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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\... 2 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)$$So the first qubit is in the state |0\rangle and the ... 2 The \dfrac{1}{\sqrt{2}} is the normalization constant to make sure the state/eigenvector is a unit vector. Note that: if |\psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}  then \bigg| \bigg| |\psi \rangle \bigg| \bigg| = |1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1 . The reason for this is because in quantum mechanics, states are always ... 1 These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent. ... 1 Upon some more reflection, the answer is probably as follows. Let \mathrm A be an observable according to the definition in the question, and assume \Omega is finite. Then any X\in\mathcal F is also some finite subset of \Omega. By definition of observable, we require the mapping \mathrm A_\psi to be additive and non-negative, and therefore$$\...

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There are no effective bounds on how small or how large the smallest non-zero eigenvalue can become upon taking the partial trace, other than the obvious constraint that it must be in $(0, 1]$. Let $\lambda_{min}(\rho)$ denote the smallest non-zero eigenvalue of operator $\rho$. Fix a real $p \in (0, \frac{1}{2}]$ and consider the pure two-qubit state  |\...

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Say you have two bases $B_1$ and $B_2$. Let $U$ be the unitary that transforms the orthonormal basis vectors in $B_1$ to the basis vectors in $B_2$. So, it is obvious that $U^{\dagger}$ will be the unitary that transforms the basis vectors in $B_2$ to that in $B_1$. For instance let $B_1 = \{|0\rangle, |1\rangle\}$ (the computational basis) and let \$B_2 = \{|...

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