Normal Ordering the $\phi^4$ interaction

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?

• I am not sure why you are dropping the terms that don't conserve particle number. Interactions create and destroy particles, that's one of the basic things in QFT. Can you explain yourself in a little bit more details? – Prof. Legolasov Oct 21 '14 at 2:57
• Because the scalar particles I am considering have lifetime more than the age of the universe, so any processes which cause decay can be neglected. – HuShu Oct 21 '14 at 3:39
• In a scalar field like what you wrote, the particles created by $a^{\dagger}_n$ are doomed to be annihilated by the quartic term $\phi^4$. This is no more than the usual phonon case: the number of phonon can not be conserved in the presence of anharmonic terms. – hyd Oct 21 '14 at 5:53
• In the lowest order there is no number violation and processes that we are talking about are not allowed due to conservation of energy and momentum, while in the higher order such processes do annihilate and conserve energy and momentum, but the rate of such processes is extremely small (time scale of decay is greater than the age of the universe) so can be neglected. – HuShu Oct 21 '14 at 17:56

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.

• I think you are right. But what I don't understand is that these terms in my case are not really infinite. The Kronecker delta makes the terms $a^{\dagger}_{\vec{n_{1}}}a_{\vec{n_{1}}}e^{-i (\vec{k}_{\vec{n_{1}}} - \vec{k}_{\vec{n_{2}}})}$ So then are we allowed to drop them? – HuShu Oct 21 '14 at 18:08
• The whole purpose of normal-ordering is dropping terms. Whether they are infinite or not. – Prof. Legolasov Oct 22 '14 at 1:09

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.