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

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 $$$$\mathcal{L} = \frac{1}{2}\left[\partial_\mu\phi(x)\right]^2 - \frac{m^2}{2}\phi(x)^2 - \frac{\lambda}{4!}\phi(x)^4,$$$$ from which we find the Hamiltonian density $$$$\mathcal{H}(x) = \underbrace{\frac{1}{2}\left[ \dot\phi^2 + (\nabla\phi)^2 +m^2\phi^2 \right]}_{\mathcal{H}_0} + \underbrace{\frac{\lambda}{4!}\phi^4}_{\mathcal{H}_1}.$$$$ Now, to expand $$$$S = T\mathrm{e}^{-i\int\mathrm{d}^4z\mathcal{H}_I(z)},$$$$ we clearly need to identify $$\mathcal{H}_I$$ in the interaction picture. This is where I start to get confused. The Hamiltonian $$H_I$$ is given by $$$$H_I = \mathrm{e}^{iH_0t}H_1\mathrm{e}^{-iH_0t},$$$$ which reduces to $$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.

• "All the textbooks I've seen simply use $\mathcal{H}_1$ for $\mathcal{H}_I$ in the S-matrix" - this isn't exactly right. $\mathcal{H}_I$ looks similar to $\mathcal{H}_1$, but it uses the interaction picture fields rather than the Schrodinger picture fields Jan 13, 2021 at 10:21
• You're right - I shouldn't generalise! I refer mostly to Lancaster & Blundell Ch 19. Jan 13, 2021 at 11:51
• I actually guessed that you did, L&B are a bit sloppy with their notation for fields Jan 13, 2021 at 11:57

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 $$H_I = e^{iH_0(t-t_0)} H_1 e^{-iH_0(t-t_0)} = e^{iH_0(t-t_0)} \phi^4_S(t_0, \vec{x})e^{-iH_0(t-t_0)} \\= e^{iH_0(t-t_0)} \phi_S(t_0, \vec{x})e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)}\phi_Se^{-iH_0(t-t_0)}...\phi_Se^{-iH_0(t-t_0)} = \phi_I^4,$$