Measuring ANC using (d,p) transfer reaction

August 24, 2021

GoLuckyRyan Basic, papers reading , Leave a comment

The reference of this post is A. M. Mukhamedzhanov, F. M. Nunes, and P. Mohr, Phys. Rev. C 77, 051601(R) (2008).

The transfer amplitude is, in a simple sense,

\displaystyle T = \left< \phi_p \phi_{B} | V | \phi_d \phi_A \right> = \left< \phi_p \psi | V | \phi_d \right>, ~~~ \left< \psi \right| = \left<B|A\right>

where \psi is the quasi-particle wave function. The function can be separated into short-range and long-range part.

\displaystyle \psi(r) \xrightarrow{r \rightarrow \infty } C \frac{W(r)}{r}

The cross section of the reaction is

\displaystyle \sigma^2 = |T|^2 = | T_{int} + T_{ext} |^2 = \left( \int_0^R \phi_p^* \psi^* V \phi_d + C \int_R^\infty \phi_p^* \frac{W^*(r)}{r} V \phi_d \right)^2

The idea is, “If a reaction is completely peripheral, it is possible to extract the nuclear ANC from the normalization to the data without any single-particle ambiguity…… For sub-Coulomb transfer reactions, one is only sensitive to the asymptotic part of the neutron wave function, and the ANC can be extracted virtually without theoretical uncertainties.“, from the reference.

Experimentally, the cross section is measured for a give reaction. The quasi-particle state is approximated by a single-particle wave function from Woods-Saxon potential.

\displaystyle \psi(r) \xrightarrow{r \rightarrow \infty } C \frac{W(r)}{r}

\displaystyle \phi(r) \xrightarrow{r \rightarrow \infty } b \frac{W(r)}{r}

Thus,

\displaystyle \psi(r) \approx \sqrt{S} \phi(r), ~~~ C(b) = \sqrt{S(b)}b

In this sense, the interior is approximated by the single-particle wave function, then, the nuclear ANC will depend on the single-particle wave function, and also the Woods-Saxon potential.

However, if the reaction is completely peripheral, the internal contribution is zero, and the nuclear ANC is independent of the Woods-Saxon parameters, or the single-particle ANC. Or, the change of the spectroscopic factor and the change of single-particle ANC will cancel out.

I think zero internal contribution and cancelation of SF and b are difference concepts. And in the DWBA framework, how to compute the internal part that it is small ? The quasi-particle state in the internal part is approximated by the single-particle wave function, and the internal part is an integral of many things, how it will not change with difference Woods-Saxon parameters?

And there is a difference between insensitive and not-contribute. If the internal part is only insensitive to the Woods-Saxon parameters, it can still be non-zero. If it is non-zero, the nuclear ANC has an offset.

By measuring the reaction at difference energy, the offset can be checked.

Using ANC to calculate the bound state radius?

August 23, 2021

GoLuckyRyan Basic, papers reading , , Leave a comment

The idea is from these papers :

  1. F. Carstoiu et al., Phys Rev. C 63, 054310 (2001)
  2. E. T. Li et al., Chinese Phys. C 40, 114104 (2016)

Here is the summary of the idea.

The mean-square radius of an orbital from nucleon transfer reaction A = B + N is

\displaystyle \left<r^2 \right> = \frac{\int_0^\infty r^4 I^2(r) dr}{\int_0^\infty r^2 I^2_{AB}(r) dr}

where I_{AB}(r) is the quasi-particle from the overlap of nuclei A and B, such that I_{AB}(r) = \left<A | B \right> . The quasi-particle state is normalized to the spectroscopic factor

\displaystyle S = \int_0^\infty r^2 I^2_{AB}(r) dr

The mean-square radius is then

\displaystyle \left<r^2 \right> = \frac{1}{S}\int_0^\infty r^4 I_{AB}^2(r) dr

We can split the integral into 2 parts, a short range and long range. The long range part should be close to the Whittaker function as the nuclear potential is essentially zero. And the quasi-particle is proportional to the Whittaker function by the ANC

\displaystyle I_{AB}(r) \xrightarrow{r\rightarrow \infty} C \frac{W(r)}{r}

The mean-square radius break down to

\displaystyle \left<r^2 \right> = \frac{1}{S} \left( \int_0^{R_n} r^4 I_{AB}^2(r) dr + C^2 \int_{R_n}^\infty r^2 W^2(r) dr \right)

Since the ANC is not sensitive to the nuclear potential and considered to be more reliable, thus, the rms radius can be extracted/calculated using the ANC and the bound state wave function, which is approximated and calculated by Woods-Saxon potential.

Approximate the normalized orbital or the bound state wave function to be

\displaystyle \phi (r) \approx \frac{1}{\sqrt{S}} I_{AB}(r)

The mean-square radius is

\displaystyle \left<r^2 \right> \approx \int_0^{R_n} r^4 \phi^2(r) dr + b^2 \int_{R_n}^\infty r^2 W^2(r) dr = \int_0^\infty r^4 \phi^2(r) dr

The last step used the fact that the bound state wave function must be approached to the Whittaker function with the single-particle ANC. The last step eliminated the need for the radius R_n . Since the bound state wave function is calculated from (usually) a Woods-Saxon potential, which reproduces the separation energy, in DWBA framework, the rms radius is directly calculated from the DWBA.

In the two references, the authors fitted the potential width and diffusiveness parameters, which reproduced the separation energy. They claimed that they found the best, model insensitive parameters, and using the ANC, to deduce the rms radius.

The formula of the rms radius still depends on the spectroscopic factor and bound state wave function. The references minimizing the difference between the bound state wave function and the Whittaker function for the long-range

\displaystyle \chi^2 = \int_{R_n}^\infty \left( I_{AB}(r) - C \frac{W(r)}{r} \right)^2 \approx \int_{R_n}^\infty \left( \sqrt{S} \phi(r) - C \frac{W(r)}{r} \right)

If one can have an independent value of the nuclear ANC C, then the minimization of the \chi^2 force the single-particle wave function, and the Woods-Saxon parameters reproduce the long-range behaviour. And give a consistent spectroscopic factor with the nuclear ANC.

For 16N,

Energy [MeV]Spin-paritysingle-particle ANCRMS radius [fm]RMS radius from reference 2 [fm]
-28.820s1/219.692.13
-15.270p3/28.552.80
-12.200p1/25.982.86
-2.340d5/20.433.763.85 +- 0.31
-2.161s1/2-2.184.814.82 +- 0.42
The Woods-Saxon parameters are V_0 = -49.57 MeV, r_0 = 1.25 fm, a_0 = a_{so} = 0.67 fm, V_{so} = 20.2 MeV, r_{so} = 1.20 fm.

In above table, I calculate the RMS radius of 16N and compare to reference 2. They are consistent, which is no surprise. In reference 2, the nuclear ANC is deduced from spectroscopic factor, which is a fatal flaw in the logic. If the nuclear ANC is not independently deduced or based on spectroscopic factor, the \chi^2 is

\displaystyle \chi^2 \approx \int_{R_n}^\infty \left( \sqrt{S} \phi(r) - C(S_0) \frac{W(r)}{r} \right)

The S_0 indicate the nuclear ANC C has implicit dependence to the spectroscopic factor S_0, which depends on some Woods-Saxon parameters. The minimization is then nothing but forcing the fitting Woods-Saxon parameters to the Woods-Saxon parameters that gave S_0 .

And I bet that the Woods-Saxon parameters that gave S_0 are based on fitting the single-particle energies.

Single-particle ANC for Square well

August 9, 2021

GoLuckyRyan Basic , Leave a comment

We studied the ANC (Asymptotic normalization coefficient) in this post. Put aside the normalization of the Whittaker function and accept the conventional from.

The nuclear ANC is defined using the quasi-single-particle wave function:

\displaystyle \phi_{nlj} = \left< A|B\right>

where \left|A \right> and \left|B\right> are the wave function of particle A and B with A = B+1. The quasi-single particle wave function is the overlap/similarity between two wave function.

The nuclear ANC C is then

\displaystyle \phi_{nlj} \xrightarrow{r \rightarrow \infty } C \frac{W_{-\eta, l + 1/2}(2 \kappa r)}{r}

where W(x) is the Whittaker function.

Note that the quasi-single-particle wave function is not necessarily normalized and the magnitude of the wave function is the spectroscopic factor

\displaystyle S(A,B) = |\phi_{nlj}|^2

The single particle wave functions of nucleus A are \psi_{nlj} and

\displaystyle \phi_{nlj} \approx \sqrt{S(A,B)} \psi_{nlj}

Thus, we have the single-particle ANC, traditionally labeled as b

\displaystyle \psi_{nlj} \xrightarrow{r\rightarrow \infty} b \frac{W_{-\eta, l+1/2}(2\kappa r)}{r}

Thus, the nuclear ANC and single-particle ANC is connected by the spectroscopic factor,

\displaystyle C = \sqrt{S} b

The single-particle ANC can be calculated with given potential. Here, we will deduce the single-particle ANC using Square well and give the sense of the value of the ANC. And the discussion is bounded for neutron only.

The potential of the square well is

\displaystyle V(r) = \begin{cases} -|V_0|, & r < a \\ 0, & r \geq a \end{cases}

The solution for r < a is the spherical Bessel function, and the solution for r \geq a is modified Bessel function divided by \sqrt{r}

\displaystyle R(r) = \begin{cases} A j_l(kr), & r < a \\ b \sqrt{\frac{2 \kappa}{ \pi r}} K_{l+1/2}(\kappa r ), & r \geq a \end{cases}

\displaystyle k^2 = \frac{2m}{\hbar^2}\left(|V_0|- |E|\right),~~~ \kappa^2 = \frac{2m}{\hbar}|E|

In the above expression, E is the binding energy, A is the normalization factor, the b is the single-particle ANC, and the factor \sqrt{2 \kappa/ \pi} is for matching the Whittaker function,

\displaystyle W_{0, l+1/2}(2\kappa r) = \sqrt{\frac{2 \kappa r}{ \pi}} K_{l+1/2}(\kappa r)

So, that the single-particle ANC is b . In fact, the function K_{l+1/2}( \kappa r) / \sqrt{r} is a valid solution of the Schrodinger solution, as the solution can be scaled up by any factor.

The inner and outer solution and their derivative must be equal at r = a , that dictated the binding energy.

There are physical constrains on the depth and width for the potential. The constrains come from the Planck’s constant and the mass of the nucleon. The factor \sqrt{2m_n}/\hbar = 0.22~\mbox{fm}^{-1} \mbox{MeV}^{-1/2} defined the possible range of the potential depth and width, because this factor should be close to half integer of pi and we will see the reason below. We calculated the energy of the 1st s-state (or the ground state) for V_0 = (-70, -20) and a = (1, 2) . The numerical solution is plotted.

For V_0 = -40~\mbox{MeV}, the minimum width of the well is ~ 1.2 fm. And the shape of the well is not scalable, i.e. the shape of V_0 = -40, a = 2 cannot scale to V_0 = -20, a = 1 . For the s-orbital the spherical Bessel function, and the decay function is

\displaystyle j_0(kr) = \frac{\sin(kr)}{kr}, ~~ \sqrt{\frac{2 \kappa}{ \pi r}} K_{1/2}(\kappa r ) = \frac{\exp(-\kappa r)}{r}

Matching the boundary condition, we have the equation for V_0, E, a of the s-orbital:

\displaystyle - \sqrt{\frac{|E|}{|V_0|-|E|}} = \cot\left( \frac{\sqrt{2m}}{\hbar} \sqrt{|V_0|-|E|} a \right)

Set \sqrt{|V_0|-|E|} = x

\displaystyle -\frac{\sqrt{|V_0|-x^2}}{x} = \cot\left( \frac{\sqrt{2m}}{\hbar} a x \right), x = (0, \sqrt{|V_0|})

I called the above equation the characteristic equation. We can plot the two sides as a function of x. The left side is infinite at x = 0 and is zero at x = \sqrt{|V_0|} . The right side is a regular cotangent function with zero at

\displaystyle x_n = \frac{\hbar}{a\sqrt{2m}}\left( \frac{\pi}{2} + n \pi \right)

Thus, the two curves will meet when x_n < \sqrt{|V_0|} , i.e.

\displaystyle |V_0| a^2 > \frac{\pi^2 \hbar^2}{8 m}, ~~ \mbox{for s-orbital}

The limit is plotted as the red curve on the above plot.

And the energy of the n-th s-orbital must lay between the n-th zero of the cotangent and \sqrt{|V_0|} .

\displaystyle \left(\frac{1}{2} +n \right)^2 \frac{\pi^2 \hbar^2}{2m a^2} - |V_0| < E_n < 0

Similarly, we write the characteristic equation for l = 0, 1, 2, 3, etc...

\displaystyle \cot(\alpha x) = f_l(x), ~~\alpha = \frac{\sqrt{2m}}{\hbar} a , ~~ x = (0, \sqrt{|V_0|})

\displaystyle f_0 (x) = - \frac{\sqrt{|V_0|-x^2}}{x}

\displaystyle f_1(x) = \frac{x}{\sqrt{|V_0|-x^2}} + \frac{|V_0|}{\alpha x (|V_0|-x^2)}

\displaystyle f_2(x) = \frac{ -x^4 \alpha^3 \sqrt{V-x^2} + V(-3 + \alpha \sqrt{V-x^2}(-3+ \alpha^2 x^2))}{ x^5 \alpha^3 - V x \alpha (3+3\alpha\sqrt{V-x^2} + \alpha^2 x^2 ) } , ~ V = |V_0|

\displaystyle f_3(x) = \frac{g_3(x)}{h_3(x)}

\displaystyle g_3(x) = \alpha ^5 \left(-x^6\right) \sqrt{V-x^2}+3 \alpha ^2 V^2 \left(2 \alpha ^2 x^2-5\right) - 45 V \\+V\alpha \left(15 \alpha x^2 \left(\alpha \sqrt{V-x^2}+2\right)-45 \sqrt{V-x^2}+\alpha ^3 x^4 \left(\alpha \sqrt{V-x^2}-6\right)\right)

\displaystyle h_3(x) = \alpha x \left(\alpha ^4 \left(x^3-V x\right)^2+15 \alpha ^2 V \left(x^2-V\right)-45 \alpha V \sqrt{V-x^2}-45 V\right)

I generate the functions using Mathematica, higher order can be generated but the function becomes very complicated.

Below is an example of the characteristic functions.

Some interesting features:

  • The 1st \cot only intercept with the s-orbital.
  • The 1st p-, d-, f- orbital meets with the 2nd cotangent.
  • The s- and d-orbitals are getting closer when x is large.
  • The p-, d-, f- solutions are near \alpha x = \pi , 3/2 \pi, 2\pi.

Ok, I think we study the finite well enough, time to solve and calculate the ANC. The ANC depends on the orbital, the binding energy, and the radius only. These 3 factors dictated the depth, and the thus, the ANC. Here, I will study the ANC for a given orbital in the function of binding energy and radius.

The depth is easily calculated for given energy and radius by using the characteristic function. We have to limit the depth not too deep or shallow, around -70 to -40 MeV should be physical.

I first study the 2nd s-orbital. here is the plot. We also plot the equal-probability contour for the nucleon stay inside the well.

The range of binding energy is (-5.01, -0.01). the range of radius is (1, 4). The data with potential depth outside (-70, -40) are rejected.

The equal-potential contours are almost perpendicular to the equal-ANC contours. The ANC is almost independent of the binding energy. And the equal-probability contours are almost parallels to the equal-ANC contours. This lead us to plot the probability / binding energy against the ANC.

This is interesting that the ANC and the probability has a clear relation. And the change of ANC is 1 for binding energy of -5 MeV, and the change is only ~0.3 for binding energy of -2 MeV. The ANC toward 0-binding energy is -0.217 and the inside probability is 0.06. It seems to be reasonable that the inside probability becomes zero as the binding energy becomes zero, as the wavefunction becomes “free”.

What is the theoretical limit of the ANC when binding energy becomes zero or unbound ?? Should it be 0 ? 1?

[2021-Aug-10] To answer the above question. The outside solution is

\displaystyle \sqrt{\frac{2 \kappa}{ \pi r}} K_{l+1/2}(\kappa r )~~, r > a

The 1st few orders of the functions are :

\displaystyle \exp(-\kappa r) \frac{1}{r}, ~\exp(-\kappa r) \frac{1+ \kappa r}{\kappa r^2}, ~\exp(-\kappa r) \frac{3+3\kappa r + \kappa^2 r^2}{\kappa^2 r^3} ...

When the binding energy becomes zero to close to zero, the \kappa \rightarrow 0 , and the functions becomes infinity, except l = 0 , which is 1/r . Others orbitals have to be normalized while the outside function becomes infinity. Thus, the ANC for zero – binding energy will becomes ZERO, except the s-orbital.

We plot a contour plot of ANC in term of radius and potential depth.

The boundary is limited by the binding energy (-5.01, -0.01).

The red line is the boundary of the ANC. For a fixed ANC, the shape of the potential has to follow a curve approximated by

\displaystyle V_0 a^2 \approx \frac{\pi \hbar}{\sqrt{2 m}} \left(\frac{1}{2} +1 + \phi \right)

Here are the potentials and the wave function ( multiplied by -50 times ) for 1s-orbitals, in which the ANCs are about -1.5,

The key massage in here is, that, when two potentials has similar V_0 a^2 values, the ANC for their corresponding orbitals should be similar. We can calculate the 0d-orbital from the above potentials, and the ANC for the 0d-orbitals are around 1.2. If we plot the ANC against V_0 a^2 , we got a quite linear line.

Since the binding energy and V_0 a^2 are closely related. In other word, for a given binding energy and orbital, the V_0 a^2 should be almost a constant, and the ANC could only variate in a finite range. Therefore, in this sense, the ANC depends on the shape of the potential, but the ANC is not sensitive to the shape of the potential for give binding energy and orbital.

Next, Lets study the 1st d-state, also the weakly bind states.

This times, the single-particle ANC becomes 0 when the binding energy becomes zero. And there is a strong correlation between ANC and inside probability.

Dimension of Wavefunction and the ANC

August 3, 2021

GoLuckyRyan Basic Leave a comment

Due to the normalization, any wave function must be normalized to 1, which is dimensionless.

\displaystyle \int_0^\infty \phi^2(r) r^2 dr \int Y_{lm}^2(\Omega) d\Omega = 1

Thus, the dimension of the square of the radial wave function \phi^2(r) is fm-3.

The relation between the single-particle ANC b , the Whittaker function and the radial wave function is

\displaystyle \lim_{r \rightarrow \infty}\phi(r) = b \frac{W_{-\eta, l+1/2}(2 \kappa r)}{r}

The Whittaker function is dimensionless (why?), so, the single-particle ANC has dimension of fm-1/2.

The relationship between the spectroscopic factor S, single-particle ANC b, and nuclear ANC C is

\displaystyle C^2 = S b^2

Thus, the dimension of the nuclear ANC also is fm-1/2, as the spectroscopic factor is dimensionless.

But for me, the dimension of the ANC is non-sense. The physical meaning of the ANC is the ratio of the single-particle wave function to the Whittaker function divided by r, or the Coulomb wave function at very far away. The Coulomb wave function has the dimension of wavefunction, so the ANC, as a ratio, has no dimension. The “dimension” of the ANC from the top session is due to, somehow, the Whittaker function is dimensionless.

Let say, in many literatures, for example, The square of ANC of 16N ground state -> 15N ground state + n, or the 0d5/2 neutron, is 0.19 fm-1. So, the invert of the square of the ANC for the 0d5/2 neutron is 5 fm. What does this 5 fm mean? a real distance from the center? The radius of the 0d5/2? the scattering length? What do we gain by giving the physical dimension of the ANC??

Asymptotic normalization coefficient

May 2, 2021

GoLuckyRyan Basic , , , , Leave a comment

The Asymptotic normalization coefficient or the ANC is thought to be an alternative to the spectroscopic factor. As the name suggested, this is the coefficient for asymptote of the radial wave function.

For a single nucleon adding/removal reaction, the reaction probability is

\displaystyle T^2 = \left|\left< \phi_b \Psi_B| V | \phi_a \Psi_A \right>\right|^2

Assume the interaction V only acts on the nucleon that being add or remove, thus

\displaystyle \left< \Psi_B | \Psi_A \right> = \phi(r)

which is the bound state wave function. Due to the nuclear interaction, the bound state wave function most probably not normalized. In other point of view, suppose the nucleus B = A + 1, the wave function of nucleus B could be

\displaystyle \left| \Psi_B \right> = \beta_{00} [\phi_{0} \times \Psi_A(0) ]_{J_B} + ... + \beta_{ij} [\phi_{i} \times \Psi_A(j) ]_{J_B}

where \phi_i is an orbital, \Psi_A(j) is the j-th excited state of nucleus A, and the square bracket is the angular coupling, antisymmetric, and normalization operator. As the wave function of nucleus B must be normalized, Thus,

\displaystyle \sum_{ij} \beta_{ij}^2 = 1 .

And the bound state can be approximated

\displaystyle \phi(r) \approx \beta_{00} \phi_{0} .

The spectroscopic factor of orbital \phi_0 in this A + 1 = B reaction is

\displaystyle \beta_{00}^2 = \int r^2 \phi^2(r) dr

At far away distance, the nucleus potential is very weak or effectively zero. The Schrodinger equation becomes

\displaystyle \left( - \frac{\hbar^2}{2m} \nabla^2 + \frac{Q_1Q_2e^2}{r} \right) \Phi = E \Phi

where e^2 = 1.44~\textrm{MeV.fm}. The Coulomb potential is still here because it is a long range force. Separate the radial and angular part,

\displaystyle \left( - \frac{\hbar^2}{2m} \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + \frac{Q_1Q_2 e^2}{r} \right) R(r) = E R(r)

This radial equation was solved before in this post. There are two solutions for u(r) = R(r)/r , one is bound (in mathematics) and one is unbound. I state the mathematically-bounded solution in here

\displaystyle F_l(r, \eta) = \frac{2^l e^{-\pi \eta /2} \left| \Gamma(l+1+i\eta) \right|}{(2l+1)!} (kr)^{l+1} e^{ikr} {_1F_1(l+1+i\eta, 2l+2, -2ikr)},

\displaystyle k^2 = \frac{2mE}{\hbar^2}

\displaystyle \eta = \frac{k}{2E} Q_1 Q_2 e^2 = \frac{1}{\hbar} \sqrt{\frac{m}{2E}} Q_1 Q_2 e^2

There is another Coulomb wave function G_l(r) , although there is no simple form of the function, the two Coulomb wave function, one behave like a sine wave (F_l(r) ) another one behave like a cosine wave (G_l(r)), they can be combined into a complex function, the Coulomb Hankel function, (it is different from the Hankel function of the first and second kind )

\displaystyle H_l^{\pm}(x, \eta) = D_l^{\pm} x^{l+1} e^{\pm i x} U(l+1\pm \eta, 2l+2, \mp 2ix)

\displaystyle D_l^{\pm} = (\mp 2i) ^{2l+1} \frac{\Gamma(l+1\pm i \eta)} {C_l }

\displaystyle C_l = \frac{2^l}{2^{\eta \pi /2}} \sqrt{\Gamma(l+1+i\eta) \Gamma(l+1-i\eta)}

where U(a,b,z) is the confluent hypergeometric function of the second kind. In Mathematica, the function is given is built-in

HypergeometrixU[l+1+i eta, 2l+2, -2 i r]

And

\displaystyle F_l(r, \eta) = \frac{1}{2i} ( H_l^+(kr, \eta) - H_l^-(kr, \eta) )

\displaystyle G_l(r, \eta) = \frac{1}{2} ( H_l^+(kr, \eta) + H_l^-(kr, \eta) )

We can see, the analogy

\displaystyle F_l(x, \eta) \sim \sin(x) = \frac{1}{2i} ( e^{ix} - e^{-ix} ) \\ G_l(x, \eta) \sim \cos(x) = \frac{1}{2} ( e^{ix} + e^{-ix} )

The long range (unbound, scattering) behaviour of the Coulomb wave function is

\displaystyle F_l(r\rightarrow \infty, \eta) = \sin \left( k r - l \frac{\pi}{2} - \eta \log(2kr) + \sigma_l \right)

\displaystyle G_l(r\rightarrow \infty, \eta) = \cos \left( k r - l \frac{\pi}{2} - \eta \log(2kr) + \sigma_l \right)

where \sigma_l = \arg(\Gamma(l+1+i\eta)) is the Coulomb phase shift.

The ANC C is the coefficient between the bound state wave function and the Coulomb wave function at r \rightarrow \infty . Since it is the bound state long range behaviour, k = i \kappa , we have to use the Coulomb Hankel function.

For neutron, \eta = 0

\displaystyle H_0^\pm(x, 0) = e^{\pm i x}

We can see that, when x = i \kappa r , the $latex H_0^\pm (i \kappa r, 0 ) = e^{-\kappa r} is a bounded and real solution.

This is the same for higher l that the decay factor e^{-\kappa r } appeared in the Hankel function.

In the following, we use the neutron and simplify the Coulomb wave function for few l . When no charge, \eta = 0 . And since it is a bound state,

In fact, the solution for \eta = 0 is the Spherical Bessel function J. As the k = i \kappa is pure imaginary, and we know that the Spherical Bessel function J and Y are unbound for pure imaginary position.

The solution is the Spherical Hankel function h_l^\pm. This is the Coulomb Hankel function divided by the radius. In Mathematica, the built-in function is

SphericalHankelH1[l, i kappa r]

In fact,

\displaystyle h_l^+(x) = \frac{-1}{x} H_l^+(x, 0) \\ h_l^-(x) = \frac{(-1)^l}{x} H_l^-(x,0)

In this paper N.K. Timofeyuk, PRC 88, 044315 (2013), in equation 3, the Whittaker function should be the Whittaker W-function. In Mathematica

WhittakerW[ - i eta, l + 1/2, 2 kappa r ]

The difference between the Whittaker W-function and the Hankel function with complex argument is the normalization factor

\displaystyle W_{-i \eta, l+1/2}(2 \kappa r ) = (2 \kappa r)^{l+1} e^{-\kappa r} U(l+1+i \eta, 2l+2, 2 \kappa r)

Since the ANC is the proportional factor between the bound state wave function and the Whittaker W-function (or the Coulomb wave function, or the spherical Hankel function ), so the normalization factor is important.

As we know that the neutron should behave as Spherical Hankel function, thus, we checked that

\displaystyle W_{0, l+1/2}(2 \kappa r) = (-1) i^l ( \kappa r ) h_{l}^+(i \kappa r) = \sqrt{\frac{2 \kappa r}{\pi}} K_{l+1/2}(\kappa r)

where K_{\alpha} is the modified Bessel function of the 2nd kind. Here are a list of the Whittaker W-function for \eta = 0

\displaystyle W_{0, 1/2}( 2 x) = e^{-x}

\displaystyle W_{0, 3/2}( 2 x) = \frac{e^{-x}}{x} (1+x)

\displaystyle W_{0, 5/2}( 2 x) = \frac{e^{-x}}{x^2} (3 + 3x+x^2)

\displaystyle W_{0, 7/2}( 2 x) = \frac{e^{-x}}{x^3} (15 + 15x+ 6x^2 + x^3)

\displaystyle W_{0, 9/2}( 2 x) = \frac{e^{-x}}{x^4} (105 + 105x+ 45x^2 + 10x^3 + x^4)

\displaystyle W_{0, 11/2}( 2 x) = \frac{e^{-x}}{x^5} (945 + 945x+ 420x^2 + 105x^3 + 15x^4+x^5)

For me, the problem of the ANC is that, the Coulomb Hankel function or the Whittaker W-function cannot be normalized. And without a proper normalization, how can we compare the magnitude of two wave functions??

For example, I calculated the 1s1/2 bound state wave function from a Woods-Saxon potential for neutron. The parameters are V_0 = -50.4, R_0 = 3.1246, a_0 = 0.67, V_{SO} = 19.6, R_{SO} = 3.0238, a_{SO} = 0.66, m_\mu = 884.297, the energy is -2.227 MeV and \kappa = 0.3181 . And here is the comparison

\displaystyle \phi(r) = C \frac{W_{0, 1/2}(2 \kappa r )}{\kappa r}

The Woods-Saxon bound state is supposed to be a pure wave function that has SF = 1 or ANC = 1. But since the Whittaker cannot be normalized, and depends on the “definition”, where the \kappa appears in the denominator or not, the ANC can be different. And for this example the ANC = 0.7, for a pure state.