# The $i\epsilon$ in non-relativistic scattering theory

When doing quantum mechanical scattering theory, we obtain the Lippman-Schwinger equation

$$|\psi\rangle=|\psi_0\rangle+(E-H_0)^{-1}V|\psi\rangle$$

Here $\psi_0$ is the unperturbed wavefunction, $\psi$ is the total wavefunction, and we define $\psi_s=\psi_0-\psi$ is the scattered wave function. When solving this, people usually replace $(E-H_0)\rightarrow E-H_0+i\epsilon$ while mumbling something about causality and $\psi_s$ not having incoming probability current. Then of course, in the final result, we indeed see that we obtain something reasonable.

However, I would like to know if there is some more robust and mathematical way to derive the $i\epsilon$ part, instead of just putting it in by hand and see that the final result is reasonable? For example, if one could start with the initial Schrodinger equation, and demand no incoming probabiltiy current on the scattered wave to see that the $i\epsilon$ comes out naturally?

• See the discussion around equation (3.1.16) in Weinberg Vol. 1. – Prahar Mitra Oct 31 '15 at 14:17
• A mathematically more precise approach is through Møller operators and associated S-operator. There the $±iϵ$ can be tied in with the behaviour of time $t→±∞$. – Urgje Nov 1 '15 at 10:55
• I think this other question will help quite a bit. In short, the $i \epsilon$ is a choice of boundary condition, and has a very nice physical motivation as well. – DanielSank Apr 15 '18 at 16:12

## 1 Answer

(This answer is pretty much just expanding on Urgje's comment.)

The time-indepent Schrödinger equation, i.e. looking at constant energy, is essentially the Fourier transform of the time-dependent Schrödinger equation. The $i\epsilon$ prescription comes from doing Fourier transforms that do not converge for scattering problems. Therefore to get something that does converge, you have to modify your Fourier transform slightly.

After this rather vague introduction, let's look at it in a bit more detail. Most of the following closely follows Newton's scattering theory bible, where the start of the 3rd chapter is probably the clearest outline of $i\epsilon$ problems you can find in the literature.

Consider the time-dependent Schrödinger equation ($\hbar=1$): $$i \dot{\psi}(t) = H \psi(t).$$

Then one can define the Green's function: $$(i \frac{\partial}{\partial t} - H ) G(t) = \mathbb{I}\delta(t) .$$

However this $G(t)$ is by far not unique, there are all kinds of initial conditions you can consider. A common choice that describes causal propagation is

$$G^+(t) = 0 \quad \textrm{for } t<0.$$

For now this is only a definition of a solution of the equation for the Green function, i.e. there are other solutions. However this solution is particularly useful to solve initial value problems, since if you know the initial state at some time $t_0$ you can write the solution of the Schrödinger equation as

$$\psi(t) = iG^+(t-t_0)\psi(t_0).$$

From there one can develop scattering theory by defining in-states in the far past and out-states in the far future and solving the wave equations.

So (almost) everything is nice in the time-dependent picture. However you would always have to specify wavepackets, which is why a time-independent/energy picture is often preferred.

To get to the energy picture we Fourier transform... pretty much everything. The Fourier transform of the time-dependent Schrödinger equation is the time-independent Schrödinger equation

$$E |\psi\rangle = H|\psi\rangle.$$

Now let's look at the Fourier transform of the Green function above. This is where the $i\epsilon$ comes in:

$$G^+(E) = \int_{-\infty}^{\infty} e^{iEt} G^+(t).$$

This clearly does not converge. As $t \rightarrow -\infty$ the integrand is zero, so that part is ok. But as $t \rightarrow +\infty$ the integrand can oscillate. Therefore we have to change our definition of $G^+(E)$ to

$$G^+(E) = \int_{-\infty}^{\infty} e^{i(Et+i\epsilon t)} G^+(t).$$

Note that there is absolutely no choice of the sign before $i\epsilon$ here. This is because above we decided that we want to solve an initial value problem. If you want to solve a final value problem instead, you will choose $G_0(t)$ to be zero in the future and you will need the opposite sign in the Fourier transform.

Where is the connection to the Lippmann-Schwinger equation? Manipulating the equations a little bit we see that

$$G^+(E) = (E-H+i\epsilon).$$

So in the Lippmann-Schwinger equations, the free Green function appears (i.e. Hamiltonian $H_0$ without interactions). So where is our choice here? The state $|\psi_0\rangle$ is not unique, either choose it as the Fourier transform of a forward travelling free wave or of a backward travelling one. So there we have our connection to choosing that we solve an initial value problem and the $i\epsilon$ is not arbitrary after all.