New answers tagged

1

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 ...


2

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} \\ ...


Top 50 recent answers are included