What does the ordering of creation/annihilation operators mean?

Question

When a system is expressed in terms of creation and annihilation operators for bosonic/fermionic modes, what exactly is the physical meaning of the order in which the operators act?

For example, for a fermionic system with states $i$ and $j$, $c_i c_j^\dagger$ is different from $c_j^\dagger c_i$ by a sign change, due to anticommutativity. I understand the mathematics of this, but what does it mean intuitively?

The former would be described as destroying a particle in state $j$ "before" creating one in state $i$, but what does "before" actually mean in this context, since there's no notion of time?

As another (bosonic) example, $a_i^\dagger a_i$ is clearly different from $a_ia_i^\dagger$, since acting the former on a vacuum state $|0\rangle$ gives zero while for the latter, $|0\rangle$ is an eigenstate, but again, what is the physical interpretation?

My normal interpretation of commutativity as a statement regarding the effect of a measurement on a state fails here since creation/annihilation are obviously not observables.

I hope the question makes sense and isn't too abstract!

Answer 1

OP is basically asking for an intuitive understanding of operator ordering. Well, the quantum world is something that us Earthlings notoriously do not understand well. Often we start with a classical model with commuting quantities. When we next want to quantize the model, we at first do not know which way we should order the corresponding non-commuting quantum operators.

Say for simplicity that the classical Hamiltonian $H=AB$ is a product of two classical quantities $A$ and $B$. And say that the corresponding two quantum operators $\hat{A}$ and $\hat{B}$ have a c-number commutator $[\hat{A},\hat{B}]~\propto~ \hbar{\bf 1}$.

There are initially many ways to choose an operator ordering and to choose a representation (ket-space), which the quantum operators act on. Say that we have chosen a specific notion of ordering that we call normal ordering $:\hat{A}\hat{B}:$, and say that we have chosen a notion of Fock space vacuum. To parametrize our ignorance we now introduce a c-number parameter $c$, and define the quantum Hamiltonian as

$$\hat{H}~=~:\hat{A}\hat{B}:~+~\hbar c{\bf 1}.$$

In this way, if we have made a wrong choice by normal ordering the operators, we can always absorb the error into the definition of the c-number parameter $c$.

One can often limit the possible choices of $c$ further by demanding Hermiticity of $\hat{H}$ and imposing other physical requirements. For instance, in (Bosonic) string theory, a similar so-called intercept parameter $a$ is completely fixed by consistency requirements (Lorentz symmetry in the light-cone formulation; nilpotency of the BRST charge in the covariant formulation), see chapter 2 and 3 in Green, Schwarz and Witten, "Superstring theory", vol. 1.

A similar story holds for Fermionic operators.

Answer 2

Because the OP asks for an interpretation of normal ordering, I would like to remark that normal ordering in QFT is the same then substracting the vacuum expectation value (which I guess is sometimes called Wick product).

Namely, if you have a free relativistic quantum field (bosonic or fermionic does not make a real difference) $\phi(x)$ which splits in a creation an annihilation part $\phi_+$ and $\phi_-$ it is easy to check using $\phi_-(x)\Omega=0$ and some formal calculation that actually $$:\phi(x)\phi(y): = \phi(x)\phi(y)- (\Omega,\phi(x)\phi(y)\Omega)$$

Note also that $\phi^2(x):=\lim_{y\to x}:\phi(x)\phi(y):$ (or a general Wick polynomial) defines a new Wightman field which lies in the same "Borchers class" i.e. is relatively local to $\phi$, while the normal square would just give infinity and makes no sense.

Answer 3

Products of these operators describe transitions of particles from one state to another, nothing else. QFT is about evolution of populations of particle states due to interactions. Consider a regular potential theory of scattering of non-relativistic particles (with no creation/annihilation operators) and then rewrite it formally with help of those operators to see what happens. Keep in mind that only the whole expression makes sense.

EDIT: In a usual description the particle momentum changes in course of scattering: $p = p(t)$. In the operator description the particle with $p(t_1)$ disappears and the particle with $p(t_1+dt)$ appears in course of interaction.