6

It's a fantastic question because the typical measurement intuition we apply no longer is sufficient - it's really necessary to formalize measurement. Specifically, we create a set of nonlinear operators $M_\psi = |\psi \rangle \langle \psi |$, where the probability of measuring $\psi$ on an arbitrary state $|\phi\rangle $ is $\langle \phi | M^\dagger M | \...


3

What is the guarantee this implementation is efficient? Is there any rule regarding when implementing such POVMs is efficient? The implementation of such a gate will only depend on the parameter $k$ (which I assume you mean to be fixed), not $n$. Since efficiency is generally phrased in terms of scaling with $n$, and you have no dependence on that, it is ...


3

Try to replace backend = BasicAer.get_backend('ibmq_santiago') with backend = provider.get_backend('ibmq_santiago') Alternativetly, you can also use this code: backend = provider.backends(name = 'ibmq_santiago')[0] You have to use backends available under you account. There are only simulators in BasicAer while the real quantum machines are under the ...


3

Here I am going to show why $\langle Z_1 Z_3 \rangle$ generally cannot be estimated from $\langle Z_1 Z_2 \rangle$ and $\langle Z_2 Z_3 \rangle$. Let's start with an arbitrary three-qubit state: \begin{align*} |\psi \rangle = c_{000} &|000\rangle + c_{001} |001\rangle + c_{010} |010\rangle + c_{011} |011\rangle + \\ +c_{100} &|100\rangle + c_{101} |...


2

Yes, the trace distance can only decrease under partial trace. One can see this via the variational characterization of the trace norm $$ \|\rho\|_1 = \max_{-I \leq M \leq I} \mathrm{Tr}[M\rho] $$ where $M$ is some hermitian operator satisfying the two operator inequalities $M \leq I$ and $M \geq - I$. This is sometimes also known as the duality between ...


2

The number of bits in the counts dictionary equals the number of qubits in the circuit. So in your first example, you have a 1-qubit circuit, therefore you're dictionary looks something like counts = {'0': 400, '1': 600} # for for 1000 shots counts = {'0': 1} # for 1 shot In the second example, the Jupyter notebook screenshot, you have three qubits. ...


2

Consider the maximal entangled state $$ |\psi \rangle = \dfrac{1}{\sqrt{2}} \big( |00\rangle + |11\rangle \big) $$ If I make a measurement on the first qubit and a zero is returned then this implies my state has collapsed into the eigenvector $|00\rangle$ and so the second qubit measurement will definitely returned a $|0\rangle$ as that the only possibility. ...


2

I assume you're happy with the idea that the state before measurement is $$|O_{out}\rangle=\frac12|0\rangle(|\phi\rangle|\psi\rangle+|\psi\rangle|\phi\rangle)+\frac{1}{2}|1\rangle(|\phi\rangle|\psi\rangle-|\psi\rangle|\phi\rangle).$$ Now you want to measure qubit 1 in the 0/1 basis. There's a couple of different ways you might approach this. Define the two ...


1

I don't know what do you mean, the error specifically says that: QiskitBackendNotFoundError: "The 'ibmq_santiago' backend is not installed in your system." This means you don't have access to this machine from your account. This machine is either dedicated to only privilege users.


1

Let $P_0=|\psi\rangle\!\langle\psi|$ and $P_1=I-P_0$. This is a projective measurement which deterministically distinguishes the two orthogonal states. More generally, consider a projective measurement with operators $\newcommand{\ket}[1]{\lvert#1\rangle}\{P_i\}_{i=1}^d$ and $\newcommand{\braket}[2]{\langle #1\rvert #2\rangle}\newcommand{\ketbra}[1]{\lvert #...


1

Yes. It seems like it is indeed adding noise to mimic the noise of Valencia. As to what kind of noise, I assuming that it takes the latest calibrated data, and incorporated the T1 errors and T2 errors along with the measurement errors too. The result will be different than the hardware run since the noise occur in the hardware is much more complicated than ...


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