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## New answers tagged mathematics

2

It is a typo as mentioned in the comments by M Stern

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These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent. ...

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Upon some more reflection, the answer is probably as follows. Let $\mathrm A$ be an observable according to the definition in the question, and assume $\Omega$ is finite. Then any $X\in\mathcal F$ is also some finite subset of $\Omega$. By definition of observable, we require the mapping $\mathrm A_\psi$ to be additive and non-negative, and therefore $$\... 0 Other important methods to check if a state is a separable or entangled are the Peres-Horodecki criterion and Schmidt decomposition. 0 CW from self-answer Reviewing Farhi et al. on quantum money from knots, one can say that the Markov chain applied by the verification algorithm that walks along the Reidemeister graph is far from ergodic, as the graph includes many individual connected components corresponding to separate knots. Each bill corresponds to a uniform superposition over ... 0 Write an ensemble as \{(p_i,\psi_i)\}_i, with p_i probabilities and \psi_i pure states. Let \mathcal I_1\equiv \{(p_i,\psi_i)\}_i and \mathcal I_2\equiv \{(q_i,\phi_i)\}_i be two such ensembles. Suppose that$$\sum_i p_i \lvert \psi_i\rangle\!\langle\psi_i\rvert = \sum_i q_i \lvert \phi_i\rangle\!\langle\phi_i\rvert$$(you can verify that this is ... 5 As is the case with ordinary multiplication, tensor product distributes over addition, so we can pull |0\rangle on the first qubit out in front$$ \begin{align} |\Psi⟩ &= \frac{1}{\sqrt{2}}|\color{red}{0}0\rangle+\frac{i}{\sqrt{2}}|\color{red}{0}1\rangle \\ &= \frac{1}{\sqrt{2}}\color{red}{|0\rangle}\otimes|0\rangle+\frac{i}{\sqrt{2}}\color{red}{|0\...

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Giving $|\psi \rangle = \dfrac{1}{\sqrt{2}}|00\rangle + \dfrac{i}{\sqrt{2}}|01\rangle$ we can see that the first qubit is in the state $|0\rangle$ so we can rewrite the state $|\psi\rangle$ as a tensor product: $$|\psi \rangle = |0\rangle \otimes \bigg( \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}}\bigg)$$ So the first qubit is in the state $|0\rangle$ and the ...

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It is normalized by dividing with the modulus or magnitude which is sqrt of (eigenvalue1 * 2 + eigenvalue2 * 2) = sqrt(2)

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The $\dfrac{1}{\sqrt{2}}$ is the normalization constant to make sure the state/eigenvector is a unit vector. Note that: if $|\psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ then $\bigg| \bigg| |\psi \rangle \bigg| \bigg| = |1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1$. The reason for this is because in quantum mechanics, states are always ...

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