From Nielsen and Chuang's Quantum Computation p.259. Even though there's an answered one on answers, I need more details.

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From Nielsen and Chuang's Quantum Computation p.259.
Even though there's an answered one on answers, I need more details.
Please explain a derivation of the equation including how the exponentials of U(t) turns to cos, sin with quarternions multiplication.
And why do they factored out 2 on second term in the equation? The formulas in the brace in second term supposed to be a rotation axis with length 1. The following texts implying that they deliberately factored out it to make the lengh of vector in the brace to 1. How did they know that it doesn't have length 1?

From Nielsen And Chuang S Quantum Computation P 259 Even Though There S An Answered One On Chegg I Need More Details 1 (66.06 KiB) Viewed 28 times

From Nielsen And Chuang S Quantum Computation P 259 Even Though There S An Answered One On Chegg I Need More Details 2 (58.04 KiB) Viewed 28 times

We have been analyzing the accuracy of the quantum simulation of the Hamiltonian (6.18) using general results on quantum simulation from Section 4.7. Of course, in this instance we are dealing with a specific Hamiltonian, not the general case, which suggets that it might be interesting to calculate explicitly the effect of a simulation step of time Δt, rather than relying on the general analysis. We can do this for any specific choice of simulation method - it can be a little tedious to work out the effect of the simulation step, but it is essentially a straightforward calculation. The obvious starting point is to explicitly calculate the action of the lowest-order simulation techniques, that is, to calculate one or both of exp(−i∣x⟩⟨x∣Δt)exp(−i∣ψ⟩⟨ψ∣Δt) or exp(−i∣ψ⟩⟨ψ∣Δt)exp(−i∣x⟩⟨x∣Δt). The results are essentially the same in both instances; we will focus on the study of U(Δt)≡exp(−i∣ψ⟩⟨ψ∣Δt)exp(−i∣x⟩⟨x∣Δt).U(Δt) clearly acts non-trivially only in the space spanned by ∣x⟩⟨x∣ and ∣ψ⟩⟨ψ∣, so we restrict ourselves to that space, working in the basis ∣x⟩,∣y⟩, where ∣y⟩ is defined as before. Note that in this representation ∣x∣(x∣=(I+Z)/2=(I+z⋅σ∗)/2, where z^≡(0,0,1) is the unit vector in the z direction, and ∣ψ⟩⟨ψ∣=(I+ψ⋅σ)/2, where ψψ​=(2αβ,0,(α2−β2) ) (recall that this is the Bloch vector representation; see Section 4.2). A simple calculation shows that up to an unimportant global phase factor, U(Δt)=​(cos2(2Δt​)−sin2(2Δt​)ψ​⋅z^)I−2isin(2Δt​)(cos(2Δt​)2ψ​+z^​+sin(2Δt​)2ψ​×z^​)⋅σ​ Exercise 6.9: Verify Equation (6.25). (Hint: see Exercise 4.15.)
−2isin(2Δt​)(cos(2Δt​)2ψ​+z^​+sin(2Δt​)2ψ​×z^​)⋅σ Exercise 6.9: Verify Equation (6.25). (Hint: see Exercise 4.15.) Equation (6.25) implies that U(Δt) is a rotation on the Bloch sphere about an axis of rotation r defined by r=cos(2Δt​)2ψ​+z^​+sin(2Δt​)2ψ​×z^​ and through an angle θ defined by cos(2θ​)=cos2(2Δt​)−sin2(2Δt​)ψ​⋅z^ which simplifies upon substitution of ψ​⋅z^=α2−β2=(2/N−1) to cos(2θ​)=1−N2​sin2(2Δt​) Note that ψ​⋅r=z^⋅r, so both ∣ψ⟩⟨ψ∣ and ∣x⟩⟨x∣ lie on the same circle of revolution about the r axis on the Bloch sphere. Summarizing, the action of U(Δt) is to rotate ∣ψ⟩⟨ψ∣ about the r axis, through an angle θ for each application of U(Δt), as illustrated in Figure 6.6. We terminate the procedure when enough rotations have been performed to rotate ∣ψ⟩⟨ψ∣ near to the solution ∣x⟩⟨x∣. Now initially we imagined that Δt was small, since we were considering the case of quantum simulation, but Equation (6.28) shows