Does the Bell's state entanglement violate the reversibility property of unitary matrices?

Question

I read unitary matrices are reversible, so when we apply a unitary operator $U$ on some input state and got an output state, then if we apply $U^\dagger$ (transpose conjugate) we get back the original input state. So, when we apply a Bell's circuit on two separate input states we get a Bell's state which is entangled and, as far as I know, can't be disentangled.

So, my question is: Bell's circuit is a unitary operator, but by applying Bell's circuit transpose conjugate onto Bell's state we don't go back to the original separable input states; doesn't it violate the property that unitary matrices are reversible? Can I say that Bell's state is an exceptional case where reversibility property of unitary matrices fails?

Answer 1

As far as I know, there is no way to violate the reversibility property because, given a valid quantum state $| \psi \rangle$ and a unitary operator $U$, you will always have that $| \psi \rangle = U^\dagger U = UU^\dagger | \psi \rangle$, since $U^\dagger = U^{-1}$ immediately follows from the definition of unitary matrix.

To show you an example by using Qiskit, if you want to disentangle a Bell state, e.g. $| \Phi_+ \rangle = \frac{1}{\sqrt{2}} \left(| 00 \rangle + | 11 \rangle \right)$, prepared by the following circuit

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector
from qiskit.visualization import array_to_latex

qc = QuantumCircuit(2)

qc.h(0)
qc.cx(0, 1)

array_to_latex(Statevector(qc))

$$ \begin{bmatrix} \tfrac{1}{\sqrt{2}} & 0 & 0 & \tfrac{1}{\sqrt{2}} \\ \end{bmatrix} $$

you simply have to apply the same quantum gates in the reverse order

qc.cx(0, 1)
qc.h(0)

array_to_latex(Statevector(qc)))

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ \end{bmatrix} $$

getting back your original state $| 0 \rangle \otimes | 0 \rangle$ used to initialize any QuantumCircuit by default.