vault backup: 2026-07-29 11:08:22

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Benjamin Neumann
2026-07-29 11:08:22 +02:00
parent b13609f695
commit 881c0eb4e2
9 changed files with 37 additions and 30 deletions
@@ -3,7 +3,7 @@ All the different possible outcomes.
`Random two bit number has deterministic possibilities 00, 01, 10, 11.`
### Probability distribution
A Probability for each possibility: Pr[x] = p where p $\in$ [0,1]
A Probability for each possibility: Pr\[x] = p where p $\in$ \[0,1]
`For a coin flip this would be Pr[Heads] = 1/2 and Pr[Tails] = 1/2`
Combine all probabilities for all possible outcomes as vector to get probability distribution+
A valid distribution d has $\sum d_i = 1$ and for all i is $d_i \geq 0$
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@@ -65,6 +65,23 @@ After measuring we get the following distribution:
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:
![[Pasted image 20260729110112.png]]
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:
![[Pasted image 20260729110515.png]]
@@ -1 +1,5 @@
1. [[Introduction]]
## 1. [[Introduction to Quantum physics]]
## 2. [[Probabilistic Systems]]
## 3. [[Quantum Systems]]
## 4. [[Observing probabilistic and measuring quantum systems]]
## 5. [[Partial observing and measuring systems]]