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From eqn. $(4.8)$ in Nielsen and Chuang, a general rotation by $\theta$ about the $\hat n$ axis is given by $$ R_\hat{n}(\theta)\equiv \exp(-i\theta\hat n\cdot\vec\sigma/2) = \cos(\theta/2)I-i\sin(\theta/2)(n_xX+n_yY+n_zZ). $$ From the qiskit textbook, a generic single-qubit gate is defined as $$ U(\theta, \phi, \lambda) = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & -e^{i\lambda} \sin\left(\frac{\theta}{2}\right) \\ e^{i\phi} \sin\left(\frac{\theta}{2}\right) & e^{i(\lambda + \phi)} \cos\left(\frac{\theta}{2}\right) \end{pmatrix}. $$ I tried to work out the matrix representation of $R_\hat{n}(\theta)$ but it looks like the first entry of that should be $\cos(\theta/2)-i\sin(\theta/2)n_z$, which is different from that of $U(\theta,\phi,\lambda)$, i.e. $\cos(\theta/2)$.

I'm wondering how does $R_\hat{n}(\theta)$ related to $U_3$ gate? In other words, given a unit vector $\hat n$ and a rotation angle $\theta$, can we represent $R_\hat{n}(\theta)$ using $U_3$?

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Any single-qubit gate can be expressed as $R_{\hat{n}}(\alpha)$ for some $\hat{n}$ and $\alpha$ and similarly any single-qubit gate can be expressed as $U(\theta, \phi, \lambda)$ for some $\theta$, $\phi$ and $\lambda$. In other words, the two generic gates provide different parametrizations for the group of single-qubit gates.

The reason that elementwise matrix comparison fails is that the two parametrizations differ by the unobservable global phase. Thus, instead of trying to solve

$$ R_\hat{n}(\alpha) = U_3(\theta, \phi, \lambda) $$

you should solve

$$ R_\hat{n}(\alpha) = e^{i\gamma}U_3(\theta, \phi, \lambda) $$

where $e^{i\gamma}$ is the unknown global phase. Optionally, you can first reduce the number of unknowns by guessing $\gamma$ from the properties of the two matrices.

For example, we observe that the two diagonal elements of $R_\hat{n}(\alpha)$ are complex conjugates of each other and so its trace is real. This suggests the guess $\gamma = -\frac{\lambda + \phi}{2}$ might work because it turns the trace of $U_3(\theta, \phi, \lambda)$ into a real number.


Note that unlike the global phase the relative phase cannot be ignored. Consequently, the above does not apply to controlled-$R_\hat{n}(\alpha)$ and controlled-$U_3(\theta, \phi, \lambda)$ gates.

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  • $\begingroup$ Thanks a lot for the answer! For controlled $R_\hat{n}(\alpha)$ and controlled $U_3$, are they also different by a global phase? $\endgroup$ – ZR- Feb 28 at 3:21
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    $\begingroup$ Controlled-$R_\hat{n}(\alpha)$ and controlled-$U_3(\theta, \phi, \lambda)$ are different gates even if (for the given parameters $\hat{n}, \alpha, \theta, \phi$ and $\lambda$) $R_\hat{n}(\alpha)$ and $U_3(\theta, \phi, \lambda)$ are the same gate. $\endgroup$ – Adam Zalcman Feb 28 at 3:30
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    $\begingroup$ Specifically, $CR_\hat{n}(\alpha) \equiv R_Z(\gamma) CU_3(\theta, \phi, \lambda)$ where $R_Z(\gamma)$ is applied to the control qubit. $\endgroup$ – Adam Zalcman Feb 28 at 3:36
  • $\begingroup$ Thanks!! Then is there a way to perform the controlled $R_\hat{n}(\alpha)$ using qiskit? or can I relate them by adding some external gates? $\endgroup$ – ZR- Feb 28 at 3:38
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    $\begingroup$ Global phase is unobservable, so yes any two unitary matrices that differ by global phase only are the same gate. $\endgroup$ – Adam Zalcman Feb 28 at 20:21

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