# Does changing the state of an entangled qubit affect its entangled partner?

I have a question about the single qubit gate effect after entanglement. Suppose we have a circuit as I draw. Qubit 0 and qubit 1 are entangled. B gate is applied to qubit 0, and C gate is applied to qubit 1 after entanglement (B and C are any single qubit gate). Would gate C affect the state of qubit 0? For example, would changing the type of gate C affect the measurement result of qubit 0?

I think this can be a bigger question. After entanglement, would changing the state of a qubit affect the state of its entangled qubit?

Imagine that after the CNOT the top qubit is sent to Alice and the bottom qubit is sent to Bob. If it were the case that Bob could apply a gate that influenced the measurement outcome probabilities of Alice's qubit, this would allow faster-then-light communication.

Gate C can affect the state of qubit 0, but can't affect the measurement result of qubit 0.

For example after the CNOT the two qubit state could be $$\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ And if gate C is a Z gate this state becomes: $$\frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$$ This is a different state, but still a maximally mixed state. If you measure a maximally mixed state along any axis, the outcome is random.

• If I understand correctly, the measurement result for qubit 0 will not change no matter what execution is applied to qubit 1 after entanglement. Also, the same phenomenon exists for the B gate's effect on qubit 1. Feb 7 at 22:52
• Yes that's correct! Feb 7 at 23:28
• "This is a different state, but still a maximally mixed state" - you might want to be more specific that you're talking about a subsystem otherwise this is not true. Feb 8 at 19:00

This depends on what you mean by "the state of qubit 0". Once your two qubits are entangled, it no longer makes much sense to talk about the state of one qubit or the state of the other. They have one collective state.

On the other hand, you might ask about being able to predict the outcomes of measurements on qubit 0. Does performing a unitary on qubit 1 affect the probabilities of different outcomes? No. One way to see this is to work through the formalism of the reduced density matrix, which gives a single-qubit description of the qubit to be measured, and still gives the best possible predictions. Through part of this formalism, you can easily prove that the reduced density matrix of one qubit does not change when you apply unitaries to another.

You then ask a more general question about changing the state of qubit 1. The other option that this introduces is measurement. When you make a measurement on one system, the overall state changes (it may still be entangled). However, if we go back to the "predicting measurement outcomes" perspective, it is a bit more subtle than that, because everything now becomes subjective, depending on who knows the measurement outcome. For anybody who knows the measurement outcome, their predictions change. However, anyone who does not know the outcome (even if they know the measurement was performed) still has the same set of predictions. This has relevance if the two qubits are distantly separated and the information about measurement outcomes on qubit 1 doesn't have enough time to propagate to the person making the measurements on qubit 0.