3
Two antipodal states in the Bloch sphere (note that $0 \leq \theta \leq \pi$):
\begin{equation}
|\psi \rangle = \cos \frac{\theta}{2} |0 \rangle + e^{i\varphi}\sin \frac{\theta}{2} |1 \rangle
\\
|\psi^\perp \rangle = \cos \frac{\pi - \theta}{2} |0 \rangle + e^{i\varphi + \pi}\sin \frac{\pi - \theta}{2} |1 \rangle = \sin \frac{\theta}{2} |0 \rangle - e^{...
3
Let's say I know that
$$
U_1|\Psi_1\rangle\otimes U_2|\Psi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle.
$$
Now, let's imagine that $U_1|\Psi_1\rangle=|\phi_1\rangle$ and $U_2|\Psi_2\rangle=|\phi_2\rangle$. So,
$$
|\phi_1\rangle\otimes |\phi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle.
$$
Now, I would just read off your desired ...
2
Let $H_1,H_2$ be two Hilbert spaces and for unit vectors $u_1,v_1 \in H_1$ and $u_2,v_2 \in H_2$ we have
$$
u_1 \otimes u_2 = v_1 \otimes v_2 \in H_1 \otimes H_2.
$$
Then it must be
$$
v_1 = \lambda u_1, ~~ v_2 = \frac{1}{\lambda} u_2,
$$
for some $\lambda \in \mathbb{C}$, $|\lambda|=1$.
To see why consider a scalar product
$$
1 = (u_1\otimes u_2, u_1\...
1
If that is all you want to do then SciPy is really the way to go. Indeed, qiskit just wraps that functionality when requesting a classical answer.
In SciPy you can use scipy.linalg.eig. You can find examples of using this function in the documentation.
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