How do local quantum gates affect an entangled state?

Question

(1) Assume we have the Bell State $$ \frac{\lvert 0_{A}0_{B}\rangle + \lvert 1_{A}1_{B}\rangle}{\sqrt{2}} $$ where A and B stand for Alice and Bob. Now say Bob applies the X gate to his qubit. I think we get $$ (I\otimes X) \frac{\lvert 0_{A}0_{B}\rangle + \lvert 1_{A}1_{B}\rangle}{\sqrt{2}} = \frac{\lvert 0_{A}1_{B}\rangle + \lvert 1_{A}0_{B}\rangle}{\sqrt{2}} $$ I would have thought that after being entangled, they would always be correlated to measuring the same value but it seems now they will always measure opposite values. Is this right?

(2) What about $$ (I\otimes H) \frac{\lvert 0_{A}0_{B}\rangle + \lvert 1_{A}1_{B}\rangle}{\sqrt{2}} = \frac{\lvert 0_{A}+_{B}\rangle + \lvert 1_{A}-_{B}\rangle}{\sqrt{2}} = \frac{\frac{1}{\sqrt{2}}\left[\lvert 0_{A}0_{B}\rangle + \lvert 0_{A}1_{B}\rangle\right] + \frac{1}{\sqrt{2}}\left[\lvert 1_{A}0_{B}\rangle - \lvert 1_{A}1_{B}\rangle\right]}{\sqrt{2}} = \frac{1}{2}\left[\lvert 0_{A}0_{B}\rangle + \lvert 0_{A}1_{B}\rangle\right] + \frac{1}{2}\left[\lvert 1_{A}0_{B}\rangle - \lvert 1_{A}1_{B}\rangle\right] $$ It seems like the Hadamard Gate just messed up entanglement.

Answer 1

All of your math is correct. Both of the states that you calculated are entangled states. An entangled state is one in which there exist measurable properties of subsystems that are correlated.

In (1), measurements of the two qubits in the computational basis are correlated; it just happens to be a negative correlation.

In (2), measurements in the computational basis are uncorrelated, but measurements in some other bases are correlated (e.g., computational for Alice and Hadamard for Bob). In an unentangled state, all measurable properties of the individual qubits would be uncorrelated.

Answer 2

To understand what entanglement is, as a crude description it means if you measure one part of the system, then knowing what the measurement outcome is will change what you expect for measurement outcomes on the remainder of the system.

This is true for both your examples, in both cases if you don't measure qubit 1, you expect to measure 0 or 1 with equal probabilities on qubit 2 (as well as equal probablities of measuring + or - on qubit 2).

• When you apply $(I\otimes X)$ and you measure qubit 1 to be $|0\rangle$ then you know qubit 2 is in state $|1\rangle$ and if you measure qubit 1 to be $|1\rangle$ then qubit 2 is $|0\rangle$
• When you apply $(I\otimes H)$ and you measure qubit 1 to be $|0\rangle$ then you know qubit 2 is in state $|+\rangle$ and if you measure qubit 1 to be $|1\rangle$ then qubit 2 is $|-\rangle$

Now at this point I'll point out that the crude description I gave you is wrong, however it will turn out to be true if we are only interested in pure states. A rigorous mathematical definition is that a pure state $|\rho\rangle$ is entangled if it cannot be written as $|\rho\rangle = |\psi_{1}\rangle \otimes |\psi_{2}\rangle$. Taking example 2 and writing general 1 qubit states $|\psi_{j}\rangle = a_{j}|0\rangle + b_{j}|1\rangle$ then you find:

$$|\rho\rangle = a_{1} a_{2} |0 0\rangle + a_{1} b_{2} |0 1\rangle + b_{1} a_{2} |1 0\rangle + b_{1} b_{2} |1 1\rangle$$

For this to represent your 2nd example you need $a_{1}a_{2} = 1$, $a_{1}b_{2} = 1/2$, $b_{1}a_{2} = 1/2$, $b_{1}b_{2} = -1/2$. These can be rearranged to find that $1 = -1$, which is not possible and so the state is entangled.