Leoragon said:
I still don't get it; I just need a brief explanation
(P.S. Note: I am assuming you have at least a basic concept of Lagrangian, Hamiltonian, and Quantum Harmonic Oscillator. If not, this will probably not make a whole lot of sense.)
It might be easier if you forget for a moment about quantum mechanics. Something very similar happens in classical systems. Consider three identical masses m in a row connected by two springs with spring coefficients of k each. They are constrained to move along one dimension for simplicity.
Lets say that positions of the masses are given by ##\small y_1##, ##\small y_2##, and ##\small y_3##. Since center of mass won't accelerate, only two of these are independent if I say that center of mass is at zero. I can then define ##\small x_1 = y_2-y_1## and ##\small x_2 = y_3-y_2##, which are spring displacements and can be treated as my actual coordinates. Because CoM isn't moving, ##\small \dot{y_1} + \dot{y_2} + \dot{y_3} = 0##. This let's me write down kinetic energy in terms of x coordinates. And all together, this yields the following Lagrangian.
L = \frac{1}{3}m (\dot{x_1}^2 + \dot{x_2}^2 + \dot{x_1}\dot{x_2}) - \frac{1}{2}k(x_1^2 + x_2^2)
Well, that isn't half as terrible as you might have expected, but notice that if you are to write down equations of motion, the differential equations you get are linked. Equations for ##\small x_1## depend on ##\small x_2## and vice versa.
Can this be helped? Well, yes it can! Let's define ##\small q_1 = x_1+x_2## and ##\small q_2 = x_1 - x_2##. This is easily visualized as ##\small q_1## stretching the edge weights apart and ##\small q_2## moving the middle mass in between. Why is this useful? Well, let's re-write the Lagrangian in terms of q variables.
L = \frac{1}{4}m \dot{q_1}^2 + \frac{1}{12}m \dot{q_2}^2 - \frac{1}{4}k(q_1^2 + q_2^2)
Would you look at that? We no longer have cross-terms. That means we'll have separate equations for ##\small q_1## and ##\small q_2##. In other words, these are the natural modes of this problem. Furthermore, each one is just a simple harmonic oscillator. Using the definition of generalized momentum.
p_i = \frac{\partial L}{\partial \dot{q_i}}
We have ##\small p_1 = \frac{m}{2} \dot{q_1}## and ##\small p_2 = \frac{m}{6} \dot{q_2}##. For completion, let us write the Hamiltonian for this system.
H = \sum_i p_i \dot{q_i} - L = \frac{p_1^2}{m} + \frac{3p_2^2}{m} + \frac{k}{4}(q_1^2 + q_2^2)
Notice that this is just sum of Hamiltonians for two independent oscillators with masses m/2 and m/6 and spring coefficients of k/2.
The most important part of second quantization is done. We have two independent harmonics. But let's imagine now that this is a QM problem. The analysis above still applies with minor editing. You still end up with the same Hamiltonian in diagonalized coordinates. And now we can treat each oscillator as quantum harmonic oscillator. That means you can add to the system energy increments of ##\small \hbar \omega_1## or ##\small \hbar \omega_2## with ##\omega_1 = \sqrt{\frac{k}{m}}## and ##\omega_2 = \sqrt{\frac{3k}{m}}##. We have two different modes, and each one can be excited with fixed quanta of energy. Reminds you of anything yet?
If you take a 3D network of masses connected by something that can be approximated as a spring, you end up with a decent model of a solid. That Hamiltonian can be similarly diagonalized, with normal modes forming plain waves. If you further quantize an effective oscillator corresponding to each plain wave, you find that you can add ##\small \hbar \omega## of energy to each such mode! These quanta of energy propagating as plain waves through the solids are the phonons. This is second quantization applied to a very similar problem.
Now you can go to a continuum case. Imagine that instead of masses and springs you have fields. You can write down a Lagrangian density for the field. If it happens to be linear, you can diagonalize it, find normal modes, apply quantum mechanics to the normal modes, and arrive at second-quantized field! Why second quantized? Because diagonalizing to normal modes already quantizes the problem. So the problem is twice quantized. Once into normal modes with discrete energies, and second time into discrete increments of these energies.
Just like normal modes of oscillations in the solid become phonons upon second quantization, the normal modes of electromagnetic field, which also happen to be electromagnetic plain waves, become photons upon second quantization.
There are several really, really exciting things that follow from this. First of all, you can use ladder operators just like you would with harmonic oscillator. These become particle creation/annihilation operators and you can describe your states as operators. Your field is now a bunch of particle states, so you can describe interactions with the same creation/annihilation operators. That let's you use the same formalism for the field theory where now you have interacting fermions. And that can all be summarized with gauge theories... But this is starting to get into more complicated stuff, and you really need a solid grasp of basic QFT formalism to dive deeper into it.
Sonderval said:
Try to get a copy of Feynman's Lectures on Gravitation. The first chapters are really fun - he treats gravity as if it were a so far unknown phenomenon and tries to describe it using QFT. The graviton then emerges quite naturally.
Absolutely. But again, you need to have solid grasp of basic QFT and preferably some QED. Otherwise, it's prohibitively difficult to follow.
