# Vacuum matrix elements

On page 87, section 7.2.3 titled Vacuum matrix elements of Quantum Field Theory and the Standard Model by Matthew Schwartz, the author writes that the vacuum state $$|\Omega>$$ is annihilated by the operators $$a_p(t)$$ of the interacting theory at time $$t = -\infty$$.

1. Then he says "to relate this to a state for which we know how the free-field creation and annihilation operators act, we need to evolve it to the reference frame $$t_0$$ where the free and interacting pictures are taken equal." Now I don't understand why must that be done as we already know that $$a_p(t)$$ annihilates $$|\Omega>$$ at time $$t= -\infty$$. Kindly explain this statement to me.

2. Next, we evolve the state to time $$t_0$$ using the operator $$S(t, t_0)$$. I thought that $$S(t, t_0)$$ takes the state at time $$t_0$$ to time $$t$$ and not the other way around. Can you please explain to me why this must be so?

3. Going along the same lines, I thought that the state $$S(t, t_0)|\Omega>$$ is annihilated by the operator $$a_p(t)$$ at time $$t= -\infty$$ and not $$a_p(t_0)$$ as written in the book.

Here is a snap from the section that I am talking about for reference.

• IMHO the way he explains is a bit confusing. I think a better description is done by properly introducing the scattering problem in terms of the so-calle Møller operators. This is done in Weinberg's The Quantum Theory of Fields Volume 1 Chapter 3 and also in Duncan's The Conceptual Framework of Quantum Field Theory. Of course this is my opinion, maybe some people will think the opposite and find this approach easier to understand.
– Gold
Oct 29, 2022 at 15:08
• @Gold Thank you for your reply. I will look into the texts that you have suggested and hopefully will get a better understanding of what is going on. Oct 29, 2022 at 16:59

I think the way to understand this is to go back to his treatment of the LSZ reduction formula. To reiterate the basics, what we want to compute is the amplitude for our state $$|\psi(t)\rangle$$ to be in some state $$|f\rangle$$ at some time $$t$$, i.e. $$\langle f|\psi(t)\rangle=\langle f|S(t, t_0)|\psi(t_0)\rangle,$$ where $$S(t, t_0)$$ is the time-evolution operator. To make life easy, we assume that the initial and final states are far removed from interactions in the far past and far future, respectively, and are multi-particle states of definite momentum. Let's try to construct these states for $$2\to n$$ scattering. Remember that we are working in the Schrodinger picture here (hence the time-evolution operator), so operators are time-independent. Our initial and final states are (note the differences to Schwartz) \begin{align*} |\psi(-\infty)\rangle&=\sqrt{2\omega_1}\sqrt{2\omega_2}\,a_{p_1}^\dagger a_{p_1}^\dagger|\Omega(-\infty)\rangle=|p_1,p_2\rangle\\|f\rangle&=\sqrt{2\omega_3}\cdots\sqrt{2\omega_n}\,a_{p_3}^\dagger\cdots a_{p_n}^\dagger|\Omega(\infty)\rangle=|p_3,\dots,p_n\rangle \end{align*} ($$|\psi(-\infty)\rangle\equiv|i\rangle$$), giving the amplitude $$\langle p_3,\dots,p_n|S(\infty, -\infty)|p_1,p_2\rangle=2^{n/2}\sqrt{\omega_1\cdots\omega_n}\langle\Omega(\infty)|\,a_{p_3}\cdots a_{p_n}S(\infty,-\infty)a_{p_1}^\dagger a_{p_2}^\dagger|\Omega(-\infty)\rangle.$$ The claim here is that, while the vacuum at asymptotic times is still technically the interacting vacuum, we are far enough removed from interactions that acting with the free-theory creation operators creates genuine single-particle states of definite momenta. Since $$a_p=a_p(t_0)$$, the annihilation operator at the reference time $$t_0$$ (when the Schrodinger and Heisenberg pictures are equal), it follows that $$\boldsymbol{a_p(t_0)}$$ annihilates $$\boldsymbol{|\Omega(\pm\infty)\rangle}$$.
Now let's move to the Heisenberg picture. In terms of the time evolved $$a$$'s, $$a_p=a_p(t_0)=S^\dagger(t_0,\infty)a_p(\infty)S(t_0,\infty).$$ Applying this to the vacuum then gives \begin{align*}a_p(t_0)|\Omega(\infty)\rangle&=S^\dagger(t_0,\infty)a_p(\infty)S(t_0,\infty)|\Omega(\infty)\rangle\\&=S^\dagger(t_0,\infty)a_p(\infty)|\Omega(t_0)\rangle,\end{align*} and similarly for $$-\infty$$. So we find that $$\boldsymbol{a_p(\pm\infty)}$$ annihilates $$\boldsymbol{|\Omega(t_0)\rangle}$$, the fixed vacuum of the Heisenberg picture that Schwartz simply refers to as $$|\Omega\rangle$$.