S-matrix expansion for the $\phi^4$ theory and the interaction picture

Question

My question is about the perturbative expansion of the S-matrix using Dyson's expansion.

Let the Lagrangian density of the $\phi^4$ theory be \begin{equation} \mathcal{L} = \frac{1}{2}\left[\partial_\mu\phi(x)\right]^2 - \frac{m^2}{2}\phi(x)^2 - \frac{\lambda}{4!}\phi(x)^4, \end{equation} from which we find the Hamiltonian density \begin{equation} \mathcal{H}(x) = \underbrace{\frac{1}{2}\left[ \dot\phi^2 + (\nabla\phi)^2 +m^2\phi^2 \right]}_{\mathcal{H}_0} + \underbrace{\frac{\lambda}{4!}\phi^4}_{\mathcal{H}_1}. \end{equation} Now, to expand \begin{equation} S = T\mathrm{e}^{-i\int\mathrm{d}^4z\mathcal{H}_I(z)}, \end{equation} we clearly need to identify $\mathcal{H}_I$ in the interaction picture. This is where I start to get confused. The Hamiltonian $H_I$ is given by \begin{equation} H_I = \mathrm{e}^{iH_0t}H_1\mathrm{e}^{-iH_0t}, \end{equation} which reduces to $H_I=H_1$ only if $[H_0,H_1]=0$. However, all of the textbooks I've seen simply use $\mathcal{H}_1$ (as above) for $\mathcal{H}_I$ in computing the S-matrix.

Here is my question: It is not clear to me that $[H_0,H_1]=0$, which would surely require $[\mathcal{H}_0,\mathcal{H}_1]=0$ (?), which implies that $[\partial_\mu\phi,\phi]=0$. However, this latter commutator does not seem to give zero when the mode expansion is used. Furthermore, I am not entirely sure how the condition $[H_0,H_1]=0$ is inferred from the Hamiltonian density. Does one need to evaluate $[\mathcal{H}_0(x),\mathcal{H}_1(x)]$, IE at the same point in spacetime?

I know that this is a small point, but I hate not knowing the logic going from one step to the next, so if anyone could shed light on the justification for using $\mathcal{H}_1$ for $\mathcal{H}_I$, I would be very appreciative.

Answer 1

In general, $[H_0, H_1] \ne 0$ and $H_1 \ne H_I$. L&B describe the interaction picture in terms of state-vectors, which I'm not the biggest fan of, so we'll work from the point of the field operators instead. The fields in the interaction picture, $\phi_I$, are defined in terms of the Schrodinger-picture fields by $$ \phi_I(t, \vec{x})= e^{iH_0(t-t_0)} \phi_S(t_0, \vec{x})e^{-iH_0(t-t_0)} \tag{1} $$ In the Heisenberg picture, the fields evolve with the full Hamiltonian $H$: the above equation is defined using $H_0$ which is the first-order approximation to the full Hamiltonian when we expand around $\lambda = 0$, since we're working perturbatively. You shouldn't think of eq. $(1)$ as an equation for time evolution, but rather as a definition for convenience.

The mode expansion for $\phi_S$ is: $$ \phi_S(\vec{x})=\int\frac{d^3 k}{(2\pi)^3}\frac1{\sqrt{2\omega_k}}\left(a(\vec{k})e^{i\vec{k}\cdot\vec{x}}+a^\dagger(\vec{k})e^{-i\vec{k}\cdot\vec{x}}\right) $$ and consequently, $\phi_I$ is (note carefully the four-vector notation, in contrast to the previous equation) $$ \phi_I(x)=\int\frac{d^3 k}{(2\pi)^3}\frac1{\sqrt{2\omega_k}}\left(a(\vec{k})e^{-ikx}+a^\dagger(\vec{k})e^{ikx}\right) $$ where $x_0 = \Delta t = t - t_0$ and $k^\mu$ is on-shell, so $\phi_I$ enjoys a free mode expansion. It is these $\phi_I$ that the interaction Hamiltonian in the interaction picture (annoyingly, both these concepts have the same name) $H_I$ in the S-matrix is expanded in terms of, so while it looks similar visually to $H_1$ (which uses $\phi_S$), the interaction picture encodes some of the time-dependence and thus hides a lot of the details. When you set up the differential equation for $U(t, t_0)$: $$ i\frac{\partial U(t, t_0)}{\partial t} = H_I(t)U(t, t_0), $$

$H_I$ is defined by wedging $H_1$ in between the same $e^{\pm iH_0(t-t_0)}$ as the fields, so you can see that $$H_I = e^{iH_0(t-t_0)} H_1 e^{-iH_0(t-t_0)} = e^{iH_0(t-t_0)} \phi^4_S(t_0, \vec{x})e^{-iH_0(t-t_0)} \\= e^{iH_0(t-t_0)} \phi_S(t_0, \vec{x})e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)}\phi_Se^{-iH_0(t-t_0)}...\phi_Se^{-iH_0(t-t_0)} = \phi_I^4,$$

with the obvious integrals and prefactors.