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0

To answer your first question, whenever you run a QPE circuit on, let's say, $n$ ancilla qubits, you will be left with the binary representation of the phase $\theta$, encoded in the ancilla qubits. Measuring these ancillary qubits provides you an estimate of the phase $\theta$ (up to n-bit precision). Note, the probabilities of the states are there to ...


2

The statement made in the research paper is right. The initialization of the $\psi$ state is more than one qubit. For n qubits, it's state is in $2^n$ dimensions. Coming back to your question, compute your state with the operator $G+A$, i.e. $\frac{1}{2}(G+A) |\psi\rangle=|\phi\rangle$, for some state $|\phi\rangle$. Now the probability of $|\phi\rangle$ is $...


1

$I_{acc}(\rho^{XA})=I(X:A)$, as the state is classical-quantum. The mutual information of product states is the sum of the mutual information of the individual states, it is additive. So since $I_{acc}(\rho^{XA})=I(X:A)$, it inherits this additive property. I will point out, however, that accessible information is not additive. It can be super-additive in ...


1

I'm assuming you are referring to this paper: Uncertainty, Monogamy, and Locking of Quantum Correlations. In proposition 6, it's not clear to me if they are referring to the same product state that you are considering. However, in that paper, where they mention the additivity of accessible information, along with the Holevo paper, they also cite the ...


2

To understand what entanglement is, as a crude description it means if you measure one part of the system, then knowing what the measurement outcome is will change what you expect for measurement outcomes on the remainder of the system. This is true for both your examples, in both cases if you don't measure qubit 1, you expect to measure 0 or 1 with equal ...


5

All of your math is correct. Both of the states that you calculated are entangled states. An entangled state is one in which there exist measurable properties of subsystems that are correlated. In (1), measurements of the two qubits in the computational basis are correlated; it just happens to be a negative correlation. In (2), measurements in the ...


0

The main idea is that when we apply measurement to qubit by default in qiskit it measures in Z bases. So applying a H gate before measurement changes the measurement bases to X bases where we have |+> and |-> bases instead of |0> and |1> bases as like in normal Z measurement. So based on what state you measure we can have the below results. For |...


2

A measurement operation includes all of the operators in the set $\{M_m\}$; we do not have control over which operator gets applied. This makes sense if we consider a number of relevant examples. Suppose we have a Stern-Gerlach device that takes our qubit and sends it on one path (call it path $0$) if it has spin up and on another path (call it path $1$) if ...


1

When a measurement is performed in a certain basis, an orthogonal projection is performed on the state in question. So given a state $|\psi\rangle$, a measurement in the hadamard basis would result in the following: $$\frac{|+\rangle\langle+|\psi\rangle}{|\langle+|\psi\rangle|}$$ with probability $|\langle+|\psi\rangle|^{2}$ or $$\frac{|-\rangle\langle-|\psi\...


0

That's the beauty of the Hadamard gate or the $\text{H}$ gate which is represented by the following matrix: It puts the qubit in what is known as superposition. So for example if we apply the $\text{H}$ gate on $|0⟩$ or $|1⟩$, we get the output qubit $|+⟩$ and $|-⟩$ respectively with $$ |+\rangle=\frac{1}{\sqrt{2}} ( |0\rangle+ |1\rangle) $$ $$ |-\rangle=\...


3

Not true. Just take Bell state $M^{AB} = |v\rangle\langle v|$, where $|v\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. It has eigenvalue 1. But $M^A \otimes I_2= \frac{1}{2}I_2 \otimes I_2 = \frac{1}{2}I_4$.


2

Here is how I think about measurements. The state $|\psi\rangle$ is in a superposition of being $|0\rangle$ or $|1\rangle$ simultaneously. If the set of measurement operators $\{M_m\}$ consists of $|0\rangle \langle 0|, |1\rangle \langle 1|$, then you are asking the following question: Is the system in the $|0\rangle$ state or in the $|1\rangle$ state? Of ...


1

I believe your problem is that you're splitting up your measurement observable too much. Yes, $M\otimes I$ has four eigenvalues, $-1,-1,1,1$. BUT, when you make the measurement, you do not get 4 difference answers. You only get 2. To see why what you've done cannot work, note that the choice of eigenvectors is not unique. So how can you sub-divide the ...


1

It is postulate or axiom of quantum mechanics that if a state $|\psi\rangle $ that is a linear superposition of eigenstates $\{ |e_i\rangle\}$ of some observable, $$ |\psi \rangle = \sum_i \alpha_i |e_i\rangle $$ then upon making measurement with respect to this observable, the state is observed in the state $|e_i\rangle$ with probability $|a_i|^2$. That is, ...


3

One idea is to do polarimetry. By using a polarizing beam splitter, the polarization qubit can have each of its polarization components directed to a different detector for photon counting (ideally a single-photon detector, here). A polarizing beam splitter might send horizontally polarized photons in one direction and vertically polarized photons in another....


2

I'll write $\rho_1 = |\psi_1\rangle \langle \psi_1|$ and $\rho_2 = |\psi_2\rangle \langle \psi_2|$. We want the discrimination to be unambiguous so we want, $$ \mathrm{tr}[\rho_1 \Pi_2] = 0 = \mathrm{tr}[\rho_2 \Pi_1]. $$ That is, when we get outcome $i\in \{1,2\}$ we know that we received $\rho_i$ as the other state has a zero probability of obtaining that ...


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Indeed, $M_i=\sqrt{E_{i}}$ is wrong. The correct relationship is $E_i = M_i^\dagger M_i$. The possible $M_i$ for a given $E_i$ are $M_i = U\sqrt{E_i}$ for any unitary $U$, as $M_i^\dagger M_i = \sqrt{E_i}U^\dagger U \sqrt{E_i} = \sqrt{E_i}\sqrt{E_i} = E_i$.


0

Lets assume that resuls of some calculation on a quantum computer is either state $|+\rangle$ or $|-\rangle$. If you measure these states in computational basis, in both cases you end up with probability of state $|0\rangle$ and $|1\rangle$ to be 50 %. This means that you are not able to distinguish between $|+\rangle$ and $|-\rangle$. However, if you put ...


2

This fully depends on the experimental implementation and the context. Generally speaking, $|\pm\rangle$ is a basis like any other, so asking "how does measuring in the $|\pm\rangle$ basis works" has the same answer as the question "how does measuring in the $|0\rangle,|1\rangle$ basis works?". However, when asking this question you ...


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