25

For a presentation from first principles, I like Ryan O'Donnell's answer. But for a slightly higher-level algebraic treatment, here's how I would do it. The main feature of a controlled-$U$ operation, for any unitary $U$, is that it (coherently) performs an operation on some qubits depending on the value of some single qubit. The way that we can write this ...


22

Computations in quantum information processing are implemented by means of unitary operations. Sometimes, we need to think not about a specific unitary operation required to execute a specific computation, but about the whole space of unitary transformations. (Examples will be given below). For a single $n$-dimensional qudit, (which can also be a tensor ...


15

Abel Molina, Thomas Vidick, and I proved that the correct answer is $c=3/4$ in this paper: A. Molina, T. Vidick, and J. Watrous. Optimal counterfeiting attacks and generalizations for Wiesner's quantum money. Proceedings of the 7th Conference on Theory of Quantum Computation, Communication, and Cryptography, volume 7582 of Lecture Notes in Computer ...


15

The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $\big(\alpha |0\rangle + \beta |1\rangle \big) \otimes \big(\alpha |0\rangle + \beta |1\rangle \big)$. This is not the state CNOT will give you. The qubits you get after applying CNOT as you ...


14

The best possible textbook reference at the moment is Coecke and Kissinger. Picturing Quantum Processes: A First Course in Quantum Theory and Diagrammatic Reasoning. Cambridge University Press, 2017. It is written by one of the two inventors of the ZX calculus (Bob Coecke), and one of the people who has contributed the most to the development of ...


14

For any matrix $A$ we can write $$ A =\sum_{i,j,k,l}h_{ijkl}\cdot \frac{1}{4}\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l, $$ where $$ h_{ijkl} = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l)^\dagger \cdot A\big) = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l) \cdot A\big) $$ ...


13

$\newcommand{\bs}[1]{{\boldsymbol #1}} \newcommand{\tildebssigma}{\tilde{\bs\sigma}} \newcommand{\bssigma}{{\bs\sigma}}$Yes, products of Pauli matrices form a basis for the set of Hermitian matrices (of dimensions that are powers of $2$). More specifically, fix an integer $n$ and let $N\equiv 2^n$, define $\bssigma\equiv(\sigma_x,\sigma_y,\sigma_y)$, and $\...


12

It is common that one refers to a density matrix (or, equivalently, a density operator) $\rho$ as acting on a particular space $\mathcal{H}$. This serves to establish the "type" of $\rho$ in computer science parlance. In particular, when there are multiple spaces under consideration, it may be helpful for a reader to know that $\rho$ corresponds specifically ...


11

Yes, the set of tensor products of all possible $n$ Pauli operators (including $I$) form an orthogonal basis for the vector space of $2^n \times 2^n$ complex matrices. To see this first we notice that the space has a dimension of $4^n$ and we also have $4^n$ vectors ( the vectors are operators in this case). So we only need to show that they are linearly ...


11

This is a good question; it's one that textbooks seem to sneak around. I reached this exact question when preparing a quantum computing lecture a couple days ago. As far as I can tell, there's no way of getting the desired 8x8 matrix using the Kronecker product $\otimes$ notation for matrices. All you can really say is: Your operation of applying CNOT to ...


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

A couple of points: The ground state is by definition the eigenvector associated with the minimum valued eigenvalue. Lets consider the Pauli Z matrix as you have. First, \begin{align*} Z = \begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}. \end{align*} As this matrix is diagonal, we can immediately see that the eigenvalues are the values on the main ...


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

The Pauli matrices form an orthogonal basis of $\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product $$ \langle A,B\rangle=\mathrm{Tr}(A^\dagger B)$$ since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to ...


9

First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "$\equiv$" symbol is taken to mean "equal up to a phase"). An easy way to see it is by computing the determinant of $H=e^{i\pi H/2}$, which gives $-1=1$ (using $\det[\exp(A)]=\exp[\operatorname{Tr}(A)] $ for all $A$ and $\operatorname{Tr}(H)=0$). Now, ...


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

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

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

First, the example that you give is not a density matrix (they must have trace 1). Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states ...


8

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ...


8

It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why $d^2$ dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are $d^2$ dimensional objects in the same sense that vectors in $\Bbb R^3$ are 3-dimensional (i.e., the cardinality of any basis set of $\...


8

Transposing a matrix is trace preserving since for $\rho = \sum_{a,b} \rho_{a,b} | a \rangle \langle b |$: $$\text{Tr}(\rho)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | a \rangle \langle b | \big) | c \rangle = \sum_{a,b,c} \rho_{a,b} \delta_{a,c} \delta_{b,c} = \sum_c \rho_{c,c}$$ $$\text{Tr}(\rho^T)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | b \...


7

"I'm looking for an explicit upper bound on the probability of successful counterfeiting ...". In "An adaptive attack on Wiesner's quantum money", by Aharon Brodutch, Daniel Nagaj, Or Sattath, and Dominique Unruh, last revised on 10 May 2016, the authors claim a success rate of: "~100%". The paper makes these claims: Main results. We show that in a ...


7

In linear algebra representation: The state of the qubit is $\alpha|0\rangle + \beta|1\rangle = \begin{bmatrix}\alpha \\ \beta \end{bmatrix}$. The Hadamard matrix is $\frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$. The state after applying a Hadamard gate can be calculated by multiplying the column vector representing the state ...


7

Calculus (e.g. $\int |\psi(x)|^2dx = 1$ ) Differential Equations (e.g. Schroedinger equation) Complex analysis Statistics/Probability theory Stochastics (especially in studying open quantum systems) Information theory Topology (e.g. topological quantum computing) Group theory (e.g. in stabilizer codes) Representation theory (e.g. in stabilizer codes) Graph ...


7

I guess you're looking at equations (130) and (131)? So, here, you have $|\psi\rangle=(|0\rangle|a\rangle+|1\rangle|b\rangle)/\sqrt{2}$ and $|\phi\rangle=|a| |0\rangle+|b| |1\rangle$. When it says to calculate $\langle\phi|\psi\rangle$, what it really means is $$ (\langle\phi|\otimes\mathbb{I})|\psi\rangle, $$ padding everything with identity matrices to ...


7

Let $g_1 \cdots g_M$ be the basic gates that you are allowed to use. For the purposes of this $\operatorname{CNOT}_{12}$ and $\operatorname{CNOT}_{13}$ etc are treated as separate. So $M$ is polynomially dependent on $n$, the number of qubits. The precise dependence involves details of the sorts of gates you use and how $k$-local they are. For example, if ...


7

An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate. (In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all ...


7

Let's represent $N^2$ as $2^a+b$, where $a$ is the greatest power of 2 that not exceeds $N^2$, and $b \ge 0$ (which is always possible to do - $a$ is just the number of bits in binary representation of $N^2$). Then $n = a+1$: $N^2 \le 2^{a+1}$, because otherwise $a$ would not be the greatest power of 2 that not exceeds $N^2$. $2^{a+1} \le 2N^2 = 2(2^a+b) = ...


7

Imagine you have a vector that can be written in the form $$ |\psi\rangle=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}c_{ij}|i\rangle|j\rangle. $$ The coefficients can be arranged as a $d_A\times d_B$ matrix $C$, with the elements $c_{ij}$ (in your special case, you're talking about setting $d_A=d_B=\sqrt{m}$). Now, if you calculate $\rho_A=CC^\dagger$, this is ...


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