# Is entanglement time-symmetric?

It is common to describe an experiment as "causing" entanglement. For example, two quantum particles that interact become entangled as a result of their interaction, so we are likely to say that the interaction "caused" the entanglement.

However, quantum mechanics is time-symmetric, so a time reversed movie of the experiment would show two entangled particles approaching each other, interacting, and then going on their way unentangled.

I suspect that this picture is incorrect, but can't put my finger on why.

• Why do you think your picture is incorrect ? I agree that reversing time it would disentangle but I am not sure to see what disturbs you ? – StarBucK Feb 15 at 17:19
• What bothers me is that in one direction it seems information would be gained, and in the other direction information would be lost. In the reverse-time direction, the original wavefunctions could only be reconstituted if the wavefunctions of the entangled particles are exactly right, so it seems the information must simply be redistributed between past and future. That being so, it seems that there must have been some sort of entanglement in the past. – S. McGrew Feb 15 at 17:58
• I am not sure to see what you mean. Reasonning on the global state at all time we have the same amount of information (you can quantify it by entropy for instance: the entropy is always $0$ in time). So could you clarify which kind of information you are talking about ? – StarBucK Feb 15 at 18:29
• If the wavefunctions of the future entangled particles could be described, it would only be necessary to describe the entangled portions for one of the two particles. But wavefunctions for two completely unentangled particles would require full description of all portions of both particles. The particle wavefunctions in the past seem to contain more information. – S. McGrew Feb 15 at 18:44
• Entanglement, likesuperposition has a part of subjectivity. It is related to the basis you choose to describe the physics. For instance if you assume your entangled state lives in a two qubit space, $|00\rangle + |11 \rangle$, you can always define a new basis of four vector in which $|00\rangle + |11 \rangle = |\psi_0 \psi_0 \rangle$ (it would be a product state in this basis). So in pure term of information you have exactly the same amount of information wether the state is entangled or not (as soon as it is pure) – StarBucK Feb 15 at 18:47

Suppose for example that two qubits initially in the product state $$|\psi_i\rangle = |+\rangle|0\rangle$$ are allowed to interact in a way that performs the CNOT gate from the first to the second qubit. Then the final state is $$|\psi_f\rangle=(|00\rangle + |11\rangle)/\sqrt{2}$$.
Now consider the above process in reverse. Two qubits begin in the Bell state $$|\psi_f\rangle=(|00\rangle + |11\rangle)/\sqrt{2}$$, interact in a way equivalent to the CNOT gate and finish in the state $$|\psi_i\rangle=|+\rangle|0\rangle$$.
In terms of spins, you can think of $$|1\rangle$$ as the "spin up" state, of $$|0\rangle$$ as the "spin down" state and of $$|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$$ as the uniform superposition of the two states with zero relative phase. A process implementing the CNOT gate is then any interaction that flips the second spin if the first spin is up.