7
$\begingroup$

So long, in my QFT courses, I've seen two definitions of the $S$-matrix:

The first, more elementary, definition is given in the interaction picture: $$S=\text T \lbrace \exp [-i\intop \text d ^4 x \,\mathscr H_{\text {int}} ^I(x)]\rbrace\qquad (1),$$where the operatore $\mathscr H _\text{int} ^I$ is the interaction hamiltonian density evaluated in the interaction picture. According to this definition, the probability amplitude for a state with in-asymptote $\vert \alpha \rangle$ to scatter into a state with out-asymptote $\vert \beta \rangle $ is given by: $$S_{\beta \alpha}=\langle \beta \vert S \vert \alpha \rangle \qquad (2).$$

The second definition is the one in terms of in/out asymptotic states: $$S_{\beta \alpha}=\langle\beta \,\text {out} \vert \alpha \,\text {in}\rangle \qquad (3).$$

In Non relativistic quantum mechanics, I understand how these two definitions are related, since there one can write the amplitude of scattering from a plane wave $\vert \mathbf k \rangle$ to a plane wave $\vert \mathbf p \rangle$ as:$$S_{\mathbf p \leftarrow \mathbf k } =\langle \mathbf p \vert S \vert \mathbf k \rangle=\langle \mathbf p -\vert \mathbf k +\rangle, $$ where $\vert \mathbf k \pm\rangle$ are the "in" and "out" solutions of the Lippmann-Schwinger equation, or generalized eigenfunctions of the full hamiltonian.

So, I guess (correct me if I'm wrong) that the "in" and "out" states in QFT are the analogues of the $\vert \mathbf k \pm \rangle $'s in potential scattering, and indeed it can be shown that $\vert \{\mathbf p_i\}\, \text {in/out}\rangle$ are eigenstates of the interacting fields hamiltonian.

However, those "in/out" states are constructed in a quite formal way, via ad-hoc constructed asymptotic in/out fields, and I'm really failing in seeing how this definition $(3)$ should relate to the more elementary $(1)$.

Just to clarify: I understand the physical idea behind the construction of "in/out" fields and, probably, if I hadn't seen def. $(1)$, I would accept $(3)$ as it stands. However I don't understand the mathematical relation beetween (1) and (3):

  • Do they agree?
  • Do they agree up to a phase?
  • Do they agree if some assumption, like "adiabatic hypothesis" is made?
  • How can I formally prove it?

Any help/thought would be appreciated, thank you for the attention.

$\endgroup$
  • $\begingroup$ Read Weinberg Vol. 1 Chapter 3. He Defines the S matrix using your equation (3) in Eq. (3.2.1) and then in Section 3.5 shows that this is equivalent to your equation (1) in Eq. (3.5.10). $\endgroup$ – Prahar Oct 20 '16 at 16:46
4
$\begingroup$

The in and out states are defined as solutions to the Lippmann-Schwinger equation, with the appropriate boundary condition prescribed by the choice of the contour ($\pm \varepsilon$). $$ | \psi^{(\pm)}_a \rangle = | \phi_a \rangle + \frac{1}{E - H_0 \pm i \epsilon} V |\psi^{(\pm)}_a \rangle $$ If you calculate the $S_{ab}=\langle \psi_b^-|\psi^+_a\rangle$ you gonna find a recursive relation for the the S-matrix. $$ S_{ab}=\delta(a-b)-2i\pi\delta(E_a-E_b)T_{ab}. $$ $$ T_{ab}=V_{ab}+\int dc\frac{V_{cb}T_{ac}}{E_a-E_c+i\varepsilon} $$ $$ T_{ab} = \langle \phi_b |V|\psi^{(\pm)}_a \rangle $$ $$ V_{ab} = \langle \phi_b |V|\phi_a \rangle $$ If you iterate this recursive relataion over and over you are left with a series. This series can be identified with a time ordering of the desired exponential if you use that: $$ \frac{1}{E_b - E_c + i \epsilon}=-i\int_0^{\infty} dt \exp\left[{i(E_b-E_c)t -\varepsilon t}\right] $$ Then you may see that adiabatic turning on and off interactions are the same thing as putting the $\pm i\varepsilon$ in the Schwinger-Lippmann equation, i.e. imposing the existence of in and out scattering states.

In LSZ approach we go to a different root. We have already a formal hamiltonian (the QFT theory). Scattering states could be build out of this hamiltonian by seeking for long-lived states with some dipersion relation. Mathematically they are poles of the two-point correlation function. The $Z$ is only requiried if you want to work with this fundamental fields inside the hamiltonian, like the fields in the correlation function. Then, you need to assume that the field create and destroy different states than the scattering ones.

The adiabatic assumption is behind all this approaches. Is ultimately related to the fact that exist scattering states behaving as a free particle. The difference is that in Lippmann-Schwinger approach you have asymptotic physical particles and in LSZ you have self-interacting bare particles giving rise to a physical one.

$\endgroup$
  • $\begingroup$ Dear Nogueira, thank you. I've recently been reading Weinberg and understand now that this is the part of the formalism which is the same as in NRQM. However, there's still something I have to figure out: if one defines the "in/out" states as those created by "in/out" fields, which are by themselves defined by the weak convergence condition (schematically: $\sqrt Z \phi (t) \to \phi _{\text{as}}(t)$), how does he/she see that this agrees with the definition given by the Lippmann-Schwinger equation that you mention? $\endgroup$ – pppqqq Oct 20 '16 at 6:33
  • $\begingroup$ Partial self-answer and clarification of the question: naively speaking, it seems to me that if the Dyson-series formalism is strictly equivalent to an adiabatic turn-off, then it should be possible to set up the whole formalism by requiring strong convergence $\phi (t) \to \phi _{\text {as}} (t)$ (that is, if the $V$ is adiabatically turned off, the interacting field becomes free in the strong convergence sense). However, when we do the LSZ thing, we get a renormalization constant $Z$ that doesn't appear anywhere in the formal Dyson series. [...] $\endgroup$ – pppqqq Oct 20 '16 at 6:53
  • $\begingroup$ [...] Ofcourse, also in the Dyson series approach one will at some point have to renormalize the infinite diagrams which inevitably appear. So, maybe, the question should be: are the LSZ and Dyson/Lippmann-Schwinger approaches two different schemes of renormalization which one simply expects (and finds out) to agree for those calculations where the adiabatic turn off should physically be irrelevant (in particular, no bound states)? Or, on the other hand, there exists a rigorous mathematical proof of the equivalence of the two approaches? $\endgroup$ – pppqqq Oct 20 '16 at 6:59
  • $\begingroup$ See the update up $\endgroup$ – Nogueira Oct 20 '16 at 16:27
  • $\begingroup$ @pppqqq The Dyson/Lippamann-Schwinger approaach don't make any assumption about the inner structure of the Hamiltonian, the exact theory. The only assumption in the approach is about the existence of the scattering state with some dispersion relation described by $H_0$, the unperturbed hamiltonian. This assumption could be translated to the fact that the interacting hamiltonian $V=H-H_0$ is adiabatically turn in and out. You can see this by the eq. that relates $\left(\Delta E +i\varepsilon\right)^{-1}$ with the exponential. $\endgroup$ – Nogueira Oct 20 '16 at 19:27
2
$\begingroup$

Nogueira's answer was really helpful, I just wanted to add a few remarks in hindsight.

When I wrote this question, one of the things which confused me was that I couldn't see how the "in" and "out" states which one defines in formal scattering theory, via e.g. the Lippmann-Schwinger equation (see Nogueira's answer), would coincide with the states created by the "in" and "out" fields from the vacuum, e.g.: $$\vert \mathbf p \text { in}\rangle=i\intop \text d^3{\mathbf x} f_{\mathbf p }^* (x) \overleftrightarrow {\partial _0} \phi _{\text {in}}(x)\vert 0 \rangle.$$

If one defines the "in" field, for the scalar case, via the linear combination: $$\phi _{\text {in}}(x)=\intop \text d ^3\mathbf p \lbrace a_{\text {in}}(\mathbf p )f_{\mathbf p}(x)+\text {h.c.}\rbrace, $$ and, in turn, defines $a(\mathbf p )$ as the destruction operator for the "in" states, then the correspondence is tautological.

However, some texts (for example, Bjorken&Drell), start from "in" and "out" operators defined by the Yang-Feldman equations:$$\phi _{\text{in}}(x)=\phi (x)-\intop \text d ^4 y \Delta _{R}(x-y)j(y),$$ where $\Delta _R$ is the retarded Green's function for the Klein-Gordon equation and $\phi$ is the (renormalized) interacting field, which satisfies: $$(\square +m^2)\phi =j.$$ In this case, the correspondence is not immediately obvious, since one has to prove that wave-packets constructed from $\phi _{\text {in}}$ converge indeed (at least weakly) to free states.

Precisely this questions are addressed in an old article by Schweber S., "On the Yang-Feldman formalism", and also (quite easier to read) in his book "Introduction to quantum field theory", sec. 17d, where he defines the in field by $$\phi _{\text {in}}(x)=e^{iHt}\Omega^+ \phi (\mathbf x ,0)(\Omega ^+)^\dagger e^{-iHt},$$ and proves that is satisfies the Yang-Feldman equation.

$\endgroup$

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.