Fields As Operators

Ray

Can anyone tell me why it is necessary to express a field as annhilation and creation operators? I just don't see why we need a field to explain the creation of particles in relativity, after all two colliding particles with enough energy produce some more.

Drakkith

I don't think relativity deals with particle creation and annihilation. That's a quantum phenomenon.

LastOneStanding

They do so because the corresponding field is an operator built out of ladder operators. Relativity just makes it possible in principle for kinetic energy to be converted into the rest mass of new particles. The actual process that does this is described by the evolution of quantum fields. Have you seen the Dyson series expansion for interactions in quantum field theory? It demonstrates how the creation and annihilation operators let you have a non-zero transition probability between incoming and outgoing states with different numbers of particles.

strangerep

Depends how much group theory you studied. :-)
The fancy-schmancy answer is that elementary particle types must correspond to unitary irreducible representations of the Poincare group, in order to be compatible with (special) relativity. But, when one delves into the technical details of this, one finds that infinite-dimensional representations are necessary -- i.e., field representations.

IOW, one is kinda forced into field representations if one wants to combine QM and relativity satisfactorily in multi-particle scenarios.

The a/c ops are a technical device, useful for working with field representations. Similar ladder operators are used when working with other dynamical groups in (say) classical dynamics, and the general theory of angular momentum. They tend to pop up wherever one uses groups in physics -- which is almost everywhere. Broadly speaking, they "move you around" between different possible solutions of the dynamical equations.

DimReg

There are less technical ways to see this, though perhaps less precise.

Take regular QM, you'll notice 1. an operator x for position, 2. no operator t for time. Position and time are given quite different roles in regular QM. But in special relativity, time and space are treated on the same footing, so this is not the correct way to proceed in constructing a relativistic theory of QM. One way to proceed (but not the only way) is to make operators depend on position in addition to time, and use this new position dependence in your operators to recover the information lost by removing the x operator.

So now you have operators that depend on position, and you use that dependence to give some notion of "where the particle is". However, there is already some object in QM that does that, although we don't call it an operator (because it isn't): the wave function. Given this, you could reasonably expect that there is some operator that now acts in a way analogous to the wave function, which we call the field operator. At this point, all that has happened is that we decided to treat space and time the same way in our operators, but as a result lost one of the tools of the operators we used to have. Then we conjecture that there is enough information in these new operators to recover the information these lost tools contained, and that this information would be in an operator that is "like" the wavefunction.

As to whether this conjectured operator exists, raising and lowering operators turn out to be a very general object: there is a theorem that states that you can construct any operator out of them (so basically, they form a "basis" just like basis vectors in a vector space). Using this theorem, and some other pieces of information (like the klein gordan equation), we can construct this field operator, which we identify as an operator that when acting on the vacuum state creates a state which acts as a one particle state. In fact, this one theorem seems to force you to have field operators: once you make your operators "relativistic" (depend on position as well as time) you end up with a space of operators that allows you to construct a field operator (because you can always use raising/lowering operators). You can try to fight it, but the field operator will be there waiting for you to use it.

Ray

Many thanks for the replies, but I still don't get it. After a collision I know the ouctome of the collision, if I know the outcome of a collision, why do I need a field?

dextercioby

The field is just a mathematical notion necessary to build the theory which furnishes predictions for experiments done with real 'things'. The collision (rather the scattering) is real, the quantum field exists only on paper.

Ray

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?

dextercioby

It has to do with the Fock representation of the commutation relations. In the finite dimensional case, they're called the 'ladder' operators, in the fields' case, they are 'creation' and 'annihilation' due to their particular interpretation for free quantum fields.

kevinferreira

It's precisely the other way around.

stevendaryl

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

[itex]\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 ... + ...)[/itex]

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 [itex]0, 1, 2, 3, ...[/itex] (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 [itex]\vert n_0, n_1, n_2, ... \rangle[/itex] be the state in which [itex]n_0[/itex] particles are in single-particle state [itex]0[/itex], [itex]n_1[/itex] particles are in single-particle state [itex]1[/itex], 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 [itex]j[/itex], rather than saying "Particles number 5, 7, 12, and 32 are in state [itex]j[/itex]". But since there are infinitely many possible states, this representation requires an infinite sequence of numbers [itex]n_j[/itex]. To make the notation manageable, we can assume that we're only going to deal with states in which all the [itex]n_j[/itex] are zero except for finitely many. So we change representations once again, as follows:

Let [itex]\vert \rangle[/itex] be the ground state, in which all particles are in the same, lowest energy level.

If [itex]\vert \Psi \rangle[/itex] is the state in which [itex]n_0[/itex] particles are in single-particle state [itex]0[/itex], [itex]n_1[/itex] particles are in single-particle state [itex]1[/itex], etc., then [itex] K a^\dagger_j \vert \Psi \rangle[/itex] is the state in which [itex]n_j + 1[/itex] particles are in single-particle state [itex]j[/itex], and the number of particles in all other states is unchanged. [itex]K[/itex] is a normalization constant, which has to be figured out, and [itex]a^\dagger_j[/itex] is a creation operator for state [itex]j[/itex].

In terms of this new representation, how would we describe a state transition in which one particle changes state from state [itex]j[/itex] to state [itex]k[/itex]? Well, if [itex]\vert \psi \rangle[/itex] is the original state, with occupation numbers [itex]n_0, n_1, n_2,[/itex], etc., then the new state will be one in which the occupation number for state [itex]j[/itex] is [itex]n_j - 1[/itex] and the occupation number for state [itex]k[/itex] is [itex]n_k + 1[/itex]. We can represent this as the state:

[itex]\vert \Psi' \rangle = C a^\dagger_k a_j \vert \Psi \rangle[/itex]

where [itex]a_j[/itex] is the annihilation operator that undoes [itex]a^\dagger_j[/itex], instead of putting an extra particle into state [itex]j[/itex], it removes one particle, and where [itex]C[/itex] 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 [itex]N_j = a^\dagger_j a_j[/itex] acts as follows: [itex]N_j \vert \Psi \rangle = n_j \vert \Psi \rangle[/itex] whenever [itex]\vert \Psi \rangle[/itex] is a state with a definite number [itex]n_j[/itex] of particles in single-particle state [itex]j[/itex]).

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 [itex]j[/itex], we want to put a particle at location [itex]x=x_0[/itex]? That one particle will have a wave function that is a [itex]\delta[/itex] function. (Strictly speaking, there is no position basis, because [itex]\delta[/itex] 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 [itex]\psi_n(x)[/itex], we can write a [itex]\delta[/itex] function as follows:

[itex]\delta(x-x_0) = \sum_n \psi^*_n(x_0) \psi_n(x)[/itex]

So putting a particle into location [itex]x=x_0[/itex] is equivalent to putting a particle into a superposition of energy levels [itex]n[/itex], weighted by the amplitude [itex]\psi^*_n(x_0)[/itex]. Inspired by this fact, we can introduce another kind of creation operator, [itex]\phi^\dagger(x)[/itex] defined by:

[itex]\phi^\dagger(x_0) = \sum_n \psi^*_n(x_0) a^\dagger_n[/itex]

There's a corresponding annihilation operator [itex]\phi(x_0)[/itex] that removes a particle from location [itex]x_0[/itex]. These position-basis operators can be normalized so that [itex][\phi(x'),\phi^\dagger(x)] = \delta(x-x')[/itex], where [itex][A,B][/itex] means the commutator [itex]AB - BA[/itex].

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 [itex]\phi(x)[/itex], and impose the commutation relations.

stevendaryl

Actually, it occurs to me that there is an assumption being made, which is implicit in the notation, which is that we are only considering states that can be represented as superpositions of states with finitely many particles. In other words, it's suitable for doing perturbation theory as perturbations of a ground state. The notation doesn't allow for a state with infinitely many particles--for instance, you can imagine a state in which there is an array of particles, one at [itex]x=0[/itex], one at [itex]x=L[/itex], one at [itex]x=2L[/itex], etc. This state cannot be described using the creation/annihilation notation, because it would require an infinite number of creation operators.

kof9595995

In a multi-particle Hilbert space, any operator can be expressed as a polynomial of creation and annhilation(c/a) operators, so if we agree we need a multi-particle physics in the first place, there's nothing wrong to use c/a operators to express everything. As for why we prefer to use it than other operators, I favor the perspective from Weinberg's qft vol1, i.e., cluster decomposition principle can be stated very simply when we use c/a operators.