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How to show that the GHZ state is absolutely maximally entangled?

I assume that you talk about the standard case of a $3$-qubit GHZ state (the generalized versions of $n$-qubit GHZ states are not absolutely maximally entangled): $$ |GHZ\rangle = \frac{1}{\sqrt{2}} \...
Ohad's user avatar
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2 votes
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Question about Nielson & Chuang Problem 9.2

I have a sketch of ideas, but haven't buttoned up all the details into a proof so use it as a set of hints. Intuition: Kraus operators have a unitary freedom, thus you can get the next set of Kraus ...
Balint Pato's user avatar
3 votes

Why is the operator $M_a |x\rangle= |a \cdot x \pmod{N} \rangle $ unitary?

Assuming $\gcd(N,a)=1$, then the operator $M_a$ mapping $x\in\mathbb Z_N$ to $a\cdot x \pmod N$ is a permutation, which means that it’s reversible, ergo unitary.
Mark Spinelli's user avatar
0 votes

Regarding the inductive proof that any Clifford gate can be made of Hadamard, phase and c-not

This question is about the second part of the cited exercise. The first part is the single qubit case. In the second part, one is basically supposed to prove by induction that with the given ...
AYS's user avatar
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1 vote

Why is a density matrix an orthogonal projector?

Suppose I have a density matrix like $\rho = \frac{1}{2}[I + \hat{n}\vec{\sigma}]$. Explicitly, we have $$ \rho = \frac{1}{2}\left(\begin{matrix} 1+n_z & n_x - in_y\\ n_x + in_y & 1 - n_z \...
hft's user avatar
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1 vote
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How to know what eigenvalue corresponds to measurements of individual qubits in a multiqubit system?

If you consider the state $$\Omega_- = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle).\tag{1}$$ you will find that there is NO violation. Let me show this. Note that the approach I will use for solving ...
Alex's user avatar
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1 vote

Why is a density matrix an orthogonal projector?

If a matrix is a projector, it squares to itself. So, you just have to verify that $\rho^2=\rho$. You'll need the assumption that the length of the Bloch vector is 1.
DaftWullie's user avatar
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1 vote

How to know what eigenvalue corresponds to measurements of individual qubits in a multiqubit system?

From your current description, it seems that you have found each eigenvector $\{|v_1\rangle,...,|v_4\rangle\}$ and its associated eigenvalue $\lambda_1,...,\lambda_4$. These eigenvalues represent the ...
Kilian's user avatar
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1 vote
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The expectation values for the values of both qubits

You can find the single-qubit expectation values by tracing out the degrees of freedom of the other qubit. Mathematically, this corresponds to $\rho_A = \text{Tr}_B(|{\psi}_{AB}\rangle\langle{\psi_{AB}...
JoJo P's user avatar
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4 votes
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What is the expression for $|\psi\rangle\!\langle\psi|$ if $|\psi\rangle=\cos(\theta/2)|0\rangle+\sin(\theta/2)e^{i\phi}|1\rangle$?

You are missing the fact that $$ \langle \psi | = \bigg( |\psi\rangle \bigg)^{\dagger} \,.$$ "$\langle \psi|$" is conjugate transpose of "$|\psi\rangle$". So, if $$|\psi\rangle = \...
FDGod's user avatar
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