$R_Z(\theta) = e^{-i\frac{\theta}{2}Z}$ and $R_{ZX}(\theta) = e^{-i\frac{\theta}{2}XZ}$

My question is, why is the rotation gate defined in an exponential format? Where does the $\frac{1}{2}$ come from? Moreover, in the exponential part of $R_{ZX}$ gate, why is it $XZ$ not $ZX$?


1 Answer 1


why is the rotation gate defined in an exponential format?

You might just put it down to convenience. An equally valid definition would be $$ R_Z(\theta)=\cos\frac{\theta}{2}I-i\sin\frac{\theta}{2}Z, $$ but it's more annoying to write out.

That said, I'd say the real origin of it is the Schrodinger equation. Remember that the interactions of physical systems are described by a Hamiltonian (typically, constant in time). This state then evolves due to the differential equation $$ \frac{d|\psi\rangle}{dt}=-iH|\psi\rangle, $$ which has the solution $$ |\psi(t)\rangle=e^{-iHt}|\psi(0)\rangle. $$

Where does the $\frac12$ come from?

This is a convention that is not really an essential part of the definition. Essentially, it accounts for the discrepancy in angles between the parametrisation of a quantum state and angles on the Bloch sphere: if you parametrise a state $$ |\psi\rangle=\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}|1\rangle, $$ then if you were plotting this state on the Bloch sphere, it would be at an angle $\theta$ (not $\theta/2$) to the vertical.

Another way that I like to look at it in terms of the $Z$ rotation is when you take into account a global phase: $$ R_Z(\theta)\equiv\begin{bmatrix} 1 & 0 \\ 0 & e^{i\theta} \end{bmatrix}, $$ so it does kind of give you the angle you'd expect.

in the exponential part of $R_{ZX}$ gate, why is it $XZ$ not $ZX$

I'm not sure where you got that definition from. It's not one I've seen before, and may just be a typo? Alternatively (and you'll only get this from context), it may be associated with the fact that when you write down a sequence of gates in a quantum circuit (e.g. $A$ followed by $B$), the product of operators has to be taken in reverse order ($U=BA$). I think this is less likely.

Edit: From the link you posted, it is clear that you missed a critical part of the information: a tensor product! So, what it actually means is $X\otimes Z$ not $XZ$. I suspect the reason for the surprising order then is that qiskit numbers its qubit wires oppositely to some standards. This is what the "note" box is referring to in your link.

  • 1
    $\begingroup$ Thanks for your comment! Could you elaborate a bit on the angle difference between quantum state and Bloch sphere? Also, the definition is found from qiskit, qiskit.org/documentation/stubs/…, I found that it has an explanation just below... $\endgroup$
    – peachnuts
    May 6 at 9:47

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