# Tag Info

10

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 ...

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

$$CNOT = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$$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,...

9

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

8

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 $A =\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ $... 8 Effectively, the Z operation (represented by the Pauli$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

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 ...

7

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, ...

7


4

If you write $Z^aX^b$, there's an implicit "add a phase $i$ to make it Hermitian if necessary", although I guess there are a couple of different conventions you might use the determine the sign used. So long as you're clear about the convention it doesn't really matter because you've got the extra $\pm 1$ freedom to add in to adjust for it. As for an ...

4

The phase of -1 generated on the ancilla is just a global phase. You can shift it to any qubit without affecting the statistics of the system. Nothing to do with entanglement. Quantum mechanics is a mathematical framework that describes reality insofar that it predicts the correct observable statistics of a system. Shifting around a global phase factor from ...

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