# Prove the parameter shift rule in case of generator is Pauli matrix

I'm studying the parameter shift rule and got stuck when doing an example with Pauli operators in https://arxiv.org/abs/1803.00745.

With $$f=e^{-i\mu\frac{\sigma_i}{2}}$$, $$\partial_\mu f=\frac{(-i\sigma_i)}{2}f=(-\frac{1}{2}if)\sigma_i$$ (1).

Because $$\frac{1}{2}\sigma_i$$ has eigenvalues $$\pm\frac{1}{2}$$, so $$r=1/2$$ and $$s=\frac{\pi}{4r}=\frac{\pi}{2}$$.

$$\rightarrow\partial_\mu f=\frac{1}{2}(f(\mu+\frac{\pi}{2})-f(\mu-\frac{\pi}{2}))$$

$$=\frac{1}{2}(e^{-i(\mu+\frac{\pi}{2})\frac{\sigma_i}{2}}-e^{-i(\mu-\frac{\pi}{2})\frac{\sigma_i}{2}})$$

$$=\frac{1}{2}(e^{-i(\frac{\pi}{4}\sigma_i)}-e^{i(\frac{\pi}{4}\sigma_i)}) f(\mu)$$

$$=-\frac{1}{2}i(2\sin(\frac{\pi}{4}\sigma_i)) f(\mu)=-\frac{1}{2}i(\frac{2}{\sqrt{2}}\sigma_i) f(\mu)$$ (2)

The results from (1) and (2) seem to not be the same, I don't know what points I missed.

• What are $r$ and $s$ here? It is also not always clear when you are saying $f(x)$ means $f$ is a function of $x$ versus $f$ is multiplied by $x$. Commented Jun 8, 2021 at 16:19
You are confusing two different quantities. Reading up on the parameter shift rule here, you will see that the gradient of an expectation value $$f(\mu) = \langle \psi | \mathcal{G}^\dagger Q \mathcal{G} | \psi \rangle$$ can be computed via $$\partial_\mu f = r[ f(\mu+s) - f(\mu-s)] \ ,$$ where $$s=\frac{\pi}{4r}$$, if $$\mathcal{G}(\mu) = e^{-i\mu G}$$ when $$G$$ has at most two eigenvalues $$\pm r$$.
You are confusing the expectation value $$f(\mu)$$ with the gate $$\mathcal{G}(\mu)$$.
• Thanks for the clear answer, I have focused so much about the gate $\mathcal{G}(\mu)$ and forgot the expectation value, maybe this case will be more complex. Commented Jun 9, 2021 at 6:51