# Tag Info

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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 ...

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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^\... 13 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 ... 12 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 ... 10 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{...

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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 ...

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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 ...

9

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, "$\ge0$" ...

8

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 $... 8 You cannot always find such a Kraus decomposition. Notice that any CPTP map$\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: $$\mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{... 7 There are several misconceptions here, most of them originate from exposure to only the pure state formalism of quantum mechanics, so let's address them one by one: All quantum operations must be unitary to allow reversibility, but what about measurement? This is false. In general, the states of a quantum system are not just vectors in a Hilbert space ... 7 The adjoint of a channel \Phi represents how observables transform (in the Heisenburg picture), under the physical process for which \Phi is the description of how states transform (in the Schrödinger picture). So, in particular, the expected value of a measurement of the observable E on a state \Phi(\rho) is equivalent to the expected value of the ... 6 Basically, it means that the correlations could be used to send a message. Or simply that Bob’s measurement outcomes can reveal some details of Alice’s actions. This is impossible when Alice and Bob each hold one qubit of a Bell pair. Despite the entanglement present, as well as contextuality, signaling in this case would result faster than light ... 6 First let me mention a minor point concerning terminology. The type of channel you are suggesting is often called a Pauli channel; the term depolarizing channel usually refers to the case where p_x = p_y = p_z. Anyway, it is not really correct to say that Pauli channels are the channel model considered for quantum error correction. Standard quantum error ... 6 Any map which is not Completely Positive, Trace Preserving (CPTP), is not possible as an "allowed operation" (a more-or-less complete account of how some system transforms) in quantum mechanics, regardless of what states it is meant to act upon. The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are ... 6 This really depends where you want to start from. For instance, you can construct the Choi state of \mathcal E, i.e.,$$ \sigma = (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle\Omega|)\ , $$with \Omega = \tfrac{1}{\sqrt{D}}\sum_{i=1}^D |i,i\rangle, and then extract the Kraus operators of \mathcal E(\rho)=\sum M_i\rho M_i^\dagger by taking any ... 6 There are several ways that you could realise the depolarising map \mathcal N_p(\rho) = (1\!-\!p)\:\!\rho + p \!\!\:\cdot\!\tfrac{1}{2}\mathbf 1 map on a quantum computer — including an idealised quantum computer, in which waiting around for the noise to do the work for you would not be an available method.\def\ket#1{\lvert#1\rangle} We start ... 6 It suffices to prove that if P and Q are positive semidefinite operators, then$$ \operatorname{im}(P) \subseteq \operatorname{im}(P+Q). $$Once you have this, the statement follows by taking P = \eta(a) and Q = \rho - \eta(a). Suppose that u is a vector with u \perp \operatorname{im}(P+Q). This implies that$$ 0 = u^{\ast} (P + Q) u = u^{\ast} ... 6 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 ... 6 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 ... 5 This is not the unitary that you have to implement: you need a two-qubit unitary $$\frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 & \omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array}\right),$$ where$\omega=e^{2i\pi/3}$, the point being that if you introduce ... 5 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 $$\mathcal{N}_{P} (\rho) = \begin{pmatrix} \rho_{00} & \rho_{01} (1-\lambda)\\ \rho_{10} (1-\lambda) & \... 5 "Church of the higher hilbert space" is a term coined by John Smolin. According to quantiki it is: for the dilation constructions of channels and states, which [...] provide a neat characterization of the set of permissible quantum operations and to quote wikipedia, it: describe[s] the habit of regarding every mixed state of a quantum system as a pure ... 5 Not exactly sure what you find confusing, but the ultimate need for Stinespring dilation theorem is that in quantum mechanics the dynamics is in general defined by a completely positive trace preserving map (CPTP) \rho \mapsto \Lambda(\rho). Now, we have a belief (rightly or wrongly) that all there is is a unitary evolution governed by Schrodinger's ... 5 A channel \Phi is said to be degradable if there exists another channel \Xi such that \Xi\Phi is complementary to \Phi. The idea here is as follows. Suppose \Phi is a channel and \Psi is complementary to \Phi. If \Phi is applied to a state \rho, then the output of the channel is \Phi(\rho) (of course), while \Psi(\rho) represents ... 5 If P is (ortho)projector, that is P^2=P=P^\dagger, then we can define unitary U = I - 2P. You can verify$$ 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 Mathematically it is a relationship between a bipartite linear operator vector space$L(X\otimes Y)$and a superoperator vector space$C(X): L(X)\to L(Y)$(maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum ... 5 "Why" is quite a nebulous concept that, ultimately, comes back to your understanding of quantum mechanics. For me, the way that I set up QM is with a set of postulates. The first postulate is that quantum states are represented by vectors in a Hilbert space of length 1. A second postulate is essentially that operations are linear. (This is not how ... 4 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 ... 4 I'll add a small bit complementing the other answers, just about the idea of measurement. Measurement is usually taken as a postulate of quantum mechanics. There's usually some preceding postulates about hilbert spaces, but following that Every measurable physical quantity$A$is described by an operator$\hat{A}$acting on a Hilbert space$\mathcal{H}\$. ...

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