20

The church of the larger (or higher, or greater) Hilbert space is just a trick that some people like (myself included) for rewriting some operations. The most general operations that you can write down for a system are described by completely positive maps, while we like describing things with unitaries, which you can always do by moving from the original ...


9

Let's start by finding a complementary channel for any channel given by a Kraus representation $$ \Phi(X) = \sum_{k=1}^n A_k X A_k^{\dagger}. $$ To make the necessary equations clear, let us assume that the channel has the form $\Phi:\mathrm{L}(\mathcal{X})\rightarrow \mathrm{L}(\mathcal{Y})$ for finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{...


6

This question is posed, and answered positively, in Nielsen & Chuang in a subsection of chapter 8 entitled "System-environment models for and operator-sum representation". In my version, it can be found on page 365. Imagine $|\psi\rangle$ is an arbitrary pure state on the space upon which you wish to enact the operators. Let $|e_0\rangle$ be some fixed ...


6

Basically, it means that the correlations could be used to send a message. Or simply that Bob’s measurement outcomes can reveal some details of Alice’s actions. This is impossible when Alice and Bob each hold one qubit of a Bell pair. Despite the entanglement present, as well as contextuality, signaling in this case would result faster than light ...


6

Definitions Denoting the Haar measure of some function $f\left(x\right)$ over $d$-dimensional unitaries as $\int_{\mathrm U\left(d\right)}f\left(x\right)d\mu\left(x\right)$, twirling some arbitrary channel $\varepsilon$ can be defined as the operation $\varepsilon \mapsto\int_{\mathrm U\left(d\right)}U^\dagger\varepsilon U dU$, which, when $\varepsilon$ is ...


5

A channel $\Phi$ is said to be degradable if there exists another channel $\Xi$ such that $\Xi\Phi$ is complementary to $\Phi$. The idea here is as follows. Suppose $\Phi$ is a channel and $\Psi$ is complementary to $\Phi$. If $\Phi$ is applied to a state $\rho$, then the output of the channel is $\Phi(\rho)$ (of course), while $\Psi(\rho)$ represents ...


5

One way to understand the relationship between the Choi representation of a channel and its possible Kraus representations is to use the vectorization map. Suppose that we have two finite-dimensional Hilbert spaces $\mathcal{X}$ and $\mathcal{Y}$, and that we have fixed a standard basis $\{|1\rangle,\ldots,|n\rangle\}$ of $\mathcal{X}$ and a standard basis $...


5

If $P$ is (ortho)projector, that is $P^2=P=P^\dagger$, then we can define unitary $U = I - 2P$. You can verify $$ UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I $$ Now we can express $P=\frac{1}{2}(I-U), Q=\frac{1}{2}(I+U)$ and calculate $$ P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)= $$ $$ = \frac{1}{4}(\rho - U\rho - \rho U +...


5

There are several ways that you could realise the depolarising map $ \mathcal N_p(\rho) = (1\!-\!p)\:\!\rho + p \!\!\:\cdot\!\tfrac{1}{2}\mathbf 1$ map on a quantum computer — including an idealised quantum computer, in which waiting around for the noise to do the work for you would not be an available method.$\def\ket#1{\lvert#1\rangle}$ We start ...


5

"Church of the higher hilbert space" is a term coined by John Smolin. According to quantiki it is: for the dilation constructions of channels and states, which [...] provide a neat characterization of the set of permissible quantum operations and to quote wikipedia, it: describe[s] the habit of regarding every mixed state of a quantum system as a pure ...


5

Any map which is not Completely Positive, Trace Preserving (CPTP), is not possible as an "allowed operation" (a more-or-less complete account of how some system transforms) in quantum mechanics, regardless of what states it is meant to act upon. The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are ...


4

First let me mention a minor point concerning terminology. The type of channel you are suggesting is often called a Pauli channel; the term depolarizing channel usually refers to the case where $p_x = p_y = p_z$. Anyway, it is not really correct to say that Pauli channels are the channel model considered for quantum error correction. Standard quantum error ...


4

This really depends where you want to start from. For instance, you can construct the Choi state of $\mathcal E$, i.e., $$ \sigma = (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle\Omega|)\ , $$ with $\Omega = \tfrac{1}{\sqrt{D}}\sum_{i=1}^D |i,i\rangle$, and then extract the Kraus operators of $\mathcal E(\rho)=\sum M_i\rho M_i^\dagger$ by taking any ...


4

Let $\mathcal{N}$ be the channels which subscripts for which conventions. $$ \mathcal{N}_{N.C.} (\rho) = \begin{pmatrix} \rho_{00} & \rho_{01} \sqrt{1-\lambda}\\ \rho_{10} \sqrt{1-\lambda} & \rho_{11} \end{pmatrix} $$ As compared to $$ \mathcal{N}_{P} (\rho) = \begin{pmatrix} \rho_{00} & \rho_{01} (1-\lambda)\\ \rho_{10} (1-\lambda) & \...


4

If you're told about an operation on a single qubit, then to convert it into an operation on both qubits, you just include the identity matrix on the other qubit. So, contrast the single qubit state $|\psi\rangle$ going through the bit-flip channel $$ |\psi\rangle\rightarrow(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X $$ with what happens on two ...


3

These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let $\Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y})$ be the channel whose capacities we are interested in and let $\Psi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Z})$ be a ...


3

Josu, You might be mis-understanding something. Your equation for the Pauli channel says that when $t\rightarrow \infty$, all operators ($X,Y,Z,I$) have an equal probability (1/4) of transforming the density matrix $\rho$. You seem to be suggesting that $t\rightarrow \infty$, the probability of the $I$ operator acting on $\rho$ whould go to 0. Keep in mind ...


3

I'll cover a slightly more general case. Let $P_k$, $k=1,...,N$ a complete set of orthogonal projectors: $\sum_k P_k=I$ and $P_j P_k=\delta_{jk}P_j$. Consider the map $\mathcal E(\rho)=\sum_k P_k \rho P_k$. We want to find a set of unitaries $\mathcal U_k$ and probabilities $p_k$ such that $\mathcal E(\rho)=\sum_\ell p_\ell\mathcal U_\ell\rho\,\mathcal U_\...


3

Let's recap a bit: In classical information theory, the analogous formula is the Shannon noisy channel coding theorem. It's charming, because it is basically just a very simple optimization of the mutual information. The quantum channel capacity is that it is given by $$ \lim\limits_{n\to\infty} \frac{1}{n}Q(T^{\otimes n}) $$ where $T$ is the quantum ...


3

Acting with the dephasing channel on the possible states of a single qubit: \begin{align}D\left(\left|0\rangle\langle0\right|\right) &= \left|0\rangle\langle0\right| \\ D\left(\left|0\rangle\langle1\right|\right) &= \left(1-p\right)\left|0\rangle\langle1\right|\\ D\left(\left|1\rangle\langle0\right|\right) &= \left(1-p\right)\left|1\rangle\...


2

Yes. Choi's Theorem a priori uses different Hilbert spaces of potentially different dimensions $d_1$ and $d_2$. Then $d_1=d_2$ is a corollory. The proof is included there.


2

As pointed out in a comment, what you wrote as $\rho$ should more precisely be written as $\rho\otimes\mathbb 1$ (although the Kraus operator decomposition can be obtained similarly with any initial ancilla state, in which case you have $\rho\otimes|\phi\rangle\!\langle\phi|$). The standard algebraic properties of tensor product spaces then tell you that $$(...


2

The channel $\mathcal{E}$ is explicitly defined in the preceding paragraph as being the depolarising channel. Thus, all you need to calculate is $$ F=\sqrt{\langle 0|\mathcal{E}(|0\rangle\langle 0|)|0\rangle}. $$


2

The general expression for the fidelity is $$ F(\rho,\sigma)=\left(\text{Tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2=(\text{Tr}|\sqrt{\rho\sigma}|)^2. $$ Assume $\rho$ and $\sigma$ are $2\times 2$ matrices. Then $\sqrt{\rho\sigma}$ is also a $2\times 2$ matrix which we assume to have eigenvalues $\lambda_1$ and $\lambda_2$. Thus, $$ F=(|\lambda_1|+|\...


2

As matrices, the natural representation and Choi representation of a map $\Phi$ have exactly the same entries, but arranged into matrices in different ways. One way to express this is like this: $$ \langle k,\ell | K(\Phi) | i,j \rangle = \langle k | \Phi(|i\rangle\langle j|) | \ell \rangle = \langle k,i | J(\Phi) | \ell,j \rangle, $$ where $K(\Phi)$ denotes ...


2

Suppose that $\mathsf{X}$ is a register that can store each possible choice for $x$, as a classical state, while $\mathsf{Y}$ is a register that can store each possible state $\rho_x$. It is then natural to associate the classical-quantum state $$ \rho = \sum_x p_x |x\rangle \langle x| \otimes \rho_x $$ with the ensemble $\{(p_x,\rho_x)\}$. Now try taking ...


1

Quantum measurement (without results recording) is just a special case of quantum operation (quantum channel). So, yes, measurement operators (as in general measurement formalism) are indeed Kraus operators. But Kraus operators are more general. For example, they can be "rectangular", while measurement operators can't.


1

This probably only addresses half of your question... Most of the time, estimates of the error probability are not so important. If you're using a non-degenerate error correcting code for a single logical qubit, all you need to know (effectively) is enough to resolve a choice between two different error corrections (i.e. you want to know that the number of ...


1

The situation of non-completely positive maps (or more generally non-linear maps) is controversial partly due to the precise definition of how you should construct the map. But it is easy to come up with an example of something that would seem to be NCP or even not linear. Non linear map. Consider a preparation device that can create a qubit in an ...


1

Here is another way to prove the equivalence of the two expressions explicitly: For the Kraus representation, $\Phi(\rho)=\sum_a A^a \rho A^{a\dagger}$, if we make the indices explicit, we get $$\Phi(\rho)_{ij}=\sum_{a,k,\ell}A^a_{ik}A^{a*}_{j\ell}\rho_{k\ell}.\tag1$$ On the other hand, unravelling the second expression we have for a generic $\sigma$ (let ...


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