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?

  • $\begingroup$ Of course Bell states can be disentangled. One way to do it is precisely applying the inverse of the Bell circuit. $\endgroup$ Commented Jan 17, 2023 at 10:30
  • $\begingroup$ Yes, applying "bell-circuit's dagger(transpose conjugate) onto bell-state" does give back the original separable input state. Why do you think that's not the case? $\endgroup$
    – glS
    Commented Jan 17, 2023 at 13:15

2 Answers 2


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.cx(0, 1)


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

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

qc.cx(0, 1)


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

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


You have to apply reverse U^ gate for vail the results. CNOT then Hadamard gate. description are given below

See the latest Inverse Unitary gates to denote the input of bipartite qubits where Bell state was formed


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