# Tag Info

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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 $\sum_{i=1}^n K_i^\... 17 Let's say we have a function$f$which maps$n$bits to$m$bits (where$m<n$). $$f: \{0,1\}^{n} \to \{0,1\}^{m}$$ We could of course design a classical circuit to perform this operation. Let's call it$C_f$. It takes in as input$n$-bits. Let's say it takes as input$X$and it outputs$f(X)$. Now, we would like to do the same thing using a quantum ... 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 What is the proof that any given unitary matrix can be converted as above? Let$U$be an arbitrary$2\times 2$unitary matrix. This is equivalent to the rows/columns of$U$forming an orthonormal system. Let us write a generic$U$as $$U=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$ The constraints imposed on the coefficients$a,b,c,d$by the requirement ... 10 Even if you only limit yourself to special-unitary operations, states will still accumulate global phase. For example,$Z = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$is special-unitary but$Z \cdot |0\rangle = i |0\rangle \neq |0\rangle$. If states are going to accumulate unobservable global phase anyways, what benefit do we get out of limiting ... 9 All operations on quantum states are unitary operations. We don't make the rules, this is just how nature seems to work. So if you want to define an operation that copies a qbit, it has to be a unitary operation. That unitary operation would look like this:$U|\psi\rangle_A|0\rangle_B=|\psi\rangle_A|\psi\rangle_B$So you have the qbit you want to copy,$|\...

9

The fact that quantum gates are unitary, is rooted in the fact that the evolution of (closed) quantum systems is by the Schrödiner equation. For a time interval in which we are trying to realise a particular unitary transformation at a constant rate, we use the time-independent Schrödinger equation: $$\tfrac{\mathrm d}{\mathrm dt} \lvert \psi(t) ... 8 Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector. Quantum states are changed/evolved according to some matrix. Given ... 8 Apply it twice:$$ O_xO_x|i\rangle|b\rangle=O_x|i\rangle|b\oplus x_i\rangle=|i\rangle|b\oplus x_i\oplus x_i\rangle=|i\rangle|b\rangle $$Hence, O_x is its own inverse, and therefore reversible. To prove unitarity, it makes more sense to prove that O_x has eigenvectors$$ |i\rangle(|0\rangle+|1\rangle)\quad\text{and}\quad|i\rangle(|0\rangle-|1\rangle) $$... 8 Your construction by gueswork in this answer is OK but not really elegant. Moreover, it's a convention to start in the state |0\rangle; we usually don't initialize a qubit with the state |1\rangle. It's better to follow the general construction which I illustrate here. The idea here is to use ancillary qubits and impose unitary evolution on the larger ... 7 There are several misconceptions here, most of them originate from exposure to only the pure state formalism of quantum mechanics, so let's address them one by one: All quantum operations must be unitary to allow reversibility, but what about measurement? This is false. In general, the states of a quantum system are not just vectors in a Hilbert space ... 7 As already mentioned in the other answers, the crucial point is that copying means implicitly that the state of the original qubit is unknown, i.e. given a qubit in an unknown state, you want to prepare a second qubit to be in exactly the same state. To make it more intelligible, there is a less mathematical argument that this should not be possible: By the ... 7 A necessary and sufficient condition is that, given an n\times n matrix M, you can construct a 2n\times 2n unitary matrix U provided the singular values of M are all upper bounded by 1. Sufficiency To see this, express the singular value decomposition of M as$$ M=RDV $$where D is diagonal and R, V are unitary. Now define$$ U=\left(\...

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Firstly simply rewrite probability amplitudes of returned states as columns of a matrix: $$U = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1-i}{2} \\ \frac{1+i}{2} & -\frac{1}{\sqrt{2}} \end{pmatrix}$$ Now do some algebra U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \frac{1-i}{\sqrt{2}} \\ \frac{1+i}{\sqrt{2}} & -1 \end{pmatrix} = \frac{1}{... 7 Take your vector \frac{1}{\sqrt{5}}(0, 1, 1, 1, 1, 1)^T and five other arbitrary ones but at the same time these vectors have to be linearly independent. After that apply Gram-Schmidt process which produces orthonormal vectors. Put these vectors to a matrix and you will get a unitary matrix with the first column equal to \frac{1}{\sqrt{5}}(0, 1, 1, 1, 1, ... 6 Born's rule states that |\psi(x)|^2 = P(x) which is the probability of finding the quantum system in the state |x\rangle after a measurement. We need the sum (or integral!) over all x to be 1: \begin{align} \sum_x P_x &= \sum_x |\psi_x|^2 = 1,\\ \int P(x)dx &= \int |\psi(x)|^2 dx= 1. \end{align} Neither of these are valid norms because they ... 6 This is actually a much easier problem. In the case of states, you're trying to use the PPT criterion, or others, to distinguish if \rho can be written in the form \rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i, $$where \sum_ip_i=1 and the \sigma^A_i and \sigma^B_i are valid states on single sites. The difficulty actually comes from the freedom that ... 5 More mathematically, because \mathbb{R}^n with an L^p norm is a Hilbert space only for p=2. 5 When writing gates for, for example, a quantum circuit diagram, you could always write them using the convention of having determinant one (from the special unitary group), but it's just a convention. It makes no physical difference to the circuit that you implement. As said elsewhere, whether what you naturally produce corresponds directly to the special ... 5 Simulating Classical "AND/NAND/OR/NOR/XOR/XNOR" Gates With the help of this answer from Blue, constructing a matrix for a classical gate is just a matter of following the steps. Here is the combined truth table for classical logic gates:$$ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Input} & \text{AND} & \text{NAND} & \text{OR} & \text{...

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Martin Vesely's answer is the way to go in general, and especially if you know more than one column. However, if you're given just one column, there's an easier trick for generating a suitable unitary. Note that $V=2|v\rangle\langle v|-I$ is a unitary ($V=V^\dagger$ and $V^2=I$). So, the question is whether you can select a $|v\rangle$ such that the first ...

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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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To answer the first part of the question (whether unitary matrix $U$ operates on $|\psi_A \rangle$ only): A unitary matrix can operate on an arbitrary number of qubits. Single-qubit gates, like Pauli X, Y and Z gates, operate on one qubit and are represented by 2x2 matrices; CNOT gate operates on two qubits and is represented by a 4x4 matrix, etc. In this ...

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A particularly efficient way is the look at the Schmidt coefficients of your target state. You know that your state can be written as $$U_1\otimes U_2(\alpha|00\rangle+\beta|11\rangle),$$ and the Schmidt decomposition tells you what $\alpha,\beta,U_1,U_2$ are. So, obviously, the problem becomes producing $$\alpha|00\rangle+\beta|11\rangle.$$ This is ...

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The simplest way to solve this problem is to work backwards from the output to the input. Suppose you have the state $a|00\rangle + b|01\rangle + b|10\rangle + b|11\rangle$. How can you reduce this to just the state $|00\rangle$ with unitary operations? Applying the inverse of those operations in reverse order will send you from $|00\rangle$ to the desired ...

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An elegant argument can be derived by asking which theories can we build which are described by vectors $\vec v = (v_1,\dots,v_N)$, where the allowed transformations are linear maps $\vec v\to L\vec v$, probabilities are given by some norm, and probabilities must be preserved by those maps. It turns out that there are basically only three options: ...

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Controlled gates are used to create entanglement. They don't have anything to do with reversibility; all unitary gates are reversible by definition, since all unitary operators have inverses. You could come up with an alternative choice of gates that you used to write algorithms in that didn't include controlled gates. For instance, for quantum computers ...

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Any quantum operation is basically a Hamiltonian $H$ acting on an isolated Hilbert space states. Now, the requirements for a Hamiltonian is to a matrix which is Unitary and Hermitian. This intrinsically implies it has an inverse, which means it is reversible because you can always find a matrix $H^{-1}$ and apply it on the state. This is where reversibility ...

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You have $U|1\rangle=e^{i\phi}(b^*|0\rangle-a^*|1\rangle)$ (and for real entries, $e^{i\phi}=\pm1$). This condition follows automatically from $$\langle 0|U^\dagger U |1\rangle=0$$ -- this is exactly the condition you describe -- together with the fact that $U|0\rangle$ and $U|1\rangle$ must have the same normalization,  \langle k|U^\dagger U |k\rangle=1 ...

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