I am highly confused with this statement and spent a lot of time understanding it.

  • $\begingroup$ Would be able to link where you found/saw this statement, so we can have more context for this? $\endgroup$
    – Mithrandir24601
    Feb 18 at 14:17

Let $\rho$ be an arbitrary single-qubit state. Since $\rho$ is Hermitian it has real eigenvalues and an orthonormal eigenbasis in which it is diagonal

$$ \rho = \lambda_1 |\psi\rangle\langle\psi| + \lambda_2 |\psi^\perp\rangle\langle\psi^\perp| $$

and since it has unit trace $\lambda_1 + \lambda_2 = 1$. Thus we can find an angle $\beta$ such that $\lambda_1 = \cos^2\frac{\beta}{2}$ and $\lambda_2 = \sin^2\frac{\beta}{2}$. The first eigenvector can be expanded in the computational basis as

$$ |\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle $$

and the other one is fixed up to global phase by orthonormality

$$ |\psi^\perp\rangle = -\sin\frac{\theta}{2}|0\rangle + e^{i\phi}\cos\frac{\theta}{2}|1\rangle. $$

Thus, we see that $\rho$ is completely described by three parameters: $\beta, \theta$ and $\phi$.

Now, recall that

$$ R_Y(\theta) = \begin{pmatrix} \cos\frac{\theta}{2} &-\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} &\cos\frac{\theta}{2} \end{pmatrix} \\ R_Z(\phi) = \begin{pmatrix} e^{-\frac{i\phi}{2}} & 0 \\ 0 & e^{\frac{i\phi}{2}} \end{pmatrix} $$

and so

$$ U(\theta, \phi) = R_Z(\phi) R_Y(\theta) \equiv\begin{pmatrix} \cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2} & e^{i\phi}\cos\frac{\theta}{2} \end{pmatrix} $$

where $\equiv$ denotes equivalence up to unobservable global phase. We see that $U|0\rangle = |\psi\rangle$ and $U|1\rangle = |\psi^\perp\rangle$.

This means that we can prepare a qubit in the $\rho$ state by first initializing a data qubit in the $|0\rangle$ state, applying $X$ gate with probability $\lambda_2$, applying $R_Y(\theta)$ and finally applying $R_Z(\phi)$. In practice, we can realize the probabilistic $X$ gate using the CNOT gate targeting the data qubit and controlled by an auxiliary qubit prepared in the state $\cos\frac{\beta}{2}|0\rangle + \sin\frac{\beta}{2}|1\rangle$. The preparation can be performed using $R_Y(\beta)$. After the CNOT gate the state of the two qubits is $\cos\frac{\beta}{2}|0\rangle|\psi\rangle + \sin\frac{\beta}{2}|1\rangle|\psi^\perp\rangle$. Finally, we discard the auxiliary qubit leaving the data qubit in the state

$$ \mathrm{tr}_1\left[\left(\cos\frac{\beta}{2}|0\rangle|\psi\rangle + \sin\frac{\beta}{2}|1\rangle|\psi^\perp\rangle\right)\left(\cos\frac{\beta}{2}\langle 0|\langle\psi| + \sin\frac{\beta}{2}\langle 1|\langle\psi^\perp|\right)\right] =\\ \cos^2\frac{\beta}{2}|\psi\rangle\langle\psi| + \sin^2\frac{\beta}{2} |\psi^\perp\rangle\langle\psi^\perp| = \rho. $$

In summary, the circuit is:

  1. Prepare a data and auxiliary qubits in $|00\rangle$ state.
  2. Apply $R_Y(\beta)$ to the auxiliary qubit.
  3. Apply CNOT with the auxiliary qubit as control and data qubit as target.
  4. Apply $R_Y(\theta)$ to the data qubit.
  5. Apply $R_Z(\phi)$ to the data qubit.
  6. Discard the auxiliary qubit.

Alternatively, we can measure the auxiliary qubit after $R_Y(\beta)$ and replace CNOT with a classically controlled $X$ gate. We discard measurement result after using it to control the $X$ gate.

  • 1
    $\begingroup$ Thank you @Adam Zalcman! $\endgroup$
    – sohamb172
    Feb 15 at 4:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.