# $S$-matrix from interacting picture

I’ve been reading a lot about the interaction picture, and I’m trying to string the ideas behind it together.

Essentially, the goal is to calculate something like $$$$. We can use states because we assume that they are free in asymptotic times, so they are eigenstates of the free Hamiltonian.

To calculate that quantity, we need to time evolve $$|i(-\infty)>$$ to $$|i(\infty)>$$ and then compare it with $$|f(\infty)>$$. We do that with the unitary operator $$U(-\infty, \infty) = Te^{i\int_{-\infty}^{\infty}H_{interaction}dt}.$$

$$H_{interaction}$$ is the interaction part of the Hamiltonian. It is a function of fields, and we plug in the free field expressions.

Now here’s the key parts that don’t make sense to me.

1. Shouldn’t we use the full Hamiltonian for the time evolution operator? Why do we only have the interacting part in the exponential?

2. Since $$\phi$$ (the interacting field), is complicated, we can try to write it in terms of the free field with $$U^{\dagger}\phi U$$ where this time $$U$$ contains the full Hamiltonian. How do we go from this relation to plugging in free fields into $$H_{interaction}$$?

I apologize in advance if this question seems unclear or redundant compared to some similar questions asked on the site. But it seems to me that answers to questions on the interacting picture just show how to get there, instead of how to connect the ideas.

Without going into Haag's theorem, I will briefly comment on why we use the interaction picture.

What we want to calculate is n-point functions: $$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle,$$ where $$\phi_H$$ is the field operator of Heisenberg representation and satisfies the EoM of the interacting theory. To calculate these n-point function, it is useful to define the partition function:

$$Z=\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_H(x)\Big)|0\rangle,$$ $$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle:=\frac{\delta^n Z[j]}{i\delta j(x_1)\cdots i\delta j(x_n)}.$$ So what we should do is replaced to calculate the partition function $$Z$$. Also, we are interested in perturbative calculations, we must separate interacting terms from free terms so that we can use a perturbative expansion. However it is quite nontrivial how to treat $$Z$$ perturbatively because the interaction term and the free term are not separated explicitly, we do not know how to handle them in a perturbative calculation.

Therefore, we split the Hamiltonian as follows, $$H=H_0+H_{\mathrm{int}}$$ with $$H_0$$ carrying the time evolution of the field operators and $$H_{\mathrm{int}}$$ carrying the time evolution of the states. Under such a division, the operator in the Heisenberg representation and the operator in the interaction representation have the following relationship: $$\phi_H(x)=U^\dagger(t,-\infty)\phi_I(t,\mathbf{x})U(t,-\infty),$$ $$U(t_1,t_2):=T\exp\Big(-i\int_{t_1}^{t_2}dt\ H_{\mathrm{int}}\Big),$$ $$U^\dagger(t_1,t_2)=U(t_2,t_1),$$ $$U(t_1,t_2)U(t_2,t_3)=U(t_1,t_3),$$ where we assume that $$\phi_I$$ and $$\phi_H$$ coincide at the far past $$t\to -\infty$$. (This coincide to choose a reference time to be $$t_0=-\infty$$.) This is natural assumption in the scattering theory because particles will be well-separated at far past.

Using the fierst equation above, we can rewrite the n-point functions as

$$\langle 0|T\phi_H(x_1)\cdots\phi_H(x_n)|0\rangle=U(-\infty,+\infty) \langle 0|T\phi_I(x_1)\cdots\phi_I(x_n) \exp\Big(i\int_{-\infty}^{\infty}dt\int d^3 x\ \mathcal{L}_{\mathrm{int}}(\phi_I(x))\Big) |0\rangle,$$ where we define the Lagrangian density as follows $$\mathcal{L}_{\mathrm{int}}:=-\mathcal{H}_{\mathrm{int}}.$$

Thus we don’t need to use the complicated expression $$Z=\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_H(x)\Big)|0\rangle,$$ but all we have to do is to evaluate $$Z= \langle 0|U(-\infty,+\infty) T\exp\Big(i\int d^4 x\ (j(x)\phi_I(x)+\mathcal{L}_{\mathrm{int}}(\phi_I(x)))\Big) |0\rangle$$ $$=\exp\Big(i\int d^4x \mathcal{L}_{\mathrm{int}}(\frac{\delta}{i\delta j(x)}) \Big) \langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$$

This expression is useful because it nicely separates the free and interacting parts. This is the reason for introducing the interaction picture. Also, since $$\phi_I$$ satisfies the free-field equation of motion, this allows $$\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$$ to be rewritten by taking the appropriate normal order as

$$\langle0|T:\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big):|0\rangle=\exp\Big(-\frac{1}{2}\int d^4x d^4y \ j(x)j(y)\Delta_0(x-y)\Big),$$ where $$\Delta_0(x-y)$$ is a free propagator.

This allows us to understand that the derivative with respect to $$j$$ contained in the interaction term acts one after the other on $$\langle0|T\exp\Big(i\int d^4x\ j(x)\phi_I(x)\Big)|0\rangle$$, thereby producing a free-field propagator. This is another simplification that arises because $$\phi_I$$ satisfies the free-field equation of motion.

In other words, the interaction picture is one prescription for successfully separating the free-field contribution from the interaction contribution for perturbation calculations.