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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!

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3 Answers 3

up vote 6 down vote accepted

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 $[\hat{A},\hat{B}]=\hbar{\bf 1}$.

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.

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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.

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I understand that they describe transitions of particles, but this to me doesn't suggest that the order of the transitions should make any difference, or indeed what that should even mean in the absence of time. Is it to do with the details of the interactions? –  Will Vousden Jun 5 '11 at 22:44
    
No, it has nothing to do with details of interactions. It is just about how we technically change the populations of states. As I said, only the whole expression makes sense, not the order of particular operators, not the color of inks we use. –  Vladimir Kalitvianski Jun 6 '11 at 8:49

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.

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