88 lines
3.2 KiB
Markdown
88 lines
3.2 KiB
Markdown
Observing = learning outcome
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![[Pasted image 20260729093733.png]]
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Because after observing all possibilities collapse to one.
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Example:
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![[Pasted image 20260729093827.png]]
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![[Pasted image 20260729093834.png]]
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The point here is that the result doesn't change if we observe at any point in the process!
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## Measuring a quantum system
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Given a quantum state $\psi \in \mathbb{C}^n$ we will
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![[Pasted image 20260729094624.png]]
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BUT!!!
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Measuring a quantum state CHANGES THE SYSTEM!!!
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![[Pasted image 20260729094710.png]]
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![[Pasted image 20260729094718.png]]
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## Elitzur-Vaidman bomb tester
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Given a box we want to determine whether it contains a bomb.
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To test, a photon can be send through the box.
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- if the bomb detects a photon it explodes!
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- if no bomb is present nothing happens
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![[Pasted image 20260729100407.png]]
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#### Beam Splitter
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![[Pasted image 20260729100435.png]]
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A beam splitter is a semi transparent mirror.
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Photons entering from up can come out on the up or down path - analog for down
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Quantum mechanically it could come in a superposition between up and down
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$\begin{pmatrix} \alpha \\ \beta \end{pmatrix}$ a = amplitude of up, b = amplitude of down
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And it would exit the beam splitter in a superposition between up and down again
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$\begin{pmatrix} \gamma \\ \delta \end{pmatrix}$ gamma = up, delta = down
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![[Pasted image 20260729101158.png]]
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The bomb tester now looks like this:
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![[Pasted image 20260729101220.png]]
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A photon in the up state $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is sent through the first beam splitter.
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Afterwards the photon is in the state $\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$
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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.
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It makes a difference whether a bomb is inside the box or not!!
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First, what happens if there is no bomb:
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Photon can pass through both paths.
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Second beam splitter is reached in any case.
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After the second splitter the photon is in the state : $B(\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix})$ = $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$
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After measuring we get the following distribution:
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![[Pasted image 20260729105142.png]]
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Now, what happens if there is a bomb:
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We effectively measure if the photon took the up/down path.
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Pr\[up] = Pr\[down] = ($\frac{1}{\sqrt{2}}$)^2 = $\frac{1}{2}$
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If the photon is in the down state the bomb explodes.
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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$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ = $\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$
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This leads to the following probability distribution:
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![[Pasted image 20260729110112.png]]
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Now, from this we can see that if the photon arrives in the down state,
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there must be a bomb present!
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Still, we have a 50% chance of getting blown to smithereens xd
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BUT having a 25% chance to observe a bomb without the photon touching it is impossible classically!!
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This setup can be improved (to much physics) to achieve the following distribution:
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![[Pasted image 20260729110515.png]]
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