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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 ... 15 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 ... 14 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 ... 13 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,$|\...

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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: $$\tfrac{\mathrm d}{\mathrm dt} \lvert \psi(t) ... 12 At risk of going off-topic from quantum computing and into physics, I'll answer what I think is a relevant subquestion of this topic, and use it to inform the discussion of unitary gates in quantum computing. The question here is: Why do we want unitarity in quantum gates? The less specific answer is as above, it gives us 'reversibility', or as ... 12 Your construction by gueswork in this answer is OK but not really elegant. Moreover, it's a convention to start in the state |0\rangle; we usually don't initialize a qubit with the state |1\rangle. It's better to follow the general construction which I illustrate here. The idea here is to use ancillary qubits and impose unitary evolution on the larger ... 11 What is the proof that any given unitary matrix can be converted as above? Let U be an arbitrary 2\times 2 unitary matrix. This is equivalent to the rows/columns of U forming an orthonormal system. Let us write a generic U as$$U=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$The constraints imposed on the coefficients a,b,c,d by the requirement ... 10 Apply it twice:$$ 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) $$... 10 Correct, unitarity is a sufficient and necessary condition. From Nielson and Chuang page 18: Amazingly, this unitary constraint is the only constraint on quantum gates. Any unitary matrix specifies a valid quantum gate! The interesting implication is that in contrast to the classical case, where only one non-trivial single bit gate exists - the NOT gate - ... 9 You cannot always find such a Kraus decomposition. Notice that any CPTP map \mathcal E which does have a decomposition as a probabilistic mixture unitaries is unital, which is to say that it maps the identity to the identity, and in particular it maps the maximally mixed state to the maximally mixed state:$$ \mathcal E(\tfrac{1}{d} \mathbf 1) = \tfrac{1}{...

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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 $... 8 Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector. Quantum states are changed/evolved according to some matrix. Given ... 8 A necessary and sufficient condition is that, given an$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 $$U=\left(\... 8 As already mentioned in the other answers, the crucial point is that copying means implicitly that the state of the original qubit is unknown, i.e. given a qubit in an unknown state, you want to prepare a second qubit to be in exactly the same state. To make it more intelligible, there is a less mathematical argument that this should not be possible: By the ... 8 Time Evolution Postulate: A pure state |\psi(t_0)\rangle in a Hilbert Space \mathbb{H} evolves to another state |\psi(t)\rangle is given by the time evolution operator U(t,t_0)$$ |\psi(t) \rangle = U(t,t_0)|\psi(t_0)\rangle $$where U(t,t_0) is the solution of the initial value problem$$ i \dfrac{d}{dt} U(t,t_0) = H(t)U(t,t_0)  U(t_0, t_0)... 8 There is a unitary that maps$\{|\psi_1\rangle,|\psi_2\rangle\}$to$\{|\phi_1\rangle,|\phi_2\rangle\}$if and only if $$\langle\psi_1|\psi_2\rangle=\langle\phi_1|\phi_2\rangle.$$ Thy "only if" direction is straightforward. Let's assume$U|\psi_i\rangle=|\phi_i\rangle$. Then $$\langle\phi_1|\phi_2\rangle=\langle\psi_1|U^\dagger U|\psi_2\rangle=\... 7 Simulating Classical "AND/NAND/OR/NOR/XOR/XNOR" Gates With the help of this answer from Blue, constructing a matrix for a classical gate is just a matter of following the steps. Here is the combined truth table for classical logic gates:$$ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Input} & \text{AND} & \text{NAND} & \text{OR} & \text{... 7 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 $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \frac{1-i}{\sqrt{2}} \\ \frac{1+i}{\sqrt{2}} & -1 \end{pmatrix} = \frac{1}{... 7 Take your vector \frac{1}{\sqrt{5}}(0, 1, 1, 1, 1, 1)^T and five other arbitrary ones but at the same time these vectors have to be linearly independent. After that apply Gram-Schmidt process which produces orthonormal vectors. Put these vectors to a matrix and you will get a unitary matrix with the first column equal to \frac{1}{\sqrt{5}}(0, 1, 1, 1, 1, ... 7 It is true that unitary evolution cannot destroy information. This is the content of the no-cloning theorem and its time reversal - the no-deleting theorem. The no-hiding theorem says something different and more subtle. Information hiding In order to understand what it says it helps to start with the observation that classical information residing in a ... 7 Right. But when you build a quantum computer, you want to have a certain set of gates that you want to implement, and all other gates (unitary matrices) can be built from that set of gates. This is known as a universal set. Surprisingly (or not), it is quite small. One example of a universal set is: G = \{H, S, T, CNOT \} where$$H = \dfrac{1}{\sqrt{2}}\... 7 No, this is impossible, because it violates the no-cloning theorem. You want to implement the function $$|A_0\rangle|A_1\rangle|i\rangle|0\rangle \mapsto |A_0\rangle|A_1\rangle|i\rangle|A_i\rangle,$$ but if you could do that you could also clone, e.g. by fixing$i=0, which would make the function be $$|A_0\rangle|A_1\rangle|i\rangle|0\rangle \mapsto |A_0\... 7 Hint: Instead of using the BCH formula in the form usually presented, for example at the top of this Wikipedia page, use this consequence of Hadamard's Lemma:$$\tag{1} e^{iHt}\hat{a}e^{-iHt} = \hat{a} + [iHt,\hat{a}] + \frac{1}{2!}[iHt,[iHt,\hat{a}] + \cdots Now substitute H into the right-hand side and evaluate the commutators between \hat{a} and ... 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, $$where \sum_ip_i=1 and the \sigma^A_i and \sigma^B_i are valid states on single sites. The difficulty actually comes from the freedom that ... 6 An isometry U:S\rightarrow S\otimes E, where S is your system and E is an environment, such that$$\mathrm{Tr}_E(U\rho U^\dagger)=\Phi(\rho)$$is called a Stinespring dilation of \Phi. An easy way to construct a Stinespring dilation from the Kraus operators is to consider \mathcal{H}_E=\mathrm{span}\{|j\rangle\}_{j=1}^r and$$U=\sum_j F_j^S\... 6 No, you can't perform anti-unitary gates. Or rather: if they are possible, then faster than light communication is possible. For example, here is a circuit that remotely detects whether or not a conjugation gate was performed: This works because conjugating sqrt(X) is equivalent to inverting it. The conjugation between the X^(1/2) and the X^(-1/2) makes ... 6 Typically this is a slight abuse of notation. One can have a unitary operatorU$chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state$|\psi_0\rangle$, a "Haar-random state" would be$|\psi_U\rangle=U|\psi_0\rangle$for$U\$ randomly chosen according to the Haar measure. If you want to find the ...

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