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# Tag Info

## Hot answers tagged mathematics

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.

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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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