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

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!

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 $$(e^Ae^B)^n = e^{i\frac{θ}{2}(X+Z) + \delta}$$.
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 $$\sin(n\phi)\frac{\tan\frac{\theta}{2n}}{\sqrt{2+\tan^2\frac{\theta}{2n}}}\sim O(\theta/n)$$.