Normal Ordering the $\phi^4$ interaction

Question

I am trying to quantize the quartic potential $(\lambda/4!)\phi^{4}$ in a box of side length $L$, with periodic boundary conditions. I have expanded the field

$$\phi = \sum \limits_{\vec{n}} \exp(i \vec{k}_{\vec{n}} \cdot \vec{x}) a_{\vec{n}}(t)+ h.c.$$

where $\vec{k}_{\vec{n}}=(2\pi \vec{n}/L)$, $a_{\vec{n}}(t)$ are the annihilation operator for state $\vec{n}$. On substituting this in $(\phi^{\dagger}\phi)^{2}$, I get many terms, and dropping all the terms that don't conserve particle number, I am left with terms like

$$aa^{\dagger}aa^{\dagger}, \quad aaa^{\dagger}a^{\dagger}$$

etc. I am trying to normal order them so that I end up getting the non relativistic result i.e.

$$\sum\limits_{\vec{n_{1}},\vec{n_{2}},\vec{n_{3}},\vec{n_{4}}}\frac{-\lambda}{4m^{2}V}\delta_{\vec{n_{1}}+\vec{n_{2}},\vec{n_{3}}+\vec{n_{4}}}a_{\vec{n_{1}}}^{\dagger}a_{\vec{n_{2}}}^{\dagger}a_{\vec{n_{3}}}a_{\vec{n_{4}}} \, .$$

My problem is when I try to normal order terms like $aa^{\dagger}aa^{\dagger}$, I get terms like $a_{\vec{n_{1}}}^{\dagger}a_{\vec{n_{2}}}\delta_{\vec{n_{3}},\vec{n_{4}}}$ in the process of normal ordering (due to equal time commutation relation). What happens to these terms, why don't they show up in the final answer? Is it because they are particle number operator and we drop them?

Answer 1

User Hindsight has already given a correct answer, but let us here try in different words.

I) Given the classical field

$$\tag{1} \phi~=~\phi_{-}+\phi_{+},$$

let us assume that the quantum field is of the form

$$\tag{2} \hat{\phi}~=~\hat{\phi}_{-}+\hat{\phi}_{+},$$

where the creation/annihilation parts $\hat{\phi}_{\pm}$ do not commute.

II) Our next problem is

How should we translate the classical interaction term $$\tag{3} \phi^4=(\phi_{-}+\phi_{+})^4~=~\phi_{-}^4 +4\phi_{-}^3\phi_{+}+6\phi_{-}^2\phi_{+}^2+4\phi_{-}\phi_{+}^3 +\phi_{+}^4 $$ into a quantum operator $\hat{V}$?

There are a priori many choices for $\hat{V}$. To be democratic, we could introduce parameters in $\hat{V}$ to parametrize all the different operator ordering possibilities consistent with the classical interaction term (3).

III) Transcribing the $\hat{V}$ operator into normal ordered form would then redistribute terms and parameters, but not fix any parameters.

IV) The parameters are next partially or fully fixed by other requirements in order to obtain a consistent theory. See e.g. my Phys.SE answer here for a simple example.

Answer 2

As far as I know, the very idea of normal-ordering implies making modifications to the quantity.

For example, consider the hamiltonian of the free real field.

$$ H = \int \frac {d^3 p} {(2 \pi)^3} \frac{\omega_p}{2} \left( a_p a^{\dagger}_p + a^{\dagger}_p a_p \right) = \int \frac {d^3 p} {(2 \pi)^3} \omega_p \left(a^{\dagger}_p a_p + \frac{1}{2} \delta^3(p) \right). $$

Note that this is not the normal-ordered Hamiltonian yet. All that I've done is used the commutation relations to write it in a nicer way.

Now the normal-ordering comes: we modify the Haimltonian by dropping the infinite constant term.

In your case this implies that terms like $a^{\dagger} a \delta$ should be present in the original result, but by definition are dropped from the normal-ordered one.