One can show that, in the interacting theory, the operators that create single-particle energy-momentum eigenstates from the vacuum are \begin{align} (a_p^{\pm\infty})^\dagger=\lim_{t\to\pm\infty}(a_p^t)^\dagger\tag*{(1)} \end{align} where \begin{align} (a_p^t)^\dagger=i\int{\rm d}^3x\,e^{-ipx}\overleftrightarrow{\partial_0}\phi(x)=\int{\rm d}^3x\,e^{-ipx}\left(\omega_{\boldsymbol{p}}\phi(x)-i\pi(x)\right).\tag*{(2)} \end{align} (see e.g. Srednicki QFT or this answer; I'm using superscript $t$ to emphasise that this operator does not undergo the usual unitary time evolution, but is instead redefined at each $t$). These are used to create the asymptotic states that go into the calculation of the $S$-matrix. For example, in $2\to n$ scattering, the "in" and "out" states would be \begin{align} |i\rangle&=|k_1k_2\rangle=(a_{k_1}^{-\infty})^\dagger (a_{k_2}^{-\infty})^\dagger|\Omega\rangle\tag*{(3)}\\ |f\rangle&=|p_1\cdots p_n\rangle=(a_{p_1}^{\infty})^\dagger\cdots (a_{p_n}^{\infty})^\dagger|\Omega\rangle,\tag*{(4)} \end{align} the inner product $\langle f|i\rangle$ of which is said to be the corresponding $S$-matrix element (Srednicki, Schwartz, et al.).

However, $\langle f|i\rangle$ is merely an inner product of products of single-particle states. How can this be right? Surely it's the Schrödinger picture states at asymptotic times that should be plane waves (or wave packets if you want to be more precise)? After all, what we want to know is the amplitude for an initial state of particles in the far past to evolve into some other state of particles in the far future (letting the limits be implicit): \begin{align} \langle f|S|i\rangle&=\langle p_1\cdots p_n|U(\infty,-\infty)|k_1\cdots k_m\rangle=\langle\Omega|a_{p_1}\cdots a_{p_n}U(\infty,-\infty)a_{k_1}^\dagger\cdots a_{k_m}^\dagger|\Omega\rangle\tag*{(5)} \end{align} But in going from the Heisenberg to the Schrödinger picture, the operators $(a_p^{\pm\infty})_S^\dagger=\lim_{t\to\pm\infty}U^\dagger(t_0,t)(a_p^t)^\dagger U(t_0,t)$ no longer create single-particle momentum eigenstates from the vacuum, so they cannot be used to construct the $S$-matrix elements $(5)$. In other words, I'm objecting to the claim that $(3)$ and $(4)$ represent the creation of particles at $t=\pm\infty$; they're applying the wrong operators!

So what gives?

Alternative construction

Just to illustrate my point further, here's an alternative construction that seems more reasonable to me (but also way too OP). Since the asymptotic operators $(a_p^{\pm\infty})^\dagger$ are time-independent anyway, we can just define operators in the Schrödinger picture by $A_p^\dagger=(a_p^\infty)^\dagger$ and create momentum eigenstates from the vacuum at asymptotic times (which still works because the vacuum is translation-invariant). In fact, it seems like this should work at whatever time we want, not just $t=\pm\infty$, then the $S$-matrix elements are \begin{align} \langle f|S|i\rangle&=\langle p_1\cdots p_n|U(t_f,t_i)|k_1\cdots k_m\rangle=\langle\Omega|A_{p_1}\cdots A_{p_n}U(t_f,t_i)A_{k_1}^\dagger\cdots A_{k_m}^\dagger|\Omega\rangle\tag*{(Schrödinger)}\\ &={_{\rm out}}\langle p_1\cdots p_n|k_1\cdots k_m\rangle_{\rm in}=\langle\Omega|A_{p_1}(t_f)\cdots A_{p_n}(t_f)A_{k_1}^\dagger(t_i)\cdots A_{k_m}^\dagger(t_i)|\Omega\rangle\tag*{(Heisenberg)} \end{align} where $A_{p}^\dagger(t)=U^\dagger(t,t_0)A_{p}^\dagger U(t,t_0)$ and I should emphasise that the Heisenberg states $|k_1\cdots k_m\rangle_{\rm in}$ and $|p_1\cdots p_n\rangle_{\rm out}$ are not products of momentum eigenstates.

  • $\begingroup$ I don't understand at all from your starting point. The asymptotic state is just an asymptotic limit of the (time-dependent) one particle (eigen) state: $\lim_{t\to\pm\infty}|p,t\rangle:=\lim_{t\to\pm\infty}(a_p^t)^\dagger|0\rangle= (a_p^{\pm\infty})^\dagger|0\rangle.$ Using this one particle state, we can naturally define the one particle wave function in the Heisenberg picture, which is equivalent to that in the Schrodinger picture, and both of them should be plane wave, so it is sufficient to define the asymptotic state in the Heisenberg picture. What on earth is the question? $\endgroup$
    – Siam
    Commented Apr 11, 2023 at 6:21
  • 2
    $\begingroup$ @Siam the state $|p,t\rangle=(a_p^t)^\dagger|0\rangle$ is not a one-particle eigenstate for general $t$. Things aren't that simple. Only $(a_p^{\pm\infty})^\dagger|0\rangle$ specifically is a one-particle eigenstate. $\endgroup$ Commented Apr 11, 2023 at 7:13
  • $\begingroup$ Either way, I still don't understand how the states that go into the $S$-matrix could be sets of plane waves in both pictures. I mean, if $a_{p_1}^\dagger\cdots a_{p_n}^\dagger|\Omega\rangle$ is a set of plane waves in the Schrodinger picture at time $t$, this equals $U(t,t_0)a_{p_1}^\dagger(t)\cdots a_{p_n}^\dagger(t)|\Omega\rangle$ in the Heisenberg picture, so $a_{p_1}^\dagger(t)\cdots a_{p_n}^\dagger(t)|\Omega\rangle$ clearly can't be a collection of plane waves at $t$... $\endgroup$ Commented Apr 11, 2023 at 7:14
  • $\begingroup$ Oh sure, you are right. But I don't think that does not cause a serious matter. In the scattering problem, we assume that each particle does not interact in the asymptotic limit, so the interaction field is a free field in the asymptotic limit, so it is enough to define the counterpart of the creation and annihilation operator of the free field by the interaction field. And since that limit must be a free field, the generated particles must also be plane waves, isn't that all? $\endgroup$
    – Siam
    Commented Apr 11, 2023 at 7:43
  • $\begingroup$ @Siam I mean, it's exactly this kind of hand wavy argumentation that has led me into all this trouble in the first place. Having tried to make things more precise, it seems to me that, no, that in fact isn't all. Hence the question. $\endgroup$ Commented May 8, 2023 at 15:53

1 Answer 1


It took me a while, but I think I've finally figured it out.

The problem was that I was writing down all these momentum eigenstates without paying attention to which momentum operators they are actually eigenstates of!

That is, the operators $(a_p^{\pm\infty})^\dagger$ don't just "create momentum eigenstates", they specifically create eigenstates of the asymptotic momentum operators $\hat{p}(\pm\infty)=U^\dagger(\pm\infty,t_0)\,\hat{p}\,U(\pm\infty,t_0)$. Likewise, the $a_p^\dagger$ create eigenstates of the Schrödinger-picture operator $\hat{p}$. Not only does this immediately invalidate my alternative construction, it explains why this whole business works in the first place.

In the Schrödinger picture, the state is evolved between asymptotic times, which makes transparent the fact that the initial and final states are particles hanging around at those asymptotic times. But it was unclear to me how fixed momentum eigenstates in the Heisenberg picture could also be interpreted as particles at asymptotic times, because I was implicitly thinking of them as being the same eigenstates of the same operator $\hat{p}$. But they're eigenstates of $\hat{p}(\pm\infty)$, and that's the point: the states in the Heisenberg picture are fixed and the operators evolve such that the predictions come out the same. In the Schrödinger picture, the initial state is a collection of eigenstates of the (fixed) momentum operator that evolves into a complicated mess during the interaction. In the Heisenberg picture the (fixed) state is a collection of momentum eigenstates of the initial momentum operator, but looks like a complicated mess to the evolved momentum operator. Ultimately, the inner product $\langle f|i\rangle$ is between collections of eigenstates of different operators, making it very non-trivial indeed.


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