Question

Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using (50) = e-a-l as a trial function with a to be varied. For a simple harmonic oscillator we have H + jmwºr? Recall that, for the variational method, the trial function (HO) gives an expectation value of H such that (016) > Eo, where Eo is the ground state energy. You may use: n! dH() ||= TH(c) – z[1 – H(r)], 8(2), dx S." arcade an+1 where (x) and (x) denote the Heaviside step function and Dirac delta function respec- tively. (Don't get confused with the Hamiltonian H and the Heaviside step function H(2).)

Answer 1

L expectation value of <H> Solution 4102 - © é alal an olaco e are o caloo <47.4197 2414 <H22 7414) = sta 2212) dre 00 - 5° 4243 dx + dx + 8 2<* da 0! + 2a [222] 2a + on (2097+0 12 1) 112 24 2x X4142 12 2 24761427 = ) 2415x214) + {mar 24/01414) da ot / marja?exp(-284) lx 222 2M w 0.227 dac+4 mar (27 fxecupfremd 2M ) 2011 tlmer (2) 6+2121 2 aylara am 28 (22)9

pll 품 Mur I 구 from eq (1) (2) and (3) 로 2414142 Fly) XH) 人. + M 3 . 222 새츠 음 Mer a로 2M