6
$\begingroup$

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!

$\endgroup$

2 Answers 2

2
$\begingroup$

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 $(e^Ae^B)^n = e^{i\frac{θ}{2}(X+Z) + \delta}$.

$\endgroup$
1
$\begingroup$

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 $\sin(n\phi)\frac{\tan\frac{\theta}{2n}}{\sqrt{2+\tan^2\frac{\theta}{2n}}}\sim O(\theta/n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.