonal elements of the input density ...


3

Peter Shor has two error correcting methods. One is the bit flip method and the other is the phase shift method. The bit flip method is similar to what you could use in classical computing, and is what I would recommend you use when comparing the two. The phase shift method is unique to quantum computing. This is a great link describing it. This circuit ...


3

It may be easier to understand how this works in terms of the computational basis if you choose $d$ to be some power of $2$. So let $d=2^n$ and then we will use $n = \log_2 d$ qubits to represent the quantum system. Then the indices $i=0,\dots, d-1$ (starting from $i=0$ for convenience) can be represented in their binary form $i := i_{n} i_{n-1} \dots i_1 $...


3

As $\rho_{ABE}$ is pure we have $\rho_{ABE} = |\psi\rangle \langle \psi|$. We'll rewrite the output of the channel $\mathcal{R}$ as $$ \rho_{ABE}' = \sum_j (P_j \otimes I_{BE}) |\psi\rangle \langle \psi| (P_j \otimes I_{BE}) $$ where $P_j = |\psi_j\rangle \langle \psi_j|$ are a collection of orthogonal rank one projections. Let $\rho_{ABE}^j = (P_j \otimes ...


3

We will use the upper bound on the entropy of a mixture (for proof see for example theorem 11.10 on p.518 in Nielsen & Chuang) $$ S\left(\sum_k p_k \rho_k\right) \leq H(p) + \sum_k p_k S(\rho_k)\tag1 $$ where $H(p) = -\sum_k p_k \log p_k$. Set $p_x := p(x)$. Note that if $|\psi_y^x\rangle$ is an eigenvector of $\rho_A^x$ associated to eigenvalue $q_y^x$ ...


3

As far as I'm aware there isn't much of a meaningful connection. The corresponding entropy for $D_{\max}$ is the min-entropy (written $H_{\min}$ or $H_{\infty}$). It measures a sort of `worst case' uncertainty whereas the Shannon or von Neumann entropies measure an average uncertainty. To answer your first question: the quantum relative entropies or ...


3

At least currently, most of the translations being made are in extraordinarily specialized areas - for example, quantum chemistry / computational chemistry. A lot of the math involves mapping domain math to quantum computers - ab initio molecular simulations need to map their traditional annihilation/creation operators to the X, Y, Z gates in quantum ...


2

Classical registers are typically used for capturing measurement results, and may also be used for conditionally applying quantum operation. See: https://github.com/Qiskit/openqasm/blob/master/spec/qasm2.rst Given the problem you described, one approach would be to have a classical program that iteratively: 1) defines and executes a quantum circuit on a ...


2

You seem to be mixing two very different concepts here. Quantum cloning is talking about the absolute limits of what is theoretically possible in a perfect world. In this absolute theoretical limit, yes we can derive how well quantum cloning can work, and we also know that classical cloning is nominally perfect. There is then a separate question of how well ...


2

In quantum information theory, the standard way to obtain what it is called reduced density operator from a quantum system composed by several quantum states is to use the so-called partial trace operation. For the case where there are two quantum states, $\rho^{AB}$ can be reduced to $\rho^A$ and $\rho ^B$ in the following way: $\rho^A=tr_B(\rho^{AB})$ $\...


2

Bra-ket notation is not necessarily tied to "quantum math," it's simply a convenient notation in many circumstances. It may seem intimidating at first, but once you understand the basics (ket = vector, bra = covector) it's straightforward to grasp, as long as you have a solid understanding of Linear Algebra. If you are shaky on Linear Algebra, different ...


2

I don't think there is a canonical "right" answer to this question as there is no universal formulation of the terminology, so let me try and pick apart a few of the things you mention, and how I understand their use within the field. The term "bit" can mean a couple of slightly different things. It can refer to how data is actually ...


2

No, this is not possible. The existence of such a hash function requires the (smooth) min-entropy of the initial state to be large enough but does not depend on its trace distance from a uniform state. For the simplest example possible let's forget about the side information $E$ and just focus on the $X$ system. The basic idea is that we can always pad $X$ ...


2

To pose a very simple answer to compete with all these complex (but also excellent) answers: the Ising model is a classical Hamiltonian because it is diagonal as it's written and therefore all of its eigenstates are classical states in the $S^z$ basis (no superposition required). Time evolution produces no mixing of the $S^z$ basis. All observables (that ...


2

While Adam's very detailed answer is probably emaculate, it's a bit long so for people that want a shorter answer, I'll give a much more compact alternative. This is not at all to challenge or try to refute Adam's answer at all. "What do they actually mean by a commuting hamiltonian?" In the specific case in your question, they are referring to ...


2

As you say, $$ \mathrm{Tr}[\rho_{XB} \log \rho_{XB}] = -S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_{B}^x \log \rho_B^x]. $$ But if you can prove the above statement, then the exact same derivation gives you $$ \mathrm{Tr}[\rho_{XB} \log \sigma_{XB}] = - S(X) + \sum_{x} p(x) \mathrm{Tr}[\rho_B^x \log \sigma_B^x]. $$ If you put both together then you get $$ \begin{...


1

Qiskit implements the parameter shift rule and the linear combination of unitaries to calculate the gradients for a QNN. These techniques are described in detail in Section 3 of this paper. If we calculate the gradients of the probabilities to measure one of the $2^n$ basis states, the circuit implementing the gradient is sampled $M$ (generally smaller than $...


1

Your description has X as a mixed state (a quantum state with classical uncertainty) and not a classical state. For example you can apply quantum gates to X but that shouldn’t be allowed if X was a classical state. However we can think of that mixed state as a classical state (see comments) and even use it as such. I’m not sure if there is some notation for ...


1

The covariance matrix is a function of the expectation values of powers of position and momentum associated to some state in a continuous-variable system. $$ \mathbf{\sigma} = \begin{pmatrix} \langle\hat{x}^2\rangle & \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\} \rangle \\ \frac{1}{2}\langle \{ \hat{x} ,\hat{p}\}\rangle & \langle\hat{p}^2\rangle\end{...


1

Why is 4! valid? We can imagine the desired operation to implement as a truth table / permutation matrix. Recognize that we may do this because none of the operations actually modify the amplitudes - they solely switch the amplitudes among basis states. Plus, because the operations provided are unitary, we know that a permutation table is valid. Let's call ...


1

This question gets right to the heart of what information does a density matrix contain about the state of a qubit. Critically, it is a subjective state of knowledge. So, if I don't know the outcome of the coin flip, my best description of the system is the same in both cases. At the moment that I learn the outcome of the coin flip, I have to update my ...


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