# Is it possible to express $U_1(\lambda)$ through the gates $R_x, R_y, R_z$ while maintaining the phase? In Qiskit for example

Is it possible to express gate $$U_1(\lambda)$$ through the gates $$R_x, R_y, R_z$$ while maintaining the phase? Both in principle and in practice (in Qiskit for example)?

The single gate $$R_z(\lambda)$$ is not suitable, because it loses phase. In qiskit the transpile function throws an error:

Cannot unroll the circuit to the given basis, ['rx', 'ry', 'rz']. No rule to expand instruction u1.

• As a side-note: In Qiskit right now, the RZ gate is equal to U1 because the global phase is ignored. So if you do circuit = QuantumCircuit(1); circuit.rz(lambda) that would implement a u1($\lambda$). – Cryoris Jun 18 '20 at 16:05
• It throws this error because u1 is already the "most basic" gate and cannot be further decomposed. – Cryoris Jun 18 '20 at 16:06
• Yes, in some cases the global phase is ignored in Qiskit: somewhere for the reason that the global phase is undetectable (| ψ⟩: = exp(iδ) | ψ⟩), and somewhere because of implementation errors, but there are also many places in Qiskit, where the global phase is not ignored. My original question is about the latter. – Aleksey Zhuravlev Jun 18 '20 at 16:25

Unless we ignore the unobservable global phase, it is not possible to express qiskit's $$U_1(\lambda)$$ gate $$\begin{pmatrix}1 & 0 \\ 0 & e^{i\lambda}\end{pmatrix}$$ for arbitrary $$\lambda$$ in terms of $$R_x$$, $$R_y$$ and $$R_z$$. Note that the three rotation gates have unit determinant
$$\det R_x(\theta) = \det \begin{pmatrix} \cos\frac{\theta}{2} & -i\sin\frac{\theta}{2} \\ -i\sin\frac{\theta}{2} & \cos\frac{\theta}{2}\end{pmatrix} = 1\\ \det R_y(\theta) = \det \begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2}\end{pmatrix} = 1\\ \det R_z(\theta) = \det \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2}\end{pmatrix} = 1$$
for all $$\theta$$. Since determinant of a product is a product of determinants, any product of $$R_x$$, $$R_y$$ and $$R_z$$ also has unit determinant. On the other hand, $$\det U_1(\lambda) = e^{i\lambda}$$.
In other words, $$R_x(\theta), R_y(\theta), R_z(\theta)\in SU(2)$$ for all $$\theta$$, but $$U_1(\lambda)\notin SU(2)$$ unless $$\lambda = 2\pi k$$ for some $$k\in\mathbb{Z}$$ (in which case $$U_1(2\pi k) = I$$).
However, if we ignore the global phase then $$U_1(\lambda) = e^{i\lambda/2} R_z(\lambda)$$ is equivalent to $$R_z(\lambda)$$.