The $|0\rangle$ represents the ground state wavefunction. If we consider the harmonic oscillator, we know that the wavefunction takes the form
$$\psi\left(x\right)=\frac{1}{2^n\sqrt{n}}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}e^{-m\omega x^2/2\hbar}\cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)$$
where $H_n(x)$ is the Hermite polynomial. The solution to the Schrdinger equation, $\hat{H}\psi=E_n\psi$, gives the energy as
$$E_n=\hbar\omega\left(n+\frac12\right)$$
Since $\psi$ has a rather complicated form, it's easier just to represent it as a ket: $\psi\to|n\rangle$, since the energy really depends on $n$. The lowest value $n$ can be is 0, so the ground state is $|0\rangle$.
The ladder operators raise and lower the index $n$ in the eigenstate:
$$
a|n\rangle = \sqrt{n}|n-1\rangle \\
a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle
$$
When you apply the bra $\langle n|$ to the above, we get
$$
\langle n|a|n\rangle = \sqrt{n+1}\,\langle n|n-1\rangle =0\\
\langle n|a^\dagger|n\rangle = \sqrt{n}\,\langle n|n+1\rangle=0
$$
because each $|n\rangle$ is orthogonal to state $|m\rangle$ if $m\neq n$. If we did a raising & lowering, we'd have
$$
\langle n|aa^\dagger|n\rangle=\sqrt{n+1}\langle n|a|n+1\rangle=\sqrt{n+1}\cdot\sqrt{n+1}\langle n|n\rangle=n+1
$$
For your assignment, you can separate the operators as
$$
aaa^\dagger a^\dagger|0\rangle=\left(aaa^\dagger\right)a^\dagger|0\rangle
$$
And then using the relation above, we find
$$
a^\dagger|0\rangle=\sqrt{0+1}|0+1\rangle=|1\rangle
$$
Now we can act the next operator on this,
$$
a^\dagger|1\rangle=\sqrt{1+1}|1+1\rangle=\sqrt{2}|2\rangle
$$
The last two give us a result of
$$
aaa^\dagger a^\dagger|0\rangle=2|0\rangle
$$
Closing this off with $\langle0|$, we get
$$
\langle0|aaa^\dagger a^\dagger|0\rangle=2\langle0|0\rangle=2
$$
First, you have a mistake in your second picture. The top tension is 5 kN, just like the other two legs. Your intuition has misled you. You don't add the tensions in the other two legs.
Second, for practical ropes, a sharp bend decreases strength. The weakest point will be where it wraps over a carabiner. Another weak point will be the knot, though to a lesser degree than if it was a knot in a single cord, or if it was an overhand knot.
Third, friction has a major effect. Loops are often made by laying two ends of a flat strap over each other and sewing. Seat belts are often done this way. The joint is far stronger than the threads that sew it together. The thread press the ends together strongly. This makes friction very large. Kind of like when a tire sits on the road.
Fourth, the knot may be hurting more than helping. Suppose you hung on the anchor and swung to the left. The left two cords would be slack. The right cord would take the full force. You should thread a single long cord under the bottom pulley, over the top left, under the bottom, over the top middle, under the bottom, over the top right, and back to the bottom, so that it makes a single continuous loop. That makes it possible for the cords to slide so the tension in all three legs stays equal. Of course, friction can mess that up.
But a single long loop that slides is not a good idea. If the loop breaks anywhere, your whole anchor fails.
You do have redundancy with your design. If the top middle carabiner or cord fails, you have two more still holding.
The best would be three independent cords to three independent top carabiners. And doubled carabiners or one extra strong carabiner (like you have) for the bottom. It is hard to tie this so that all three have tension. One generally will be slack. But it is redundant.
The figure 8 knot is a good idea if you do use one long cord for this anchor. It turns the single loop into three. One leg can break and the other two will hold.
Best Answer
Breaking strength refers to the maximum tension in the cord. Now, from the sounds of this problem, you've probably been doing force diagrams involving cords. What happens when you attach two cords to a single 100N object (and keep it stationary)? Is the tension in both of those cords 100N? Or is the combined force 100N, so that each just has 50N?
Put another way, most ropes you see will be made of many individual little threads. Each one of them is much weaker than the whole rope. See what I'm getting at?