# Fields As Operators

1. Feb 5, 2013

### 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.

2. Feb 5, 2013

### Staff: Mentor

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

3. Feb 5, 2013

### VantagePoint72

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.

4. Feb 5, 2013

### 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.

5. Feb 6, 2013

### 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.

6. Feb 9, 2013

### 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?

7. Feb 9, 2013

### 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.

8. Feb 10, 2013

### 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?

9. Feb 10, 2013

### 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.

10. Feb 10, 2013

### kevinferreira

It's precisely the other way around.

11. Feb 10, 2013

### stevendaryl

Staff Emeritus
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.

12. Feb 10, 2013

### stevendaryl

Staff Emeritus
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 $x=0$, one at $x=L$, one at $x=2L$, etc. This state cannot be described using the creation/annihilation notation, because it would require an infinite number of creation operators.

13. Feb 10, 2013

### 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.

14. May 22, 2013

### TrickyDicky

Maybe this is a too naive question, but if in QFT the quantum field is the fundamental object and particles are just the quanta of that field, wich is an important conceptual departure from the QM picture, why are multi-particle physics and particle number issues that seem linked to the "particle as fundamental object" conceptual framework still dragging on in the quantum field picture?
wouldn't this "dragging" lead one to think the transition to a truly quantum field theory is not finished?

Last edited: May 22, 2013
15. May 22, 2013

### TrickyDicky

And what is not "just a mathematical notion" in quantum physics? What are the "real things" then, particles as tiny bullets that collide(scatter)? How would you go about showing that with certain rigour if not using mathematical notions?

16. May 22, 2013

### DarMM

I would would take strangereps point of view.

The combination of special relativity and quantum mechanics demands that the most basic states be unitary irreps of the Poincaré group. In current particle physics this is what particles are, quantum states which for an irrep of the Poincaré group.

Now obviously once you have these states $\vert p, s\rangle$ you can immdiately define a creation operator:
$a^{*}_{p,s} \vert 0\rangle = \vert p, s\rangle$

Of course the irreps of the Poincaré group are labelled by momentum, so you are working in momentum space. If you want to work in position space rather than momentum space you must work with the (Lorentz invariant) Fourier transform of this operator. This Fourier transform turns out to be a free field basically.

So free fields are just the creation and annihilation operators for particles in position space.

Interacting fields are known, via the Källén–Lehmann spectral represntation or more rigorously Haag-Ruelle theory, to asymptote to free fields. Hence their states asymptote to free field states, i.e. particles. Although finite-time interacting field states are not particles and hence interacting fields are not composed of creation and annihilation operators.

17. May 22, 2013

### strangerep

M

Let's distinguish between (1) the quantum field representation of an elementary system, and (2) that of a composite system.

Case 1: A (fully relativistic) elementary system must correspond to a unitary irreducible representation ("unirep") of the Poincare group. However, there are no finite-dimensional unireps of the Lorentz group, and this means we need an infinite-dimensional unirep to describe the elementary system. (This contrasts with the non-rel QM case where we can get by with finite-dimensional unireps.)

An "infinite-dimensional unirep" can also be called a "field unirep". For an elementary system characterized by mass M and total spin 0, I'll write it in 4-momentum space as $\Phi_M(p)$, where $p$ is a 4-momentum vector constrained to the mass hyperboloid $p^2 = M^2$. I.e., $\Phi_M(p)=0$ off the mass hyperbloid. Thus, $\Phi_M$ has only 3 degrees of freedom, not 4, so if we Fourier-transform back to position-time space, we have a field with correlations (in general) between different spacetime points, expressed via a propagator. This must be so, since the field has only 3 degrees of freedom, not 4. (There's some other contraints, since we also want positive-energy representations only, but I'll omit those extra complications for now.)

Case 2: A (fully relativistic) composite system involving two non-interacting elementary systems. In ordinary non-rel QM, we already learned that composite systems must be described via a tensor product of Hilbert spaces, hence a tensor-product wavefunction, and not by a more complicated wavefunction that somehow lives in the familiar 3-position space. (This is a lesson learned from Hamiltonian dynamics where we describe N-particle systems using a 3N-dimensional phase space.) Therefore, in the (non-interacting) relativistic case, we construct a tensor product of two field representations, e.g., $\Phi_M(p_a)\Phi_M(p_b)$, where $a,b$ label the (independent) momentum spaces of the elementary systems, and the product should be suitably symmetrized (details omitted) since we're dealing with a spinless boson field here. Continuing this process, we construct the usual Fock space.

In the interacting case, we Fourier-transform back to a (product of) position-time spaces, impose the constraint that there's only 1 time parameter, and try to define a Hamiltonian on the resulting space such the interaction term is only non-zero for $x_a=x_b$ (i.e., a "local" interaction). Of course, this is not the way that interacting fields are usually introduced in QFT textbooks, which typically start from a Lagrangian, decompose the basic field into free modes, and construct a Fock (tensor product) space accordingly on which the full interacting Hamiltonian supposedly acts. One is then punished by various divergences, requiring regularization+renormalization as workarounds. Hence, more rigorous treatments of interacting fields use a more general framework.

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But anyway, the point I really wanted to make in this post is this: even for a single elementary (relativistic) system it's best to think in terms of a field unirep of the Poincare group, rather than the old notion of "particle". Hence, in the modern era, it's better to think in terms of "elementary" and "composite" field unireps instead of using the anachronistic terms "single-particle" and "multi-particle", even though describing a composite system involving interacting fields at finite times exactly is still an open problem. As to why the term 'particle' is still "dragging on the quantum field picture", well it's just the usual phenomenon of human inertia.

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Edit: The betting window is now open for how many ways my post will be misunderstood.

Last edited: May 22, 2013
18. May 23, 2013

### TrickyDicky

Thanks for this didactic reply, strangerep.

There is a quote by Max Born that I find deep and fitting here:"It would indeed be remarkable if Nature fortified herself against further advances in knowledge behind the analytical difficulties of the many-body problem."