Physics Asked by Sito on January 14, 2021
I am currently reading Niklas Beisert’s lecture notes on QFT, Chapter 10, on the scattering matrix $S$.$^1$ My main confusion lies in the construction of $vert rm in rangle$ and $vert rm out rangle$ states and in what picture these states are in.
Let $phi(x)$ be the interacting field, that can be decomposed into
$$phi(x) = sqrt{Z}phi_{rm as}(x) + …tag{10.8},$$
where $phi_{rm as}$ is a canonically normalized free field of the physical mass $m$ with creation and annihilation operators $a(vec{p})$ and $a^dagger (vec p)$. We define now
$$H_{rm as} := frac{1}{2} int frac{d^3p}{(2pi)^3}a^dagger(vec p)a(vec p).tag{10.10}$$
For the scattering setup we define two asymptotic regions of spacetime, one in thedistant past $t_{rm i}to -infty$ and one in the distant future $t_{rm f} to infty$. On the initial timeslice we construct a wave packet in the form of the initial state $vert irangle$ and evolve this in time to the final state
$$vert frangle = exp(-iH(t_{f}-t_i))vert irangle .tag{10.16}$$
If I’m not misunderstanding, we are working here in the Schrödinger picture, i.e. states are time independent and we use the interacting Hamiltonian $H$ to perform the time evolution. The lecture notes state:
It is hard to compare them to see what the effect of scattering is.
I don’t understand the logic behind this statement. The Hilbert space in question is the multi-particle Hilbert space (Fock space). What does this space care about the time parameter in out theory? We just pic two random states of this space $vert f rangle$ and $vert i rangle$ and compare them.
The author then constructs the states:$^2$
$$vert {rm out}rangle := exp(iH_{rm as} t_f)vert frangle quad text{and}quad vert {rm in}rangle := exp(iH_{rm as} t_i)vert irangle tag{10.19}$$
and claims
The $vert {rm out}rangle$ and $vert {rm in}rangle$ states are both defined at time $t= 0$. Consequently,they are elements of the same Hilbert space and can be compared directly.
What does this “shift in time” using the free Hamiltonian do and why can I now compare states?
Last but not least, it seems the $vert {rm out}rangle$ and $vert {rm in}rangle$ states don’t seem to be in the Schrödinger picture anymore.. Are they in the Interaction picture?
$^1$ All references with equations will be with respect to the above lecture notes.
$^2$ Is there a typo in this equation? I would have expected $vert {rm in}rangle := exp(-iH_{rm as} t_i)vert irangle$…
The "in" and "out" states are defined in the Heisenberg picture where the states are time independent. The "in" states have a simple interpretation as a collection of widely separated particles in the distant past and the out states have a simple description as widely separated particles in the distant future. Although they are usually considered as plane-waves, we should really assemble wavepackets to keep the particles well separated.
Answered by mike stone on January 14, 2021
The issue here is not specific to QFT or to using Heisenberg/Schrödinger picture. Rather it is the difference in the boundary conditions and the two rather different types of problems studied in quantum mechanics - let us call them eigenvalue and scattering problems.
In eigenvalue problems one is determining the allowed energies and states of the system by solving the Schrödinger equation subject to specific boundary conditions. These problems, the approximation methods for solving them, and the associated mathematical issues are extensively covered in QM textbooks. One typically solves the solvable part of the problem in the whole space and then looks for corrections to this solution. Problems with continuous spectrum, of course, can be also treated as eigenvalue problems, although boundary conditions are trickier.
Scattering problems are a very different view at solving Schrödinger equation - we typically know the energy and the solution in some regions of space and need to relate these solutions via a scattering matrix, characterizing the region of space where the exact solution of the Schrödinger equation is not possible. Introductory QM books and courses usually cover this topic only partially when discussing tunneling or delegate it to later chapters (like in Landau&Livshits), while when studying QFT one is already expected to have a good grip of the boundary conditions, scattering matrix formalism, optical theorem, etc.
$|inrangle$ and $|outrangle$ states refer to the set of boundary conditions frequently employed in scattering problems, designed to mimic a classical scattering of a particle from an object - which is where the terminology in terms of behavior at $trightarrow pm infty$ comes from, although one may be actually working in Schrödinger picture. Using Heisenberg picture makes this boundary conditions more intuitive, and most naturally they probably come when using the Keldysh formalism (but by then one already feels comfortable with the scattering theory).
Answered by Vadim on January 14, 2021
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