Measuring One Qubit: When you measure entangled qubits, all related qubits collapse.
If I have a qubit and want you to decide which option the qubit is:
either |0\rangle, |1\rangle: 100% probability
either |+\rangle, |-\rangle: 100% probability
either |Rot_\theta\rangle, |Rot_{\theta+90\degree}\rangle: 100% probability
anything choice that is not orthogonal will result in error, but we can choose 2-sided error or 1-sided error
Distinguish Between Qubits
you cannot repeat test by copying cubit
you cannot re-test one cubit
you can do one sided error or two sided error for copies of cubit
you can measure in both basis for copies of cubit waiting for 100% answer in one sided error.
If you got the mystery rotation machine: to distinguish rotation \theta = 0 between \theta = \frac{\pi}{n}, you only need to send the mystery cubit into the machine by \frac{n}{2} times.
Say we want you to tell the difference between |Rot_{0}\rangle and |Rot_{\pi/n}\rangle where n is big.
Strategy: measure in |0\rangle, |1\rangle basis and only guess |Rot_{\pi/n}\rangle if result is |1\rangle
Given n cubits to test with, the probability of guessing correct with one sided error is: n\sin(\pi/n)^2 \simeq \frac{\pi^2}{n} which is bad.
Given n^2 cubits to test with, the expectorated number of time we guess correct is \simeq n^2 \cdot \frac{\pi^2}{n^2} = \pi^2 which is big. The probability of failure is (1 - \sin(\frac{\pi}{n})^2)^{n^2} \simeq (1 - \frac{\pi^2}{n^2})^{n^2} \leq (e^{-\pi^2/n^2})^{n^2} = e^{-\pi^2} which is small.
With 2-sided error, the asymptotic complexity is the same.
Qubit Generator: given |0\rangle, generate a fixed mystery |test\rangle qubit. We know the machine either generate |0\rangle or |Rot_\frac{\pi}{n}\rangle.
We can feed the generated qubit back to the machine \frac{n}{2} times to get either 0\rangle or 1\rangle
Elitzur-Vaidman Bomb Problem: We got a box, potentially with a bomb fused to a horizontal filter. We want to know whether the box has a bomb in it. We can only do this by sending photon passing through box. However, if there is a filter and the filter generates heat, the bomb will explode.
Bomb | Nothing | |
---|---|---|
Send 0 | No Info | No Info |
Send 1 | Explode | No Bomb |
Observation:
Sending |1\rangle will immediately explode if there is bomb
Sending |0\rangle is useless and give us no information.
Sending |+\rangle
If we send |Rot_{\pi}{n}\rangle instead of |+\rangle, to know the final answer, we need distinguish between |Rot_{\pi}{n}\rangle from |0\rangle. We need to a know that takes \frac{n}{2} many tries. Since the probability of explode is \frac{\pi^2}{n^2}, by union bound, the probability of explode is \frac{n}{2} \cdot \frac{\pi^2}{n^2} = \frac{\pi^2}{2n}. Therefore, if we choose a large n, we can almost make it certain that the bomb will not explode before we know the status of the bomb for certain.
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