- #1

UnreliableObserver

- 4

- 4

In scattering theory, the quantity of interest is the amplitude for the system—initially prepared as a collection of (approximate) momentum eigenstates—to evolve into some other collection of momentum eigenstates. For example, for ##m\to n## scattering, the amplitude we're interested in is $$\bra{p_1\cdots p_n}U(t_f,t_i)\ket{k_1\cdots k_m},\tag*{(1)}$$ where ##t_i## is the time at which we prepare the state, ##t_f## is the time at which we observe the result of the scattering process, and ##U(t_2,t_1)## is the time evolution operator.

We can imagine finding operators ##a_p## capable of creating such states from the vacuum, in which case we could write ##(1)## as $$\bra{\Omega}a_{p_1}\cdots a_{p_n}U(t_f,t_i)a_{k_1}^\dagger\cdots a_{k_m}^\dagger\ket{\Omega},\tag*{(2)}$$ or, in the Heisenberg picture, $$\bra{\Omega}a_{p_1}(t_f)\cdots a_{p_n}(t_f) a_{k_1}^\dagger(t_i)\cdots a_{k_m}^\dagger(t_i)\ket{\Omega},\tag*{(3)}$$ where ##a_p(t)=U^\dagger (t, t_0)a_p U(t, t_0)##. I hope this is all rather uncontroversial so far.

Now, it can be shown (see e.g. Srednicki QFT) that the operators that create single-particle momentum eigenstates from the vacuum are ##(a_p^{\pm\infty})^\dagger=\lim_{t\to\pm\infty}a_p^\dagger(t)## where

$$a_p^\dagger(t)=\int{\rm d}^3x\,e^{-ipx}\left(\omega_{\boldsymbol{p}}\phi(x)-i\pi(x)\right).\tag*{(4)}$$ However after this point things start to fall apart for me. According to everything I've read, these operators are used to create the "initial" and "final" states ##\ket{i}=(a_{k_1}^{-\infty})^\dagger\cdots(a_{k_m}^{-\infty})^\dagger\ket{\Omega}## and ##\ket{f}=(a_{p_1}^{+\infty})^\dagger\cdots(a_{p_n}^{+\infty})^\dagger\ket{\Omega}##, with which the corresponding S-matrix element is given as ##\langle f|i\rangle##. But how can this be right? If these operators do indeed create single-particle momentum eigenstates, then isn't this inner product very trivial? If we refer back to the Heisenberg picture expression ##(3)##, we are free to write it as ##\langle f|i\rangle##, but these states are clearly not products of momentum eigenstates, but rather whatever complicated mess results from the action of ##a_p(t)## on the vacuum (as they must be to give a non-trivial amplitude). The products of eigenstates are those in ##(1)##, so shouldn't the operators ##(a_p^{\pm\infty})^\dagger## instead assume the role of those in ##(2)##?

We can imagine finding operators ##a_p## capable of creating such states from the vacuum, in which case we could write ##(1)## as $$\bra{\Omega}a_{p_1}\cdots a_{p_n}U(t_f,t_i)a_{k_1}^\dagger\cdots a_{k_m}^\dagger\ket{\Omega},\tag*{(2)}$$ or, in the Heisenberg picture, $$\bra{\Omega}a_{p_1}(t_f)\cdots a_{p_n}(t_f) a_{k_1}^\dagger(t_i)\cdots a_{k_m}^\dagger(t_i)\ket{\Omega},\tag*{(3)}$$ where ##a_p(t)=U^\dagger (t, t_0)a_p U(t, t_0)##. I hope this is all rather uncontroversial so far.

Now, it can be shown (see e.g. Srednicki QFT) that the operators that create single-particle momentum eigenstates from the vacuum are ##(a_p^{\pm\infty})^\dagger=\lim_{t\to\pm\infty}a_p^\dagger(t)## where

$$a_p^\dagger(t)=\int{\rm d}^3x\,e^{-ipx}\left(\omega_{\boldsymbol{p}}\phi(x)-i\pi(x)\right).\tag*{(4)}$$ However after this point things start to fall apart for me. According to everything I've read, these operators are used to create the "initial" and "final" states ##\ket{i}=(a_{k_1}^{-\infty})^\dagger\cdots(a_{k_m}^{-\infty})^\dagger\ket{\Omega}## and ##\ket{f}=(a_{p_1}^{+\infty})^\dagger\cdots(a_{p_n}^{+\infty})^\dagger\ket{\Omega}##, with which the corresponding S-matrix element is given as ##\langle f|i\rangle##. But how can this be right? If these operators do indeed create single-particle momentum eigenstates, then isn't this inner product very trivial? If we refer back to the Heisenberg picture expression ##(3)##, we are free to write it as ##\langle f|i\rangle##, but these states are clearly not products of momentum eigenstates, but rather whatever complicated mess results from the action of ##a_p(t)## on the vacuum (as they must be to give a non-trivial amplitude). The products of eigenstates are those in ##(1)##, so shouldn't the operators ##(a_p^{\pm\infty})^\dagger## instead assume the role of those in ##(2)##?

Last edited: