Quantum Mechanics-II

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Chapter#4: Scattering Theory

All exercise S.Qs of Chapter#4: Scattering Theory of Quantum Mechanics-II. We have arranged all S.Qs for BS/MSc Physics students.

Q.4.1 why differential area d\sigma and solid angle d\Omega directly proportional to each other?

Answer: Particles incident within an infinitesimal patch of cross- section area  will scatter into a correspond infinitesimal solid angle .

So;

    \[d\sigma\propto d\mathrm{\Omega}\]

 d\mathrm{\Omega} is the vector solid angle subtended by surface area  about the target particles incident in the area  scattering into solid angle

Q.4.2 what is role of impact parameter b in scattering?

Answer: Impact parameter b is small then scattering angle θ will be large.

Q.4.3 Show that total cross section area  depends over all solid angle?

Answer: The total scattering cross- section is the integral of D(θ) over all solid angles.

    \[D\left(\theta\right)=\ \frac{d\sigma}{d\mathrm{\Omega}}\]

    \[d\sigma=D\left(\theta\right)d\mathrm{\Omega}\]

    \[\int{d\sigma=\ \int D\left(\theta\right)d\mathrm{\Omega}}\]

    \[\sigma=\ \int D\left(\theta\right)d\mathrm{\Omega}\]

Q.4.4 Define incident and scattered density.

Answer: Incident density:    

    \[density= \frac{mass}{volume}\]

Number of incident particle per unit area per unit time.

    \[J_{inc}=\ \frac{dN}{d\sigma}\]

Scattered density:

Number of scattered particles per unit area per unit time;

    \[J_{sca}=\ \frac{dN}{d\mathrm{\Omega}}\]

Q.4.5 Define the relation between differential scattering and scattering amplitude.

Answer: Differential scattering is D(\theta) and scattering amplitude is f(\theta) .

    \[D\left(\theta\right)=\ \frac{d\sigma}{d\mathrm{\Omega}}\ \ \ \ \ \ \ ;\ \ \ \left|f(\theta)\right|2 =\ \frac{d\sigma}{d\mathrm{\Omega}}\]

    \[\left|f(\theta)\right|2=D(\theta)\]

Differential scattering cross- section is equal to modulus square or absolute square of scattering wave amplitude f(\theta).

Q.4.6 Differentiate between incident and scattered wave function.

Answer: Incident wave function:

    \[\mathrm{\Psi}_{inc\ }=\ {Ae}^{ikz}\]

Scattered wave function:

    \[\mathbf{\mathrm{\Psi}_{sca}=Af\left(\theta\right)\frac{e^{ikr}}{r}}\]

Q.4.7 Differentiate between spherical Bessel and Hankel functions.

Answer: Spherical Bessel function:

\mathbf{\frac{d^2R}{{dp}^2}+\ \frac{2}{p}\ \ \frac{dR}{dp}+\left[1-\frac{l(l+1)}{p^2}\right]\ R=0} \because p=kr

This is known as the spherical Bessel equation, solution to this equation and first kind is spherical Bessel function.

    \[j_{l\ }\left(P\right)=({-P)}^l\ \left(\frac{1\ }{P\ \ }\ \frac{d}{dp}\right)\ ^l (\frac{sinP}{P})\]

    \[U\left(r\right)=\ {Ae}^{ikr}=A_rj_{l\ }(kr)\]

Hankel function:

The combination of sine function and cosine function of the 1D free particle.

    \[h_l^1\left(kr\right)=j_l\left(kr\right)+in_l(kr)\]

This is known as spherical hankel function.

Q.4.8 Differentiate between partial wave and Born approximation.

Partial wave:

    \[\mathrm{\Psi}\left(r,\ \theta,\ \emptyset\right)=A\left[e^{ikz}+\sum_{l=0}^{\infty}{C_lh_l^1(kr)Y_l^m(\theta,\ \emptyset)}\right]\]

As V(r) is spherically symmetric potential, scattered wave function must not depend on  \emptyset,\ hence\ m=0\ and\ Y_l^m\left(\theta,\ \emptyset\right)=\sqrt{\frac{2l+1}{4\pi}\ } P_l\left(cos\theta\right).

Also, for every large r;

    \[\mathrm{\Psi}\left(r,\ \theta\right)=A\left[e^{ikz}+\sum_{l=1}^{\infty}{C_l({-i)}^{l+1}\frac{e^{ikr}}{kr}}\ \sqrt{\frac{2l+1}{4\pi}}\ \ P_l\ (cos\theta)\right]\]

This equation is a partial wave.

Born approximation:

In born approximation, V(r) is localized about r = 0 (Potential drops to zero outside some finite region) and the incoming plane wave is changed slightly by the weak potential (incident and scattering particles or wave are in plane wave states) because the potential is weak.

Q.4.9 Discuss the validity of born approximation.

Answer: The born approximation is valid for large incident energies or sufficiently weak scattering potentials. That is, when the average interaction energy between the incident particle and the scattering potential is much smaller than the particle incident kinetic energy, the scattered wave can be considered to be a plane wave.

Q.4.10 Differentiate between Lab and CM frame.

Answer: Lab frame:

Thelaboratory frame is the frame whereby the positions and velocities are measured with respect to the C.M frame.

C.M frame:

The C.M frame is the frame whereby the positions and velocities are measured with respect to the C.M frame.

Q.4.11 Why scattered spherical wave must carry a factor 1/r? 

Answer: In \psi_{sc}={A\ f\left(\theta\right)\ e}^{ikr} as  approaches to zero ( ) the scattering probability also approaches to zero. That’s why there must be a factor \frac{1}{r} to avoid the scattering probability to become zero.

    \[\psi_{sc}=A\ f\left(\theta\right)\frac{e^{ikr}}{r}\]

Q.4.12 What are the limitations of born approximation?

Answer: Born approximation is only valid for weak scattering potentials and large incident energies. Born approximation cannot be applied for strong potentials.

Q.4.13 What is modulation factor?

Answer: Modulation factor also called scattering amplitude, is amplitude of the outgoing spherical wave depending on scattering angle . It tells the probability of scattering in each direction , and is related to the differential scattering cross section .

Q.4.14 What is meant by elastic scattering?

Answer: In this, the kinetic energies of the particles remain conserved in CM (center of mass) frame but due to interaction with other particle their direction of propagation must be changed.

Q.4.15 What is difference between classical and quantum scattering?

Answer:

Classical scatteringQuantum scattering
It deals with particlesIt deals with the fields of quantum particles
Particles collide with each other exchanging energy or momentum Fields interact with each other, but there is always a separation between then, called impact factor.
For example, collision of billiard ballsFor example, electron diffraction, Compton scattering

Q.4.16 Prove that U_{(r)}\approx\ e^{ikr}

Answer: By Schrödinger wave equation:

    \[-\frac{\hbar}{2m}\frac{d^2U}{dr^2}\ +\left[V_{\left(r\right)}+\frac{\hbar^2}{2m}\frac{l\left(l+1\right)}{r^2}\right]U=EU\]

At large , potential becomes weak also centrifugal term (\frac{\hbar^2}{2m}\frac{l\left(l+1\right)}{r^2}) becomes negligible

    \[\frac{d^2U}{dr^2}\approx-\frac{2mE}{\hbar^2}U\]

    \[\frac{d^2U}{dr^2}\approx{-k}^2U\]

The solution of this 2nd order homogenous differential equation

    \[U_{(r)}=Ce^{ikr} + De^{-ikr}\big m\]

as D\rightarrow 0 for incoming wave

    \[U_{\left(r\right)}\approx Ce^{ikr}\]

    \[U_{(r)}\approx\ e^{ikr}\]

Since  is calculated by normalization, it is also negligible

Q.4.17 At large incident energies, scattered wave can be a plane wave why?

Answer: When the average interaction energy between the incident particle and the scattering potential is much smaller than the particle’s incident kinetic energy, the scattered wave can be a plane wave.

Q.4.18 Why we use Hankel function to solve Bessel function equation?

Answer: To solve the equation

    \[\frac{d^2U}{dr^2}-\frac{l\left(l+1\right)}{r^2}U=-k^2U\]

This equation is solved by using spherical Hankel function, because we need linear combinations analogous to (outgoing) and  (incoming)

    \[h_l^{\left(1\right)}\left(kr\right)=j_l\left(kr\right)+in_l\left(kr\right)\]

    \[h_l^{\left(2\right)}\left(kr\right)=j_l\left(kr\right)-in_l\left(kr\right)\]

When  becomes large, 1st and 2nd kind of Hankel functions goes like  \frac{e^{ikr}}{r} and \frac{e^{-ikr}}{r} .


Q.4.19 Separate radial and spherical part in the given equation   \psi_{(r,\theta,\phi)}=R\left(r\right)\ Y_l^{m_l}(\theta,\phi)         

Answer:

  1. R(r) is radial part
  2.  Y_l^{m_l}(\theta,\phi)is spherical harmonic part

Q.4.20 What do you mean by differential cross-section, mathematically?

Answer: Differential cross section is the ratio of infinitesimal patch of cross section area  to the corresponding infinitesimal solid angle .

    \[d\sigma\propto d\Omega\]

    \[d\sigma=D\left(\theta\right)d\Omega\]

    \[D\left(\theta\right)=\frac{d\sigma}{d\Omega}\]



Differential cross section has the dimensions of cm2/rad.

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