5
$\begingroup$

Consider a scalar field theory with a $\phi^{4}$ interaction term $$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4},$$ where $\lambda\ll 1$.

I am confused about the following statement, given in p.49 of David Tong's QFT notes:

'We can get a hint for what the effects of this extra term will be. Expanding out $\phi^{4}$ in terms of $a_{\mathbf{p}}$ and $a^{\dagger}_{\mathbf{p}}$, we see a sum of interactions that look like $a^{\dagger}_{\mathbf{p}}a^{\dagger}_{\mathbf{p}}a^{\dagger}_{\mathbf{p}}a^{\dagger}_{\mathbf{p}}$ and $a^{\dagger}_{\mathbf{p}}a^{\dagger}_{\mathbf{p}}a^{\dagger}_{\mathbf{p}}a_{\mathbf{p}}$ etc. These will create and destroy particles.'

My questions are:

  1. In the free field theory, we can expand $\phi$ in terms of $a_{\mathbf{p}}$, $a^{\dagger}_{\mathbf{p}}$ as $$\phi(\mathbf{x},t)=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}(a_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}+a^{\dagger}_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}).$$ However, once the interaction has been added, this solution is no longer correct (the Klein-Gordon EOM has an extra non-linear term). So why do we use this to expand out the interaction term?

  2. How do we actually do the expansion? Where have the integrals gone?

  3. How do those terms create and destroy particles any more so than the quadratic terms from $\phi^{2}$ would?

$\endgroup$
3
$\begingroup$

I think, the sentence you are citing is meant to give a rough idea about the intuitive meaning of the $\phi^4$ term, not to be verified computationally.

  1. The solution of the free theory (without the interaction term) is what defines things like "particles" and propagators. $a^\dagger$ as defined in the free theory's expansion creates a free particle. Usually, if you consider a scattering process, before the actual scattering occurs the particles are considered free.

  2. They are still there. Usually you do the expansion not in the Lagrangian but when you write down a scattering amplitude. Then contractions of creator and annihilator operators will lead to commutators which will produce delta distributions which consume the momentum integrals.

  3. Not sure if I understand the question correctly but if you have a look at the Feynman rules for the different terms in the Lagrangian, you will see that a $\phi^4$ term corresponds to a vertex with four legs while the free Lagrangian (which is quadratic in the field) corresponds to a propagator with two legs. So the number of legs in a Feynman rule corresponds to the power of the field in the corresponding term.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Let's look at Tong's comment in context, by which I mean start reading Chapter 3 from the start.

He's discussing interacting fields, and begins by writing a Lagrangian that features arbitrarily high powers of $\phi$. However, since in a $4$-dimensional spacetime the energy dimension of $\lambda_n$ is $4-n$, it follows that $\lambda_n$ with $n>4$ are supressed at low energies, viz. $\lambda_n=g_n\Lambda^{4-n}$ with $\Lambda$ a new-physics energy scale. Thus at an energy $E\ll \Lambda$ we get $\lambda_n\propto \left(\frac{E}{\Lambda}\right)^{n-4}$, which is small for $n>4$. This observation motivates a classification of coefficients as relevant, marginal or irrelevant if their energy dimensions are respectively positive, zero or negative. (The term "relevant" means "dominant at the scales we've probed"; "marginal" means "scale-independent, thus dominant at neither high nor low energies".) Tong notes that the fact only finitely many terms are relevant or marginal simplifies QFT.

Then we get to the comment you asked about, in which he discusses how several theories ($\phi^4$ being the first) behave under small perturbations. He's thrown away irrelevant couplings, but $\phi^4$ has been retained, and is the only such term that's not also present in the case of free fields. We recover the free-field case as $\lambda_4\to 0$, so the $\phi^4$ theory with small $\lambda_4$ is a perturbation around the theory of a free field, making your integral representation of $\phi$ approximately valid. The exact expression for $\phi$ therefore adds on some $\lambda_4$-dependent terms that, in comparison to the original integral, are quite small. It is therefore still reasonable to think of what happens to ladder-operator monomials' matrix elements.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$
  1. You switch to the quantum mechanical 'interaction picture', in which operators are given in terms of static Schrodinger operators as ($H_0$ is the free Hamiltonian, $H_I$ is the interaction term, $H=H_0+H_I$ is the full Hamiltonian):

$$ A_I = e^{iH_0t}A_Se^{-iH_0t} $$

and states are given in terms of static Heisenberg states as:

$$ |\psi_I(t)\rangle = e^{iH_0t}e^{-i(H_0+H_I)t}|\psi_H\rangle. $$

You can then show that $U_{int} \equiv e^{iH_0t}e^{-i(H_0+H_I)t}$ is given by a Dyson series expansion in $H_I$:

$$ U_{int} = \exp\left(-i\int e^{iH_0t}H_Ie^{-iH_0t}\ dt\right) $$

and we generally say $H_{int} \equiv e^{iH_0t}H_Ie^{-iH_0t}$. It's in Section 3.1 of David Tong's notes, but Section 4.2 of Peskin and Schroeder has a more detailed discussion I found useful.

  1. You expand out $U_{int}$ as a 'Dyson series':

$$ U_{int} = 1 - i\int dt' H_{int}(t') + (-i)^2 \int dt' \int dt'' H_{int}(t') H_{int}(t'') + \cdots $$

The $n$th term in this series corresponds to a family of Feynman diagrams all with $n$ vertices.

  1. They don't, they just result in a non-linear equation of motion. $\phi^2$ terms give linear terms in the Klein Gordon equation and so you can just look at the plane wave solutions (and you just find that these $\phi^2$ terms essentially give rise to a mass term in the dispersion relation). If you want to study more complex behaviour that we generally call 'interaction', you need higher powers.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.