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qft-lsn-04008
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I Session Overview
Goals
To compute scattering cross section for particle;
Second quantized Schrodinger field in
External field (potential) \(V(x)\)
To learn application of first order perturbation theory.
Skills needed
The skill to compute matrix elements of product of field operators
between Fock states; Properties of normal product
Statement of the Problem
Assume that the interaction of Schrodinger field with an external source is given by
\[\Hsc{'}(t) = \int d\mathbf x \psi^\dagger (\mathbf x,t) V(\mathbf x) \psi(\mathbf x,t)\] Show that differential scattering cross section is give by \[\sigma(\theta)= |f(\theta, \phi)|^2\] where \(f\) is the scattering amplitude \[f(\theta,\phi)= \frac{m}{2\pi\hbar^2}\int d\mathbf x e^{i(\mathbf k_i-\mathbf k_f)\cdot\mathbf x} V(\mathbf x)\]
II Recall formulae needed
Expansion of fields
In all the examples here we work in the interaction picture. The field operator will be expanded as
\begin{equation}
u_n(\mathbf x,t)= u_n(\mathbf x) e^{-iE_nt/\hbar}
\end{equation}
First order perturbation result
\[\text{Transition amplitude } c_{fi} =-\frac{1}{i\hbar}\int_{t_1}^{t_2} \bra{f}H_I{'}(t)\ket{i} dt\]
Some Commutators
\begin{eqnarray}
[a,\psi^\dagger(x,t)] &=& u^*_\mathbf k(\mathbf x) \Rightarrow a\psi^\dagger(\mathbf x) = u_\mathbf k^*(\mathbf x) + \psi^\dagger(\mathbf x) a\\{} [\psi(x,t), a^\dagger] &=& u_\mathbf k(\mathbf x)\Rightarrow \psi(\mathbf x) a^\dagger= u_\mathbf k(\mathbf x) + a^\dagger \psi(\mathbf x) \end{eqnarray}
The initial and final states
Initial and final states have one particle. Corresponding wave functions are
\[u_i(\mathbf x)=\frac{e^{\mathbf k_i\dot{\mathbf x}}}{(2\pi)^{3/2}}, \quad u_f(\mathbf x)=\frac{e^{\mathbf k_f\dot{\mathbf x}}}{(2\pi)^{3/2}}.\]
The final states are the states with energy \(E_f\) and momentum \(\mathbf k_f\), with \(E_f=\frac{\hbar^2 k_f^2}{2m}\), and momentum vector in the solid angle \(d\Omega\).
\[u_i(\mathbf x)=\frac{e^{\mathbf k_i\dot{\mathbf x}}}{(2\pi)^{3/2}}, \quad u_f(\mathbf x)=\frac{e^{\mathbf k_f\dot{\mathbf x}}}{(2\pi)^{3/2}}.\]
The final states are the states with energy \(E_f\) and momentum \(\mathbf k_f\), with \(E_f=\frac{\hbar^2 k_f^2}{2m}\), and momentum vector in the solid angle \(d\Omega\).
Cross section
The differential cross section for potential scattering is given by
\begin{equation} \sigma(\theta,\phi) d\Omega = \frac{w_{fi}}{J_\text{inc}} \end{equation} where \(w_\text{fi}\) is the transition probability per unit time for scattering into solid angle \(d\Omega\).\\ \(\vec{J}_\text{inc}\) is calculated from probability current density for wave function \(u_{\mathbf k i}(x)\) and is given by \begin{equation} \vec{J}_\text{inc} = \frac{1}{(2\pi)^{3/2}} \frac{\hbar \mathbf k_i}{m}. \end{equation}
III Solution
Scattering-external-source-Schrodinger-field
V Useful Tips
It will be seen that squaring gives
\[(2\pi) \delta(\omega_{fi}) \stackrel{\text{square}}{\longrightarrow}(2\pi)^2 \delta(0) \delta(\omega_{fi})\] and computing transition probability per unit volume amounts to dropping the factor \((2\pi) \delta(0)\): \[\boxed{(2\pi) \delta(\omega_{fi}) \stackrel{\underbrace{\text{square}}}{\longrightarrow} (2\pi)^2 \delta(\omega_{fi}) \delta(\omega_{fi})= (2\pi)^2 \delta(0) \delta(\omega_{fi}) \stackrel{\underbrace{\text{per unit time}}}{\longrightarrow}(2\pi) \delta(\omega_{fi})}\]
This replacement can be intuitively understood as follows
\begin{eqnarray} (2\pi) \delta(\omega) &=& \int _{-\infty}^{\infty} e^{i\omega t} \,\, dt\\ &=& \lim_{T\to\infty} \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\\ (2\pi) \delta(0) &=& \lim_{T\to\infty} \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\big|_{\omega=0}\\ \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\big|_{\omega=0}&=& \int_{T/2}^{T/2} \,dt\\ &=& T. \end{eqnarray}
Hence taking per unit time amounts to replacement
\[ (2\pi)\delta(\omega)\big|_{\omega=0} \to 1\] }
The above sequence of statement cannot be
justified, a complete argument has however been given earlier.
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