When studying the quantization of a field theory with free fields, the creation operators $a^\dagger(k)$ are independent of time. In an interacting theory, they are time-dependent, and therefore $a^\dagger(k, -\infty)$ is not equal to $a^\dagger(k, +\infty)$. I would like to understand better what is the form taken by this difference : how exactly do they differ ? What are the physical consequences ?


Let us consider a real, scalar field theory for simplicity (and metric signature $+ - - -$). In the free theory, one can use the mode expansions of the field $\phi(x)$ and its canonical conjugate momentum $\pi(x)$ to derive the following expressions for the creation and annihilation operators: $$ a(p)=i\int d^3x\ e^{ipx}\overset{\leftrightarrow}{\partial_0}\phi(x),\ a^\dagger(p)=-i\int d^3x\ e^{-ipx}\overset{\leftrightarrow}{\partial_0}\phi(x)$$ where $g\overset{\leftrightarrow}{\partial_0}f:=g\partial_0f-(\partial_0 g)f$. This relation is always true (and exact) in the free field theory.

In the interacting theory, one makes the assumption - usually accompanied by some hand-waving argument about 'turning off interactions in the distant past and future' - that these relationships still hold asymptotically (i.e. as $t\to \pm \infty$). Now, we want to find out how much $a$ and $a^\dagger$ change with time, so we consider the following: $$\Delta a(p)=a(p,\infty)-a(p,-\infty)=\int\partial_0 a(p,t)\ dt=i\int d^4x\ \partial_0 \left(e^{ipx}\overset{\leftrightarrow}{\partial_0}\phi(x)\right)$$ One simply writes out these derivatives, and uses an additional assumption about the behavior of $\phi(x)$ at infinity to cancel some terms, finally arriving at $$\Delta a(p)=i\int d^4x\ e^{ipx}(m^2+\partial^2_t-\partial_i^2)\phi(x)=i\int d^4x\ e^{ipx}(\square+m^2)\phi(x) $$ This immediately shows that $\Delta a(p)$ is zero in the free theory, since the free field obeys the Klein-Gordong equation $(\square+m^2)\phi=0$, reflecting the fact that $a(p)$ is time-independent in this case. However, in an interacting theory the equation of motion is different and (generally) non-linear (e.g. $(\square+m^2+\lambda\phi^3)\phi=0$ in the case of $\phi^4$-theory), therefore $\Delta(p)$ will not vanish in the interacting case. It is easy to derive a very similar expression for $\Delta a^\dagger(p)$, which can be given the same interpretation.

Addendum (18-12-2014):

It turns out that it is not completely straightforward to find a good interpretation for the main expressions. However, I believe I have now found a good mental picture. To make it a little more clear, however, we have to do some more work. From the expressions for $\Delta a(k)$ and $\Delta a^\dagger(k)$ it is quite straightforward to derive the LSZ reduction formula for $N$ incoming particles with momenta $p_i$ and $M$ outgoing particles with momenta $k_i$ (in position space). If one then performs a Fourier transform, we find :

\begin{align}\langle k_1\dots k_M|S|p_{1}\dots p_{N}\rangle &=(2\pi)^4\delta^{(4)}\Big(\sum P\Big) \prod_{i=1}^M i(k_i^2+m^2)\prod_{j=1}^Ni(p_i^2+m^2)\\ &\hspace{1cm}\times i\tilde G_{M+N, c}(p_1,\dots,k_{M-1})\\ &=(2\pi)^4\delta^{(4)}\Big(\sum P\Big) \Big(i\tilde G_{2, c}^{(0)}(k_1)\Big)^{-1} \dots\Big(i\tilde G_{2, c}^{(0)}(p_N)\Big)^{-1}\\ &\hspace{1cm}\times i\tilde G_{M+N, c}(p_1,\dots,k_{M-1}) \end{align}

Here, it is more simple to give an interpretation. The final 'big' propagator basically takes the $N$ initial particles and lets them time-evolve, including the interaction terms, to $M$ particles in the 'out'-state - it's basically a black box. Then, the product of inverse free (note the superscript $(0)$) propagators can be interpreted as subtracting off the free time evolution, of all the particles individually.

With this in mind, we can give the same interpretation to the expression for $\Delta a$ and its hermitian conjugate: We can interpret the $(\square+m^2)$ as 'subtracting off' the 'free' time evolution, keeping only the part that is due to the interaction (as we know, the free creation/annihilation operators should be time-independent). I'm aware that this is very handwavey, but I think that's the best I can do in terms of providing an intuition for what 'it means' - of course one should mostly just shut up and calculate ;)

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    $\begingroup$ Why do you guys so often answer simple questions by referring to really complex frameworks like relativistic QFT? This is just a question about interacting Hamiltonians and has absolutely nothing to do with path integrals. Adding relativity on top of path integration just to answer a question about interacting Hamiltonians seems weird. I really would like to know the answer to this, I'm not just complaining :) $\endgroup$ – DanielSank Dec 15 '14 at 22:15
  • $\begingroup$ @DanielSank It's because that's where we learn most of our field theory ;) Feel free to add a better answer if you can! And honestly, I don't think relativistic QFT is harder; maybe it's you who's biased ;) $\endgroup$ – Danu Dec 15 '14 at 22:17
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    $\begingroup$ @DanielSank Also, I'm not sure where you're getting 'path integration' in this answer. There is no path integration going on here; this is canonical quantization. $\endgroup$ – Danu Dec 15 '14 at 22:21
  • $\begingroup$ Ooof, you're right about the path integration. Sorry, my mistake. Oh man, that first comment of mine doesn't make any sense, sorry. $\endgroup$ – DanielSank Dec 15 '14 at 22:41
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    $\begingroup$ @JamalS: While it is true that a real scalar is the simplest field theory, it is definitely not the simplest case in which you can discuss the evolution of raising/lowering operators in an interacting system. Furthermore, it is precisely because the answers are for the whole community that I encourage them to be as simple as possible. As I am a member of the community, I don't see how it makes sense to tell me that I have no right to complain :) $\endgroup$ – DanielSank Dec 16 '14 at 7:04

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