1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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($\lambda$). $\endgroup$ – Cryoris Jun 18 '20 at 16:05
  • $\begingroup$ It throws this error because u1 is already the "most basic" gate and cannot be further decomposed. $\endgroup$ – Cryoris Jun 18 '20 at 16:06
  • 1
    $\begingroup$ 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. $\endgroup$ – Aleksey Zhuravlev Jun 18 '20 at 16:25
1
$\begingroup$

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 $R_z(\lambda)$.

$\endgroup$

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.