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Lippman-Schwinger Equation with Outgoing Solutions

+ 2 like - 0 dislike
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I'm reading about Green's functions and how the Lippmann-Schwinger equation eventually leads to the textbook expression for the form of scattered wavefunctions by a central potential in the far radiation zone.

Why does adding $i\epsilon$ to the energy in the Green's function ensure that the L-S equation will give us outgoing solutions? Where does that intuition come from?

This post imported from StackExchange Physics at 2014-05-04 11:41 (UCT), posted by SE-user eqb
asked Apr 25, 2014 in Theoretical Physics by eqb (10 points) [ no revision ]

2 Answers

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One way to look at it is simply a mathematical trick that encodes the boundary conditions of the Schroedinger equation. An alternative and only slightly more intuitive view is the following. In order to obtain only outgoing solutions it is essential to assume that the scattering potential is slowly switched on adiabatically. Formally one can achieve this by treating the scattering potential as a function of time of the form $V(r,t) = V(r)e^{\epsilon t}$, so that $V\to 0$ for large negative $t$. Working through the algebra, you find that exactly the same effect is achieved by adding a small imaginary term $i\epsilon$ to the energies instead. This is discussed in Landau & Lifshitz QM, section 43, and also mentioned somewhere in the extensive sections on scattering theory in the same text. As you probably already know, this procedure shifts the pole of the Green function in the right direction so that you pick up the outgoing spherical wave after contour integration in the complex momentum plane.

So long as the interaction is switched on very slowly, the adiabatic theorem tells you that the system will remain in the same eigenstate, even though the form of that eigenstate itself changes as the Hamiltonian is altered. It is not so hard to believe that the adiabatic deformation of a coherent plane wave state at momentum $\mathbf{k}$ is again a wave at momentum $\mathbf{k}$, but now with a small outgoing spherical wave component. If you instead switch the scattering potential on abruptly, the sudden 'jolt' this gives to the system will excite modes at all possible momenta. The interference between these different modes can lead to both incoming and outgoing spherical wave components, in general.

Strictly speaking, adiabaticity requires the interaction to be switched on slower than any inverse frequency difference. For scattering in a large volume $L^3$, this is essentially impossible since the frequency differences are proportional to $1/L^2$. However, as long as the interaction is not switched on infinitely fast (which of course is also impossible), one expects the initial transients to eventually average to zero, leaving only the outgoing spherical wave. And of course this is indeed what happens in the experiment.

This post imported from StackExchange Physics at 2014-05-04 11:41 (UCT), posted by SE-user Mark Mitchison
answered Apr 26, 2014 by Mark Mitchison (270 points) [ no revision ]
I like this idea of encoding boundary conditions into Schrodinger's equation. Thanks for that reference in L&L's QM. I didn't tie the idea with the adiabatic theorem as you did when asking the question, so thank you for that additional point! :)

This post imported from StackExchange Physics at 2014-05-04 11:41 (UCT), posted by SE-user eqb
+ 1 like - 0 dislike

First, as I understand the Lippman-Schwinger equation, there are actually two different cases, denoted by the $\pm$'s in the standard form of the equation: \begin{equation} | \psi^{(\pm)} \rangle = | \phi \rangle + \frac{1}{E - H_0 \pm i \epsilon} V |\psi^{(\pm)} \rangle. \ \end{equation}

In this way, we're talking about not just the outgoing solution (the $+$ equations), but also an incoming solution (the $-$). Furthermore, the $i\epsilon$ term exists in the Lippman-Schwinger equation before you introduce the Green's function (you might introduce the Green's function while solving the equation).

As described on the Wikipedia page (here) the complex term you're referring to is simply introduced to avoid making the expression blow up for $(E-H_0)\rightarrow 0$. You can contour integrate around the pole then, in the usual fashion. Perhaps I'm wrong (and I'd be happy to understand this in a deeper way), but I don't think of the $i\epsilon$ term as being physically interesting; instead, I see it as more of a math trick to allow us to solve.

This post imported from StackExchange Physics at 2014-05-04 11:41 (UCT), posted by SE-user Sam
answered Apr 25, 2014 by sam (45 points) [ no revision ]

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