Is it essential to apply Quantum Singular Value transformation twice for Hamiltonian simulation?

Question

I have been reading the paper A Grand Unification of Quantum Algorithms and I need clarification on the Hamiltonian simulation algorithm provided in the paper on page 23. . In procedure part point 2 says apply QSVT twice. I am confused here. Does that mean I have do QSVT for cos 2 times and for sine 2 times or does that mean 1 time cos and 1 time sine?. I have observed greater success when done 2 times each. This can be a coincidence because I don't have a clarification for it. Also anyone knows from where we are getting the -i for sine part in LCU picture and not for cosine?

Answer 1

You can find the answer to your question in page 22:

How might this problem be solved with QSVT? Naively, one may try to employ QSVT with a polynomial approximation to $e^{ −ixt}$ [...]. However, because the exponential function does not have definite parity, this function does not satisfy the constraints on $Poly(a)= \langle +|U_{\vec{\phi}}|+\rangle$ discussed in Section II A and Appendix A). To circumvent this issue, one can instead apply QSVT twice - once with an even polynomial approximation to $\cos(xt)$, and once with an odd polynomial approximation to $\sin(xt)$, both of which have definite parities. Then, using the circuit illustrated in Figure 17, one can sum together the results of these two QSVT executions to obtain $\cos^{(SV)}(\mathcal{H}t) − i\sin^{(SV)}(\mathcal{H}t) = e^{−i\mathcal{H}t}$, as desired.

So,

1. Applying QSVT twice means once with $\cos(xt)$ and once with $\sin(xt)$
2. We are getting the $-i$ for $\sin$ part from the equality $e^{−ix} = \cos(x) \color{red}{− i}\sin(x)$