Lecture 012 - Elitzur-Vaidman Bomb

Video: introduction to Elitzur-Vaidman bomb

Guessing about One Qubit

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:

Making Copies

Distinguish Between Qubits: summary

Given n Cubits

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:

n\sin(\pi/n)^2 \simeq \frac{\pi^2}{n}

Given n^2 cubits to test with, the expectorated number of time we guess correct is good:

\simeq n^2 \cdot \frac{\pi^2}{n^2} = \pi^2

The probability of failure is small

(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}

With 2-sided error, the asymptotic complexity is the same.

Obtain the Qubit Generator

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

Elitzur-Vaidman Bomb Problem

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


If we send |Rot_\frac{\pi}{n}\rangle instead of |+\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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