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relationship between helstrom operators acting on different pairs of quantum states

tl;dr: In the single-qubit case the explicit formula $$P(|0\rangle\langle0|-\rho)=\begin{cases} 0&\rho=|0\rangle\langle 0|\\ \frac{|0\rangle\langle0|-\rho}{\sqrt{|\langle 0|\rho|1\rangle|^2+(\...
Frederik vom Ende's user avatar
1 vote

How to calculate the Schmidt decomposition of a state without SVD

When a question says "without using the SVD" it may be implying that there are easier ways, at least in this specific case. It's always worth just having a stare at the state for a minute (...
DaftWullie's user avatar
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2 votes

What is meant with "different ensembles can give rise to the same density matrix?"

An ensemble in this context is a set of states with attached probabilities. In your example the ensembles would be written as $\{(\frac12,|a\rangle\langle a|),(\frac12,|b\rangle\langle b|)\}$ and $\{(\...
glS's user avatar
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1 vote

What is meant with "different ensembles can give rise to the same density matrix?"

The two ensembles are $|0\rangle, |1\rangle$ and $|a\rangle, |b\rangle$. The point of this example is to show you that the same density matrix $\rho = \begin{pmatrix} 3/4 & 0 \\ 0 & 1/4 \end{...
David Dentelski's user avatar
4 votes
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How to calculate the Schmidt decomposition of a state without SVD

Why do you need to calculate the Schmidt decomposition without SVD? Doing it in other ways will just lead to the same result. That aside, the problem with the calculation here is that knowing the ...
glS's user avatar
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0 votes

Interpretation of a circuit that yields the same result for initializations $|+\rangle$ and $|-\rangle$

To see why $|+\rangle$ and $|-\rangle$ states are not the same consider $|\psi\rangle = |i\rangle$ and the measurement outputs $|1\rangle$. The two states only differ by a global phase, but they are ...
Daniele Cuomo's user avatar
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Interpretation of a circuit that yields the same result for initializations $|+\rangle$ and $|-\rangle$

One way of seeing this is that the difference between the circuits is a $Z$-gate applied to the first qubit. This won't affect the measurement, and $Z$-gates are not 'transposed' by the CNOT gate to ...
JoJo P's user avatar
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1 vote

Interpretation of a circuit that yields the same result for initializations $|+\rangle$ and $|-\rangle$

The way to think about the circuit you showed (and it can generalise to some multi-qubit cases) is that when you measure, you collapse the qubit into a particular state, 0 or 1. You can now run the ...
DaftWullie's user avatar
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1 vote
Accepted

What is the density matrix of a GHZ state when onle a qubit is in a decoherence channel?

Suppose Rob's channel applies a bit flip with probability $p$. It means that with probability $1-p$, the state is: $$\rho_0=\frac{|000\rangle\!\langle000|+|000\rangle\!\langle111|+|111\rangle\!\...
Tristan Nemoz's user avatar
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1 vote

Can different density matrices have 100% fidelity with a given pure state?

If $|\psi\rangle$ is a pure state and $\rho$ is a density matrix, then $F(|\psi\rangle\!\langle\psi|,\rho)=\langle\psi|\rho|\psi\rangle=1$ iff $\rho=|\psi\rangle\!\langle\psi|$. Or more generally, ...
glS's user avatar
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4 votes
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Exponential Quantum Speedup for the Traveling Salesman Problem - where is the catch?

The catch is the parameter $\kappa$, the condition number of the embedding matrix. I can't comment on the correctness of the whole work. This will require a thorough examination of the content. But ...
Manish Kumar's user avatar
4 votes

What's the Schmidt decomposition of $|\psi\rangle = 1/ \sqrt{3}( |0\rangle| 0\rangle + |0\rangle |1\rangle + |1\rangle |1\rangle)$?

One way of computing the decomposition is through density matrices, but then you will have to diagonalize those density matrices. This requires the eigenvalue decomposition of each density matrix. ...
MonteNero's user avatar
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4 votes
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Why is the trace distance between two density matrices not always $0$?

The problem in your analysis is the fourth equality: unless $\rho_1 - \rho_2$ is zero it will not be positive semidefinite, so it will not be the case that $\sqrt{(\rho_1 - \rho_2)(\rho_1 - \rho_2)} = ...
John Watrous's user avatar
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1 vote

Analyzing the composition of a channel with its adjoint in relation with an identical composition obtained for the channel's complement

The short answer is: yes the inclusion in question is true for all maps $\Phi$ which are completely positive and completely co-positive. In fact, it turns out to hold for the slightly more general ...
Frederik vom Ende's user avatar
2 votes
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If states are close together does there always exist a channel close to the identity mapping one to the other?

We shall see that for general mixed states no such upper bound can exist in the following precise sense: For no continuous function $c:[0,2]\to[0,2]$ with $c(0)=0$ does it hold that for all $\|\rho-\...
Frederik vom Ende's user avatar
2 votes
Accepted

How to prove the inclusion relation $\text{Im} (\rho) \subseteq \text{Im} (\rho[X] \otimes \rho[Y])$ about density operators?

Let's consider some examples first to see what we're talking about: Start thinking about a simple case: what does this mean if $\sqrt2|\Psi\rangle=|00\rangle+|11\rangle\in\mathbb{C}^2\otimes\mathbb{C}...
glS's user avatar
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2 votes
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Efficient Clifford simulation and entropy of reduced density matrices

TL;DR: Yes, we can efficiently compute von Neumann entropy $S(\rho)$ of a $k$-qubit mixed state $\rho$ obtained by tracing out $n-k$ qubits from a pure $n$-qubit stabilizer state $|\psi\rangle$ using \...
Adam Zalcman's user avatar
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