3.2 KiB
Observing = learning outcome
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Because after observing all possibilities collapse to one.
Example:
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The point here is that the result doesn't change if we observe at any point in the process!
Measuring a quantum system
Given a quantum state \psi \in \mathbb{C}^n we will
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BUT!!! Measuring a quantum state CHANGES THE SYSTEM!!!
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Elitzur-Vaidman bomb tester
Given a box we want to determine whether it contains a bomb. To test, a photon can be send through the box.
- if the bomb detects a photon it explodes!
- if no bomb is present nothing happens
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Beam Splitter
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A beam splitter is a semi transparent mirror.
Photons entering from up can come out on the up or down path - analog for down
Quantum mechanically it could come in a superposition between up and down
\begin{pmatrix} \alpha \\ \beta \end{pmatrix} a = amplitude of up, b = amplitude of down
And it would exit the beam splitter in a superposition between up and down again
\begin{pmatrix} \gamma \\ \delta \end{pmatrix} gamma = up, delta = down
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The bomb tester now looks like this:
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A photon in the up state \begin{pmatrix} 1 \\ 0 \end{pmatrix} is sent through the first beam splitter.
Afterwards the photon is in the state \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}
So the photon is in a superposition between up and down, so between passing through the box with maybe a bomb and passing through empty air.
It makes a difference whether a bomb is inside the box or not!!
First, what happens if there is no bomb: Photon can pass through both paths. Second beam splitter is reached in any case.
After the second splitter the photon is in the state : B(\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
After measuring we get the following distribution:
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Now, what happens if there is a bomb:
We effectively measure if the photon took the up/down path.
Pr[up] = Pr[down] = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}
If the photon is in the down state the bomb explodes.
If the photon is in the up state \begin{pmatrix} 1 \\ 0 \end{pmatrix} post-measurement after no explosion it reaches the second splitter after which it is in the state B$\begin{pmatrix} 1 \ 0 \end{pmatrix}$ = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}
This leads to the following probability distribution:
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Now, from this we can see that if the photon arrives in the down state, there must be a bomb present!
Still, we have a 50% chance of getting blown to smithereens xd BUT having a 25% chance to observe a bomb without the photon touching it is impossible classically!!
This setup can be improved (to much physics) to achieve the following distribution:
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