Scattering theory in quantum mechanics

Question

I am currently reading through Weinsteins Lectures on Quantum Mechanics. In Chapter 8 he deals with general scattering theory.

Let

$$H = H_0 + V(\vec{x})$$

where $V(\vec{x})$ is a central potential, that satisfies $V(\vec{x}) \rightarrow 0$ for $\vert \vec{x} \vert \rightarrow \infty$. In addition to that, we introduce $\Psi^{\pm}_{\alpha}$ (interpreted as in- and out-states) as eigenstates of the full Hamiltonian

$$H\Psi^{\pm}_{\alpha} = E_{\alpha}\Psi^{\pm}_{\alpha},$$

as well as $\Phi_{\alpha}$, which are defined as eigenstates of the free-particle Hamiltonian

$$H_0 \Phi_{\alpha} = E_{\alpha}\Phi_{\alpha}$$

With that being said, the following causes some trouble to me:

The definition of $\Psi_{\alpha}^{+}$ and $\Psi_{\alpha}^{-}$ can be made more precise by specifying that if $g(\alpha)$ is a sufficiently smooth function of the momenta in the state $\alpha$, then

$$\int d\alpha \, g(\alpha) \,\Psi^{\pm}_{\alpha}\,\exp\left(-\frac{iE_{\alpha}t}{\hbar}\right) \rightarrow \int d\alpha \, g(\alpha) \,\Phi_{\alpha} \, \exp\left(-\frac{iE_{\alpha}t}{\hbar}\right)$$

for $t \rightarrow \mp \infty$.

I am not really sure, what to make out of this. I think, that Weinberg wants to emphasise, that the in- and out-states $\Psi^{\pm}_{\alpha}$ look like states of the free-particle Hamiltonian, if the scattering event either will take place in the (far) distant future or did already happen in the long-gone past. However in this case, I would have either expected

$$\int d\alpha \, g(\alpha) \,\Psi^{\pm}_{\alpha}\,\exp\left(-\frac{iE_{\alpha}t}{\hbar}\right) \rightarrow \int d\alpha \, g(\alpha) \,\Phi_{\alpha}$$

for $t \rightarrow \mp \infty$ or something like

$$\int d\alpha \, g(\alpha) \,\Psi^{\pm}_{\alpha,t_0}\,\exp\left(-\frac{iE_{\alpha}t}{\hbar}\right) \rightarrow \int d\alpha \, g(\alpha) \,\Phi_{\alpha} \, \exp\left(-\frac{iE_{\alpha}t}{\hbar}\right)$$

for $t_0 \rightarrow \mp \infty$ ($t_0$ marks the starting point of our observation)

An explanation of why the expression from Weinstein makes sense, would be much appreciated.

Answer 1

I think Weinberg makes this much clearer in his book on quantum field theory. The point is that with the full Hamiltonian $H=H_0+V$, there is in general no way of defining particle states. To define scattering, we assume that at in the far past we have particles defined by the free Hamiltonian $H_0$, which are so far apart that they cannot interact. As time evolves, these particles can come together and interact in nontrivial ways as long as in the far future, they separate far enough that they don't interact and we can identify them with particles again.

In this description, there are two regimes where particles are well-defined, i.e. at $t\to\mp\infty$. Hence why we define the states $\Psi_{\alpha}^{\pm}$. These are states which are eigenstates of the full Hamiltonian $$H\Psi_{\alpha}^{\pm}=E_{\alpha}\Psi_{\alpha}^{\pm}$$

But somehow has to correspond to the free particle states of $H_0$ at $t\to\mp\infty$. Accordingly, we define these free particle states $\Phi_{\alpha}$ to be eigenstates of $H_0$

$$H_0\Phi_{\alpha}=E_{\alpha}\Phi_{\alpha}$$

Notice that implicit here is the assumption that $H$ and $H_0$ have the same spectrum.

Now, an observer $\mathcal{O}$ seeing a state $\Psi$ will see the state $\exp(-iH\tau/\hbar)\Psi$ after $\tau$ time has elapsed. Of course, since $\Psi_{\alpha}^{\pm}$ is an energy eigenstate, it is not localized in time. So in order to make a statement about how these states behave at $t\to\mp\infty$, we must consider a wave packet $\int d\alpha g(\alpha)\Psi_{\alpha}^{\pm}$ to localize it.

$$\exp(-iH\tau/\hbar)\int d\alpha g(\alpha)\Psi_{\alpha}^{\pm}=\int d\alpha g(\alpha)\Psi_{\alpha}^{\pm}e^{-iE_{\alpha}\tau/\hbar}$$

Now that we have a state which is localized in time, we define $\Psi_{\alpha}^{\pm}$ so that as we take $\tau\to\mp\infty$, the state is a wave packet of free particle states

$$\int d\alpha g(\alpha)\Psi_{\alpha}^{\pm}e^{-iE_{\alpha}\tau/\hbar}\to\int d\alpha g(\alpha)\Phi_{\alpha}e^{-iE_{\alpha}\tau/\hbar}$$

You should understand this to mean that the r.h.s. is the leading contribution as $|\tau|$ becomes large, not the actual limit.

Hopefully this clears up your confusion, let me know if you still are confused I can try to accommodate.