8
$\begingroup$

A single-qubit state can be parametrized with real $\theta$ and $\phi$ as follows: $$|\psi(\theta, \phi)\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i \phi} \sin\frac{\theta}{2}|1\rangle.$$

I would like to know how to parameterize an arbitrary two-qubit state in a similar way.

$\endgroup$
1

1 Answer 1

7
$\begingroup$

TL;DR: The logic behind the equation $|\psi\rangle=\cos\frac{\theta}{2}|0\rangle+e^{i\phi}\sin\frac{\theta}{2}|1\rangle$ can be iterated to obtain a real parameterization for states of arbitrary finite dimension.

Iterative procedure

We parameterize an expansion of $|\psi\rangle$ in basis $|0\rangle,\dots,|N\rangle$ in $N$ iterations. The $k$th iteration for $k=1,\dots,N$ consists of two steps. First, we introduce $\theta_k\in[0,\pi]$ to apportion the unit norm between the magnitudes of the amplitudes of $|k-1\rangle$ and the projection $|\psi_{k\dots N}\rangle$ of $|\psi\rangle$ onto the subspace spanned by $|k\rangle,\dots,|N\rangle$. Next, we stick a relative phase $\alpha_k\in[0,2\pi)$ on $|k\rangle$. Thus, the $k$th iteration rewrites $|\psi_{k-1\dots N}\rangle$ as $$ |\psi_{k-1\dots N}\rangle=\cos\frac{\theta_k}{2}|k-1\rangle + e^{i\alpha_k}\sin\frac{\theta_k}{2}|\psi_{k\dots N}\rangle.\tag1 $$ The final result can be simplified by setting $\phi_k:=\alpha_1+\dots+\alpha_k\mod 2\pi$.

Each step introduces a single real parameter, for a total of $2N$ real parameters, as expected for an $N+1$ dimensional system.

Two-qubit case

For a two-qubit state the above procedure yields $$ \begin{align} |\psi\rangle&=\cos\frac{\theta_1}{2}|00\rangle\\ &+ e^{i\phi_1}\sin\frac{\theta_1}{2}\cos\frac{\theta_2}{2}|01\rangle\\ &+ e^{i\phi_2}\sin\frac{\theta_1}{2}\sin\frac{\theta_2}{2}\cos\frac{\theta_3}{2}|10\rangle\\ &+ e^{i\phi_3}\sin\frac{\theta_1}{2}\sin\frac{\theta_2}{2}\sin\frac{\theta_3}{2}|11\rangle \end{align}\tag2 $$ where $\theta_1,\theta_2,\theta_3\in[0,\pi]$ and $\phi_1,\phi_2,\phi_3\in[0,2\pi)$.

This isn't the only way to parameterize a two-qubit state. An alternative approach uses Schmidt decomposition and single-qubit parameterization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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