Some thoughts on the quenching of spectroscopic factor (II)

January 1, 2024

GoLuckyRyan Basic Leave a comment

The nature of the spectroscopic factor is discussed here (from the point of view of mean-field) and here (from an experimental point of view). Here are some updates and additional discussions on the topic.

Up to now, I still believe that the quenching of the SF simply reflects the mismatch between experiment and theory, due to experimental conditions that limited configurations can be observed (so that the sum of SF is less than 1), and also the reaction theory does not take into account the correlation (so that the SF of each state is quenched). If both limitations are removed, there is no quenching. Although the SF is quenched, careful normalization gives us a great deal about the nuclear structure.

Is the spectroscopic factor or the cross-section quenched?

It seems that since the spectroscopic factor and the cross-section are closely related by

\displaystyle \sigma_{exp} = C^2S \sigma_{th}

so, the question is meaningless, because when SF is quenched from unity, the cross-section is also quenched from the therotical cross-section. However, since cross-section is an observable while SF is not, people consider that it is more proper to say quenching of cross-section.

Whether SF is an observable or not, an article Spectroscopic Factors: Observability and Measurability by L. D. Blokhintsev stated that “Spectroscopic factors are non-invariant under the unitary transformations of nuclear forces“, and Non-observability of Spectroscopic Factors by B. K. Jennings said that “At the level of microscopic calculations, they can be varied by unitary transformations and we expect hard and soft nucleon-nucleon potentials to generate quite different values for them. At the phenomenological level, the spectroscopic factor can be varied by field redefinitions.“. Thus, the spectroscopic factor is a pure construction that is fixed on a specific basis and interaction. Another article Nonobservable nature of the nuclear shell structure: Meaning, illustrations, and consequences by T. Duguet, H. Hergert, J. D. Holt, and V. Somà is worth reading.

However, I think SF is an observable because there should be a natural basis and fixed interaction. when the basis is fixed, and the 1st order of the nuclear interaction is well-known, the SF is well-defined, just like the coefficient of fractional percentage. Another argument is that there are many articles on extracting the spectroscopic factors, and although the uncertainty can be as large as 50% due to various experimental conditions and analysis methods, the community has a consensus on the quenching. That is, the community reached a consensus on the basis and range of interaction. I think that means, we are reaching the natural basis and 1st order understanding of the interaction.

The situation is like if we have the theoretical framework but lack some fundamental parameters fixed. The theory can produce all kinds of phenomena. Say, neutrino mass, without that, we don’t know the neutrino mass hierarchy. But we cannot say the neutrino mass hierarchy should fall in either the normal or inverted hierarchy and therefore neutrino mass hierarchy is not well-defined. My point is once the nuclear interaction and the basis are fixed, the SF is uniquely determined. So, some people would say, SF is an quasi-observable and that it is not an observable yet, but it will. May be, we should call SF is an extractable/deductible, which is a extractable/deductible quantity with some will-be-fixed parameter sets.

The last comment is that, if SF is not an observable, and is a pure arbitrary construct, all the works on it become meaningless. But, I believe SF is not aether, it reveals the structure and nucleon configuration of a nucleus. Also, the effective single-particle energy (ESPE) is an SF weighted energy of an orbital. If SF is arbitrarily constructed and can be anything, so the ESPE. but the ESPE very much agrees with the energy levels from Woods-Saxon potential which is a 1st order approximation of the nuclear mean field. A similar thing to the asymptotic normalization coefficient (ANC), ANC is a ratio of the wavefunction to the Coulomb wave function outside the nuclear potential, and the wavefunction is not an observable, so ANC is also not. But the wavefunction is surely can be deduced/extracted from the experimental data by fixed some nuclear parameters.

So, if SF is an observable, it is meaningless to ask is SF quenched or cross-section quenched.

If, the theoretical cross-section is perfectly calculated, the SF would be unity and no quenching for either SF or cross-section, and the nucleon configuration is embedded into the coefficient of fractional percentage.

[Papers Reading] On the recent progress of the quenching of spectroscopic factor

May 20, 2023

GoLuckyRyan papers reading , Leave a comment

The journey started as early as 1993. J.M. Udías et. al used (e,e’p) to study the spectroscopic factors (SFs) of 40Ca and 208Pb [ PRC 48, 2731 (1993) ]. The experiment used the Coulomb force (which is well-understood) to extract the SF and found that it is quenched, i.e. about 30% smaller than the SF calculated from the Shell model calculation. Although there were many experiments that also found the quenching of SF using nuclear force, for example, (d,n), (d,p) experiments, people had generally concerned that the hadron probe would introduce some quenching, for example, the sensitivity of the optical potentials. So, using (e,e’p) experiment with fully relativistic analysis confirmed the quenching.

In 2001, G.J. Kramer, H. P. Blokb, and L. Lapikás published a study [ NPA 679, 267 (2001) ] and found that the SF quenching from (e,e’p) and (d,3He) experiments are consistent. Later, J. Lee, M. B. Tsang, W. G. Lynch published another study [ PRC 75, 064320 (2007) ] that confirmed the quenching is consistent with (d,p) and (p,d) transfer reactions from Z = 3 – 24. Thus the SF quenching was considered a fact for stable nuclei.

The figure is taken from NPA 679, 267 (2001). It shows that the SF quenching is consistent for (e,e’p) and (d,3He) reactions for various stable nuclei. This founding confirmed the hardon probe is also a good probe given that a consistent analysis framework is used.

Meanwhile, many theoretical studies were published to understand the cause of the quenching. For example, W. H. Dickhoff and C. Barbieri [ Prog. part. Nucl. Phys. 52, 377 (2004) ] pointed out that the quenching is due to Short-range correlation and Long-range correlation. The correlation will virtually excite the nucleon out of its orbital, creating high-momentum pairs, and occupation in “unoccupied” orbitals. Another example is I. Sick, [ Quasi-free knockout reaction, QFS workshop at ECT, Trento, 2008 ] studied 208Pb. There are many recent progresses on nuclear correlation, but I am not so familiar.

  • Taken from Prog. Part. Nucl. Phys. 52, 377 (2004).
  • Taken from I. Sick, Quasi-free knockout reaction, QFS workshop at ECT, Trento, 2008.

In 2008, A. Gade et. al published a systematic study on the SF quenching [ PRC 77, 044306 (2008), updated on PRC 90, 057602 (2014) ] with decades of experimental data on unstable nuclei. They found that when plotting the SF vs the “boundness” of the knockout nucleon, the quenching is small for weekly bound nucleons and the quenching can be as large as 60% for deeply bounded nucleons. They suggested that the weekly bounded nucleons are more “free” and like “independent-particle” than those that are deeply bound. As the SF is a concept from the independent particle model, (the 1st order interaction on a nucleon is from a mean-field generated by other nucleons, a 2nd order interaction is the fine detail of NN-interaction, and effective non-1st order interaction is called residual interaction. This model formed the basis of the Shell model calculation.) that, a weekly bound nucleon has weaker residual interaction so it is mainly influenced by the mean field, and the quenching is small. And the metric of the boundness is from the difference between the proton-neutron Fermi surface, or more precisely, the difference between the separation energies.

Taken from PRC 90, 057602 (2014)

This result drew much attention and leaded 2 studies using (p,2p) knockout experiment on the oxygen isotope chain from GSI, Germany [ PRL 120, 052501 (2018) ], and RIKEN, Japan [ Prog. Theor. Exp. Phys. 2018, 021D01 (2018) ]. Both of them found that the quenching does almost not depend on the boundness from 14O ( \Delta S = -19 MeV ) to 24O ( \Delta S = 23 MeV ). The discrepancy between the (p,2p) studies and Gade et al. results in many debates and discussions. One focus is on the reaction mechanism. In the Gade plot, the data were produced from nucleon removal reaction at 100 MeV/u on 9Be target. Many suspected that the reaction theory using 9Be target may be not complete, while others suspect that the reaction theory for (p,2p) is based on the impulse approximation, which is still unclear. On top of that, the direct comparison with different reactions is also in question, for example, the treatment of final state interaction.

  • Taken from Prog. Theor. Exp. Phys. 2018, 021D01 (2018).
  • Taken from Phys. Rev. Lett. 120, 052501 (2018)

Recently, there are 2 studies using (d,p) reaction at 10 MeV/u [ Kay et. al PRL 129, 152501 (2022) ] and (p,pN) at 100 MeV/u [ Pohl et. al PRL 130, 172501 (2023) ] report that the quenching almost not depends on the boundness.

Kay et. al used a 14N and 14C mixed beam to perform (d,p) reaction simultaneously. The reaction mechanism for (d,p) reaction is well understood and is considered to be a reliable way to extract the SF. And the simultaneous reactions eliminated most of the systematic uncertainty. Pohl et. al used a proton target to perform nucleon knockout on 14O. The study found that when including the inelastic that the 14O is excited above the proton threshold and emits a proton, the quenching almost does not depend on the boundness. However, if the inelastic channel was ignored, the quenching does depend on the boundness. In other words, the study suggests that the reaction theory used in the Gade plot is not complete.

  • Taken from PRL 129, 152501 (2022)
  • Taken from PRL 130, 172501 (2023)

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.

Spectroscopic factor and the sum-rule

April 5, 2020

GoLuckyRyan Basic Leave a comment

In the last 2 posts (here and here), we discussed the issue with spectroscopic factor and its quenching. Especially in the previous post, the discussion is from a generic basis.

However, there is a “constrain” on what kind of basis can be used. i.e. the one that is spherical symmetric. Because the total wave function has to be spherical, or to say, the total Hamiltonian must be commuted with total spin J. Also, the experimentally observed quantities must be spherical symmetric. Thus, it is “natural” to use spherical harmonics ( or the hyper-spherical harmonic, spherical-spin harmonic) to represent the angular part of the “single-particle state”. In this sense, only the radial part can be chosen arbitrary, but I doubt that the physics would be changed, except gaining computation convenience.

And because the radial part can be chosen freely, the use of Woods-Saxon could be a standard. And the problem of quenching, become the choice of Woods-Saxon parameters, or the “mean-field”. Once we defined or self-consistently calculated the mean field, the single-particle state is well defined. Or in the worst case, the spin must be well defined for a single-particle state.

In this sense, and also with the residual interaction that perturb the single-particle state. The quenching could mean two things:

  1. The residual interaction is large. i.e., the two-body interaction has a large contribution to the nucleon motion that this two-body interaction is partially captured by One-body mean-field approximation, like ~60%.
  2. The consistency between the optical potential and the bound state mean-field is not good.

Currently, the community tends to lean on the 1st point and the quenching is due to short- and long- range correlation.

The short-range correlation lead to high excitation that is inaccessible for experiments. But the long-range correlation that couples the single-particle state with nearly by excited states of the core nucleus, is < 20 MeV. And in some case, the long-range correlation create fragmentation of the states and can be observed clearly. So, the quenching of SF should be ~80%? It seems that the present DWBA calculation has to be improved as well.

At last, about the sum-rule. From the calculation, we can see that the sum-rule is integrating all the internal degree of freedom, and acts like finding the “magnitude” of the wave function. In fact, the only external degree of freedom for a wave function is the number of nucleons. And the sum-rule is telling us, the number of nucleons is “conserved”. In this sense, we can imagine that the sum-rule is independence of the choice of basic. An analogy is that, the magnitude of a vector, is independence of the rotation (i.e., change of basis). Thus, it may be no surprise that in the paper, “Test of Sum Rules in Nucleon Transfer Reactions”, J. P. Schiffer et al., PRL 108, 022502 (2012), that, the normalized SF reproduce the sum-rule, although the SF is ~ 0.55.

However, if the short-range correlation block experimental access to high excitation states. The sum-rule should also be quenched.

However, if the short-range correlation is universal effect that affect all nucleons indiscriminately, a renormalization could be OK.

Spectroscopic factor (again)

April 3, 2020

GoLuckyRyan Basic , , Leave a comment

In the last posts on SF, my mind have been biased to the mean field model. I started everything from the mean field. Now, lets forget the mean filed, just starting from experimental view point and any complete basis.

Suppose an arbitrary complete single-particle basis, say, a Woods-Saxon numerical basis, 3D spherical harmonic oscillator, or spherical Gaussian basis, and donate it as \phi_i . Using this basis,, we can form the Slater determinate of N-nucleons as

\displaystyle \Phi_i = \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_1(1) & ... & ... \\ ... & \phi_r(s) & ... \\ ... & ... & \phi_N(N) \end{vmatrix} .

The solution of any N-body Hamiltonian can be calculated by solving the eigen-system with the matrix elements

\displaystyle H_{ij} = \left< \Phi_i |H| \Phi_j \right>

Say, the solution is

\displaystyle  H \Psi_i = E_i \Psi_i

Experimentally, for the A(d,p)B reaction, we are observing the transition probability

\displaystyle T = \left< \chi_p \Psi_i(B)  |V|\chi_d \Psi_0(A) \right>,

where \chi_k is the distorted plane wave of  particle k and V is the interaction during the reaction. The integration summed all internal degree of freedom, particularly,

\displaystyle  \left<\Psi_i(B) |V_{BA} | \Psi_0(A) \right > \approx  \left<\Psi_i(B) | \Psi_0(A) \right > = \sum_i \alpha_i \phi_i = \psi

where \psi is called the quasi-particle wave function and it is expressed in the complete basis of \phi_i .

And then, the quasi-particle wave function will also be integrated.

\displaystyle T = \left< \chi_p \psi |V'|\chi_d \right>

In fact, it is the frame work of many calculations, that require 4 pieces of input,

  1. the optical potential for the incoming channel to calculate the distorted wave of the deuteron,
  2. the optical potential for the outgoing channel to calculate the distorted wave of the proton,
  3. the quasi-particle wave function, and
  4. the interaction.

If the Woods-Saxon basis is being used, then we can have shell model picture that, how the quasi-particle “populates” each Woods-Saxon states, and gives the spectroscopic factor.

The Woods-Saxon basis is somewhat “artificial”, it is an approximation. the nature does not care what the basis is, the transition probability is simple an overlap between two nuclei (given that the interaction is not strong, i.e. in the case of Direct reaction). Since the spectroscopic factor is always in reference for a basis, i.e., the quasi-particle wave function has to be projected into a basis, so, the spectroscopic factor is model dependence. If a weird basis is used, the shell model picture is lost. In this sense, SF is pure artificial based on shell model.

In experiment, we mainly observe the scattered proton, and from the angular distribution, the orbital angular momentum is determined. And from the conservation of angular momentum, we will know which orbital the neutron is being added to nucleus A.

But it is somehow weird. Who said the quasi-particle must be in an orbital angular momentum but not many momenta? If the spin of nucleus A is zero, because of conservation of angular momentum, the spin of nucleus B is equal to that of added neutron. But if the spin of nucleus A is not zero,  the neutron can be in many difference spins at once. In general, the conservation of angular momentum,

\displaystyle \left < \Psi_i(B) | \phi \Psi_0(A) \right >

could be difference than that of

\displaystyle \left < \chi_d \Psi_i(B) | V | \chi_p \Psi_0(A) \right >

We discussed that the spectroscopic factor is artificial. However, if we use a self-consistence basis, i.e. the basis that describes both nucleus A and B very well (is such basis exist?) , then the quasi-particle wave function is being described using the basis of nucleus B. To be explicit, lets say, we have a basis can describe nucleus B as

\Psi_0(A) = 0.8 \Phi_0 + 0.6 \Phi_1

\Psi_i(B) = \sum_{jk} \beta_{jk} \left( \phi_j \otimes  \Psi_k(A) \right)

where the single particle basis capture most of the nucleus A. is the spectroscopic factor meaningful in this sense?

So, is such basis exist? i.e, can it describe a nuclear wave function (almost) perfectly? In ab initial calculation, it can expend the nuclear wave function using some basis and give the overlap. Can the quasi-particle wave function be projected into the “single-particle state” ? How to define the “single-particle state” in ab initial calculation?

The experimental spectroscopic factor is calculated by compare the experimental cross section with that from theory. In the theory, the cross section is calculated by assuming the bound state is from a Woods-Saxon potential without correlation. Thus. the spectroscopic factor is with respect to the Woods-Saxon potential, an approximation.

Some thoughts on the quenching of spectroscopic factor

March 31, 2020

GoLuckyRyan Basic , , , , , Leave a comment

Spectroscopic factor plays the central role in unfolding the nuclear structure. In the simplest manner, the total Hamiltonian of the nucleus is transformed into a 1-body effective potential and the many-body residual interaction, i.e.,

\displaystyle H = \sum_i^N \frac{P_i^2}{2m_i} + \sum_{i \neq j}^N V_{ij} = \sum_i^N  \left( \frac{P_i^2}{2m_i} + U \right) + \sum_{i\neq j}^N \left( V_{ij} - U \right) \\ = \sum_{i}^N h_i + H_R = H_0 + H_R

The effective single-particle Hamiltonian has solution:

\displaystyle h_i \phi_{nlj}(i) = \epsilon_{nlj} \phi_{nlj}(i)

where \epsilon_{nlj} is the single-particle energy. The solution for H_0 is

\displaystyle H_0 \Phi_k(N) = W_k \Phi_k(N)

\displaystyle \Phi_k(N)= \frac{1}{\sqrt{N!}}\begin{vmatrix} \phi_{p_1(k)}(1) & \phi_{p_1(k)}(2) & ... & \phi_{p_1(k)}(n) \\ \phi_{p_2(k)}(1) & \phi_{p_2(k)}(2) & ... & ... \\ ... & ... & ... & ... \\ \phi_{p_n(k)}(1) & \phi_{p_n(k)}(2) & ... & \phi_{p_N(k)}(N) \end{vmatrix}

\displaystyle W_k = \sum_i^N \epsilon_{p_i(k)}

where p_i(k) is the set of basis for state k from \phi_{nlj} , and W_k is the eigenenergy.

The residual interaction is minimized by adjusted the mean-field U. Thus, the residual interaction can be treated as a perturbation. This perturbs the nuclear wave function

H \Psi_k(N) = W_k \Psi_k(N),    \Psi_k(N) = \sum_i \theta_{i}(k) \Phi_i(N) .

The normalization requires \sum_i \theta_{i}^2(k) = 1 .

In the Slater determinant \Phi_k, a single-particle wave function for a particular orbital can be pull out.

\displaystyle \Phi_k(N) = \phi_{\mu} \otimes \Phi_{k}(N-1)

where \otimes is anti-symmetric, angular coupling operator. Thus,

\displaystyle \Psi_k(N) = \sum_{\mu i} \theta_{\mu i}(k) \phi_{\mu} \otimes \Phi_i(N-1)

The \theta_{\mu i}^2 (k) is the spectroscopic factor. There are another sum-rule for adding and removing a nucleon. so that the sum is equal to the number of particle in a particle orbital.

I always imagine the quenching is because we did not sum-up the SFs from zero energy to infinity energy (really???), thus, we are always only observing a small fraction of the total wave function. For example,  the total wavefunction would look like this:

\displaystyle \Psi_k(N) =  \phi_{0} \otimes \left(\theta_{00}(k) \Phi_0(N-1) + \theta_{01}(k)\Phi_1(N-1) +....  \right) \\ + \phi_{1} \otimes \left( \theta_{10} \Phi_0(N-1) +...   \right) +...   .

In experiment, we observe the overlap between ground-state to ground-state transition

\displaystyle  \left<  \phi_0 \Phi_0(N-1) | \Phi_0(N-1) \right> = \theta_{00}(0)

for ground-state to 1st excited state transition for the same orbital is

\displaystyle  \left<  \phi_0 \Phi_1(N-1) | \Phi_0(N-1) \right> = \theta_{01}(0)

And since we can only observed limited number of excited states, bounded by either or boht :

  • experimental conditions, say incident energy
  • the excited states that are beyond single-particle threshold.
  • finite sensitivity of momentum

Thus, we cannot recover the full spectroscopic factor. This is what I believe for the moment.

Experimentally, the spectroscopic factor is quenched by 40% to 50%. The “theory” is that, the short-range interaction quench ~25%, the long-range interaction quench ~20%. The long- and short-range interaction correlate the single-particle orbital and reduce the degree of “single-particle”.

Annotation 2020-03-31 010604.png

The short-range interaction is mainly from the “hard-core” of nucleon, i.e., the interaction at range smaller than 1 fm. The long-range interaction is coupling with nearby vibration states of the rest of the nucleus.

For example, from the 19F(d,3He) reaction, the spectroscopic factor for 19F 1s1/2 state is ~0.4, and 0d5/2 is ~0.6.

Annotation 2020-03-31 011328.png

Thus, the wavefunction of 19F is

\left|^{19}\textrm{F}\right> \approx  \sqrt{0.4} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.6} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right>

It is worth to note that the above SFs is not re-analysised and the “quenching” is not shown. Many old data had been re-analysised using global optical model and the SF is reduced and show that the sum of SFs is ~ 0.55.

If it is the case for 19F, the wavefunction would become,

\left|^{19}\textrm{F}\right> \approx  \sqrt{0.2} \left|1s_{1/2}\right> \otimes \left|^{18}\textrm{O}_{g.s.} \right> + \sqrt{0.3} \left|0d_{5/2} \right> \otimes \left|^{18}\textrm{O}(1.98) \right> + \sqrt{0.5} \Psi_k

Here I use \Psi_k for the “correlated wavefunction” that the single-particle orbital cannot simply pull out. Nevertheless, if x and y are correlated,

f(x,y) \neq  g(x) h(y)

Am I misunderstood correlation?

My problem is, What does a correlated wave function look like?

In my naive understanding, the Slater determinant \Phi_k is a complete basis for N-nucleon system. A particular single-particle orbital can ALWAYS be pull out from it. If it can not, therefore, the Slater determinant is NOT complete. The consequence is that all theoretical calculation is intrinsically missed the entire CORRELATED SPACE, an opposite of Slater determinant space (of course, due to truncation of vector space, it already missed somethings).

If the theory for correlation is correct, the short-range interaction is always there. Thus, the spectroscopic factor for deuteron 0s1/2 orbital is ~0.8, assuming no long-range correlation. However, we already knew that 96% of deuteron wavefunction is from 0s1/2 and  only 4% is from 1d5/2 due to tensor force. Is it not mean the spectroscopic factor of deuteron 0s1/2 state is 0.96?  Is deuteron is a special case that no media-modification of nuclear force? But, if the short-range correlation is due to the hard core of the nucleon, the media-modification is irrelevant. Sadly, there is no good data such as d(e,e’p) experiment. Another example is 4He(d,p)5He experiment. What is the spectroscopic factor for g.s. to g.s. transition, i.e. the 0p3/2 orbital? is it ~0.6 or ~ 1.0?

Since the experimental spectroscopic factor has model dependency (i.e. the optical potential). Could the quenching is due to incomplete treatment of the short- and long-range correlation during the interaction, that the theoretical cross section is always bigger?

In the very early days, people calibrate their optical potential using elastic scattering for both incoming and out-going channel, and using this to produce the inelastic one. At that time, the spectroscopic factors are close to ~1. But since each optical potential is specialized for each experiment. It is almost impossible to compare the SF from different experiments. Thus, people switch to a global optical potential. Is something wrong with the global optical potential? How is the deviation?

Let me summarize in here.

  1. The unperturbed wave function should be complete, i.e. all function can be expressed as a linear combination of them.
  2. A particular single-particle orbital can be pull out from the Slater determinate \Phi_k.
  3. The residual interaction perturbs the wave function. The short-/long-range correlation should be in the residual interaction by definition or by construction of the mean field.
  4. The normalization of wave function required the sum of all SF to be 1.
  5. Another sum rule of SFs equals to the number of particle.
  6. By mean of the correlation, is that many excited states have to be included due to the residual interaction. No CORRELATED space, as the Slater determinant is complete. (pt. 1)
  7. Above points (1) to (7) are solid mathematical statements, which are very hard to deny.
  8. The logical result for the quenching of the observed SF is mainly due to not possible to sum up all SFs from all energy states for all momentum space.
  9. The 2-body residual interaction can create virtual states. Are they the so called collective states?
  10. But still, collective states must be able to express as the Slater determinant (pt. 1), in which a particular single-particle orbital can be pull out (pt. 2).
  11. May be, even the particular single-particle orbital can be pull out, the rest cannot experimentally observed ? i.e. \Phi_k(N-1) is not experimentally reachable. That go back to previous argument for limitation of experiments (pt. 8).
  12. For some simple systems, say doubly magic +1, deuteron, halo-nucleon, very weakly bounded exited state, resonance state, the sum of SF could be close to 1. Isn’t it?
  13. The theoretical cross section calculation that, the bound state wave function is obtained by pure single-particle orbital. I think it is a right thing to do.
  14. The use of global optical potential may be, could be not a good thing to do. It may be the METHOD to deduce the OP has to be consistence, instead of the OP itself has to be universal. Need more reading from the past.

My PhD thesis

December 20, 2019

GoLuckyRyan Basic , , , , , Leave a comment

Here https://doi.org/10.15083/00075194

or Here

The title is : Quasi-free proton knockout reaction of 23,25F

Abstract

The change of the neutron dripline from oxygen to fluorine indicates the 1d5/2 proton affects the neutron shell structure. We aim to know how the neutron sd-shell structure is changed by the 1d5/2 proton in neutron-rich 23F and 25F nucleus using proton spectroscopy. The spectroscopy is free from the effects of the proton shell structure, because the 1d5/2 proton in 23F or 25F is a single-particle state due to the 𝑍=8 magicity and even neutron number. If the neutron shell structure is not changed by the proton in 23,25F, after the sudden removal of that proton, the spectroscopic factor of that proton should be unity and not fragmented. Therefore, the effect on the neutron-shell from the proton will be shown on the spectroscopy.

The quasi-free 23,25F(p,2p) direct knockout reactions in inverse kinematics were performed in RIBF, RIKEN Nishina Center. Secondary beams of 23F and 25F were produced at ~280A MeV. The missing four-momentum of the residual oxygen (22O or 24O) was reconstructed using coincidence measurement of the incident nucleus and the two scattered protons. The excitation energy of the residue was then deduced.

thesis1.png
From the experimental results, the occupation number of the 1d5/2 proton of 25F was 0.1 ± 0.3 and the proton is indeed in single-particle state. Meanwhile, the spectroscopic strength of the 1d5/2 proton of 23F or 25F were fragmented. These pointed that the change of the sd-shell neutron structure due to the 1d5/2 proton is the reason of the fragmentation. The change of neutron shell suggests the disappearance of 𝑁=16 magicity. The nuclear structures of the 25F and 23F demonstrated the Type-1 shell evolution. The comparison with the present shell model interactions (SFO, USDB, and SDPF-MU) indicated that the tensor force should be stronger. Also, the spectroscopic strength of the p-orbit was ~0.8 in 23,25F, this shows that the short-range correlation in neutron-rich nuclei is as same as stable nuclei.

Sum rule of Spectroscopic factors

April 7, 2019

GoLuckyRyan Basic , , , Leave a comment

For transfer reaction, the wave function of nucleus B = A + 1 can be written as

\displaystyle \Psi_{J_B m_B}(A+1) = \sum_{A'nlj} \beta_{nlj}(B,A) A[\Psi_{J_A'}(A) \phi_{nlj}(r)]_{J_B m_B} \\ = \sum_{A' nlj} \beta_{nlj}(B,A') \frac{1}{N_A} \sum_{m_A' m}C_{J_A M_A j m}^{J_B M_B} \Psi_{J_A' m_A'}(A) \phi_{nljm}(r)

where the N_A = _{A+1}C_1 due to anti-symetrization between the single nucleon and the core nucleus A, the Clebsh-Gordon coefficient is come from the angular coupling, and \beta_{nlj}(B,A) is the square root of the spectroscopic factor for orbital nlj between nuclei B and A.

We can see that, to find out the spectroscopic factor, we can integrate out the core nucleus A

\displaystyle \beta_{nlj}(B,A) = \langle \Psi_{J_B mB} | A [\Psi_{J_A} \phi_{nlj}]_{J_B m_B} \rangle \\  = \sum_{m_A m} C_{J_A m_A j m}^{J_B m_B} \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle \\ = \sum_{m_A m} (-1)^{J_A - J + m_B} \sqrt{2J_B+1} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix} \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle

In the last step, we write the Clebsh-Gordon coefficient in Winger 3j-symbol. Using the Wigner-Eckart theorem

\displaystyle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle \\ = \frac{(-1)^{J_B-m_B}}{\sqrt{2J_B+1}} \begin{pmatrix} J_B & j & J_A \\ -m_B & m &  m_A \end{pmatrix} \langle J_B || a_{nljm}^\dagger ||J_A \rangle \\ = \frac{(-1)^{-m_B - J_A - j }}{\sqrt{2J_B+1}} \begin{pmatrix} J_A & j & J_B \\ m_A & m &  -m_B \end{pmatrix} \langle J_B || a_{nljm}^\dagger ||J_A \rangle

Thus,

\displaystyle \beta_{nlj}(B,A) = \sum_{m_A m} (-1)^{-2J} \sqrt{2J_B+1} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix}^2 \langle J_B|| |a_{nljm}^\dagger ||J_A\rangle

using the identity

\displaystyle  \sum_{m_A m} \begin{pmatrix} J_A & j & J_B \\ m_A & m & -m_B \end{pmatrix}^2 = \frac{1}{2J_B+1}

we have

\displaystyle \beta_{nlj}(B,A) = \frac{(-1)^{2J}}{\sqrt{2J_B+1}} \langle J_B|| |a_{nljm}^\dagger ||J_A\rangle

replacing the reduced matrix element in the Wigner-Eckart theorem with previous result,

\displaystyle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle = \frac{(-1)^{J_B-m_B}}{\sqrt{2J_B+1}} \begin{pmatrix} J_B & j & J_A \\ -m_B & m &  m_A \end{pmatrix} \frac{\sqrt{2J_B+1}}{(-1)^{2j}} \beta_{nlj}(B,A)

multiple the adjoint

\displaystyle (2J_B+1)\begin{pmatrix} J_B & j & J_A \\ -m_B & m &  m_A \end{pmatrix}^2 \beta_{nlj}^2(B,A) \\=\langle J_A m_A |a_{nljm} |J_B m_B \rangle \langle J_B m_B |a_{nljm}^\dagger |J_A m_A \rangle

For A(d,p) B reaction, the nucleus A is in the ground state, we can sum all the m_B and m states, then sum all the J_B , notice that

\displaystyle \sum_{J_B m_B} |J_B m_B \rangle \langle J_B m_B | = 1

we have

\displaystyle \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) =\langle J_A m_A |\sum_{m} a_{nljm}a_{nljm}^\dagger |J_A m_A \rangle

recall

a_{nljm}a_{nljm}^\dagger = 1 - a_{nljm}^\dagger a_{nljm} \\ \langle J_A m_A | \sum_{m} a_{nljm}^\dagger a_{nljm}|J_A m_A \rangle = n_{nlj}(A)

The last equality mean the number of nucleon in orbital nlj in nucleus A. Thus, we obtain

\displaystyle \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) = 2j+1  - n_{nlj}(A)

For B(p,d)A reaction, the nucleus B is in the ground state and we can sum all J_A, m_A, m,

\displaystyle \sum_{J_A} \beta_{nlj}^2(B,A) = n_{nlj}(B)

Thus, the adding A(d,p)B and removing A(p,d)C reaction from nucleus A, the sum of spectroscopic factors

\displaystyle  \sum_{J_B} \frac{2J_B+1}{2J_A+1} \beta_{nlj}^2(B,A) + \sum_{J_C} \beta_{nlj}^2(A,C) = 2j+1

 

 

2-state system (an invert problem)

January 25, 2019

GoLuckyRyan Basic , , Leave a comment

Given 2 states with energy E_1 < E_2 , and interaction energy between the two states V , the Hamiltonian of the system is

\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E_2 \end{pmatrix}

This system is already discussed in this post. The solution state in here again as:

The eigen-energies

\epsilon_{\pm} = \bar{E} \pm \sqrt{dE^2 + V^2 }

the eigen-states are

\Phi_{+} = \alpha \phi_1 + \beta \phi_2

\Phi_{-} = -\beta \phi_1 + \alpha \phi_2

where

\displaystyle \alpha = \frac{dE + \sqrt{dE^2 + V^2 }}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}, \beta = \frac{V}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}

The \phi_{1,2} are the wavefunction of pure state of E_{1,2} .

This can be easily extended to 3-state system or n-state system. The number of coupling constants is n(n-1)/2.

Experimentally, we observed the mixed states and the spectroscopic factors originated from one of the pure state. For example, in a neutron transfer reaction, the transferred neutron may coupled with the 2+ state of the core. The neutron is sitting in a pure orbital (\phi_1), say 1d5/2 orbital, it couple with the 2+ state (\phi_2) with interaction energy V . The final states will have energy \epsilon_{\pm} and we extract the spectroscopic factors (\alpha^2, \beta^2) for the 1d5/2 orbital $.

To state more clear, in a transfer reaction, we may observed two excited states A_+, A_- with transfer of |0p_{1/2}\rangle neutron. This neutron assumed to couple with a core state |C\rangle and a core excited state |C^+\rangle . And we supposed that the 2 observed states are:

|A_+ \rangle = \alpha |0p_{1/2}\rangle |C\rangle + \beta |0p_{1/2} \rangle |C^+\rangle  = \alpha |\phi_1\rangle + \beta |\phi_2\rangle

|A_- \rangle = \beta  |0p_{1/2}\rangle |C\rangle - \alpha |0p_{1/2} \rangle |C_+\rangle = \beta |\phi_1\rangle - \alpha |\phi_2\rangle

We then want to find out the un-perturbed energy E_{1,2} and the interaction V . And this is actually an easy problem by the diagonalization process.

The Hamiltonian was diagonalized into

\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E_2 \end{pmatrix} =  P \cdot D \cdot P^{-1}

where

\displaystyle P = \begin{pmatrix} \alpha && -\beta \\ \beta && \alpha \end{pmatrix} , det(P) = 1

\displaystyle D = \begin{pmatrix} \epsilon_+ && 0 \\ 0 && \epsilon_- \end{pmatrix}

Thus,

E_1 = \alpha^2 \epsilon_+ +\beta^2 \epsilon_-

E_2 = \beta^2 \epsilon_+ +\alpha^2 \epsilon_-

\displaystyle V = \alpha \beta (\epsilon_+ - \epsilon_-)  = \frac{1}{2}\sqrt{d\epsilon^2 - dE^2}

 

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