5
$\begingroup$

As I understand it the Pauli-X, Y and Z gates are the same as their rotational gates with a rotation of $\pi$. But given the expression for those gates, I find that there is a factor of $-i$ in each of them. Can someone explain why?

For example the X-gate:

$R_X(\theta)=\begin{align}\begin{bmatrix} \cos\frac{\theta}{2}&-i\;\sin\frac{\theta}{2}\\ -i\;\sin\frac{\theta}{2}&\cos\frac{\theta}{2} \end{bmatrix}\end{align}\\ $ $R_X(\pi)=\begin{align}\begin{bmatrix} 0&-i\\ -i&0 \end{bmatrix}\end{align}\\ $ $X=\begin{align}\begin{bmatrix} 0&1\\ 1&0 \end{bmatrix}\end{align}\\ $

$\endgroup$
2
$\begingroup$

Let's start from the basics. Any arbitrary single qubit state can be written as

$$|\Psi\rangle = e^{i\gamma} \left(\cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle\right),$$ where $\theta, \phi, \gamma \in \Bbb R$. $0\leq \theta \leq \pi$ and $0\leq \phi\leq 2\pi$. $e^{i\gamma}$ is a global phase here. Qubit states with arbitrary values of $\gamma$ represent the same quantum state as global phases don't have any observable effects. So the standard notation for qubit states is

$$|\Psi\rangle = \cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle $$

where we neglect the global phase factor $e^{i\gamma}$.

Remember that you are working on a vector space structure where $|0\rangle$ and $|1\rangle$ are your basis elements. The rotation operators $R_X, R_Y$ and $R_Z$ are simply linear maps in this context (in the standard $\{|0\rangle, |1\rangle\}$ basis). And so are the Pauli $X, Y$ and $Z$.

Now, note that you can write $R_X(\pi)$ as $-iX$. When $X$ acts on an arbitrary state $|\Psi\rangle$ it will produce $X|\Psi\rangle$. If $-iX$ acts, it will produce $-iX|\Psi\rangle$ instead. However, the states $-iX|\Psi\rangle$ and $X|\Psi\rangle$ simply differ by a global phase of $-i$, or in other words by a factor of $e^{i\gamma}$ where $\gamma = (2n+1)\frac{\pi}{2}$ and $n\in \Bbb N$. So, as you can see, $R_X(\pi)$ and $X$ are equivalent mappings as global phases don't matter in quantum mechanics!

As for why the Pauli operators and rotation operators differ by a phase factor, you'll have to check out the derivations for the following. You'll find an outline here.

$$R_X(\theta) = e^{-i \frac{\theta}{2}X},$$

$$R_Y(\theta) = e^{-i \frac{\theta}{2}Y},$$

$$R_Z(\theta) = e^{-i \frac{\theta}{2}Z}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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