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8

First, the example that you give is not a density matrix (they must have trace 1). Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states ...

8

The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $\big(\alpha |0\rangle + \beta |1\rangle \big) \otimes \big(\alpha |0\rangle + \beta |1\rangle \big)$. This is not the state CNOT will give you. The qubits you get after applying CNOT as you ...

8

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ...

8

For any matrix $A$ we can write $$A =\sum_{i,j,k,l}h_{ijkl}\cdot \frac{1}{4}\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l,$$ where $$h_{ijkl} = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l)^\dagger \cdot A\big) = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l) \cdot A\big)$$ ...

7

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 ...

7

Hint: To make your induction work, write \eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)} Then, use triangle inequality and finally the fact that ...

6

This is actually a much easier problem. In the case of states, you're trying to use the PPT criterion, or others, to distinguish if $\rho$ can be written in the form $$\rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i,$$ where $\sum_ip_i=1$ and the $\sigma^A_i$ and $\sigma^B_i$ are valid states on single sites. The difficulty actually comes from the freedom that ...

6

A density matrix $\rho$ has the properties of being Hermitian, non-negative and has trace 1. Any $2\times 2$ matrix can be written in the form $$\rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}.$$ The trace being 1 fixes that $n_0=1$, while the Hermitian property imposes that $\vec{n}\in\mathbb{R}^3$, where $\vec{\sigma}$ is the vector of the 3 Pauli ...

6

In the paper I called it the Burnside decomposition, but it looks like the standard name is the Wedderburn decomposition. That might simply have been a mistake in terminology on my part. Anyway, there are two good ways to get the summands to be $V \otimes V^*$. (Of course they are closely related.) 1) You can interpret $\mathbb{C}[G]$ as an associative ...

6

Yes, since the trace norm is the sum of the absolute value of the singular values, and the singular values can be found for each of the $a$ blocks independently.

6

Your reasoning is correct if your two Hamiltonians commute. But, as you say, it doesn't work if they don't commute. In that case, the trick is to find something that approximates the the thing you want. So, what you should really be thinking about is taking terms in the opposite order: $$e^{iH_1t/2}e^{iH_2t/2}e^{iH_1t/2}e^{iH_2t/2}\approx(e^{i(H_1+H_2)t/2})... 6 First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "\equiv" symbol is taken to mean "equal up to a phase"). An easy way to see it is by computing the determinant of H=e^{i\pi H/2}, which gives -1=1 (using \det[\exp(A)]=\exp[\operatorname{Tr}(A)]  for all A and \operatorname{Tr}(H)=0). Now, ... 6 \newcommand{\bs}{{\boldsymbol #1}} \newcommand{\tildebssigma}{\tilde{\bs\sigma}} \newcommand{\bssigma}{{\bs\sigma}}Yes, products of Pauli matrices form a basis for the set of Hermitian matrices (of dimensions that are powers of 2). More specifically, fix an integer n and let N\equiv 2^n, define \bssigma\equiv(\sigma_x,\sigma_y,\sigma_y), and \... 5 The idea is to apply a unitary transformation which will map these states to a different pair of orthogonal states which are easy to distinguish by measuring in computational basis. Let's construct a unitary U which prepares |\psi_0\rangle starting from |000\rangle state. The first step is to prepare a W_3 = \frac{1}{\sqrt{3}}\left(|100\rangle + |... 5 Alice receives a quantum state |\psi\rangle, which is an element of some basis \mathcal{B}, though she does not know what \mathcal B is. She then teleports this to Bob, who is told by someone else what \mathcal B is. Furthermore, it seems that specifically either \mathcal{B} = \{ \lvert 0 \rangle, \lvert 1 \rangle \} or \mathcal{B} = \{ \lvert + \... 5 Your calculations are correct, however, there are two issues: first, note that the promise on the dual bent function is actually not$$ f^\prime(x\oplus s) = 2^{-\frac{n}{2}} \sum_{y \in \mathbb{F}_2^n} (-1)^{xy} f(y) , \; \forall x \in \mathbb{F}_2^n $$as stated. Instead you really want to define here the dual as in my paper https://arxiv.org/abs/0811.... 5 This is not the unitary that you have to implement: you need a two-qubit unitary$$ \frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 & \omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array}\right), $$where \omega=e^{2i\pi/3}, the point being that if you introduce ... 5 There are a couple variants of the HOG test. "Old HOG" computed the proportion of unique samples whose probability is larger than the median probability of the distribution. It then compares that proportion to a threshold, e.g. 2/3. If you have enough larger-than-median outputs, you pass the test. "New HOG" instead computes the mean of the probabilities of ... 5 You just need to do a bit more algebra: Note that$$ \sum_{i=0}^n (\overline{x_i+y_i})(x_i+y_i)=\langle x+y|x+y\rangle$$and then you can distribute the right-hand side to get$$\langle x|x\rangle+\langle x|y\rangle+\langle y|x\rangle+\langle y|y\rangle.$$Since | x\rangle and | y\rangle are normalized, we know that \langle x|x\rangle=\langle y|y\... 5 It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why d^2 dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are d^2 dimensional objects in the same sense that vectors in \Bbb R^3 are 3-dimensional (i.e., the cardinality of any basis set of \... 5 A matrix function f(A) for normal matrix A is defined as follows \begin{equation} f(A)=\sum_{i=1}^{n}f(\lambda_i)v_iv_i^T \end{equation} where \lambda_{i} is an eigenvalue and v_{i} is coresponding eigenvector (note: transposed vector v_{i} is a row vector). In your case: f(A) = \mathrm{e}^A and A = -i\frac{\phi}{2}\sigma_{2}. 5 Quantum computing, in general, does not use Heisenberg's matrix mechanics or Dirac's interaction picture. I suggest that you carefully read the Wikipedia page on dynamical pictures. The differences between the Heisenberg picture, the Schrödinger picture and Dirac (interaction) picture are well summarized in the following chart. The development of matrix ... 5 All tensor products of n Pauli operators \{I,X,Y,Z\} (that is 4^n combinations) form an orthogonal basis for the vector space of 2^n \times 2^n complex matrices. Hence, for every matrix there is a unique decomposition as a linear combination of tensor products of Pauli unitaries. Same is true if we fix some other unitary basis. If we not fix the ... 4 I think the idea of the proof is that if |h\rangle can be shown to be orthogonal to |\mathcal H\rangle then it would imply that that h \not\in \mathcal{H}. Otherwise, h\in \mathcal{H}. Not really as far as the method shown in the linked video is concerned. The algorithm described there uses a controlled-unitary operation of the form$$\newcommand{\...

4

The partial transpose is not the only positive but not completely positive operation that is possible on 2x2 and 2x3 systems. Trivially, any completely positive operation (such as a local unitary) combined with the partial transpose is a different positive operation. The point is that, as wikipedia puts it every such map $\Lambda$ can be written as $... 4 Shor's algorithm relies on determining the period of$a^x\bmod N$. If you only evaluate up to$N$, then you are undersampling, in much the same way that you would classically be below the Nyquist criteria. For example, if you measure the second register and get$y$, the first register collapses to all$x$such that$a^x\bmod N =y$. These$x$collide at$...

4

$I^{\otimes 23} = I\otimes I\otimes I\otimes \cdots \otimes I$ (containing 23 identity operators, each presumably being $2\times 2$) The $\otimes$ operator is just the Left Kronecker Product. Assuming that $y$ and $x$ represent qubits, then $|yx\rangle\langle yx|$ is some $4\times 4$ matrix, which can be calculated as the outer product of the column ...

4

Matrix just encodes linear operation that transforms basis vectors to some other vectors. For example, matrix $M$ can transform vector $|0\rangle$ to vector $m_{11}|0\rangle + m_{12}|1\rangle$ and vector $|1\rangle$ to $m_{21}|0\rangle + m_{22}|1\rangle$. In this case, this matrix is written as $\left(\begin{matrix} m_{11} & m_{21} \\ m_{12} & m_{22} ... 4 By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of$n$non-negative eigenvalues$\lambda_i$with$n$corresponding mutually orthogonal eigenvectors$|v_i\rangle$such that $$\rho = \sum_{i=1}^n{\lambda_i |v_i\rangle \langle v_i|}$$ This ... 4 In the paper that you refer to, they are essentially asking "when can we implement the partial transpose map$\Theta=I_2\otimes\Lambda$?". So, that means the SPA of this map must be positive. What you have calculated, by comparison, is to ask when the SPA of the transpose map$\Lambda\$ can be made positive. It might sound like these ought to be the same ...

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