# Tag Info

## Hot answers tagged projection-operator

### Confusion regarding projection operator

Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication? You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as ...
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### What does "measuring a state" mean?

Look like a lot of misunderstanding. If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information ...
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### What's the observable when measuring multiple qubits in the computational basis?

Note that your current definitions of the projection matrices $\{P_{1},P_{2},...,P_{n}\}$ are actually not projection matrices, since $P_{i}^{2} = I \not= P_{i} \,\, \forall i$. What works 'better' is ...
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### Implement a projection operator as a quantum circuit

Consider the states $|\psi\rangle = a|0\rangle|\psi_0\rangle + b|1\rangle|\psi_1\rangle$, and $|\psi'\rangle = a|0\rangle|\psi_0'\rangle + b|1\rangle|\psi_1\rangle$, where $a$ and $b$ are non-zero and ...
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### Quantum error correction using bit-flip code for the amplitude damping channel

Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined ...
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### Find the unitary implementing the transformation $|0\rangle\to\frac1{\sqrt2}(|0\rangle+|1\rangle),|1\rangle\to\frac1{\sqrt2}(|0\rangle-|1\rangle)$

Another way to solution: Hadamard gate changes $|0\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |+\rangle$ and $|1\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |-\rangle$. In ...
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### Relation between symmetric subspaces and $n$-exchangeable density matrices

The answer is: no, it is not true that any $n$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace (that is supported on the symmetric subspace). ...

### When discussing error correction, what are the objects in the expression $PE_i^\dagger E_j P=\alpha_{ij} P$?

Generally speaking, if a quantum channel $\Phi$ sends operators in $\mathcal X$ into operators in $\mathcal Y$, and its Kraus representation reads $\Phi(X)=\sum_a A_a X A_a^\dagger$, then the Kraus ...
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Yes. Clearly, ${\rm Ker}(P) \subset {\rm Ker}(P\rho P)$. On the other hand, since $\rho$ is a density operator and has full rank, we have $\rho > 0$, that is, $\langle x | \rho | x \rangle > 0$ ...
Firstly, I'm presuming that when you write $E_0^3$ it corresponds to $E_0^{q_1} \otimes E_0^{q_2} \otimes E_0^{q_3}$. In question 2. you've written out how the Superoperator of the Kraus operators ...