The approximation of the time evolution operator U, using Trotter formula, can't hold anymore taking a time step of $\Delta t = \pi$

Question

The problems arises from a consideration written on the book "Quantum computation and quantum information" from Michael A. Nielsen and Isaac L. Chuang on page 259.

In this chapter it's studied the Grover search algorithm as a quantum simulation and it's taken as a sample hamiltonian $H =|x\rangle\langle x| + |\psi\rangle \langle\psi|$ where $| x \rangle$ is the solution and $|\psi\rangle$ is the state we start from which is in the form: $$ |\psi\rangle=\sum_{x}\frac{|x\rangle}{\sqrt{N}} $$

The time evolution operator $e^{-i(|x\rangle\langle x| +|\psi\rangle\langle \psi|)t}$ is approximated for small $\Delta t$ as $$ U(\Delta t) = e^{-i|\psi\rangle\langle\psi|\Delta t}e^{-i| x\rangle\langle x|\Delta t} $$

and choosing $|y\rangle$ such that $\{|x\rangle, |y\rangle\}$ is an orthonormal base we take

$$ |\psi\rangle\doteq \begin{pmatrix} \alpha \\ \beta \end{pmatrix},|x\rangle\doteq\begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ Then we represent the projectors $|x\rangle\langle{x}|$, $|\psi\rangle\langle\psi|$ in this base as $$ |\psi\rangle\langle\psi|=\frac{\mathbb{I}+\vec{\psi}\cdot\vec{\sigma}}{2} \hspace{1cm}\text{where}\hspace{1cm}\vec{\psi}= \begin{pmatrix} 2\alpha\beta \\ 0 \\ \alpha^2-\beta^2 \end{pmatrix} $$

$$ |x\rangle\langle x|=\frac{\mathbb{I}+\hat{z}\cdot\vec{\sigma}}{2} \hspace{1cm}\text{where}\hspace{1cm}\hat{z}= \begin{pmatrix} 0\\ 0 \\ 1 \end{pmatrix} $$

Calculating $U(\Delta t)$ in this base we obtain $$ U(\Delta t) = \left( \cos^2\left(\frac{\Delta t}{2}\right) - \sin^2\left(\frac{\Delta t}{2}\right)\vec{\psi}\cdot \hat{z} \right)\mathbb{I} - 2i\sin\left(\frac{\Delta t}{2}\right)\left(\cos\left(\frac{\Delta t}{2}\right)\frac{\vec{\psi}+\hat{z}}{2} + \sin\left(\frac{\Delta t}{2}\right)\frac{\vec{\psi}\, \times \, \hat{z}}{2} \right)\cdot\vec{\sigma} $$ And this relation should be valid for small $\Delta t$. Then the book says that a smart choice for a step would be $\Delta t = \pi$, because this would lead to the Grover iteration.

BUT doesn't choosing $\Delta t=\pi$ violate the approximation assumed in the beginning?

PS:

I tried to see if $$ U(\Delta t_1)U(\Delta t_2) = U(\Delta t_1 + \Delta t_2) $$ on a computer but it does not.

Answer 1

As you allude to, Nielsen and Chuang say on pp. 259-260,

Now initially we imagined that $\Delta t$ was small, since we were considering the case of quantum simulation, but Equation (6.28) shows that the smart thing to do is to choose $\Delta t = \pi$, in order to maximize the rotation angle $\theta$.

Note that Equation (6.28) is your equation for $U(\Delta t)$, and $\theta$ is the angle that this unitary operator rotates by, along an axis $\vec{r}$ in the Bloch sphere spanned by $|x\rangle$ and $|y\rangle$.

There is nothing incorrect about the mathematics here. N&C are not claiming that $U(\Delta t) = e^{-i\Delta t (|x\rangle\langle x| + |\psi\rangle\langle \psi|)}$, nor that they are even approximately equal for large $\Delta t \sim \pi$. Rather, the point of this section more pedagogical: they want to give some intuition about how the Grover search procedure might be obtained from a physical perspective (i.e., from performing some sort of Hamiltonian simulation). The expression for $U(\Delta t)$ gives us an idea of how, under this intuitive picture, the Grover process manages to take the initial state $|\psi\rangle$ to the solution $|x\rangle$, i.e., by rotating it in the Bloch sphere about an appropriate axis. Their point about setting $\Delta t = \pi$ is to get to this punchline faster.