# Tag Info

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$, and you have no dependence on that, it is ... 3 Try to replace backend = BasicAer.get_backend('ibmq_santiago') with backend = provider.get_backend('ibmq_santiago') Alternativetly, you can also use this code: backend = provider.backends(name = 'ibmq_santiago') You have to use backends available under you account. There are only simulators in BasicAer while the real quantum machines are under the ... 3 Here I am going to show why$\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 $$\|\rho\|_1 = \max_{-I \leq M \leq I} \mathrm{Tr}[M\rho]$$ whereM$is some hermitian operator satisfying the two operator inequalities$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$\newcommand{\ket}{\lvert#1\rangle}\{P_i\}_{i=1}^d$and$\newcommand{\braket}{\langle #1\rvert #2\rangle}\newcommand{\ketbra}{\lvert #...