Video: introduction to Elitzur-Vaidman bomb
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^\circ}\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: summary
you cannot repeat test by copying cubit
you cannot re-test one cubit, since it will collapse
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 generate as many mystery qubit as you want: 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. Why? We will see later.
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 bad:
Given n^2 cubits to test with, the expectorated number of time we guess correct is good:
The probability of failure is small
With 2-sided error, the asymptotic complexity is the same.
Deterministic Mystery 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.
In this case, we know n, we just don't know whether the machine always generate |0\rangle or always generate |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
When we measure the final qubit, it is 1, then it generates |Rot_\frac{\pi}{n}\rangle.
When we measure the final qubit, it is 0, then it generates |0\rangle.
Elitzur-Vaidman Bomb Setting: We got a box, with a small hole sealed with 2 horizontal filter at both sides. There might be a bomb in it that can be triggered if the filter generate heat by block lights.
Elitzur-Vaidman Bomb Problem: 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_\frac{\pi}{n}\rangle instead of |+\rangle:
If there is a bomb:
If there is no bomb: photon come out as |Rot_\frac{\pi}{n}\rangle.
To know the final answer, we need distinguish between |Rot_\frac{\pi}{n}\rangle from |0\rangle. It takes \frac{n}{2} many tries of repeatedly sending the "mystery" photon comming out of the box back to box again. 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}. Only measure once.
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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