I am reading the QFT book by M. Schwartz. More specifically, I have issues with the section about LSZ. I am puzzled with the way the creation and annihilation operators from the free theory act there.

Harmonic oscillator

To explain my problems, let me compare the QFT case with the usual harmonic oscillator deformed with interaction terms. For the harmonic oscillator one has the lowest-energy state, which can be referred to as the vacuum state $|0\rangle$. Then there is a whole other tower of energy eigenstates and the convenient way to generate them is to use the creation and the annihilation operators $a$ and $a^\dagger$. These are defined in terms of $x$ and $p$ by the well-known formulas. In particular, one has $a|0\rangle =0$ and higher energy eigenstates are generated from $|0\rangle$ by applying the creation operator.

Let us now add a small correction to the Hamiltonian. First of all, to the best of my understanding the Hilbert space remains unchanged: it is spanned by square-integrable functions. So, we can, in principle, use the previous basis for the Hilbert space -- $|0\rangle$, $a^+|0\rangle$, etc. The problem is that it is not very useful as in most cases it does not diagonalise the new Hamiltonian.

If we are interested in diagonalising the Hamiltonian, we may start from the ground state. In most cases one can only express it in terms of the previous basis perturbatively. We will not care about this relation and just denote it $|\Omega \rangle$. There are also other energy eigenstates $|\Omega_1\rangle$, $|\Omega_2\rangle$ and so on. To the best of my understanding, in general, \begin{equation} a|\Omega\rangle \ne 0, \qquad a^\dagger|\Omega \rangle \ne |\Omega_1\rangle, \qquad \dots . \qquad (1) \end{equation} Here by $a$ and $a^\dagger$ I mean just the very same creation and annihilation operators as for harmonic oscillator before we introduced interactions.

In fact, if we assume that (1) is correct (in the sense of replacing inequalities with equalities), we can realise $a$ and $a^\dagger$, say, in the coordinate representation and find that the equations defining the energy eigenstates in the harmonic and in the deformed cases are the same. So these are actually the same states \begin{equation} |\Omega \rangle = |0 \rangle, \qquad |1 \rangle = |\Omega_1\rangle, \qquad \dots ! \end{equation} Thus the only effect of deforming the Hamiltonian was that energies of eigenstates have changed. This is true only for a very limited set of deformations, more specifically, $[H,H_{int}]=0$.

QFT case

Now we go back to the QFT case. When switching on interactions one says that the vacuum state has changed from $|0\rangle$ to $|\Omega\rangle$. But then (here I mean the book by Schwartz) one uses that \begin{equation} a_p |\Omega\rangle=0 \qquad (2) \end{equation} and that the single-particle states are \begin{equation} |p\rangle = a^\dagger_p |\Omega\rangle. \qquad (3) \end{equation} Similar things are done with asymptotic multi-particle states as well. In these formulas $a_p$ and $a_p^\dagger$ are the same as in the free theory (they evolve in time with different exponents, but this is due to the fact that they may have different energies, while in the Schroedinger picture these are just the free theory operators).

As I illustrated above with the quantum mechanical case, I do not think that (2) and (3) are true.

How things work, to the best of my understanding

There should be a ground energy state $|\Omega\rangle$ in interacting QFT, which is, in general, different from $|0\rangle$. It is not annihilated by $a_p$ from the free theory. There are still exact energy eigenstates, which are analogous to $|\Omega_1\rangle$. These are the single-particle states $|p^{int}\rangle$ associated with stable particles. Indeed, considering that these are stable and have no other particles to interact with (it is a single-particle state), they just travel in time with energy defined by their effective mass and spatial momentum.

Then it gets more tricky. One also defines states $|p_1^{int},p_2^{int}\rangle$, etc. I do not know how this thing is defined exactly, but, apparently, it is something like: if interactions between these particles could have been ignored (e.g. when they are far away), it would be a two-particle state. It is also not an exact eigenstate of energy, unless interactions between particles are ignored.

Assuming that we have defined $|\Omega\rangle$, $|p^{int}\rangle$, $|p_1^{int},p_2^{int}\rangle$, etc, one can think of a way of defining $a_p$ and $a^\dagger_p$, which connect these states as in the free theory. But these $a_p$ and $a^\dagger_p$ are not at all the same as in the free theory.


  1. Am I write that (2) and (3) is not, strictly speaking, correct?

  2. Is what I wrote in the last part correct?

  3. How does one define $a_p$ and $a^\dagger_p$ in the interacting theory more precisely? Are these just defined by their natural action on the asymptotic states?

  4. Can anyone add anything more specific on how the asymptotic states are defined (except single-particle states, which I believe, I understand how to define)? I guess, if particles are well separated, one can safely think that their interaction is negligible. However, naively a state with definite momentum is evenly smeared over the whole space. How can one separate such states spatially? Why do people say that $t=\pm \infty$ when speak of asymptotic states? Why can't one separate particles at finite time?

  • $\begingroup$ There is a closely related discussion here physics.stackexchange.com/q/210119 I do not, however, find the top-voted answer there completely convincing. Is there any reference on textbooks or peer-reviewed sources? $\endgroup$
    – Dr.Yoma
    Commented Mar 9, 2023 at 18:02

1 Answer 1


Indeed, (2) and (3) are not correct, if $a$ and $a^\dagger$ are understood as in the free theory. Besides the arguments presented with the question, one can also look into the discussion of the Kallen-Lehmann representation (following, say, the book by Peskin and Schroeder). In that discussion one encounters \begin{equation} \langle \Omega|\phi(0)|\lambda\rangle, \end{equation} where $\Omega$ is the interacting vacuum, $\phi(0)$ is the same field operator as in the free theory and $\lambda$ is any of the asymptotic states in the interacting theory. In particular, $|\lambda\rangle$ include the vacuum itself, the one-particle asymptotic states $|p\rangle$ of the interacting theory, the two-particle asymptotic states, etc. From explicit computations of $\langle \Omega| T\{\phi(x)\phi(y)\}|\Omega\rangle$ one knows that all \begin{equation} \langle\Omega|\phi(0)|p\rangle, \qquad \langle\Omega|\phi(0)|p_1,p_2\rangle, \qquad \dots \end{equation} are typically non-vanishing. This means that $\langle\Omega|\phi(0)$, which is a superposition of $\langle \Omega|a$ and $\langle \Omega|a^\dagger$ with $a$ and $a^\dagger$ from the free theory produce all possible multi-particle state in the free theory. Thus, (2) and (3) are not correct.

Still, there is a way to define $a$ and $a^\dagger$ in the interacting theory so that (2) and (3) are true. This discussion can be found e.g. in the book by Weinberg (cluster decomposition section). The idea is very simple. We have the Hilbert space of the asymptotic states of the interacting theory and we introduce interacting $a$ and $a^\dagger$ so that they act in the same way as in the free theory. Since, by this we define the action of $a$ and $a^\dagger$ on any state in the Hilbert space this is equivalent to saying that we have defined these operators. The problem, however, is that these $a$ and $a^\dagger$ are not connected in a simple way to the field that enters the Lagrangian (the field operator from the free theory).


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