# Tag Info

## Hot answers tagged quantum-information

5

After several discussions on that topic, I came to think that it is more appropriate to separate those different types of "bits" into four instead of two classes: A deterministic classical bit is an ordinary bit can be either in state $\left|0\right>$ or in state $\left|1\right>$. No other states allowed. A random classical bit can be in a "unknown" ...

3

Use that the trace is invariant under cyclic permutations of its arguments, i.e. $$\mathrm{Tr}(ABC) = \mathrm{Tr}(CAB) = \mathrm{Tr}(BCA)$$

3

For a freely falling observer the local geometry of spacetime is always flat i.e. described by the Minkowski metric. So the freely falling observer can never observe themself to fall through an event horizon, because that contradicts the requirement that spacetime be locally flat. In fact the freely falling observer will observe an apparent horizon that ...

3

Following http://arxiv.org/abs/quant-ph/0110082, for two qudits the manifold of all product states is $4(d-1)$ dimensional, while the manifold of maximally entangled states has $d^2-1$ dimensions. Thus, with the possible exception of $d=3$, these two sets cannot be mapped onto each other by any kind of "nice" mapping (and in particular not by a linear map). ...

3

The ground state of the toric code can be understand as a superposition of all loop configurations in the $z$ basis. The fact that these loops fluctuate at all length scales (and thus around the torus) leads to the topological order in the system. The $\sigma_z$ terms lead to a "tension" in the loops, penalizing long loops. Eventually, this tension will ...

2

Coherent state is one thing and decoherence is something else. The coherent state has the form $(\text I) \ |\alpha \rangle = e^{-\alpha ^2/2} \sum _n \frac {\alpha ^n}{\sqrt {n!}} |n \rangle.$ where $|n \rangle$ is a Fock state of n identical particles. This state is a coherent superposition of Fock states, and its density matrix has diagonal and ...

2

For one thing, toric codes (and other error-correcting codes) are really about ways to store quantum information(logical qubits) in several physical qubits, so there is not much point in asking for a continuous limit. On the other hand, if you view it as a topological quantum phase of matter, then surely there are continuous versions. For example, the ...

1

There are essentially two questions here: (mathematical) stability under perturbation (i.e. changing $\tau$ to $\tau+\varepsilon \eta$ for some small $\varepsilon$). channel stability under perturbation. Let's fix notation: We define the Choi matrix $\tau$ of a completely positive map $T$ from $n\times n$ matrices to $n\times n$ matrices via ...

1

Here, the idea behind introducing the subspace $S$ and its orthogonal complement $S^\bot$ was to show that all vectors on the RHS of equation 6.5.1 and 6.5.2 form a vector space of dimension $n-1$. As you say, indeed, the zero vector $\mathbf{0}$ is a vector that we need, but you are wrong in saying that it is 'forbidden' because it appears in the subspace ...

1

If you are interested in how these sets look like, I can provide their characterization. A state on two $d$-dimensional systems is maximally entangled if and only if it can be written as follows: $$\frac{1}{\sqrt{d}} \sum_{i=1}^d |u_i\rangle |v_i\rangle$$ where $|u_i\rangle$ is the $i$-th column of some $d \times d$ unitary matrix $U$ and similarly ...

1

This is exactly analogous to the procedure for finding matrix elements of normal operators. Let's first recall how this works in the familiar case. You choose an orthonormal basis of vectors, say $|n\rangle$, with $n = 1,2,\ldots D$, where $D$ is the dimension of the Hilbert space, such that $\langle n\rvert m\rangle = \delta_{mn}$. Now the matrix elements ...

1

If you want to write a super-operator representing left- or right-multiplication, there is a distinct method which is simpler and more elegant. Let us define the left-multiplication superoperator by $$\mathcal{L}(A)[\rho] = A\rho,$$ and the right-multiplication superoperator by $$\mathcal{R}(A)[\rho] = \rho A.$$ It should be clear that these operations ...

1

I'll leave out the math, partly because I'm not sure I remember it precisely enough. And then there's the formatting problem. A bit (per Shannon's information theory) is the smallest amount of information. So in digital circuitry, it is a voltage (or lack thereof) on a TTL (or similar) gate. Where it came from (ie, random sequence or one bit of an ASCII ...

1

It does not matter whether or not you look. What matters is that it is possible to look in principle, because the quantum particle interacts with the LED and therefore becomes entangled with it (your unitary matrix acts on both the particle and LED). So the outcome will always be the same as experiment 1. The technical term for this is "which-way" ...

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