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  • 原子核结构壳模型:粒子空穴转换

    1. 准自旋( quasi-spin )的SU(2)代数

    对于每个单 (j) 壳,定义两个算符(S_+, S_-)

    [S_+(j)=sum_{m>0}(-1)^{j-m}a^+_ma^+_{-m},S_-(j)=sum_{m>0}(-1)^{j-m}a_{-m}a_m, ]

    其中 (a_m)(a_{jm})的缩写,因为这里的讨论都局限于一个单 (j) 壳内,所以这样简略表达。
    (S_pm)是 S 对产生算符和湮灭算符。通过费米子产生湮灭算符的反对易子,可以推导 (S_pm) 的对易子:

    [[S_-,S_+]=Omega_j - hat{N}_j, ]

    其中(hat{N}_j)(j) 壳上的粒子数算符,(Omega_j = (2j+1)/2)
    另外有

    [[S_pm, hat{N}_j] = mp 2, ]

    ( 因为对产生湮灭算符会增加减少粒子数 2 )。
    定义

    [S_z= frac{1}{2}(N_j-Omega_j) ]

    会得到

    [[S_pm, S_z] = pm 1, [S_-, S_+] = -2S_z, ]

    所以可以定义

    [S_x = frac{1}{2}( S_+ + S_-), S_y = frac{1}{2i}(S_+ - S_-), ]

    由于(S_+, S_-)互为共轭,所以(S_x, S_y)都是厄米算符,(S_z)显然也是厄米算符。利用上面的对易子,很容易得到

    [[S_alpha, S_eta] = i epsilon_{alpha eta gamma} S_gamma, ]

    即 SU(2) 代数,与自旋算符的代数一样,所以叫做准自旋。显然,(S_z) 与粒子数相关,如果波函数是S对condensate,那么(S_z)量子数就是”超过一半容量的对数“。

    [S^2 = S_+ S_- - S_z + S^2_z, ]

    在给定粒子数的情况下,是一个衡量波函数中有多少配成(S)对的算符。

    2. 粒子空穴共轭算符:(Gamma = e^{ipi S_y})

    2.1 (Gamma)是幺正算符

    因为 (S_y) 是一个厄米算符,所以(Gamma^dagger = e^{-ipi S_y}), 所以有

    [Gamma Gamma^dagger = Gamma^dagger Gamma = 1. ]

    2.2 (Gamma ilde{a}_m Gamma^dagger = - a^dagger_{m}, Gamma a^dagger_m Gamma^dagger = ilde{a}_m)

    约定 ( ilde{a}_m = (-1)^{j+m}a_{-m})。对于任意算符(S, Q),定义

    [f(x) = e^{xS} Q e^{-xS}, onumber ]

    则有偏导数

    [frac{ d f}{ d x} = [S, f(x)], ~~~~~ frac{ d^2 f}{ d x^2} = [S,[S,f(x)]], ~~~~~ frac{ d^n f }{ d x^n } = [S, [S, cdots[S,f(x)]cdots]]. ]

    现在取 (x = ipi, Q = ilde{a}_m),由泰勒展开式,得到

    [Gamma ilde{a}_m Gamma^dagger = ilde{a}_m + i pi [S_y, ilde{a}_m] + frac{(ipi)^2}{2!} [S_y,[S_y, ilde{a}_m]]+cdots. onumber ]

    由费米子产生湮灭算符的反对易子,可以推导得到

    [[S_-, a^dagger_m] = - ilde{a}_m, ~~ [S_-, ilde{a}_m]=0, ~~ [S_+, a^dagger_m]=0, ~~ [S_+, ilde{a}_m] = - a^dagger_m, ]

    所以有

    [[S_y, ilde{a}_m] = frac{i}{2} a^dagger_m, ~~~~ [S_y, a^dagger_m] = - frac{i}{2} ilde{a}_m. ]

    代入(Gamma ilde{a}_m Gamma^dagger)得到

    [ Gamma ilde{a}_m Gamma^dagger = ilde{a}_m ( 1 + (ipi/2)^2/2! + (ipi/2)^4/4! + cdots ) + i a^dagger_m ( ipi/2 + (ipi/2)^3/3! + (ipi/2)^5/5! + cdots ) ~~~~~~~~ = ilde{a}_m cos frac{pi}{2} - a^dagger_m sin frac{pi}{2} = - a^dagger_m. ]

    类似地,还能得到

    [Gamma a^dagger_m Gamma^dagger = ilde{a}_m. ]

    进一步

    [Gamma S_+ Gamma^dagger = - S_-, ~~~ Gamma S_- Gamma^dagger = - S_+. ]

    2.3 (Gamma |0 angle = frac{1}{Omega!}S^Omega_+ |0 angle, Omega = (2j+1)/2)

    显然,(frac{1}{Omega!}S^Omega_+ |0 angle)表示占满整个 (j) 壳的状态。
    (n/2 leq Omega) 时,利用(S_pm)的对易子,可以推导出

    [langle 0 | S^{n/2}_- S^{n/2}_+ | 0 angle = frac{ n!! }{ 2^n } frac{ (2j+1)!! }{ (2j+1-n)!! } = frac{ (n/2)! Omega! }{ (Omega-n/2)! }, ]

    所以立即得到

    [langle 0 | S^Omega_- S^Omega_+ | 0 angle = (Omega ! )^2 ]

    所以,(frac{1}{Omega!}S^Omega_+ |0 angle) 是归一化的,是归一化的满占据波函数。下面证明(Gamma |0 angle = frac{1}{Omega!}S^Omega_+ |0 angle)
    首先,如果定义

    [S_+(m) = (-1)^{j-m} a^dagger_m a^dagger_{-m}, ~~~ S_-(m) = (-1)^{j-m} a_{-m} a_m,~~~ S_z(m) = frac{1}{2}( hat{N}_{m,-m} - 1 ) ]

    也可以在 ((j,m), (j,-m)) 这两个轨道上定义一个迷你的准自旋 SU(2) 代数。
    那么,(Gamma = e^{ipi S_y} = e^{ipi sum S_y(m) } = prod Gamma_m)

    [Gamma_m | 0 angle = e^{frac{pi}{2} [S_+(m) - S_-(m) ]} | 0 angle = sum_k frac{ (pi/2)^k }{ k! } [S_+(m) - S_-(m) ]^k | 0 angle. ]

    容易看出

    [(S_+(m) - S_-(m))^2 |0 angle = - S_-(m) S_+(m) | 0 angle = - | 0 angle, ]

    所以,计算得到 (Gamma_m | 0 angle = S_+(m) | 0 angle)

    [Gamma | 0 angle = prod_m Gamma_m | 0 angle = prod_m S_+(m) | 0 angle = frac{1}{Omega!} S^Omega_+ | 0 angle. ]

    最后一个等号是因为 (S^Omega_+ | 0 angle) 会产生 (Omega!)(prod_m S_+(m)|0 angle)

    2.4 (Gamma Psi_{IM u} (j^n))正比于(Psi_{IM u})的对偶空穴态

    如果把(Psi_{IM u}(j^n))中的所有粒子都拿走,所有没被占据的轨道都填上,会得到一个角动量为((I,-M))(2Omega-n)粒子波函数,咱们暂且把这个波函数叫做(Psi_{IM u})的对偶空穴态。
    这里假设了seniority是好量子数,即(Psi_{IM u})可以表示为

    [Psi_{IM u}(j^n) = (frac{(Omega -q - u)!}{q!(Omega - u)!})^{1/2} phi_{IM}(j^ u)S^q_+ |0>, ]

    即一部分是归一化之后的 seniority 为 0 的((frac{(Omega -q - u)!}{q!(Omega - u)!})^{1/2} S^q_+ |0>),另一部分是

    [phi_{IM}(j^ u) = sum_k alpha_{Ik} phi_M(k), phi_M(k) = a^dagger_{m_1 } cdots a^dagger_{m_ u} | 0 angle, ]

    那么,

    [Gamma Psi_{IM u} (j^n) = Gamma phi_{IM}(j^ u) Gamma^dagger Gamma (frac{(Omega -q - u)!}{q!(Omega - u)!})^{1/2} S^q_+ Gamma^dagger frac{1}{Omega!} S^Omega_+ | 0 angle, ]

    最后一行使用了 (Gamma | 0 angle = frac{1}{Omega!} S^Omega_+ | 0 angle).
    利用 (Gamma S_+ Gamma^dagger = - S_-)(S_pm)的对易子,以及 (Gamma a^dagger_m Gamma^dagger = ilde{a}_m),得到

    [Gamma Psi_{IM u} (j^n) = (-1)^q (frac{q!}{(Omega-q- u)!(Omega- u)!})^{1/2} ilde{phi}_{IM} S_+^{Omega-q- u}. |0> ]

    (phi_{IM})占据了 ( u)((j,pm m)) 轨道对中的一半,如果这一半空出,而另一半占据则正比于( ilde{phi}_{IM})
    除了这些对偶轨道之外的轨道能容纳 (Omega - u) 个对,(Phi_{IM u})占据了(q)个,剩下(Omega- u-q)个”对轨道“,若这些被占据则构成((frac{q!}{(Omega-q- u)!(Omega- u)!})^{1/2} S_+^{Omega-q- u} |0 angle)。 
    所以,(Gamma Psi_{IM u})正比于(Phi_{IM u})的对偶空穴态。

    2.5 粒子空穴变换

    所以,任何矩阵元的计算,都可以插进几个(Gamma^dagger Gamma),进行粒子空穴转换:

    [langle Phi | hat{H} | Psi angle = langle Phi Gamma^dagger Gamma hat{H} Gamma^dagger Gamma Phi angle. ]

    如果空穴数少于粒子数,就把哈密顿量做个粒子空穴变换(hat{H} ightarrow Gamma hat{H} Gamma^dagger),以 (Gamma | Phi angle) 为基矢做计算。
    注意,如果 (Phi) 没有好的 seniority,(Gamma | Phi angle)并不是严格的 (Phi) 的空穴对偶态,而是各个不同seniority组分的空穴对偶态乘以一个相因子以后的叠加。具体相因子与相应组分中的 (S) 对个数、角动量和第3角动量有关,即上面公式中的 ((-1)^q (-1)^{I +M})

    3. 壳模型相互作用的粒子空穴转换

    3.1 壳模型 1+2body 相互作用形式

    在pn格式下(与之相对的是isospin格式),壳模型1+2body哈密顿量为

    [H = H_{pp} + H_{nn} + H_{pn}, ]

    三个部分分别是质子、中子、质子-中子三个部分。

    [H_{pp} = sum_{a in pi} epsilon_a sum_{m_a} hat{c}^dagger_{ j_a, m_a } hat{c}_{j_a, m_a} + sumlimits_{abcd in pi} frac{ sqrt{ (1+delta_{ab})(1+delta_{cd}) } }{4} sum_I V_{pp}(abcd;I) sum_M hat{A}^dagger_{IM}(ab) hat{A}_{IM}(cd), ]

    其中(hat{A}^dagger_{IM}(ab) = ( a^dagger otimes b^dagger )_{IM}),是不可约张量算符。
    中子部分格式与质子部分是相似的,

    [H_{nn} = sum_{a in u} epsilon_a sum_{m_a} hat{c}^dagger_{ j_a, m_a } hat{c}_{j_a, m_a} + sumlimits_{abcd in u} frac{ sqrt{ (1+delta_{ab})(1+delta_{cd}) } }{4} sum_I V_{nn}(abcd;I) sum_M hat{A}^dagger_{IM}(ab) hat{A}_{IM}(cd), ]

    质子-中子部分略有不同

    [H_{pn} = sumlimits_{a,cin pi; b,d in u; I} V_{pn}(abcd;I) sum_M hat{A}^dagger_{IM}(ab) hat{A}_{IM}(cd). ]

    3.2 (H_{pp} Rightarrow Gamma H_{pp} Gamma^dagger)

    3.2.1 单体项

    单体项很简单,因为

    [Gamma hat{c}^dagger_{j_a, m_a} hat{c}_{j_a, m_a} Gamma^dagger = hat{c}_{j_a, -m_a} hat{c}^dagger_{j_a, -m_a} = 1 - hat{c}^dagger_{j_a, -m_a} hat{c}_{j_a, -m_a} , ]

    所以有

    [Gamma sum_{a in pi} epsilon_a sum_{m_a} hat{c}^dagger_{ j_a, m_a } hat{c}_{j_a, m_a} Gamma^dagger = sum_{a in pi} epsilon_a (2j_a+1) - sum_{ainpi} epsilon_a sum_{m_a} hat{c}^dagger_{ j_a, m_a } hat{c}_{j_a, m_a}, ]

    即满占据的总单粒子能减去空穴的“单粒子能”,空穴的单粒子能是粒子的单粒子能的负数。

    3.2.2 两体项

    利用前面推导的(Gamma)算符的性质,得到,

    [Gamma sum_M hat{A}^dagger_{IM}(ab) hat{A}_{IM}(cd) Gamma^dagger = - sum_M sum_{alpha eta gamma delta} C^{IM}_{a alpha, b eta} C^{IM}_{c gamma, d delta} a_alpha b_eta c^dagger_gamma d^+_delta, ]

    利用 Wick 定理,即任意算子都可以表示成一系列正则排序的算子之和,

    [abc^dagger d ^dagger = delta_{bc} delta_{ad} - delta_{ac} delta_{bd} - delta_{bc} d^dagger a^dagger + delta_{ac} d^dagger b + delta_{bd} c^dagger a - delta_{ad} c^dagger b + c^dagger d^dagger a b, ]

    经过亿点繁琐的计算,得到

    [Gamma H_{pp} Gamma^dagger = sum_{abin pi, I} frac{ 1+delta_{ab}}{2} (2I+1) V_{pp}(abab;I) - sum_{adinpi} delta_{j_a j_d} ( sum_{binpi, I} sqrt{(1+delta_{ab})(1+delta_{bd})} ) frac{2I+1}{2j_a+1} V(abdb;I) ) sum_alpha d^dagger_alpha a_alpha + sumlimits_{abcd in pi} frac{ sqrt{ (1+delta_{ab})(1+delta_{cd}) } }{4} sum_I V_{pp}(abcd;I) sum_M hat{A}^dagger_{IM}(ab) hat{A}_{IM}(cd), ]

    3.3 (H_{nn} Rightarrow Gamma H_{nn} Gamma^dagger)

    (Gamma H_{nn} Gamma^dagger)也是完全类似的,把上面的公式中的轨道全部换成中子轨道即可。

    3.4 (H_{pn} Rightarrow Gamma H_{pn} Gamma^dagger)

    3.4.1 质子、中子都做粒子空穴转换

    经过相似的一点推导,得到

    [Gamma H_{pn} Gamma^dagger = sum_{ainpi, bin u, I} (2I+1) V_{pn}(abab;I) - sum_{ainpi, bdin u, I} V_{pn}(abad;I) frac{2I+1}{2j_b+1}delta_{j_bj_d} sum_eta d^dagger_eta b_eta - sum_{acinpi, bin u, I} V_{pn}(abcb;I) frac{2I+1}{2j_a+1}delta_{j_aj_c} sum_alpha c^dagger_alpha a_alpha + sum_{acinpi, bdin u, I} V_{pn}(abcd;I) sum_M hat{A}^dagger_{IM}(cd) hat{A}_{IM}(ab). ]

    所以变换以后的两体部分不变,多了常数项、质子单体项、中子单体项。

    3.4.2 只有中子做粒子空穴转换

    [Gamma_n H_{pn} Gamma^dagger_n = sum_{ac in pi, b in u, I} delta_{j_a j_c} V_{pn}(abcb;I) (2I+1)/(2j_a+1) sum_alpha a^dagger_alpha c_alpha - sum_{ac in pi, bd in u, L} sum_I (2I+1) V_{pn}(abcd;I) left{ egin{array}{ccc} a & b & I \ c & d & L end{array} ight} sum_M A^dagger_{LM} (ad) A_{LM} (cb) ]

    3.4.3 只有质子做粒子空穴转换

    [Gamma_p H_{pn} Gamma^dagger_p = sum_{a in pi, bd in u, I} delta_{j_b j_d} V_{pn}(abad;I) (2I+1)/(2j_b+1) sum_eta b^dagger_eta d_eta - sum_{ac in pi, bd in u, L} sum_I (2I+1) V_{pn}(abcd;I) left{ egin{array}{ccc} a & b & I \ c & d & L end{array} ight} sum_M A^dagger_{LM} (cb) A_{LM} (ad) ]

    4.计算验证

    我用 jun45 相互作用(使用 scaling),分别用 PandaWarrior 和 Bigstick 进行计算:1) (p2n2) 2) (p ar{2} n ar{2}) 3) (p2nar{2}) 4) (par{2}n2) 四种情况,其中 (p2) 表示 (2) 质子,(par{2}) 表示 2 质子空穴。

    4.1 (p2n2)(^{60}_{30}Zn)

    前20个能谱如下,PandaWarrior 和 Bigstick 的结果完全一样(虽然PandaWarrior没有isospin)。

    BigStick States:
      State      E        Ex         J       T 
        1    -50.42585   0.00000     0.000  -0.000
        2    -49.43000   0.99585     2.000  -0.000
        3    -47.78510   2.64075     4.000   0.000
        4    -46.30908   4.11676     2.000  -0.000
        5    -46.24605   4.17980     0.000   0.000
        6    -45.97339   4.45246     2.000  -0.000
        7    -45.85816   4.56769     6.000   0.000
        8    -45.69710   4.72875     1.000  -0.000
        9    -45.25258   5.17327     4.000  -0.000
       10    -45.22663   5.19922     3.000  -0.000
       11    -45.00065   5.42520     1.000  -0.000
       12    -44.98837   5.43748     2.000   1.000
       13    -44.80961   5.61624     2.000  -0.000
       14    -44.70301   5.72284     3.000   0.000
       15    -44.61498   5.81087     2.000   1.000
       16    -44.56598   5.85987     3.000   0.000
       17    -44.56188   5.86397     0.000  -0.000
       18    -44.51231   5.91354     1.000   1.000
       19    -44.47028   5.95556     2.000   0.000
       20    -44.43870   5.98715     1.000   1.000
       
    PandaWarrior States:
    States   E       Ex      J pi    ?th
    1    -50.4258    0        0+  1th
    2    -49.43    0.99585        2+  1th
    3    -47.7851    2.64075        4+  1th
    4    -46.3091    4.11677        2+  2th
    5    -46.246    4.17981        0+  2th
    6    -45.9734    4.45246        2+  3th
    7    -45.8582    4.56769        6+  1th
    8    -45.6971    4.72875        1+  1th
    9    -45.2526    5.17327        4+  2th
    10    -45.2266    5.19922        3+  1th
    11    -45.0006    5.4252        1+  2th
    12    -44.9884    5.43748        2+  4th
    13    -44.8096    5.61624        2+  5th
    14    -44.703    5.72284        3+  2th
    15    -44.615    5.81087        2+  6th
    16    -44.566    5.85987        3-  1th
    17    -44.5619    5.86397        0+  3th
    18    -44.5123    5.91354        1+  3th
    19    -44.4703    5.95557        2+  7th
    20    -44.4387    5.98715        1+  4th
    

    4.2 (p20n20)(^{96}_{48}Cd)

    BigStick states:
      State      E        Ex         J       T 
        1   -528.99745   0.00000    -0.000   0.000
        2   -528.09635   0.90111     2.000   0.000
        3   -527.01004   1.98741     4.000   0.000
        4   -526.96621   2.03125    -0.000   0.000
        5   -526.33770   2.65975     2.000   0.000
        6   -526.31427   2.68319     5.000   0.000
        7   -526.06750   2.92996     3.000   0.000
        8   -525.97606   3.02140     6.000   0.000
        9   -525.51433   3.48312     8.000   0.000
       10   -525.33979   3.65767     4.000   0.000
       11   -524.98870   4.00875     4.000   0.000
       12   -524.96354   4.03391     8.000   1.000
       13   -524.85037   4.14709     2.000   1.000
       14   -524.75086   4.24659     6.000   0.000
       15   -524.71605   4.28141     8.000   0.000
       16   -524.64199   4.35546     5.000   0.000
       17   -524.61983   4.37762     6.000   0.000
       18   -524.61334   4.38412     2.000   0.000
       19   -524.58394   4.41351     1.000   1.000
       20   -524.53035   4.46711     4.000   0.000
    
    PandaWarrior States:
    States   E       Ex      J pi    ?th
    1    -528.998    0        0+  1th
    2    -528.096    0.901055        2+  1th
    3    -527.01    1.98744        4+  1th
    4    -526.966    2.03116        0+  2th
    5    -526.338    2.65966        2+  2th
    6    -526.314    2.68317        5-  1th
    7    -526.068    2.92988        3-  1th
    8    -525.976    3.02133        6+  1th
    9    -525.514    3.4831        8+  1th
    10    -525.34    3.65767        4-  1th
    11    -524.989    4.0087        4-  2th
    12    -524.964    4.03392        8+  2th
    13    -524.85    4.14708        2+  3th
    14    -524.751    4.24655        6-  1th
    15    -524.716    4.2814        8+  3th
    16    -524.642    4.35541        5-  2th
    17    -524.62    4.3776        6+  2th
    18    -524.613    4.38411        2+  4th
    19    -524.584    4.4135        1+  1th
    20    -524.53    4.46708        4+  2th
    

    4.3 (p2n20)(^{78}_{30}Zn)

    BigStick States:
      State      E        Ex         J       T 
        1   -198.39336   0.00000     0.000   9.000
        2   -197.34820   1.04516     2.000   9.000
        3   -196.60879   1.78457     4.000   9.000
        4   -196.51383   1.87952     2.000   9.000
        5   -195.98205   2.41131     4.000   9.000
        6   -195.90511   2.48824     0.000   9.000
        7   -195.70748   2.68587     3.000   9.000
        8   -195.67498   2.71837     2.000   9.000
        9   -195.65626   2.73710     5.000   9.000
       10   -195.52211   2.87124     6.000   9.000
       11   -195.48953   2.90383     0.000   9.000
       12   -195.48833   2.90502     2.000   9.000
       13   -195.46283   2.93052     1.000   9.000
       14   -195.45651   2.93684     4.000   9.000
       15   -195.43577   2.95759     1.000   9.000
       16   -195.43262   2.96074     8.000   9.000
       17   -195.40300   2.99036     4.000   9.000
       18   -195.25651   3.13685     3.000   9.000
       19   -195.14086   3.25249     6.000   9.000
       20   -195.00435   3.38901     2.000   9.000
    
    PandaWarrior States:
    States   E       Ex      J pi    ?th
    1    -198.393    0        0+  1th
    2    -197.348    1.04516        2+  1th
    3    -196.609    1.78457        4+  1th
    4    -196.514    1.87952        2+  2th
    5    -195.982    2.41131        4+  2th
    6    -195.905    2.48821        0+  2th
    7    -195.707    2.68587        3+  1th
    8    -195.675    2.71837        2+  3th
    9    -195.656    2.73707        5-  1th
    10    -195.522    2.87124        6+  1th
    11    -195.49    2.90381        0+  3th
    12    -195.488    2.90503        2+  4th
    13    -195.463    2.93053        1+  1th
    14    -195.456    2.93684        4+  3th
    15    -195.436    2.95758        1+  2th
    16    -195.433    2.96073        8+  1th
    17    -195.403    2.99034        4-  1th
    18    -195.256    3.13684        3+  2th
    19    -195.141    3.25249        6+  2th
    20    -195.004    3.389        2+  5th
    

    4.4 (p20n2)(^{78}_{48}Cd)

    结果应该与 (^{78}_{30}Zn) 相同。

    BigStick States:
      State      E        Ex         J       T 
        1   -198.39335   0.00000     0.000   9.000
        2   -197.34820   1.04515     2.000   9.000
        3   -196.60878   1.78457     4.000   9.000
        4   -196.51381   1.87954     2.000   9.000
        5   -195.98205   2.41131     4.000   9.000
        6   -195.90512   2.48824     0.000   9.000
        7   -195.70748   2.68588     3.000   9.000
        8   -195.67498   2.71837     2.000   9.000
        9   -195.65626   2.73709     5.000   9.000
       10   -195.52212   2.87124     6.000   9.000
       11   -195.48953   2.90382     0.000   9.000
       12   -195.48833   2.90503     2.000   9.000
       13   -195.46283   2.93052     1.000   9.000
       14   -195.45651   2.93684     4.000   9.000
       15   -195.43576   2.95759     1.000   9.000
       16   -195.43261   2.96074     8.000   9.000
       17   -195.40298   2.99037     4.000   9.000
       18   -195.25651   3.13684     3.000   9.000
       19   -195.14086   3.25250     6.000   9.000
       20   -195.00434   3.38901     2.000   9.000
    
    PandaWarrior States:
    States   E       Ex      J pi    ?th
    1    -198.393    0        0+  1th
    2    -197.348    1.04516        2+  1th
    3    -196.609    1.78457        4+  1th
    4    -196.514    1.87952        2+  2th
    5    -195.982    2.41131        4+  2th
    6    -195.905    2.48821        0+  2th
    7    -195.707    2.68587        3+  1th
    8    -195.675    2.71837        2+  3th
    9    -195.656    2.73707        5-  1th
    10    -195.522    2.87124        6+  1th
    11    -195.49    2.90381        0+  3th
    12    -195.488    2.90503        2+  4th
    13    -195.463    2.93053        1+  1th
    14    -195.456    2.93684        4+  3th
    15    -195.436    2.95758        1+  2th
    16    -195.433    2.96073        8+  1th
    17    -195.403    2.99034        4-  1th
    18    -195.256    3.13684        3+  2th
    19    -195.141    3.25249        6+  2th
    20    -195.004    3.389        2+  5th
    

    所以,PandaWarrior 内部的处理代码给出了正确的结果。
    PVPC/src/class.cpp 中的 xpn_int::Pandya 中有一个bug, Ln2588
    mononew[i][i] += nspe[i];//(mono)[i][i];
    应改为
    mononew[i][i] += nspe[ i - jspace.nj_p ];//(mono)[i][i];
    更正以后,用 PVPC 做 Pandya,然后用 PandaWarrior 重复了上面四种情况的检验,验证通过了。

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  • 原文地址:https://www.cnblogs.com/luyi07/p/14603628.html
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