0
$\begingroup$

When I think about scattering process, I reach to slightly another definition to the S-matrix. because I understand my reasoning I hope someone could refine it to a correct one so that I can have a clear idea about the S-matrix.

In QM if $A$ be an observable with eigenstates $|A_\alpha\rangle$ you can set system at time $t_1$ to state $|A_1\rangle$. system at time $t_2$ will be at $e^{-ih(t_2-t_1)}|A_1\rangle$. now if you measure $A$ the probability of collapsing to $A_2$ is $\left<A_2|e^{-iH(t_2-t_1)}|A_1\right>$.

Now in scattering experiment $t_1=-\infty$, $t_2=+\infty$ and $A=H_0$. So $$ S_{21}= \lim_{t_i=\mp} \left<\Phi_2|e^{-iH(t_2-t_1)}|\Phi_1\right> $$ where $\Phi_\alpha$s are eigenstates of $H_0$. but conventional S-matrix reads: (Weinberg Vol.1 P.114) $$ S_{21}= \lim_{t_i=\mp}\left<\Phi_2|e^{iH_0t_2}e^{-iH(t_2-t_1)} e^{-iH_0t_1} |\Phi_1\right>. $$ Is my reasoning refinable?

$\endgroup$
  • 1
    $\begingroup$ What's your "argument"? You just omitted the $H_0$ factors, the reason for which you should find e.g. in the very Weinberg book you quote. What's the question here? $\endgroup$ – ACuriousMind May 30 '16 at 19:04
  • $\begingroup$ Weinberg's approach to defining S-matrix is quite different from what I wrote here. when I try to consider situation more reasonable for myself the very $H_0$ factors disappear! $\endgroup$ – moshtaba May 30 '16 at 19:25
  • 1
    $\begingroup$ To get the proper QM analogon, you need to consider the Dirac/interaction picture, not the Schrödinger one you seem to be assuming implicitly here. $\endgroup$ – ACuriousMind May 30 '16 at 19:45
  • $\begingroup$ yes I'm trying the Schroedinger picture. but transition probability what have to do with pictures? Weinberg's discussion is in Heisenberg picture and Greiner's (Vol.5 P.216) is in Dirac picture and S-matrix are the same. then why same S-matrix can't be derived in Schroedinger picture? $\endgroup$ – moshtaba May 30 '16 at 20:31
  • $\begingroup$ @moshtaba: see my answer to physics.stackexchange.com/q/336599 $\endgroup$ – Abdelmalek Abdesselam Feb 25 '18 at 3:14
0
$\begingroup$

I consider a related question but where I consider a pair of Schrodinger equations for unperturbed and perturbed states. The occurrence of time is an inconvenience, so instead work with the Hamiltonians and Green' function. This is much the same, in that the S-matrix is concerned with the asymptotic scattering states and their differences in eigenspectrum.

The ordered S-matrix is constructed to that a set of vertices, or particles, transition. Suppose we have the state $$ |\phi\rangle = |p_1, p_2, \dots, p_n\rangle $$ that transitions to the state $$ |\phi'\rangle = |q_1, q_2, \dots, q_n\rangle. $$ The evaluation of the overlap of this state with the S-matrix $$ \langle\phi |S|\phi'\rangle = \langle p_1, p_2, \dots, p_n|S|q_1, q_2, \dots, q_n\rangle. $$ This operation is unitary and a part of this is that $S = 2\pi iT$ for $T$ the transition matrix

The transition matrix can be evaluated explicitly. The S-matrix describes the transition of a state that obeys $H|\phi\rangle = E\phi\rangle$ to $H'|\phi'\rangle = E\phi'\rangle$. The state $|\phi'\rangle = |\phi\rangle + V|\phi'\rangle$, where $V$ is the perturbing or potential that induces scattering. We are looking at a process that propagates a scattering influence, and propose the Green's function $(E - H)G = 1$ for $G = G(x,x')$. We can then see $$ (H' - E)|\phi'\rangle = (H' - E - V)|\phi'\rangle + (E - H)GV|\phi'\rangle $$ $$ (E - E)(|\phi'\rangle - GV|\phi'\rangle) = (H - E)|\phi\rangle = 0 $$ makes this consistent. The relevant term is $ |\phi'\rangle - GV|\phi'\rangle$ that gives the scattered portion of the wave $|psi\rangle = |\phi'\rangle - |\phi\rangle$ $ = GV|\phi'\rangle$. The transition matrix is then found by the expansion of the state $|\phi'\rangle$ $$ |\phi'\rangle = (1 + GV + GVGV + \dots)|\phi\rangle = (1 - GV)^{-1}|\phi\rangle $$ so the transition matrix $$ T = \langle\phi|T|\phi\rangle = \langle\phi|V|\phi'\rangle = \langle\phi|V(1 - GV)^{-1}|\phi\rangle. $$ So the transition matrix is $T = V(1 - GV)^{-1}$.

The unitarity of the S-matrix can be verified by computing $SS^\dagger$. $$ SS^\dagger = (1 + 2\pi iT)(1 - 2\pi iT^\dagger) = 1 + 2\pi i(T - T^\dagger) + 4\pi^2TT^\dagger + O(T^3) $$ The first term $T - T^\dagger = T^\dagger(G^\dagger - G)T$ where for $G = (E - H - i\epsilon)^{-1}$ gives $T - T^\dagger = 2\pi TT^\dagger$ and the unitarity of the S-matrix is proven.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.