Prove that $E(R_n(\alpha),R_n(\theta)^n)<\epsilon/3$ from $E\big(R_n(\alpha),R_n(\alpha+\beta)\big)=|1-\exp(i\beta/2)|$

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This is from my reference Page 196, Quantum Computation and Quantum Information by Nielsen and Chuang.

How do we prove that for any $\epsilon>0$ there exists an $n$ such that $E(R_n(\alpha),R_n(\theta)^n)<\epsilon/3$ given $E(U,V)=\max_{|\psi\rangle}||(U-V)|\psi\rangle ||=||U-V||$ is the error when $V$ is implemented instead of $U$ ?

Note: $\theta$ is defined such that $\cos(\theta/2)\equiv\cos^2(\pi/8)$, i.e., $\theta$ is an irrational multiple of $2\pi$

My Attempt

Step 1

We have taken, $\delta>2\pi/N$ and $\theta_k=(k\theta)\;mod\;2\pi$ such that $\theta_k\in[0,2\pi)$, i.e., $k\theta=2\pi t_k+\theta_k$

Assume that, $\theta_{k+1}-\theta_k>\frac{2\pi}{N}$ for all $1\leq k\leq N$

$$ \theta_N>(N-1)\frac{2\pi}{N}\\ \theta_1+2\pi-\theta_N=2\pi-\theta_N<2\pi-(N-1)\frac{2\pi}{N}=\frac{2\pi}{N} $$ $\implies $ there are distinct $j$ and $k$ in the range $1,\cdots,N$ such that $|\theta_k-\theta_j|\leq \frac{2\pi}{N}<\delta$

Assuming $k>j$ we have $|\theta_k-\theta_j|=|\theta_{k-j}|<\delta$

Step 2

It is defined that the error when $V$ is implemented instead of $U$ is $E(U,V)=\max_{|\psi\rangle}||(U-V)|\psi\rangle ||=||U-V||$,

i.e., $E(U,V)=||(U-V)|\phi\rangle ||$ for some $|\phi\rangle$

$$ E\big(R_n(\alpha),R_n(\alpha+\beta)\big)=|(R_n(\alpha)-R_n(\alpha+\beta))|\phi\rangle|=|(R_n(\alpha)-(R_n(\alpha)R_n(\beta))|\phi\rangle|=\Big|R_n(\alpha)\big[1-R_n(\beta)\big]|\phi\rangle\Big|=\sqrt{\langle\phi|\big[1-R_n(\beta)\big]^\dagger R_n^\dagger(\alpha)R_n(\alpha)\big[1-R_n(\beta)\big]|\phi\rangle}\\ =\sqrt{\langle\phi|\big[1-R_n^\dagger(\beta)\big] \big[1-R_n(\beta)\big]|\phi\rangle}=\sqrt{\langle\phi|\big[1-R_n(-\beta)\big] \big[1-R_n(\beta)\big]|\phi\rangle}=\sqrt{\langle\phi|\big[1-R_n(-\beta)-R_n(\beta)+1\big]|\phi\rangle}=\sqrt{2-2\cos(\beta/2)}=|1-\exp(i\beta/2)| $$