Finite Spherical Square Well

Hello, everyone, in order to calculate deuteron by Hartree-Fock method, I need a basis. The basis of infinite spherical square well is too “rigid”, that it has to “extension” to non-classical region. Beside of the basis of Wood-Saxon potential. The finite spherical square well is a good alternative. The radial equation is basically the same as infinite spherical square well.

The potential is

$\displaystyle V(r) = -|V_0|, r\leq a$

$\displaystyle V(r) = 0, r > a$

Within the well, the wave vector is

$\displaystyle k = \sqrt{\frac{2m}{\hbar^2} (|V_0 |-|E|) }$

, outside the well, the wave vector is

$\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2} |E| }$

The solution is spherical Bessel function. Since the Bessel function of the first and second kind are oscillating like sin or cosine function. To form a decay function when $r > a$, we need the Hankel function with complex argument.

$\displaystyle h_n( i \kappa r) = h_n(x) = - (i x)^n \left( \frac{1}{x} \frac{d}{dx}\right)^n \left(\frac{\exp{(-x)}}{x} \right)$

To make it real, we need a factor $i^n$.

The boundary conditions are continuity and differential continuity.

$\displaystyle j_l(ka) = A i^l h_l(i\kappa a)$

$\displaystyle \left(\frac{d}{dr}j_l(kr)\right)_{r=a} = A i^l \left(\frac{d}{dr}h_l(i\kappa a) \right)_{r=a}$

These two conditions solved for two parameters $A$ and $E$. However, I cannot find an analytical solution to the energy $E$.

If the potential depth is 60 MeV for proton. Radius is 1 fm for a light nuclei. Set $\hbar = 1, m = 1$.

The result is follow,

The 1st column is 1s, 1p, 1d, 1f, and 1g. The 2nd column is 2s, 2p, 2d, 2f, and 2g. The 3rd column is 3s, 3p, and 3d.

By compare with infinite square well,

The energies are lower in finite well, because the wave functions can spread-out to non-classical region, so that the wave length is longer and energy is lower.

In this example, the 3d and 2g orbits are bounded (of course, all orbits in infinite well are bound). This is not because of the depth of the well, but the boundary of the well. In other word, to bring down an orbit, the wave function has to spread out, that is connected with the neutron-halo.

Wave function in momentum space

The wave function often calculated in spatial coordinate. However, in experimental point of view, the momentum distribution can be extracted directly from the experimental data.

The conversion between momentum space and position space is the Fourier transform

$\displaystyle \phi(\vec{k}) = \frac{1}{\sqrt{2\pi}^3} \int Exp\left(-i \vec{k}\cdot \vec{r} \right) \phi(\vec{r}) d\vec{r}$

Using the plane wave expansion

$\displaystyle \exp(i k\cdot r) = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l(\hat{k}\cdot\hat{r})$

or

$\displaystyle \exp(i k\cdot r) = 4\pi \sum_{l=0}^\infty \sum_{m=-l}^{l} i^l j_l(kr) Y_{lm}(\Omega_k) Y_{lm}^{*}(\Omega_r)$

Thus,

$\displaystyle \phi(\vec{k}) = \frac{4\pi}{\sqrt{2\pi}^3} \sum_{l=0}^\infty (-i)^l \sum_{m=-l}^{l} \int j_l(k r) Y_{lm}(\Omega_k) Y_{lm}^*(\Omega) \phi(\vec{r}) r^2 dr d\Omega$

where $j_l (x)$ is spherical Bessel function. Usually, due to conservation of angular momentum, the angular part can be separated from the spatial part. Let assume the wave function in position space is

$\phi(\vec{r}) = \psi(r) Y_{l_r m_r}(\Omega)$

$\phi(\vec{k}) = \psi(k) Y_{l_k m_k}(\Omega_k)$

Then we have

$\displaystyle \psi(k) = \frac{4\pi}{\sqrt{(2\pi)^3}} \int j_l(k r) \psi(r) r^2 dr$

$\displaystyle \phi(\vec{k}) = \psi(k) (-i)^l Y_{lm}(\Omega_k)$

where $l = l_r = l_k, m=m_r = m_k$, due to the orthogonality of spherical harmonics.

For s, p, d, f-state, the spherical Bessel function is

$\displaystyle j_0(x) = \frac{\sin(x)}{x}$

$\displaystyle j_1(x) = \frac{\sin(x)}{x^2} - \frac{\cos(x)}{x}$

$\displaystyle j_2(x) = \sin(x)\frac{2-x^2}{x^3} - \cos(x)\frac{3}{x^2}$

$\displaystyle j_3(x) = \sin(x)\frac{15-6x^2}{x^4} - \cos(x)\frac{15-x^2}{x^3}$

For Hydrogen-like wave function, the non-normalized momentum distribution is

$\displaystyle \psi_{10}(k) = \frac{4 Z^{5/2}}{(k^2 + Z^2)^2}$

$\displaystyle \psi_{20}(k) = \frac{16 \sqrt{2}Z^{5/2}(4k^2-Z^2)}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{21}(k) =\sqrt{\frac{2}{3}}\frac{64 k Z^{7/2}}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{30}(k) = \frac{36 \sqrt{3} Z^{5/2} (81k^3-30k^2Z^2+Z^4)}{(9k^2 + Z^2)^4}$

$\displaystyle \psi_{31}(k) = \frac{144 \sqrt{6} Z^{7/2} (9k^3-kZ^2)}{(9k^2 + Z^2)^4}$

Here is the plot for momentum distribution $(\psi_{nl}(k))^2 k^2$.

It is interesting that, the number of node decrease with higher angular momentum. But be-aware that it is only in atomic case, not a universal.

The higher the principle quantum number, the smaller of the spread of momentum. This is because, the spread of position wave function getting larger, and the uncertainty in momentum space will be smaller. This is a universal principle.

We also plot the Hydrogen radial function in here $\psi(r)^2 x^2$, for reference,

Fermi and Gamow-Teller Transition

The beta decay is caused by the weak interaction. The weak interaction is very short range, because the mediate particles, the $W^{\pm}$ and $Z^0$ bosons are 80 GeV and 91 GeV respectively. The effective range is like $10^{-3}$ fm. So, the interaction can assumed to be a delta function and only the coupling constant matter. The Fermi coupling constant is

$1.17 \times 10^{-11} (\hbar c)^2~ \mathrm{MeV^2}$

The fundamental process of beta decay is the decay of quark.

$\displaystyle u \xrightarrow{W^+} d + e^+ + \nu_e$

Since a pion is made from up and down quark, the decay of pion into position and electron neutrino is also due to weak interaction.

The Hamilton of the beta decay is

$\displaystyle H_w(\beta^{\pm})=G_V \tau_{\mp} + G_A \sigma \tau_{\mp}$

where $G_V$ is the vector coupling constant, the term is called Fermi transition. The $\tau_{\pm}$ is the isospin ladder operator. The beta+ decay changes the isospin from +1/2 (neutron) to -1/2 (proton). The $G_A$ is the axial coupling constant, the term is called Gamow-Teller transition. $\sigma$ is spin operator. Because of this operator, the Gamow-Teller transition did not preserve parity.

The $G_A$ is different from $G_V$, which is caused by the effect of strong interaction. The Goldberger-Trieman relation

$\displaystyle g_A = \frac{G_A}{G_V} = \frac{f_\pi g_{\pi N}}{M_N c^2} = -1.3$

where $f_\pi \sim 93~\textrm{MeV}$ is the pion decay constant. $g_{\pi N} \sim 14 \times 4\pi$ is the coupling constant between pion and nucleon.  This, we can see the effect of the strong interaction, in which pion is the meson for strong nuclear force.

The transition probability can be estimated by Fermi-Golden rule

$\displaystyle W(p_e)=\frac{2\pi}{\hbar}|\left< \psi_f|H|\psi_0\right> |^2 \rho(E_f)$

the final state wavefunction

$\displaystyle \left|\psi_f\right> = \frac{1}{\sqrt{V}} e^{ik_e r} \frac{1}{\sqrt{V}} e^{ik_{\nu}r} \left|j_f m_f\right>$

$\displaystyle e^{ikr} = \sum \limits_{L}\sqrt{4\pi (2L+1)} i^L j_L(kr) Y_{L0}(\theta)$

using long wavelength approximation, the spherical Bessel function can be approximated by the first term.

$\displaystyle j_L(kr) \sim \frac{(kr)^L}{(2L+1)!!}$

$\displaystyle \left| \psi_f\right>=\frac{1}{V}(1 + i \sqrt{\frac{4\pi}{3}} Y_{10} + ...) \left|j_f m_f\right>$

The first term 1, or L=0 is called allowed decay, so that the orbital angular momentum of the decayed nucleus unchanged. The higher order term, in which the weak interaction have longer range has very small probability and called L-th forbidden decay.

The density of state is

$\displaystyle \rho(E_f) = \frac{V}{2\pi^2 \hbar^7 c^3} F(Z,E_e)p_e^2 (E_0-E_e) ( (E_0-E_e)^2-(m_{\nu} c^2)^2)^2$

where the $F(Z, E_e)$ is the Fermi function.

The total transition probability is the integration with respect to the electron momentum.

$\displaystyle W = \int W(p_e) dp_e = \frac{m_e^5 c^4}{2 \pi^3 \hbar^7} f(Z,E_0) |M|^2$

where $f(Z,E_0)$ is the Fermi integral. The half-life

$\displaystyle T_{1/2} = \frac{\ln{2}}{W}$

To focus on the beta decay from the interference of the density of state, the ft-value is

$\displaystyle ft = f(Z,E_0) T_{1/2} =\frac{2\pi^3\hbar^7}{m_e^5 c^4} \frac{\ln{2}}{|M|^2}$

The ft-value could be difference by several order.

There is a super-allowed decay from $0^{+} \rightarrow 0^{0}$ with same isospin, which the GT does not involve. an example is

$\displaystyle ^{14}\mathrm{O} \rightarrow ^{14}\mathrm{N} + e^+ + \nu_e$

The ft-value is 3037.7s, the smallest of known.

Fermi Gamow-Teller
$\Delta S=0$ $\Delta S=1$
$J_f=J_i + L$ $J_f=J_i + L+1$
$T_f=T_i + 1$

transition L $\log_{10} ft_{1/2}$ $\Delta J$ $\Delta T$ $\Delta \pi$
Fermi GT
Super allowed 3.1 ~ 3.6 $0^+ \rightarrow 0^+$ not exist 0 no
allowed 0 2.9 ~ 10 0 (0), 1 0, 1 ; $T_i=0\rightarrow T_f=0$ forbidden no
1st forbidden 1 5 ~ 19 (0),1 0, 1, 2 0,1 yes
2nd forbidden 2 10 ~18 (1), 2 2, 3 no
3rd forbidden 3 17 ~ 22 (2), 3 3, 4 yes
4th forbidden 4 22 ~ 24 (3), 4 4, 5 no

The () means not possible if either initial or final state is zero. i.e $1^{-} \rightarrow 0^+$ is not possible for 1st forbidden.

Scattering phase shift

for a central potential, the angular momentum is a conserved quantity. Thus, we can expand the wave function by the angular momentum wave function:

$\sum a_l Y_{l , m=0} R_l(k, r)$

the m=0 is because the spherical symmetry. the R is the radial part of the wave function. and a is a constant. k is the linear momentum and r is the radial distance.

$R_l(k,r) \rightarrow J_{Bessel} (l, kr )$

which is reasonable when r is infinite and the nuclear potential is very short distance. when r goes to infinity,

$J_{Bessel} (l,kr) \rightarrow \frac {1}{kr} sin( k r - \frac{1}{2} l \pi )$

for elastic scattering, the probability of the current density is conserved in each angular wave function, thus,

the effect of the nuclear potential can only change the phase inside the sin function:

$\frac{1}{kr} sin( k r - \frac {1}{2} l \pi +\delta_l )$

with further treatment, the total cross section is proportional to $sin^2(\delta_l)$.

thus, by knowing the scattering phase shift, we can know the properties of the nuclear potential.

for more detail : check this website