Observing = learning outcome

![[Pasted image 20260729093733.png]]

Because after observing all possibilities collapse to one.

Example:

![[Pasted image 20260729093827.png]]
![[Pasted image 20260729093834.png]]

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
![[Pasted image 20260729094624.png]]

BUT!!!
Measuring a quantum state CHANGES THE SYSTEM!!!

![[Pasted image 20260729094710.png]]
![[Pasted image 20260729094718.png]]


## 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

![[Pasted image 20260729100407.png]]

#### Beam Splitter

![[Pasted image 20260729100435.png]]

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

![[Pasted image 20260729101158.png]]

The bomb tester now looks like this:

![[Pasted image 20260729101220.png]]

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:
![[Pasted image 20260729105142.png]]


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]]





