Consider the Lagrangian density:

$$\mathcal{L} = \mathcal{L}_0 - \frac{\lambda}{4!}\left(\vec{\phi}^2\right)^2$$

Where $\mathcal{L}_0$ contains the standard mass and kinetic terms for $N$ scalar fields. This Lagrangian is invariant under $\mathrm{SO(N)}$ rotations.

I am trying to show the following identity for the two point correlation function: $$S_{ad}\langle T\phi_d(x)\phi_c(y)\rangle S_{bc} = \langle T\phi_a(x)\phi_b(y)\rangle$$ Where $S$ is a group element of $\mathrm{SO(N)}$. My progress so far:

I wrote $S$ in its infinitesimal form: $S_{ad} = \delta_{ad} + \epsilon~ T_{ad}$, with $\epsilon$ being an infinitesimal parameter and $T_{ad}$ an anti symmetric matrix.

Doing this for the two S elements in the expression yields four terms: 1 vanishes as it has a second order infinitesimal and another is the desired result, which means that the proof boils down to showing that the remaining two vanish:

$$T_{bc}\langle T\phi_a(x)\phi_c(y)\rangle + T_{ad}\langle T\phi_d(x)\phi_b(y)\rangle = 0$$

From here I'm unsure of how to proceed. If I write the expression for the correlation functions explicitly I can put all the terms in the same integral, and requiring the integrand to be 0 yields the condition:

$$T_{bc}\phi_a(x)\phi_c(y) + T_{ad}\phi_d(x)\phi_b(y) = 0$$

But I still don't know how to show this. Plus, I don't know if this is the right way, as maybe it is possible to show that the integral is 0 without the integrand being 0?

I'd appreciate any help!

EDIT: Forgot to mention this, but the integration measure in the path integral is assumed to be invariant.


1 Answer 1


I believe that the most elegant way to show the above identity is using the path integral formula for the correlators, as follows (I'll omit the time ordering as it's automatic in this formalism).

Let $S\in SO(N)$, then the fields transform as $\vec{\phi'}(x)=S\vec{\phi}(x)$, or in components $\phi_a'(x)=S_{ab}\phi_b(x)$. Now we expand the left side:

$\displaystyle S_{ad}\langle\phi_d(x)\phi_c(y)\rangle S_{bc}=S_{ad}\frac{\int \mathcal{D}\vec{\phi}\;\phi_d(x)\phi_c(y)\;\exp{iS[\vec{\phi}]}}{\int \mathcal{D}\vec{\phi}\;\exp{iS[\vec{\phi}]}}S_{bc}$

We can put the $S$ factors inside the integral as they're $\phi$ independent, and use the transformation $\phi_a'(x)=S_{ab}\phi_b(x)$ to obtain:

$\displaystyle \frac{\int \mathcal{D}\vec{\phi}\;\phi_a'(x)\phi_b'(y)\;\exp{iS[\vec{\phi}]}}{\int \mathcal{D}\vec{\phi}\;\exp{iS[\vec{\phi}]}}$

The integration measure transforms as $\mathcal{D}\vec{\phi}=f(S)\mathcal{D}\vec{\phi'}$, where $f(S)$ just depends on $S$, and the action is by construction invariant: $S[\vec{\phi}]=S[\vec{\phi'}]$. The function $f$ is both in the numerator and the denominator so it cancels, and we are left with:

EDIT: The integration measure transforms as $\mathcal{D}\vec{\phi}=|J(\phi',\phi)|\mathcal{D}\vec{\phi'}=|\det S^{-1}|\vec{\phi'}$ just like in regular integration, but as $S\in SO(N)$ the determinant is 1, and the action is by construction invariant: $S[\vec{\phi}]=S[\vec{\phi'}]$. Thus we are left with:

$\displaystyle \frac{\int \mathcal{D}\vec{\phi'}\;\phi_a'(x)\phi_b'(y)\;\exp{iS[\vec{\phi'}]}}{\int \mathcal{D}\vec{\phi'}\;\exp{iS[\vec{\phi'}]}}$

However, the primes are redundant as they're just integration variables, hence we can relabel $\phi'\rightarrow\phi$ and then we have:

$\displaystyle \frac{\int \mathcal{D}\vec{\phi}\;\phi_a(x)\phi_b(y)\;\exp{iS[\vec{\phi}]}}{\int \mathcal{D}\vec{\phi}\;\exp{iS[\vec{\phi}]}}$

but this is by definition just $\displaystyle \langle\phi_a(x)\phi_b(y)\rangle$, thus completing the proof.

  • $\begingroup$ Thanks for the answer, it seems really simple now. I have just one question: why does the integration measure transform just as a function of S? It doesn't actually matter for this case, as I'm told the integration measure is invariant (should probably put that in the question), but I'd still like to know. $\endgroup$
    – StackUFS
    Commented Dec 3, 2016 at 15:52
  • $\begingroup$ Ah, right, my bad, haven't thought that part through: it's not a function of $S$, but rather the measure must transform as $\mathcal{D}\vec{\phi'}=|J(\phi',\phi)|\mathcal{D}\vec{\phi}$ just like in ordinary integration, but as the Jacobian determinant is just the determinant of $S$, it's just 1 as $S\in SO(N)$, and the measure is indeed invariant. $\endgroup$
    – blueshift
    Commented Dec 3, 2016 at 16:51

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