TDHF for frequency-dependent operators

[benroberts999]

• TDHF method doesn't work for frequency-dependent E1v operator, though diagram method does
• Possibly linked to #3 (possibly unrelated though)
• Neither method is correctly symmetric for E1v (probably $|\omega|$ vs. $\omega$)
• "Basis" method performs the same as TDHF - which is somewhat interesting
  • "Basis" method is equivalent to TDHF, but instead of solving TDHF equations, expands solutions using a basis
  • nb: "basis" method is extremely slow, and should only be used for testing

Length form (independent of $\omega$)	E1(L)	$\omega$	HF	TDHF	Diagram	Basis
6p- - 6s+	0.0418	-5.2777	-4.9747	-4.9747	-4.9747
6p+ - 6s+	0.0436	7.4264	7.0137	7.0137	7.0137
6s+ - 6p-	-0.0418	-5.2777	-4.9747	-4.9747	-4.9747
6s+ - 6p+	-0.0436	-7.4264	-7.0137	-7.0137	-7.0137

Velocity form ($\omega$ dependent)	E1(v)	$\omega$	HF	TDHF	Diagram	Basis
6p- - 6s+	0.0418	-5.0371	1.9758	-4.9747	1.9755
6p+ - 6s+	0.0436	7.0662	-1.9837	7.0137	-1.9834
6s+ - 6p-	-0.0418	5.0371	12.0498	5.0994	12.0495
6s+ - 6p+	-0.0436	7.0662	16.1161	7.1187	16.1157

Full code input/output: E1_LvsV.txt

[benroberts999]

Note: Dzuba code has same issue with TDHF (though gives different number again):

w = 0.04175221

Iteration #   1 Delta=0.1442E-10 eps=0.1000E-09
matv:  -0.816497  <-1  6|V| 1  6>=  0.503706E+01  0.584510E+01
matv:  -0.816497  < 1  6|V|-1  6>=  0.503706E+01  0.422902E+01

w = 0.04358263

Iteration #   1 Delta=0.1717E-10 eps=0.1000E-09
matv:   1.154701  <-1  6|V|-2  6>= -0.706618E+01 -0.803144E+01
matv:  -1.154701  <-2  6|V|-1  6>=  0.706618E+01  0.610092E+01

[benroberts999]

At the very least, issue probably has something to do with incorrect sign for omega (that seems to be part, but not the fill problem): See output: e1-v.txt