New answers tagged nielsen-and-chuang
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} \\
...
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