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 ...


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 ...


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 ...


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 ...


Only top voted, non community-wiki answers of a minimum length are eligible