# Tag Info

21

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^\... 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 ... 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 ... 11 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 ... 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{...

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 $... 6 Basically, it means that the correlations could be used to send a message. Or simply that Bob’s measurement outcomes can reveal some details of Alice’s actions. This is impossible when Alice and Bob each hold one qubit of a Bell pair. Despite the entanglement present, as well as contextuality, signaling in this case would result faster than light ... 6 It suffices to prove that if$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 $$0 = u^{\ast} (P + Q) u = u^{\ast} ... 6 This question is posed, and answered positively, in Nielsen & Chuang in a subsection of chapter 8 entitled "System-environment models for and operator-sum representation". In my version, it can be found on page 365. Imagine |\psi\rangle is an arbitrary pure state on the space upon which you wish to enact the operators. Let |e_0\rangle be some fixed ... 5 This is not the unitary that you have to implement: you need a two-qubit unitary$$ \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 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 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}. ... 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 ... 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 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 ... 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 The partial transpose is not the only positive but not completely positive operation that is possible on 2x2 and 2x3 systems. Trivially, any completely positive operation (such as a local unitary) combined with the partial transpose is a different positive operation. The point is that, as wikipedia puts it every such map \Lambda can be written as ... 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 In the paper that you refer to, they are essentially asking "when can we implement the partial transpose map$\Theta=I_2\otimes\Lambda$?". So, that means the SPA of this map must be positive. What you have calculated, by comparison, is to ask when the SPA of the transpose map$\Lambdacan be made positive. It might sound like these ought to be the same ... 3 Acting with the dephasing channel on the possible states of a single qubit: \begin{align}D\left(\left|0\rangle\langle0\right|\right) &= \left|0\rangle\langle0\right| \\ D\left(\left|0\rangle\langle1\right|\right) &= \left(1-p\right)\left|0\rangle\langle1\right|\\ D\left(\left|1\rangle\langle0\right|\right) &= \left(1-p\right)\left|1\rangle\... 3 For amplitude damping,\gamma$is something like$e^{-\Delta t/T_1}$where$\Delta t$is how long the Kraus operator is supposed to act. But be very careful, Kraus evolution assumes your system has no initial correlations, that every qubit interacts with identical baths and that every qubit is identical. All the assumptions are most likely violated and so ... 3 The state$\mid \psi \rangle$is fixed. You can write it as$a |0 \rangle + b |1 \rangle$. If you write that in the other basis and get$ (a,b) = (1,0) \implies (c,d) = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) $They are the same state so it is still true that if you do a measurement in the$|0 \rangle$,$|1 \rangle$basis, you will surely get$\mid 0 \...

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

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

This gate, which I'll denote as $U$ is essentially the square root of not gate, whose decomposition has already been discussed elsewhere. Hence, we only need the conversion: Specifically, if we perform the corresponding matrix multiplications, we have  \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & ...

3

In quantum mechanics, not all operators are observables. Many operators are observables, and in the first year or two of treatments in some physics courses you will only see operators which are observables; but not all operators of interest are observables. The operators you have mentioned all happen to be Hermitian, and could therefore be interpreted as ...

3

I agree with the main points that Niel makes: not all operators are observables, and the purpose of the ones you list is typically to transform states, not to be measured. However, the operators you list happen to be hermitian (allowing them also to represent observables) as well as unitary (allowing them to represent transformations), so in this case we ...

3

Those maps are linear, so if it preserves the trace of trace 1 matrices then it preserves the trace of any other matrix. We just don't need to restrict ourselves by considering only trace 1 matrices (from mathematical point of view).

Only top voted, non community-wiki answers of a minimum length are eligible