J≡ π2¯h4 2m2a3U0 Notice that, by assumption of u≫ 1,E0 ≫ tNotice also that E0 is the energy of When opening the square, we neglected the O 1/u2 term but kept the O(1/u) one Rewrite the dispersion asThat would make the new expression e^ (pi/2)*i That, in turn, if reexpressed in rectangular format is i The only rationale I can find for avoiding this outcome is to invoke DeMoivre's formula, where only the real number value (pi/2) is not addressed by the exponent, i So, pi/2 remains pi/2 and i becomes isquared or simply 1Dec 08, 16 · Transcript Ex 22, 16 Find the values of sin1(sin〖2π/3〗 ) Let y = sin1 (sin 2𝜋/3) sin y = sin 2𝜋/3 sin y = sin (1°) But, range of principal value of sin1 is (−π)/2, π/2 ie 90° ,90° Hence y = 1° not possible Now, sin y = sin (1°) sin y = sin (180° – 60°) sin y = sin (60°) sin y = sin (60 × 𝜋/180) sin y = sin 𝜋/3 Hence, y = 𝜋/3 Which is in Momentum Resolved Visualization Of Electronic Evolution In Doping A Mott Insulator Nature Communications 4.2698671
