# Tag Info

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. ...

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 $|+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$, $-|+\rangle = \frac{1}{\sqrt2}(-|0\... 7 Yes, the set of tensor products of all possible$n$Pauli operators (including$I$) forms an orthogonal basis for the vector space of$2^n \times 2^n$complex matrices. So 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 ... 6 A unitary$U$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 ... 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 CNOT gate is a 2-qubit gate, and consequently, its operation cannot be expressed by the tensor product of two one-qubit gates as the example you gave with the Hadamard gates. An easy way to check that such matrix cannot be expressed as the tensor product of two other matrices is to take matrices$A =\begin{pmatrix}a & b \\ c & d\end{pmatrix}...

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 ...

4

This looks essentially similar to the property of non-commutativity of the Kronecker product: $X\otimes \lambda_6\neq \lambda_6\otimes X$: $$X\otimes\lambda_6 = \begin{pmatrix}0&1 \\1&0\end{pmatrix}\otimes \begin{pmatrix}0&0&0 \\0&0&1 \\0&1&0\end{pmatrix} = \begin{pmatrix}0&0&0&0&0&0 \\ 0&0&0&... 4 There is a very closely related representation of the tableau representation of Aaronson (and Gottesman), which works not only for qubits but for qudits of arbitrary finite dimension, which works particularly well for purely Clifford circuits (i.e. at most one terminal measurement). In this alternative representation, one has tableaus describing how ... 4 This phenomenon is sometimes known as a discretization of errors. It is a property of certain error correcting codes that allows it to work. It is described (somewhat briefly) in Section 10.2 of Nielsen and Chuang. Suppose that we have an arbitrary error that affects just one qubit, and suppose that we represent this error by a channel \Phi mapping one ... 4 This does indeed work because this construction works for any matrix H where H^2=\mathbb{I}.$$ 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 ... 3 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(... 3 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)$$since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to ... 3 Let's say you have a Hamiltonian of the form$$ H=\sigma_1\otimes\sigma_2\otimes\sigma_2\otimes\ldots\otimes\sigma_n $$There's a straightforward circuit construction that lets you implement its time evolution e^{-iHt}. The trick is basically to decompose the state that you're evolving into the components that are in the \pm 1 eigenspaces of H. Then, ... 2 When I talked about a controlled-phase gate, I meant the standard gate that has unitary matrix$$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right) $$Note that this is related to the controlled-not via the action of a Hadamard on the target qubit. ... 2 Pick an element a_i of A and find its position in B, disregarding changes in phase. The shift in position uniquely identifies the series of X or Y applications needed for the transformation. The relative phase of (a_0, b_0) tells you, in steps of -i rotations, how many Y gates you need for the transformation. The relative phase of (a_1,b_1) to ... 2 The way that I thought to start was to look at the reduced density matrices of the individual qubits. If they cannot be interconverted using Pauli matrices, then the whole thing can't. The only problem is that this idea breaks down as soon any of the reduced density matrices are maximally mixed. At that point, you could ask "are the two states equivalent ... 2$$ CNOT = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} $$But what does this matrix mean? The above matrix means: on a two qubit system (such as \left|00\right>, \left|10\right>, \left|11\right>, etc.) if the first qubit is a one,... 2 I like to use the projectors$$ 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. $$Here you can ... 2 Let's start from the basics. Any arbitrary single qubit state can be written as$$|\Psi\rangle = e^{i\gamma} \left(\cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle\right),$$where \theta, \phi, \gamma \in \Bbb R. 0\leq \theta \leq \pi and 0\leq \phi\leq 2\pi. e^{i\gamma} is a global phase here. Qubit states with arbitrary ... 2 To draw out Aaronson and Gottesman's techniques a bit more explicitly: you can set up each stabilizer as a bit string of length 2N (for N qubits). The first N bits specify the locations of Z operators, and the second set of N specify the locations of X operators (so, X_1Z_2 for N=2 is 0110). For your circuit on four qubits, the transformation ... 2 Your expressions for \langle X_0X_1 \rangle and \langle Y_0Y_1 \rangle are correct under the assumption that the two qubits are independently random. In the case that they are correlated, these expressions will not yield the right answer. This is because you have to think of X_0X_1, for example, as an operator in its own right, rather than just a ... 2 For your more general question in the title, the answer is yes. This is more of a lesson in linear algebra than in quantum computing. What you found out is that actually the set of eigenvectors of a matrix A corresponding to a given eigenvalue \lambda forms a vector subspace of the space upon which A acts, this is called the eigenspace of \lambda, ... 1 Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of \pm i Y have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of \pm i Y. This is only enough to simulate real stabiliser circuits, which is what you're noticing.... 1 Quantum algorithms provide a computational speedup by orchestrating constructive and destructive interference of the amplitudes. It is as if there must be a "minus" sign somewhere in the matrices - otherwise we merely work in the classical world, and would not see a computational speedup. Let's consider the following gates as controlled Pauli matrices: \... 1 Here Preskill is using a physics convention that the states |x\rangle are the eigenstates of the X operator. So |x\rangle with x=0 actually means |+\rangle and with x=1 actually means |-\rangle. 1 Yes if you work with general phase shift there would be \phi in the final answer. In fact you would be able to take \phi=0 and just get A back. Try \phi=\frac{\pi}{4}. Looks like notational mismatch of what's called a phase gate/phase shift gate. Whether it means the entire 1 parameter family or just one specific value of \phi. EDIT: Incorrect, ... 1 With the IBM Q Experience you can implement that transformation with what they call the U3 gate, which takes three input arguments and looks like this:$$ U3(\theta,\phi,\lambda) = \begin{bmatrix} \cos(\theta/2) & -e^{i\lambda}\sin(\theta/2) \\ e^{i\phi}\sin(\theta/2) & e^{i\lambda+i\phi}\cos(\theta/2) \end{bmatrix}.  So in this case you could ...

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