# 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. ...

8

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

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 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}...

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 ...

6

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, ...

6


3

This is not an attempt to answer the question directly, but might help with the intuition. They key point is that multi-qubit gates can propagate the effect of a single-qubit rotation to other qubits. So, there's the technicality of what effect is propagated that I'll leave to other answers, but just to convince you that it can propagate: Take a very simple ...

3

Your mistake is computing exponent of matrix; use the formula $$\exp(i\theta\sigma_2)=\cos(\theta)\cdot I+i\cdot \sin(\theta)\cdot\sigma_2$$

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