Investigating the scaling of the error of a Trotter-Suzuki-approximation

Question

I am doing an assignment and I am being asked to investigate the scaling of the error with the number of repetions $n$ of a approximation of the Hadamard with $R_x$ and $R_y$. This is the approximation, where $\theta = \frac {\pi} {\sqrt2}$: $$ H \equiv \lim_{n\rightarrow\infty} \left( ~R_x\left(\frac{\theta}{n}\right) ~~R_z \left(\frac{\theta}{n}\right) ~\right)^n = e^{i \frac{\theta}2 (X+Z)}$$

I am not sure how to approach this problem. I know that the error $\delta$ is polynomial to $n$ here, but I don't know how to get the scaling more specifically: $$U = \left(e^{i\frac\theta{2n}P}e^{i\frac{\theta}{2n}P'}\right)^n + \delta$$

I appreciate the help!

Answer 1

You can use the Baker–Campbell–Hausdorff formula that states that for $e^Ae^B = e^C$ (assuming $e^A,e^B \approx I$) $C$ is given by: $$C = A + B + \frac12[A, B] + \frac1{12}[A, [A, B]] + \frac1{12}[B, [B, A]] + O(K^4) + \cdots$$

Setting $A = i\frac{θ}{2n}X$ and $B=i\frac{θ}{2n}Z$:

$$C = i\frac{θ}{2n}(X + Z) + \frac12\frac{θ^2}{2n^2}iY - \frac1{12}\frac{θ^3}{2n^3}i(X+Z) + O(\frac1{n^4})Y + O(\frac1{n^5})(X+Z) + \cdots$$

So $$(e^Ae^B)^n = e^{nC}$$ $$nC = i\frac{θ}{2}(X + Z) + \frac12\frac{θ^2}{2n}iY - \frac1{12}\frac{θ^3}{2n^2}i(X+Z) + O(\frac1{n^3})Y + O(\frac1{n^4})(X+Z) + \cdots$$

Therefore, $$-\delta = \frac12\frac{θ^2}{2n}iY - \frac1{12}\frac{θ^3}{2n^2}i(X+Z) + O(\frac1{n^3})Y + O(\frac1{n^4})(X+Z) + \cdots$$ where $\delta$ is $(e^Ae^B)^n = e^{i\frac{θ}{2}(X+Z) + \delta}$.

Answer 2

For this particular calculation, you can keep your results exact for quite a long time. To see this, start with the exact thing $$ H_0=e^{i\pi/2(X+Z)/\sqrt{2}}=i\frac{X+Z}{\sqrt{2}}. $$ Now for the approximation. We have one step is $$ e^{i\theta X/(2n)}e^{i\theta Z/(2n)}. $$ If you expand this out, you'll find it's equivalent to a rotation $$ e^{i\phi \vec{n}\cdot\vec{\sigma}} $$ where $\cos\phi=\cos^2\frac{\theta}{2n}$ and $$ \vec{n}=\frac{(1,\tan\frac{\theta}{2n},1)}{\sqrt{2+\tan^2\frac{\theta}{2n}}}. $$ So, the $n^{th}$ power is just $e^{in\phi \vec{n}\cdot\vec{\sigma}}$. Now we just need to calculate this distance $$ \left\|I\cos(n\phi)+i\sin(n\phi)\vec{n}\cdot\vec{\sigma}-i\frac{X+Z}{\sqrt{2}}\right\|. $$ It's only at this point that you want to start approximating $\phi=\frac{\theta}{\sqrt2n}+O(\frac{\theta^2}{n^2})$. I believe that you'll find the accuracy is $O(\theta/n)$ when you work out the details: observe that there is a $Y$ term with coefficient $\sin(n\phi)\frac{\tan\frac{\theta}{2n}}{\sqrt{2+\tan^2\frac{\theta}{2n}}}\sim O(\theta/n)$.