Ray said:
Still not not there, obviously an operator valued field is a mathematical construction, that's the original point. Why are the creation and anhilation operators lodged in a field?
I might get criticism for this, because this is not the usual way that the subject is introduced, but to me, the following is a way to motivate quantum field theory to someone already understands quantum mechanics of several particles. It's unorthodox because it doesn't start with a Lagrangian, and it doesn't mention relativity at all, which are two key ingredients in the usual development of quantum field theory. But this is the way I think about it.
In many-particle quantum mechanics with a fixed finite number of particles, you can represent the total state as a symmetric (for identical bosons) or anti-symmetric (for identical fermions) product of one-particle states. (For simplicity, let's assume that we only have one type of particle). In the boson case, with N particles
\vert \Psi \rangle = K (\vert \psi_1 \rangle \vert \psi_2 \rangle \vert \psi_3 \rangle ... + \vert \psi_2 \rangle \vert \psi_1 \rangle \vert \psi_3 \rangle ... + ...)
where K is a normalization constant. You have to sum over all possible permutations of the N particles. If the one-particle states are discrete, with quantum numbers 0, 1, 2, 3, ... (for example, if we're talking about harmonic oscillators), then this description in terms of products of single-particle states is equivalent to a description in terms of occupation numbers:
Let \vert n_0, n_1, n_2, ... \rangle be the state in which n_0 particles are in single-particle state 0, n_1 particles are in single-particle state 1, etc. If the original set of single-particle states were complete, then this new representation gives a complete basis for multiparticle states.
What's nice about this representation is that it's immediately obvious that it correctly treats the particles as indistinguishable: we only count the number of particles in state j, rather than saying "Particles number 5, 7, 12, and 32 are in state j". But since there are infinitely many possible states, this representation requires an infinite sequence of numbers n_j. To make the notation manageable, we can assume that we're only going to deal with states in which all the n_j are zero except for finitely many. So we change representations once again, as follows:
Let \vert \rangle be the ground state, in which all particles are in the same, lowest energy level.
If \vert \Psi \rangle is the state in which n_0 particles are in single-particle state 0, n_1 particles are in single-particle state 1, etc., then K a^\dagger_j \vert \Psi \rangle is the state in which n_j + 1 particles are in single-particle state j, and the number of particles in all other states is unchanged. K is a normalization constant, which has to be figured out, and a^\dagger_j is a creation operator for state j.
In terms of this new representation, how would we describe a state transition in which one particle changes state from state j to state k? Well, if \vert \psi \rangle is the original state, with occupation numbers n_0, n_1, n_2,, etc., then the new state will be one in which the occupation number for state j is n_j - 1 and the occupation number for state k is n_k + 1. We can represent this as the state:
\vert \Psi' \rangle = C a^\dagger_k a_j \vert \Psi \rangle
where a_j is the annihilation operator that undoes a^\dagger_j, instead of putting an extra particle into state j, it removes one particle, and where C is again some normalization constant, which we can work out.
(The normalization constants for creation and annihilation operators are chosen so that the number operator N_j = a^\dagger_j a_j acts as follows: N_j \vert \Psi \rangle = n_j \vert \Psi \rangle whenever \vert \Psi \rangle is a state with a definite number n_j of particles in single-particle state j).
This is all just notation, so far. I haven't introduced any new physics. It's just a different notation for doing many-particle quantum mechanics.
Now, let's make a transition to a different basis, a position basis. Suppose instead of putting a particle into state j, we want to put a particle at location x=x_0? That one particle will have a wave function that is a \delta function. (Strictly speaking, there is no position basis, because \delta functions are not square-integrable. However, physicists being sloppy can act as if there were a position basis without getting into too much trouble.) In terms of single particle states \psi_n(x), we can write a \delta function as follows:
\delta(x-x_0) = \sum_n \psi^*_n(x_0) \psi_n(x)
So putting a particle into location x=x_0 is equivalent to putting a particle into a superposition of energy levels n, weighted by the amplitude \psi^*_n(x_0). Inspired by this fact, we can introduce another kind of creation operator, \phi^\dagger(x) defined by:
\phi^\dagger(x_0) = \sum_n \psi^*_n(x_0) a^\dagger_n
There's a corresponding annihilation operator \phi(x_0) that removes a particle from location x_0. These position-basis operators can be normalized so that [\phi(x'),\phi^\dagger(x)] = \delta(x-x'), where [A,B] means the commutator AB - BA.
As I said earlier, all of this is simply notation. There is no new physics involved beyond many-particle quantum mechanics (plus the requirement of bose or fermi statistics---I've only mentioned bose statistics here). However, the notation can be used in a more general setting than many-particle quantum mechanics. Once we've introduced creation and annihilation operators, we can easily talk about interactions that change the total number of particles.
An alternative approach (the standard approach, of course) to the same end is to start with a description of physics in terms of field operators \phi(x), and impose the commutation relations.
