I
X
outcome to Bob. If Bob wants his qubit in a|+> + b|->, what gate does he apply in the following two cases? Choose your answer from I,X,Z,H.
Alice's outcome is 0:
Alice's outcome is 1:
X
Alice's outcome is 1: X,H
The state |+> is called the superposition state and is defined as |+> = (|0> + |1>) / sqrt(2).
The state |-> is another superposition state defined as |-> = (|0> - |1>) / sqrt(2).
Now, let's consider the two cases:
1. Alice's outcome is 0:
This means that when Alice measures her qubit in the standard basis, she obtains 0.
When Alice measures her qubit in the standard basis, the state of the qubit collapses to the measurement outcome. So, Alice's measurement outcome being 0 implies that the state after her measurement is now |0>.
To transform the state of Bob's qubit to a|+> + b|->, where Alice's measurement outcome is 0, Bob needs to apply the X gate. The X gate flips the state of the qubit, so applying the X gate to the qubit in state |0> would give us |1>. Therefore, applying the X gate to the qubit would give us a|+> + b|->.
Answer: X gate.
2. Alice's outcome is 1:
This means that when Alice measures her qubit in the standard basis, she obtains 1.
Similarly, Alice's measurement outcome being 1 implies that the state after her measurement is now |1>.
To transform the state of Bob's qubit to a|+> + b|->, where Alice's measurement outcome is 1, Bob needs to apply the Z gate. The Z gate introduces a phase flip to the qubit. Applying the Z gate to the qubit in state |1> would give us -|1>. Therefore, applying the Z gate to the qubit would give us a|+> - b|->.
Answer: Z gate.