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Hot answers tagged pauli-gates

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$$CNOT = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$$But what does this matrix mean? The above matrix means: on a two qubit system (such as $\left|00\right>$, $\left|10\right>$, $\left|11\right>$, etc.) if the first qubit is a one,...

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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 $$h_{ijkl} = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l)^\dagger \cdot A\big) = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l) \cdot A\big)$$ ...

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I suggest two different ways of trying to solve this, which will give you experience of different bits of the formulation of Quantum Information Theory. I'll give examples that are closely related to the question you asked, but are not what you asked so that you still get the value of answering the question yourself. Long-hand Method Represent the kets as ...

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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 ...

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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 ...

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I call this the "Paulinomial decomposition" as you are writing the matrix $H$ as a polynomial of Pauli matrices: $H=a_{XX}X_1X_2 + a_{XY}X_1Y_2 +a_{XZ}X_1Z_2 + a_{XI}X_1 + a_{YY}Y_1Y_2 + \cdots$ (for the 2-qubit case). To get the coefficients, you can use this formula: $a_{AB}=\frac{1}{4}\textrm{tr}\left((A_1\otimes B_2)H\right)$ For example, here ...

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While Hadamard gate is defined as $$H= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},$$ $y$-rotation by $\pi/2$ leads to gate $$Ry(\pi/2)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.$$ So, there is a difference in position of -1 in the second column. Application of the $X$ gate returns the -1 in $... 9 One way order to perform Z rotations by arbitrary angles is to approximate them with a sequence of Hadamard and T gates. If you need the approximation to have maximum error$\epsilon$, there are known constructions that do this using roughly$3 \lg \frac{1}{\epsilon}$T gates. See "Optimal ancilla-free Clifford+T approximation of z-rotations" by Ross et al. ... 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, ... 9 Qiskit uses "little endian" bit ordering. That means, if A and B are$2 \times 2$unitary matrices then$B \otimes A$(note the order) is equivalent to applying$A$to first qubit and$Bto second qubit. Hence, $$CNOT = I \otimes P_0 + X \otimes P_1$$ where $$P_0 = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right) , P_1 = \... 8 Effectively, the Z operation (represented by the Pauli Z matrix) applies a rotation about the Z-axis. As you note, rotations can also be written in the form e^{-i Z t}. To see that, you can use a trick pretty similar to the one used to derive Euler's identity (e^{i \theta} = \cos(\theta) + i \sin(\theta)) to rewrite the Taylor series that you quoted ... 7 The quantum states that differ by a global phase (i.e., by a complex number multiple which has absolute value of 1) are considered the same quantum state, since they can not be distinguished using any operations or measurements. Thus, the eigenstates for hω are |+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle), -|+\rangle = \frac{1}{\sqrt2}(-|0\... 7 What \sigma^z_i means is that you've got a Pauli-Z applied to qubit i, and nothing else on the other qubits (i.e. the identity). So, you could expand it as$$ I^{\otimes(i-1)}\otimes\sigma^z\otimes I^{n-i} $$if your system has n qubits. A term such as \sigma^z_i\sigma^z_j is then a product of two of these, which is equivalent to the tensor product ... 7 As a general rule, you wouldn't bother constructing this: it is just a global phase that has no observable consequence. If you really insist on doing this, introduce an ancilla qubit in the |1\rangle state and apply a Z gate to it. PS "inverse identity gate" is a really bad name for it. The identity operation is its own inverse. 7 Starting with the state |\psi_0 \rangle = |0\rangle, and we want to get to the state |\psi_f \rangle = \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} then we must realize that we need to create some sort of a superposition between the state |0\rangle and the state |1\rangle. This is where the Hadamard gate will come into play. The Hadamard gate which ... 7 We can transform the left hand side as follows$$ \begin{align} \sum_{l=x,y,z}\langle J_l^2\rangle &= \sum_{l=x,y,z}\langle\psi|J_l^2|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}J_l^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}\left(\sum_{i=1}^N\frac12\sigma_l^i\right)^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,... 6 A unitaryU$and$e^{i\phi}U$, which differs from it by a phase, act exact identically on any quantum state. Thus, they should really be considered the "same" unitary in terms of their action. You can therefore use$X$instead of your unitary, which is$-X$. This will have exactly the identical action in any circuit. (Why is this? There are different ... 6 For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define$\vec \sigma = (\sigma_1,\sigma_2,\sigma_3)$to represent the three Pauli matrices $$\sigma_1 = X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \;\;\; \sigma_2 = Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, \;\;\; \... 6 A matrix function f(A) for normal matrix A is defined as follows $$f(A)=\sum_{i=1}^{n}f(\lambda_i)v_iv_i^T$$ where \lambda_{i} is an eigenvalue and v_{i} is coresponding eigenvector (note: transposed vector v_{i} is a row vector). In your case: f(A) = \mathrm{e}^A and A = -i\frac{\phi}{2}\sigma_{2}. 6 \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} There is no reason to factor out the i, it just make thing more cumbersome. I think what you are trying to do or assuming is that you can ignore the phase factor i since it is just an overall phase factor. That would be okay, as long as you ... 6 As noted by @KAJ226 in another answer, the global phase factor i is unobservable and can be ignored, unless we are considering a controlled gate in which case the phase factor i becomes a relative phase which may not be ignored. Consequently, we can choose to represent the single-qubit gate known as Y using the real matrix$$ Y=\begin{pmatrix} 0 & -... 5 You get two decompositions for your matrix (let's call it$A$) because you are using two different operatorial bases. In the first case you are considering the matrix as acting in a space of dimension$3\times 2$, that is, using the operatorial basis$\{\lambda_i\sigma_j\}_{ij}\equiv\{\lambda_i\otimes\sigma_j\}_{ij}$. In other words, you are computing the ... 5 I like to use the projectors $$P_0=|0\rangle\langle 0|\equiv\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\qquad P_1=|1\rangle\langle 1|\equiv\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right)$$ to express the intuition of the controlled-not when constructing it: $$CNOT=P_0\otimes\mathbb{I}+P_1\otimes X.$$ Here you can ... 5 Let's say you have a Hamiltonian of the form $$H=\sigma_1\otimes\sigma_2\otimes\sigma_2\otimes\ldots\otimes\sigma_n$$ There's a straightforward circuit construction that lets you implement its time evolution$e^{-iHt}$. The trick is basically to decompose the state that you're evolving into the components that are in the$\pm 1$eigenspaces of$H$. Then, ... 5 This does indeed work because this construction works for any matrix$H$where$H^2=\mathbb{I}$. $$R_\theta(H)=e^{-i\theta H/2}=\cos\frac{\theta}{2}\mathbb{I}-i\sin\frac{\theta}{2}H$$ There are several ways that you could prove this. I think the easiest way is to realise that because$H^2=\mathbb{I}$, then the eigenvalues of$H^2$are 1, and hence the ... 5 Expanding on Jalex Look at what happens on the possible terms. \begin{eqnarray*} \mid 0 \rangle \otimes \mid + \rangle &\to& \mid 0 \rangle \otimes \mid + \rangle\\ \mid 0 \rangle \otimes \mid - \rangle &\to& \mid 0 \rangle \otimes \mid - \rangle\\ \mid 1 \rangle \otimes \mid + \rangle &\to& \mid 1 \rangle \otimes \mid + \rangle\\ \... 5 The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of$\pm i\$ that are missing. In lieu of a tikz commutative diagram \begin{align} <X,Y,Z> & \hookrightarrow & U(2)\\ \downarrow & & \downarrow\\ <[X],[Z]> & \hookrightarrow & PU(...

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