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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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