HW3 - Problem 1
Problem 1: Bell States and Partial Measurements
We’ve discussed the Bell pair in class, which (like any state) can be part of an orthonormal basis. Since these vectors are length four, the basis should have four states. There is a called the Bell basis, formed of the four canonical Bell states. The four states are
- \(\left|\Phi^+\right> := \frac{1}{\sqrt{2}}(\left|00\right> + \left|11\right>)\)
- \(\left|\Phi^-\right> := \frac{1}{\sqrt{2}}(\left|00\right> - \left|11\right>)\)
- \(\left|\Psi^+\right> := \frac{1}{\sqrt{2}}(\left|01\right> + \left|10\right>)\)
- \(\left|\Psi^-\right> := \frac{1}{\sqrt{2}}(\left|01\right> - \left|10\right>)\)
- Design a two qubit circuit that starts in the state \(\left|00\right>\), and ends with \(\ket{\Psi^-}\).
- For the remaining parts for this problem, we will use the following notation.
\(\left|\theta\right> := \cos(\theta)\left|0\right> + \sin(\theta)\left|1\right>.\)
Consider the basis \(\{\left|\pi/6\right>, \left|4\pi/6\right>\}\). Write down \(\left|0\right>\) and \(\left|1\right>\) as a weighted sum of \(\left|\pi/6\right>\) and \(\left|4\pi/6\right>\) in the above basis. Recall that
\(\cos(\pi/6) = \sqrt{3}/2 \;\;\;\; \sin(\pi/6) = 1/2\)
\(\cos(4\pi/6) = -1/2 \;\;\;\;\sin(4\pi/6) = \sqrt{3}/2\)
- Suppose Alice has the first qubit and Bob has the second qubit of a \(\left|\Psi^-\right>\) state. If Alice measures her qubit in the standard basis, what are the probabilities of each outcome, and the state of the two qubits after the measurement?
- If Alice instead chooses to measure in the \(\{\left|\pi/6\right>, \left|4\pi/6\right>\}\) basis, at are the probabilities of each outcome, and the state of the two qubits after the measurement?
- Verbally describe what happens to the second qubit when the first qubit of a \(\left|\Psi\right>\) state gets measured.