TF DerivationOfCoulombForce

Poisson's Equation

$\nabla^2 \phi(\vec{\xi}) = - \frac{\rho}{\epsilon_0} =- \frac{e}{\epsilon_0} \delta(\vec{\xi})$

Fourier Transform of Poisson's Equation

$\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \nabla^2 \phi(\vec{\xi})dV = - \frac{1}{(2 \pi)^{3/2}} \frac{e}{\epsilon_0} \int e^{-i \vec{k} \cdot \vec{\xi}}\delta(\vec{\xi}) dV$

$\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \cdot (\vec{\nabla} \phi(\vec{\xi}))dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)$

Product rule for dervatives

$\frac{1}{(2 \pi)^{3/2}} \int \left \{ \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) - (\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi) \right \} dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)$

Gauss' Theorem:

$\int \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) dV = \oint_S e^{-i \vec{k}\cdot \vec{\xi}} \vec{\nabla}\cdot d\vec{A}$

Definition of derivative:

$(\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi ) = \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k}}) - \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}}$

Substituting

$\frac{1}{(2 \pi)^{3/2} } \left \{ \oint_S e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi \cdot d\vec{A} - \int \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV + \int \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}} dV \right \} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}$

Gauss' Low:

$\int \vec{\nabla}\cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV = \oint_S \phi \vec{\nabla} e^{-i k \xi } \cdot d\vec{A}$

$\frac{1}{(2 \pi)^{3/2} } \left \{\oint_S \left \{ e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi - \phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} \right \} \cdot d\vec{A} + \int \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}} dV \right \} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}$

$\frac{1}{(2 \pi)^{3/2} } \int \phi (-ik) (-ik) e^{-i \vec{k} \cdot \vec{\xi}} dV = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}$

$-k^2 \frac{1}{(2 \pi)^{3/2} } \int \phi(\xi) e^{-i \vec{k} \cdot \vec{\xi}} dV_{xi} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}$

$-k^2 \phi(k) = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}$

1.) Coulomb $\phi(k) = \frac{e}{(2 \pi)^{3/2} \epsilon_0} \frac{1}{k^2}$ = potential in "k"(momentum) space

To find the potential in "coordinate" $(\xi)$ space just inverse transform

$\phi (\xi) = \frac{1}{(2 \pi)^{3/2} } \int e^{+ i \vec{k} \cdot \vec{\xi}} \phi (k) dV_k$

$= \frac{1}{(2 \pi)^{3/2} } \int e^{i \vec{k} \cdot \vec{\xi}} \frac{e}{2 \pi)^{3/2} \epsilon_0} \frac{1}{k^2} dV_k$

$= \frac{e}{(2 \pi)^{3} \epsilon_0} \int \frac{e^{i \vec{k} \cdot \vec{\xi}}}{k^2} dV_k$

$dV_k=k^2 sin{\theta}_k d{\theta}_k d{\phi}_k dk$

$=\frac{e}{(2 \pi)^{3} \epsilon_0} {{\int}_0}^{2\pi} d{\phi}_k {{\int}_0}^{\pi} d{\theta}_k {{\int}_0}^{\infty} dk \times k^2 sin{\theta}_k e^{i \vec{k} \cdot \vec{\xi}}$

$=\frac{e}{(2\pi)^2 \epsilon_0} {{\int}_0}^{\pi} {{\int}_0}^{\infty} sin{\theta}_k e^{ik \xi cos{\theta}_k} k^2 dk$

$u=cos\theta$

$du=sin\theta d\theta$

$\phi(\xi) = \frac{e}{4 {\pi}^2 \epsilon_0} {{\int}_0}^infty {{\int}_{-1}}^1 \frac{e^{ik\xi u}}{k^2} du k^2 dk$

$=\frac{e}{4 {\pi}^2 \epsilon_0} {{\int}_0}^infty \frac{e^{ik \xi} - e^{-ik\xi}}{ik\xi} dk$

$=\frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{i\xi} (i\pi) = \frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{\xi}$

$=\frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{|\vec{r} - \vec{r}^'|} =$ Coulomb potential

2) Nuclear potential

Consider the force field generated by a point source (nucleon) at location $\vec{r}$ from the origin of a coordinate system.

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Assume a particle of mass m is e charged to generate the field (In Coulomb force particle was m=o photon).

Definition of relativistic Energy:

$E^2=(mc^2)^2 + (cp)^2$

In terms of Hamiltonian

$-\hbar \frac{d^2}{dt^2} \phi(\vec{r}) = [ (mc^2)^2 + (\frac{c\hbar \vec{\nabla}}{i})^2 ] \phi (r)$

In a static case

$[(mc^2)^2 - c^2 {\hbar}^2 {\nabla}^2]\phi(r)=0$

$[ {\nabla}^2 - (\frac{mc}{\hbar})^2]\phi(r)=0$

Lets $\mu = \frac{mc}{\hbar} = \frac{(140 \frac{MeV}{c^2})c}{6.6 \times 10^{-16}eV \cdot s} (\frac{10^6 eV}{MeV})$

$=\frac{2.1 \times 10^23}{(3\times 10^8 m)} (\frac{10^{-15}m}{fm}) =\frac{0.7}{fm} (200 MeV fm) = 140 MeV$

$\hbar c=(6.6 \times 10^{-16} eV \cdot s) (3 \times 10^8 \frac{m}{s}) (\frac{10^{15} fm}{m}) (\frac{1 MeV}{10^6 eV}) = 198 MeV \cdot fm \approx 200 MeV \cdot fm$

$\mu = \frac{mc^2}{\hbar c} = \frac{mc^2}{200 MeV \cdot fm}$ $\Longrightarrow$ if $mc^2\approx 200 MeV$ then interaction length $\sim 1 fm$.

With the source term

$[{\nabla}^2 - {\mu}^2]\phi(\xi) = -{{\xi}_0}^' \delta(\xi)$

As seen before for Coulomb force

$\frac{1}{(2\pi)^{3/2}} \int e^{-ik \xi} [{\nabla}^2 - {\mu}^2] \phi (\xi) dV = -\frac{{{\xi}_0}^'}{(2\pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \delta (\xi) dV$

$[-k^2 -{\mu}^2]\phi(\vec{k}) = -\frac{{{\xi}_0}^'}{(2\pi)^{3/2}}$

$\Longrightarrow \phi(\vec{k}) = \frac{{{\xi}_0}^'}{(2\pi)^{3/2}} \frac{1}{k^2 + {\mu}^2}$

$\phi(\vec{\xi}) = \frac{1}{(2 \pi)^{3/2}} \int \frac{e^{+i \vec{k} \cdot \vec{\xi}}}{(2 \pi)^{3/2}} \frac{{{\xi}_0}^'}{k^2 + {\mu}^2} dV_k$ : inverse fourier transform

$= \frac{{{\xi}_0}^'}{(2\pi)^3} \int \frac{e^{+i \vec{k} \cdot \vec{\xi}}}{k^2 + {\mu}^2} k^2 sin\theta d\theta d\phi dk$

$\phi(\xi) = \frac{{{\xi}_0}^'}{(2\pi)^2} \int \frac{e^{i\vec{k} \cdot \vec{\xi}}}{(k^2 + {\mu}^2)} k^2 d[cos\theta] (2\pi) dk$

$= \frac{{{\xi}_0}^'}{(2\pi)^2} {{\int}_0}^{\infty} \frac{k^2}{k^2 + {\mu}^2} {{\int}_0}^{\pi} e^{ik\xi cos\theta} d[cos\theta] dk$

$= \frac{{{\xi}_0}^'}{(2\pi)^2} {{\int}_0}^{\infty} \frac{k^2}{k^2 + {\mu}^2} dk \frac{e^{ik\xi \mu}}{ik\xi} {|_{-1}}^1$

$= \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{1}{i\xi} {{\int}_0}^{\infty} \frac{k}{k^2 + {\mu}^2} (e^{ik\xi} - e^{-ik\xi})$

$= \frac{{{\xi}_0}^'}{(2\pi)^2}\frac{1}{i\xi} {{\int}_{-\infty}}^{\infty} \frac{e^{ik\xi}kdk}{k^2 + {\mu}^2}$

${{\int}_{-\infty}}^{\infty} \frac{e^{ik\xi}kdk}{k + i{\mu}}{k - i{\mu}} = i\pi e^{-\mu \xi}$

$\phi(\xi) = \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{e^{-\mu \xi}}{\xi}$

$\Longrightarrow$

$\phi(\vec{r}) = \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{1}{|\vec{r} - {\vec{r}}^'|}$

${\phi}_{EM}(\vec{r}) = \frac{e}{4\pi \epsilon_0} \frac{1}{|\vec{r} - {\vec{r}}^'|}$

Coupling constants are:

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Summary

There are now at least two forces which act between Nucleons, the Coulomb force and the Nucleon force. We can write the force in terms of a potential

$V_{EM} = \frac{e}{4\pi \epsilon_0} \frac{1}{|\vec{r} - {\vec{r}}^'|}$

$V_{Nuc} = \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{e^{-\mu |\vec{r} - {\vec{r}}^'|}}{|\vec{r} - {\vec{r}}^'|}$

$V_{tot} = \frac{1}{4\pi} \left \{ \frac{Q_1 Q_2}{\epsilon_0 e } - \frac{{\xi}_0}{\pi} e^{-\mu |\vec{r} - {\vec{r}}^'|} \right \} \frac{1}{|\vec{r} - {\vec{r}}^'|}$

$\mu=\frac{m_{\pi}c}{\hbar}=\frac{1}{R} \sim (1.5 fm)^{-1}$

C.) Deuteron

$({H_1}^2)$, $(D^2)$, (d) = a proton-neutron bound state

General properties: L=0 - orbital angular momentum

$J^{\pi} = 1^+$ - (Nuclear spin)

$\sqrt{r^2}=2.095 fm$ - Mean radius.

$B.E.=2.2246 MeV$ - Binding energy.

Non-relativistic Schrodiger solution

$B.E. = 2.2246 MeV \lt \lt 1.8 GeV m(H^2)c^2$

$\Longrightarrow$ weakly bound system

Instead of Dirac equation try 3-D Square Well Schrod. Eq. approximation for Deuteron wavefunction.

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$V(r) = -V_0$ when $r \leq R = 2.2$

$V(r) = 0$ $r\gt R$

Assume $\psi = Y_{00} \frac{V(r)}{r}$ : No angular dependence, only radial dependence.

Schrod. Equation

$-\frac{\hbar^2}{2m} {\nabla}^2 \psi + V(r)\psi = E \psi$

${\nabla}^2 = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} (sin\theta \frac{\partial}{\partial \theta}) + \frac{1}{r^2 sin^2 \theta} \frac{{\partial}^2}{\partial {\phi}^2}$

$-\frac{\hbar^2}{2m} Y_{00} \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial (U(r)/r)}{\partial r} + \frac{1}{r^2 sin\theta}\frac{U(r)}{r}(\frac{\partial}{\partial \theta} sin \theta \frac{\partial Y_{00}}{\partial \theta})$

$+ \frac{U(r)}{r^3 sin^2 \theta} \frac{{\partial}^2}{\partial {\phi}^2} Y_{00} + \frac{V(r)Y_{00} U(r)}{r} = \frac{E U(r)}{r}$

$\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} \frac{U(r)}{r} = \frac{1}{r^2} \frac{\partial}{\partial r} r^2 [\frac{1}{r}\frac{\partial U(r)}{r} - \frac{U(r)}{r^2}]$

$= \frac{1}{r^2} [\frac{\partial U(r)}{\partial r} + r \frac{\partial^2 U(r)}{\partial r^2} - \frac{2rU(r)}{r^2} - \frac{r^2}{r^2} \frac{\partial U(r)}{\partial r} - r^2 U(r)\frac{-2}{r^3}]$

$= \frac{1}{r}\frac{\partial^2 U(r)}{\partial r^2} + \frac{1}{r^2} \frac{\partial U(r)}{\partial r} - \frac{1}{r^2} \frac{\partial U(r)}{\partial r} - \frac{2U(r)}{r^2} + \frac{2U(r)}{r^2}$

$= \frac{1}{r}\frac{\partial^2 U(r)}{\partial r^2}$

$\frac{U(r)}{r^3 sin\theta}(\frac{\partial}{\partial \theta} sin\theta \frac{\partial}{\partial \theta})Y_{00} = \frac{U(r)}{r^3 sin\theta} l(l+1)Y_{00}$

$=\frac{U(r)}{r^3 sin\theta}[\frac{\partial}{\partial \theta} sin\theta \frac{\partial}{\partial \theta}]\frac{1}{\sqrt{4\pi}}=0$

$\frac{U(r)}{r^3 sin\theta} \frac{\partial^2}{\partial {\phi}^2} Y_{00} = 0$ : $Y_{00} = \frac{1}{\sqrt{4\pi}}$

Schrod. Equation becomes

$-\frac{\hbar^2}{2m} \frac{Y_{00}}{r} \frac{\partial^2 U(r)}{\partial r^2} + V(r) Y_{00}\frac{U(r)}{r} = E \frac{U(r)Y_{00}}{r}$

$\Longrightarrow$ for $r\leq R$ : $V(r)= -V_0$

$\frac{\partial^2 U(r)}{\partial r^2} + \frac{2m}{\hbar^2}(E+V)U(r) = 0$

for $r\gt R$ : $V(r)=0$

$\frac{\partial^2 U(r)}{\partial r^2} + \frac{2m}{\hbar^2} E U(r) = 0$

File:ImageTF 1.jpg

$\psi_{I}$

$\frac{\partial^2 U_I (r)}{\partial r^2} + {k_1}^2 U(r) = 0$ : ${k_1}^2 = \frac{2m}{\hbar^2}(E + V)$

$\Longrightarrow$ $U_I (r) = A sin(k_1 r) + Bcos(k_1 r)$ : spring simple harmonic motion

Boundary condition:

$\psi_I (r=0)=0$ $\Longrightarrow$ $B=0$

$\psi_{II}$

$\frac{\partial^2 U_{II} (r)}{\partial r^2} + {k_2}^2 U(r) = 0$ : ${k_2}^2 = \frac{2m}{\hbar^2} E \lt 0$

E<0 for bound states. Taking out " - " sign in ${k_2}^2$

$\frac{\partial^2 U_{II} (r)}{\partial r^2} - {k_2}^2 U(r) = 0$

New definition of ${k_2}^2$ : ${k_2}^2=\frac{2m}{\hbar^2}|E|$

$\Longrightarrow$ $U_{II}(r) = C e^{-k_2 r} + D e^{k_2 r}$

Boundary condition:$U_{II} (r=\infty)$ - finite $\Longrightarrow$ D=0

$U_I (r) = A sin(k_1 r)$

$U_{II} (r) = C e^{-k_2 r}$

Bounding condition

$U_I (r=R) = U_{II} (r=R)$

$\frac{\partial U_I}{\partial r} {\mid}_{r=R} = \frac{\partial U_{II}}{\partial r} {\mid}_{r=R}$

$\Longrightarrow$ $A sin(k_1 R) = C e^{-k_2 R}$

$A k_1 cos(k_1 R) = -C k_2 e^{-k_2 R}$

Dividing two equations $\Longrightarrow$

$\frac{tan(k_1 R)}{k_1} = - \frac{1}{k_2}$

$\Longrightarrow$ $tan(k_1 R) = - \frac{k_1}{k_2} = -\sqrt{\frac{\frac{2m}{\hbar^2}(V + E)}{\frac{2m}{\hbar^2}|E|}}$

But E<0 for bound states

$tan(k_1 R) = - \sqrt{\frac{(V - |E|)}{|E|}} = -X = -\frac{k_1}{k_2}$

$k_1 R = X k_2 R = X\beta$

$-X = tan (\beta X)$

Solving the ??? 59 eqution:

$X=-tan(\beta X)$

$\beta X = k_1 R$ $\Longrightarrow$ $\beta = \frac{k_1}{X}R = k_2 R = \sqrt{\frac{2m|E|}{\hbar^2}} R$

m = reduced mass

$\beta = \sqrt{\frac{2(\frac{938.98}{2} MeV) |E|}{(\hbar c)^2}} R$ :

$:=\sqrt{\frac{(938.98) MeV (2.224 MeV)}{ (197.3 \cdot MeV \cdot fm)^2}} R$

$=\sqrt{\frac{(938.98) \cdot (2.224) MeV^2}{(197.3)^2 MeV^2 \cdot fm^2}} (2.095 fm)$

$= 0.4853$

$|E| = |-2.224 MeV|$

Find X s.t. $X = -tan (0.4853 X)$

Using ?? 59 of graphing $\Longrightarrow$ $X=3.91 = \frac{k_1}{k_2}$

$k_2 = \sqrt{\frac{2m |E|}{\hbar^2}} = \frac{\beta}{R} = \frac{0.9853}{2.095 \cdot fm} = \frac{1}{fm}$

$k_1 = X k_2 = (3.931) (\frac{0.231}{fm}) = \frac{0.91}{fm} = \sqrt{\frac{2m (V - |E|)}{\hbar^2}}$

$V_0 = 36 MeV$

Spin and Parity

$(I^{\pi})$

66-78 pages

The shrodinger equation for this scattering:

$-\frac{\hbar^2}{2m} \nabla^2 \psi + \nabla \psi = E \psi$

In spherical coordinates this may be written as:

Let $\psi = \frac{U_l (r) Y_{lm}(\theta, \phi)}{r}$

$-\frac{\hbar^2}{2m} \frac{\partial^2 U_l (r)}{\partial r^2} + \frac{l(l+1) \hbar}{2m} \frac{U_l (r)}{r^2} + V U_l(r) = E U_l (r)$

or

$-\frac{\hbar^2}{2m} \frac{\partial^2 U_l (r)}{\partial r^2} + ( V + \frac{l(l+1) \hbar^2}{2m} \frac{1}{r^2}) U_l (r) = E U_l(r)$

General solution:

$\frac{U_l(r)}{r} = A_l J_l (kr) + B_l N_l (kr)$

where

$J_l(kr)$ = Bessel function

$N_l(kr)$ = Neiman function

1.) Distant scattering: r is large such that neutron "glances" off.

$J_l (kr) \approx \frac{sin(kr - l \pi /2)}{kr}$

$N_l (kr) \approx - \frac{cos (kr - l\pi /2)}{kr}$

$\Longrightarrow$ $\frac{U_l(r)}{r} = \frac{A_l sin(kr - \frac{l \pi}{2}) - B_l cos(kr - \frac{l \pi}{2})}{kr}$

$\frac{\psi_I (r)}{r} = A_{lI} \frac{sin(k_I r - \frac{l \pi}{2})}{k_I r}$

$k_I = \sqrt{\frac{2m (V + E)}{\hbar^2}}$

$U_I (r=0)$ = finite, no cosine term.

$\frac{U_{II}(r)}{r} = A_{l II} \frac{sin(k_{II}r - l \pi /2)}{k_{II}r} - B_{l II} \frac{cos(k_{II}r - l \pi /2)}{k_{II}r}$

$k_{II} = \sqrt{\frac{2mE}{\hbar^2}}$

Normalizing and simplifying $U_{II}(r)$

$\frac{U_{II}(r)}{r} = \sqrt{{A_{l II }}^2 + {B_{l II}}^2} [ \frac{A_{l II}}{\sqrt{{A_{l\pi}}^2 + {B_{l II}}^2}} \frac{sin(k_{II}r - l \pi /2)}{k_{II}r} - \frac{B_{l II}}{\sqrt{{A_{l II}}^2 + {B_{l II}}^2}} \frac{cos(k_{II}r - l \pi /2)}{k_{II}r} ]$

$cos (\delta_l) = \frac{A_{l II}}{\sqrt{{A_{l II}}^2 + {B_{l II}}^2}}$

$sin (\delta_l) = \frac{B_{l II}}{\sqrt{{A_{l II}}^2 + {B_{l II}}^2}}$

$\Longrightarrow$

$\frac{U_{II}(r)}{r} = \frac{\sqrt{{A_{l II}}^2 + {B_{l II}}^2}}{k_{II}r} \left \{ cos (\delta_l) sin(k_{II}r - l \pi /2) + sin ( \delta_l)cos(k_{II}r - l \pi /2) \right \}$

$+/- cosA sinB + sinA cosB = sin(A +/- B)$

$\frac{U_{II}(r)}{r} = \frac{\sqrt{{A_{l II}}^2 + {B_{l II}}^2}}{k_{II}r} sin (k_{II} r - l \pi /2 + \delta_l)$

$=C_l \frac{sin(k_{II} r - l \pi /2 + \delta_l)}{k_{II}r}$

$A_{l II}$ and $C_l$ are found by applying Boundary conditions:

$\psi_I (r=R) = \psi_{II} (r=R)$

$\frac{\partial \psi_I}{\partial r} \mid_{r=R} = \frac{\partial \psi_{II}}{\partial r} \mid_{r=R}$

Example

l=0 special case

$\psi = \frac{U_l (r)}{r} Y_{lm}(\theta, \phi)$ $\Longrightarrow$ $\psi_{l=0} = \frac{U(r)}{r}$

$\frac{U_I (r)}{r} = A_I \frac{sin(k_I r - 0 \pi /2)}{k_I r}$

${k_I}^2 = \frac{2m(V + E)}{\hbar^2}$

$\frac{U_{II} (r)}{r} = C \frac{sin(k_{II} r - 0 \pi /2 + \delta_0)}{k_{II} r}$

$k_{II}^2 = \frac{2m E}{\hbar^2}$

Apply Boundary Conditions:

$U_I (r=R) = U_{II} (r=R)$

$A_I sin(k_I R) = C sin(k_{II} R + \delta_0)$

$\frac{\partial \psi_I}{\partial r} \mid_{r=R} = \frac{\partial \psi_{II}}{\partial r} \mid_{r=R}$

$A_I k_I cos(k_I R) = C k_{II} cos(k_{II} R + \delta)$

$k_I$, $k_{II}$ and R are known. $\delta$ is unknown.

$\frac{U_I (R)}{ \frac{\partial U_I (r)}{\partial r} } = \frac{U_{II} (R)}{ \frac{\partial U_{II] (r)}{\partial r} }$

$\Longrightarrow$

$k_I cot(k_I R) = k_{II} cot (k_{II} R + \delta)$

$k_{II}^2 = \frac{2m E}{\hbar^2}$

$k_{II}^2 = \frac{2m E}{\hbar^2}$

If V=0 $\Longrightarrow$ $k_I = k_{II}$ $\Longrightarrow$ $\delta = 0$: Free neutron; No target; No phase shift.

image70_1

If V>0 $\Longrightarrow$ $k_I \gt k_{II}$:

image70

If V<0 $\Longrightarrow$ $k_I \lt k_{II}$ $\Longrightarrow$ $\delta \lt 0$: E>0; Neutron not bound.

If you measure the phase shift you can determine "sign" of V.

Cross-section:$\frac{d \sigma}{d \Omega} =$ differential X-section = $\frac{#scattered/ solid-angle}{#in / Area} [m^2]$

=probability of scattering into the solid angle $( d\Omega )$

$\sigma = \int (\frac{d\sigma}{d\Omega}) d\Omega =$ probability of scattering in any direction.

Let$j \equiv \frac{#particles}{area}$

$j_{scattered} = \frac{#Nucleons scattered}{area}$

From Q.M. $j = \frac{\hbar}{2mi}(\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) =$ particle current density

This comes from the continuity equation

$\frac{\partial \ro}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0$

Time dependent Shrodinger equation \Longrightarrow

$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar} H \psi$

$\frac{\partial \psi^*}{\partial t} = \frac{i}{\hbar} H \psi^*$

$\frac{\partial}{\partial t}\psi^* \psi =$ time derivative of the particle density

$\psi^* \frac{\partial \psi}{\partial t} + \psi \frac{\partial \psi^*}{\partial t}$

$\psi^* (-\frac{iH}{\hbar}\psi) + \psi (\frac{iH}{\hbar} \psi^*)$

$\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* H \psi - \psi H \psi^*] = 0$

If free particle $H = \frac{p^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial X^2}$

Cross-section

$\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* (- \frac{\hbar^2}{2m}) \frac{\partial^2 \psi}{\partial X^2} - \psi (-\frac{\hbar^2}{2m})\frac{\partial^2 \psi}{\partial X^2} ] = 0$

$\frac{\partial \psi^* \psi}{\partial t} - \frac{i\hbar}{2m} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0$

Check: $\frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) = \frac{\partial \psi^*}{\partial X} \frac{\partial \psi}{\partial X} + \psi^*\frac{\partial^2 \psi}{\partial X^2} - \frac{\partial \psi}{\partial X} \frac{\partial \psi^*}{\partial X} - \psi \frac{\partial^2 \psi^*}{\partial X^2}$