Why the cross section can be obtained directly from the stationary scattering states?

Question

I'm currently studying scattering theory in the book Quantum Mechanics, Vol. 2 by Cohen-Tannoudji. In the book the author deduces that to find the number of particles detected far from the target at a position described by $(\theta,\phi)$ within the area $d\Omega$ we just need to find the scattering cross section $\sigma(\theta,\phi)$, so that $dn = F_i \sigma(\theta,\phi) d\Omega$, being $F_i$ the intensity of the beam.

The author then states:

In order to describe in quantum mechanical terms the scattering of a given incident particle by a potential $V(\mathbf{r})$, it is necessary to study the time evolution of the wave packet representing the state of the particle.

Actually, to simplify the calculations, we are going to base our reasoning directly on the stationary states and not on wave packets.

In that case he considers a fixed energy $E$ and looks for an eigenstate $|\varphi\rangle$ of $H$ with this energy. The evolution of this state is obviously $|\varphi(t)\rangle = e^{-iE/\hbar t}|\varphi\rangle$. The author states that this is the same kinetic energy of the incident particle before it reaches the zone of influence of the target.

The author then states regarding this procedure:

Therefore, we shall specify, using wave packet properties in an intuitive way, the conditions that must be imposed on the solutions of $H|\varphi\rangle = E|\varphi\rangle$ if they are to be used in the description of a scattering process. We shall call the eigenstates of the Hamiltonian which satisfy these conditions stationary scattering states.

Finally to compute $\sigma$ and hence $dn$ the author uses the probability currents. He consider $\mathbf{J}_i$ the incident current, considering the wavefunction to be $e^{ikz}$.

The whole point of my question is that the author then simply picks the wavefunction $\varphi(\mathbf{r})=\langle \mathbf{r}|\varphi\rangle$ associated to the stationary scattering state with energy $E$, calls the corresponding probability current $\mathbf{J}_d$ and says

Similarly, the number $dn$ of particles which strike the opening of the detector per unit time is proportional to the flux of the vector $\mathbf{J}_d$ across the surface $dS$ of this opening:

$$dn = C\mathbf{J}_d\cdot dS.$$

From this the author obtains $\sigma$. My whole point here is:

Before reading this, the most natural approach to the problem in my opion would be: knowing the initial state, we evolve it in time according to the Hamiltonian $H = H_0 + V$ in the zone of influence of the potential. With knowledge of the time evolution of the initial state we compute the probability of finding the incident particle inside the area $d\Omega$ located at $(\theta,\phi)$ far from the target.

That is: we follow the evolution of the particle to see what is happening with it.

The author's approach, which is as I found out, the standard approach, bases itself just on the so called stationary scattering states. That is, instead of following what happens to the state of the particle, we end up finding the solution on certain eigenvectors of the Hamiltonian.

What is the reasoning behind this? Why knowledge of just some eigenstates of the Hamiltonian is enough to find the scattering cross section?

Answer 1

To understand the method referred above one can consider the following:

I. one can note that the scattering process being treated is time-independent, and can give steady-state scattering currentsindependent of time.

II.For elastic scattering the particle energy is fixed and well defined.

III.The evolution of the system is completely governed by the positive energy solutions to the energy eigenvalue equation .

This naturally drives us to ask whether or not there exist stationary solutions to the energy eigenvalue equation which have asymptotic properties ?corresponding to the experimental situation of interest ?

The answer is yes and the solutions of interest are referred to as stationary scattering states of the associated potential V (r):

To understand these states one can consider the one dim. analogy of a free particle incident upon a potential barrier.

For this situation, there exist solutions in which the wave function to the left of the barrier is a linear combination of a right-going (incident) and left-going (scattered) wave,

while the wave function to the right of the barrier contains a part that corresponds to the transmitted or “forward scattered” part of the wave.

We note that experimentally, the wave function in the barrier region is inaccessible, and the only information that we can obtain is by measuring the relative magnitudes of the forward and backward scattered waves.....

Here the form of scattered wave function satisfies the asymptotic nature of the wave function with the scattering amplitude as coefficient representing the scattered intensity . Its a modified plane wave.

The fact is that asymptotically the radial component is the main contributor the (Phi , theta ) part of J(s) current density become negligible compared to it.

The modulus square of scattering amplitudes are straight way related to differential scattering cross sections.(as we would get in partial wave analysis of the scattering process)

The details can be seen at >http://web.mst.edu/~parris/QuantumTwo/Class_Notes/Scattering.pdf

Answer 2

The answer to this question fundamentally boils down to "linear algebra," or perhaps "signals and systems."

It is a well known result of, say, Fourier analysis, that a signal can be broken into components that depend only on the context and medium of the signal and not on the signal itself. For example, the classic example of breaking a violin note down into a sum of resonant frequencies that depend on the shape and length of the instrument, or in a more complex example, frequency-dependent attenuation across a reactant wire.

It so happens that in quantum mechanics, we can give these base resonances a physical meaning - particles. In the end, however, the principle is the same: a function of probability in respect to time can be broken down to a specific combination of general vibrational modes, which are much simpler to manipulate.