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Unitary operations are only a special case of quantum operations, which are linear, completely positive maps ("channels") that map density operators to density operators. This becomes obvious in the Kraus-representation of the channel, $$\Phi(\rho)=\sum_{i=1}^n K_i \rho K_i^\dagger,$$ where the so-called Kraus operators $K_i$ fulfill $\sum_{i=1}^n K_i^\... 23 The church of the larger (or higher, or greater) Hilbert space is just a trick that some people like (myself included) for rewriting some operations. The most general operations that you can write down for a system are described by completely positive maps, while we like describing things with unitaries, which you can always do by moving from the original ... 17 Definitions Denoting the Haar measure of some function$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$is ... 16 One way to understand the relationship between the Choi representation of a channel and its possible Kraus representations is to use the vectorization map. Suppose that we have two finite-dimensional Hilbert spaces$\mathcal{X}$and$\mathcal{Y}$, and that we have fixed a standard basis$\{|1\rangle,\ldots,|n\rangle\}$of$\mathcal{X}$and a standard basis$...

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Short Answer Quantum operations do not need to be unitary. In fact, many quantum algorithms and protocols make use of non-unitarity. Long Answer Measurements are arguably the most obvious example of non-unitary transitions being a fundamental component of algorithms (in the sense that a "measurement" is equivalent to sampling from the probability ...

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At risk of going off-topic from quantum computing and into physics, I'll answer what I think is a relevant subquestion of this topic, and use it to inform the discussion of unitary gates in quantum computing. The question here is: Why do we want unitarity in quantum gates? The less specific answer is as above, it gives us 'reversibility', or as ...

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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 $\mathcal{... 11 It appears that the statement is not true in general. Suppose$X = Y = \{0,1\}$,$\mathcal{H}$is the Hilbert space corresponding to a single qubit, and$Wis defined as \begin{align} W(0,0) & = | 0 \rangle \langle 0 |,\\ W(0,1) & = | 1 \rangle \langle 1 |,\\ W(1,0) & = | 1 \rangle \langle 1 |,\\ W(1,1) & = \frac{1}{2} | 0 \rangle \langle 0 |... 10 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$be some fixed ... 10 Matrix inequalities of the form$A\ge B$should be read as $$A-B\ge 0\ ,$$ which in turn means that all eigenvalues of$A-B$are larger or equal than zero. In the given case,$M\le I$means that all eigenvalues of$M$are smaller or equal than one. (Note that this convention for$\ge$used on matrices depends on the field. In other fields, "$\ge0$" ... 9 You cannot always find such a Kraus decomposition. Notice that any CPTP map$\mathcal E$which does have a decomposition as a probabilistic mixture unitaries is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state: $$\mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{... 9 No, not necessarily. For example, take \rho to be a GHZ state and let \sigma be the completely mixed state of one qubit. We then have \lambda=4 and \mu=2. 9 The existence of the inverse of a linear map is independent of the way the map affects the trace. Moreover, if an invertible map preserves a property then its inverse necessarily also preserves the property. Since depolarizing channel preserves the trace, so does its inverse. Inverse of depolarizing channel We can derive the formula for the inverse \mathcal{... 8 Any map which is not Completely Positive, Trace Preserving (CPTP), is not possible as an "allowed operation" (a more-or-less complete account of how some system transforms) in quantum mechanics, regardless of what states it is meant to act upon. The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are ... 8 There are several misconceptions here, most of them originate from exposure to only the pure state formalism of quantum mechanics, so let's address them one by one: All quantum operations must be unitary to allow reversibility, but what about measurement? This is false. In general, the states of a quantum system are not just vectors in a Hilbert space ... 8 "Church of the higher hilbert space" is a term coined by John Smolin. According to quantiki it is: for the dilation constructions of channels and states, which [...] provide a neat characterization of the set of permissible quantum operations and to quote wikipedia, it: describe[s] the habit of regarding every mixed state of a quantum system as a pure ... 8 Let \mathcal{N} be the channels which subscripts for which conventions.$$ \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) & \... 8 The adjoint of a channel$\Phi$represents how observables transform (in the Heisenburg picture), under the physical process for which$\Phi$is the description of how states transform (in the Schrödinger picture). So, in particular, the expected value of a measurement of the observable$E$on a state$\Phi(\rho)is equivalent to the expected value of the ... 8 An elementary method is to simply carry out the integration \begin{align} \overline{F} &= \int\langle\psi|\mathcal{N_\gamma}(|\psi\rangle\langle\psi|)|\psi\rangle d\psi\\ &=\int\langle\psi|K_0|\psi\rangle\langle\psi|K_0^\dagger|\psi\rangle + \langle\psi|K_1|\psi\rangle\langle\psi|K_1^\dagger|\psi\rangle d\psi\\ & =\frac{1}{4\pi}\int_0^\pi\... 8 Every quantum channel has many Kraus representations that may differ in the number of Kraus operators. For example, for any positive integer n and numbers p_i with i=1,\dots,n and \sum_{i=1}^np_i=1 the matrices E_i=\sqrt{p_i}I form a valid, if impractical, Kraus representation of the identity channel with n Kraus operators. This example also ... 7 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 ... 7 Not exactly sure what you find confusing, but the ultimate need for Stinespring dilation theorem is that in quantum mechanics the dynamics is in general defined by a completely positive trace preserving map (CPTP) \rho \mapsto \Lambda(\rho). Now, we have a belief (rightly or wrongly) that all there is is a unitary evolution governed by Schrodinger's ... 7 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. 7 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 ... 7 There is an ambiguity in the choice of Kraus operators: If \{E_a\} is a set of Kraus operators for a channel \mathcal E, so is \{F_b\} with F_b=\sum_a v_{ab} E_a, with (v_{ab}) an isometry. In particular, you can choose a (v) which diagonalizes the matrix X_{ac}=\mathrm{tr}[E_a^\dagger E_b], in which case \{F_b\} satisfies your ... 7 For the specific linear function you are interested in, the solution turns out to be trivial: you can take the channel to be N_{X\rightarrow Y}(\rho) = \operatorname{Tr}(\rho) |\psi\rangle\langle \psi| for |\psi\rangle being an eigenvector of \sigma_Y having the largest possible eigenvalue. More generally, however, you can optimize any real-valued ... 7 ( I copied some text from a previous answer of mine) Defining the Choi and \chi matrix The Choi matrix is a direct result of the Choi-Jamiolkowski isomorphism. Some intuition on what this is can be found in this previous answer. Consider the maximally entangled state |\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle, where ... 7 No, this is not always possible. A counterexample is given by \sigma=I/d' and \Phi(X)=\mathrm{tr}(X)|0\rangle\langle0|. To see this, note that for X=I/d, \begin{align} 2(1-1/d) & = \|\,|0\rangle\langle0|-I/d\|_1 \\ &= \|\Phi(X)-I/d\|_1 \\ &\le \left\|U\left(X\otimes \frac{I}{d'}\right)U^\dagger-U\left(\frac{I}{d}\otimes\frac{I}{d'}\right)... 7 No. The minimal size of the environment is just the rank of the Choi matrix of \mathcal E, call it J(\mathcal E). Since J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n} and \text{rank}(A \otimes B) = \text{rank}(A)\text{rank}(B), the minimal size of the environment is just \text{rank}\big(J(\mathcal E)\big)^n. 7 No, that doesn't work. It's fine to use an arbitrary pure state because the unitary U can always be used to take it to any pure state you want. This argument doesn't work for a mixed state, as unitaries cannot take mixed states to pure states. As a concrete example, consider the CPTP map \Lambda(\rho) = |0\rangle\langle 0| \operatorname{tr}(\rho),\$ ...

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