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3

Suppose $\lambda_0 = 1$ and the rest are $0$. $$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \...


2

If I understand what you are asking for (I don't know much about tensor networks), both equations are singular value decompositions of $U$, just with respect to different indices (and in the latter case highlighting the singular values, which in the first equation is "hidden" in $A,B$). There are two things to notice here. First of all, given an arbitrary ...


3

In the most widespread convention, the Bloch sphere uses $\theta = 0$ radians latitude to indicate the north pole $|0\rangle$, $\theta = \pi$ to refer to the south pole $|1\rangle$ and $\theta = \pi/2$ to refer to the equator, which includes the superpositions $(1+i)/\sqrt 2$ and $(1-i)/\sqrt 2$ as well as $i|1\rangle$ and $-i|1\rangle$. If two great ...


2

If $|z\rangle$ are orthogonal to each other, then $$ \log(\sum_z |z\rangle\langle z| \cdot b_z) = \sum_z |z\rangle\langle z| \cdot \log(b_z) $$ So $$ \mathrm{trace}(\sum_z |z\rangle\langle z| \cdot a_z \cdot \log(\sum_{z^\prime} |z^\prime\rangle\langle z^\prime| \cdot b_{z^\prime})) $$ $$ =\mathrm{trace}(\sum_z \sum_{z^\prime} |z\rangle\langle z| \cdot |z^...


6

Imagine you have a vector that can be written in the form $$ |\psi\rangle=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}c_{ij}|i\rangle|j\rangle. $$ The coefficients can be arranged as a $d_A\times d_B$ matrix $C$, with the elements $c_{ij}$ (in your special case, you're talking about setting $d_A=d_B=\sqrt{m}$). Now, if you calculate $\rho_A=CC^\dagger$, this is ...


4

The reason is relatively straightforward. Consider an $m$ dimensional vector space $V$ with basis $\lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace$, and an $n$ dimensional vector space $W$ with basis $\lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace$. As your intuition suggests, we can naturally express any element $A \in V \otimes W$ in the ...


1

Here is a possible, though expensive, way. First, find all prime factors of the dimension d of your vector. In your example, the dimension is 9, and the only prime factor is 3. Next for each prime factor $p$, try a tensor product of a vector of dimension p and another vector of dimension $d/p$. Then you need to solve $d$ quadratic equations with $p+d/p$ ...


1

Outer product is a mapping operator. You can use it to define quantum gates, just sum up outer products of input and (desired) output basis vectors. For example, $$\vert{0}\rangle\rightarrow\vert{1}\rangle,\vert{1}\rangle\rightarrow\vert{0}\rangle$$ $$\vert{0}\rangle\langle{1}\vert+\vert{1}\rangle\langle{0}\vert=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{...


1

Measurements cannot produce an imaginary result. So if you want to measure an imaginary part, you need a suitable transformation before you measure. I haven't looked into the details of the mentioned operations but I'm sure that is what they do. On the last part of your question: the operation $R_x (\pi /2)$ can be visualized on the Bloch sphere by a ...


1

In the first summation for $U$: $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\cos(\Omega t)\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}(\text{This is wrong})$$ $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\...


2

There's a mistake. It's incorrect even for $a=0,b=0,U=I$. In this case the correct formula is $$ |\phi\rangle \otimes |B_{00}\rangle = \frac{1}{2} \bigg(|B_{00}\rangle \otimes |\phi\rangle + |B_{01}\rangle \otimes Z|\phi\rangle + |B_{10}\rangle \otimes X|\phi\rangle + |B_{11}\rangle \otimes XZ|\phi\rangle\bigg) $$ but their formula swaps $X$ and $Z$ in the ...


4

You seem to be overcomplicating this somewhat! You are right to split it up into the two terms $H_1$ and $H_2$. So, we have $$ e^{-i(H_1+H_2)t}=e^{-iH_1t}e^{-iH_2t}. $$ Now, straightforwardly, $$ e^{-iH_1t}=I+(e^{-i\delta t}-1)|00\rangle\langle 00|. $$ Next, we need to think about the $e^{-iH_2t}$ term. Of course, it maps $|00\rangle$ to $|00\rangle$. So, ...


3

If you wish to distinguish two states $|\psi\rangle$ and $|\phi\rangle$, you can only guarantee to do this if $\langle\psi|\phi\rangle=0$. You do this by measuring in a basis defined by the two states (alternatively, you apply a unitary $U$ such that $$ U|\psi\rangle=|0\rangle,\qquad U|\phi\rangle=|1\rangle, $$ and then measure in the standard $Z$ basis. ...


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