# How does a general rotation $R_\hat{n}(\theta)$ related to $U_3$ gate?

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 $$U_3$$?

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.

• Thanks a lot for the answer! For controlled $R_\hat{n}(\alpha)$ and controlled $U_3$, are they also different by a global phase? – ZR- Feb 28 at 3:21
• 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. – Adam Zalcman Feb 28 at 3:30
• Specifically, $CR_\hat{n}(\alpha) \equiv R_Z(\gamma) CU_3(\theta, \phi, \lambda)$ where $R_Z(\gamma)$ is applied to the control qubit. – Adam Zalcman Feb 28 at 3:36
• 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? – ZR- Feb 28 at 3:38
• Global phase is unobservable, so yes any two unitary matrices that differ by global phase only are the same gate. – Adam Zalcman Feb 28 at 20:21