Lecture 012 - 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

Making Copies

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.

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 $|1\rangle$

• 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 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

Observation:

• Sending $|1\rangle$ will immediately explode if there is bomb

• Sending $|0\rangle$ is useless and give us no information.

• Sending $|+\rangle$

• If there is bomb:
• 50% explode
• 50% no explode and photon come out as $|0\rangle$, we know there is bomb
• If there is no bomb: photon come out as unmodified $|+\rangle$, but when we measure it we either get 1-sided error with $\frac{1}{2}$ or two sided error $15\%$

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

• If there is a bomb:

• $|\langle 1 | Rot_\frac{\pi}{n}\rangle|^2$ (very small) chance explode
• $|\langle 0 | Rot_\frac{\pi}{n}\rangle|^2$ (very big) chance come out as $0\rangle$, we know there is 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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