# Tag Info

Let's start by considering specific density matrices $\rho=|i\rangle\langle i|$. This immediately tells you that $$\langle i|\sum_kE_k^\dagger E_k|i\rangle=1,$$ and hence all diagonal elements of $\sum_kE_k^\dagger E_k$ are 1. Next, consider a more general $\rho$, which we divide into diagonal and off-diagonal components, $$\rho=\rho_d+\rho_o.$$ We ...
Suppose that $$\mathrm{tr}\left(\sum_k E_k\rho E_k^\dagger\right) = \mathrm{tr}(\rho)$$ for all $\rho$. Then $$\mathrm{tr}\left(\sum_k E_k^\dagger E_k\rho\right) = \mathrm{tr}(I\rho)$$ for all $\rho$. The last equation can be rewritten in terms of Hilbert-Schmidt inner product as $$\left\langle \sum_k E_k^\dagger E_k,\rho\right\rangle_{HS} = \left\... 2 For every matrix A=\begin{pmatrix}a & x-iy\\x+iy & b\end{pmatrix}(hermitian here) with real number a,b,x,y. And A satisfy Tr(A\rho) and Tr(\rho)=1, let's consider two by two matrix for example, we can choose \rho=|0\rangle\langle 0| to make sure Tr(A\rho)=I. Then A_{11} must be 1. For the same reason we can get A_{22}=1. Now A=\... 1 The proof consists in connecting together two arguments. The first, covered by the exercise, reduces the problem of approximating the rotation gate R_\hat{n}(\alpha) to the problem of approximating the rotation angle \alpha. The second, described in the quoted text from Nielsen & Chuang, shows that one can achieve arbitrarily fine approximations of ... 0 As a general rule, if a bipartite pure state \Psi\in\mathbb C^n\otimes\mathbb C^m can be written as$$\Psi\equiv\sum_k (u_k\otimes v_k)$$for some collection of orthogonal vectors \{u_k\}_k, then$$\operatorname{Tr}_1[\mathbb P(\Psi)] = \sum_k \|u_k\|^2 (v_k v_k^\dagger), \qquad \mathbb P(\Psi)\equiv \Psi\Psi^\dagger.$$This is precisely the case at hand.... 1 The approach I found the best: On page 380 of Nielsen and Chuang, B|0\rangle(a|0\rangle+b|1\rangle)=a|00\rangle+b \cos\theta|01\rangle + b \sinθ|10\rangle. So, to find the element, just use \langle 0|(B|0\rangle|\psi\rangle)=\langle 0(B|0\rangle(a|0\rangle+b|1\rangle))=a|0\rangle+b \cos\theta|1\rangle which is then equal to E_0 |\psi\rangle. 1 I don't think you need the statements about general measurements ≡ unitary dynamics + projective measurements as it's not used/needed to explain the first circuit you discuss. If I understand correctly, the problem with using the circuit U|ψ⟩ is that upon measuring you destroy the state in the sense that you get the outcome m but you don't obtain a state ... 8 In the situation described in the book, Alice and Bob share the state$$ |\psi\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}}. $$Using the definition |\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2 and simple algebra we can see that |\psi\rangle can also be written$$ |\psi\rangle = \frac{|{++}\rangle+|{--}\rangle}{\sqrt{2}}.  Now, if Alice measures $|\... 2 Forget about the whole$n(1-p)$for a minute. For simplicity, let$k$be even and think of the product$\tfrac{(|0\rangle+|1\rangle)}{\sqrt{2}}^{\otimes k}$like it's the binomial$(x+y)^k$with commuting indeterminates$x$and$y$. When you expand the binomial$(x+y)^k$the monomial$x^jy^{k-j}$with the largest coefficient i.e. the monomials that appears ... 1 If you set$P_y(t)=0$for a particular length of time, then you just get an$X$rotation. Similarly, if you set$P_x(t)=0$for a certain length of time, you just get a$Y\$ rotation. So, you do a sequence that looks something like \begin{align*} P_X(t)=\left\{\begin{array}{cc} J & 0\leq t\leq \frac{d}{2J} \\ 0 & 0<t-\frac{d}{2J}\leq \frac{c}{2J} \\ ...