Quantum Field Question I am watching a video on QFT (https://www.youtube.com/watch?v=jlEovwE1oHI).
And have kind of a fundamental question(s)...


*

*Why are the fields that require hundreds of MeV (or more than 174 GeV like the top field) so unstable and those that require only a few MeV (down quark) so stable?

*My next question is how does the energy from one field transferred to a lower field?  For instance when a Strange quark ---> Up + W- ---> e- + anti Ve
How does the energy transfer to the Up field?


*From the example above, how doers the residual energy transfer to the electron and electron neutrino fields?

 A: Simplified speaking a quantum field can be considered as a number of infinite harmonic oscillators, and one of the most important quantum properties of a quantum oscillator compared to its classic counterpart is that the ground state of the quantum oscillator has non-zero energy. 
$$E^{oszi}_0 = \frac{1}{2}\hbar\omega$$
This can be understood as a consequence of uncertainty-relation for energy: A for long time $\Delta t$ observed oscillator in the ground state is allowed to fluctuate in energy $\Delta E=\hbar/\Delta t$. Furthermore, if we consider the main degree of freedom of a quantum harmonic oscillator $q$, its average value and its fluctuations, we find:
$$\langle0 | q |0\rangle = 0\quad\quad \text{whereas}\quad \quad \langle0 | q^2 |0\rangle = \frac{\hbar}{2m\omega}$$
The variable $m$ stands for the mass of the oscillating particle and $\omega$ for its angular frequency. 
If we now (from now on we set $\hbar=1$) consider a quantum field $\phi$, for simplicity a real scalar field (the reasoning for non-scalar fields is similar) the most striking difference is that it has an infinite number of degrees of freedom of its fourier components characterized by the 4-vector $k$
(and $E_k = \sqrt{\mathbf{k}^2 +m^2}$ its energy):
$$\phi(x) = \int \frac{d^3k}{\sqrt{(2\pi)^3 2E_k}}\left(a(k) e^{ikx} + a^\dagger(k) e^{-ikx}\right)$$
where $a(k)$ and $a^\dagger(k)$ are annihilation and creation operators with the rules
$$\langle n_k-1 | a(k)| n_k\rangle = \sqrt{n_k} \quad \text{and} \quad \langle n_k | a^\dagger(k)| n_k-1\rangle = \sqrt{n_k}$$
and with commutation rules: 
$$[ a(k), a^\dagger(l)] = \delta^3 (\mathbf{k}-\mathbf{l})$$
We can tentatively compute the average value of the quantum field $\phi$:
$$\langle 0 | \phi(x) |0\rangle = 0 $$
according the rules of the creation and annihilation operators $a(k)$ and $a^\dagger(k)$ and its fluctuations: 
$$\langle 0|\phi(x)\phi(y)|0\rangle =\int  \frac{d^3k d^3 l}{(2\pi)^3 \sqrt{2E_k 2E_l}} e^{-ikx + ily} \langle 0 | a(k) a^\dagger(l) |0\rangle =  \int \frac{d^3k}{(2\pi)^3 2E_k} e^{-ik(x-y)} $$
For $y\rightarrow x$ we get for the fluctuations an infinite value:
$$\langle 0|\phi(x)^2|0\rangle =\int\frac{d^3k}{(2\pi)^3 2E_k}=\infty$$ 
So due to the infinite number of freedom of a continous quantum field its fluctuations in the ground state is infinite. That explains the used picture of the numerous fluctuations shown in the video. In difference to classical physics where the vacuum is considered as empty, 
the quantum vacuum is not really empty, quantum oscillations are ubiquitous inspite the average value of the quantum field is zero. 
The particles electron, (by the way the forgotten neutrino), up- and down-quark are very stable, simply because they cannot decay (according to the rules of the Standard Model) in more lighter particles simply because  those lighter particles don't exist. They belong --- as it is well-known --- to the first generation of elementary particles in the Standard Model.
Whereas the top-quark belongs to the third generation and is a lot of more heavier and can decay (as single particle) via the weak interaction mostly in b-quarks, those continue to decay in other particles. 
