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.

Thanks for reading!

  • $\begingroup$ 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 $x$. $\endgroup$ Jun 8, 2021 at 16:19
  • $\begingroup$ @QuantumMechanic Thanks for the answer, seem to I be confused between these quantities in this problem. $\endgroup$
    – Monad
    Jun 9, 2021 at 6:46

1 Answer 1


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 $\mathcal{G}(\mu)$.

  • $\begingroup$ 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. $\endgroup$
    – Monad
    Jun 9, 2021 at 6:51

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