vault backup: 2026-07-29 11:08:22
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"title": "Partial observing and measuring systems"
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@@ -3,7 +3,7 @@ All the different possible outcomes.
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`Random two bit number has deterministic possibilities 00, 01, 10, 11.`
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### Probability distribution
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A Probability for each possibility: Pr[x] = p where p $\in$ [0,1]
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A Probability for each possibility: Pr\[x] = p where p $\in$ \[0,1]
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`For a coin flip this would be Pr[Heads] = 1/2 and Pr[Tails] = 1/2`
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Combine all probabilities for all possible outcomes as vector to get probability distribution+
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A valid distribution d has $\sum d_i = 1$ and for all i is $d_i \geq 0$
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+17
@@ -65,6 +65,23 @@ After measuring we get the following distribution:
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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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@@ -1 +1,5 @@
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1. [[Introduction]]
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## 1. [[Introduction to Quantum physics]]
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## 2. [[Probabilistic Systems]]
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## 3. [[Quantum Systems]]
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## 4. [[Observing probabilistic and measuring quantum systems]]
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## 5. [[Partial observing and measuring systems]]
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