Question

-ax²12 directly into the Schroedinger equation, as broken down in the following steps. Show that the energy of a simple harmonic oscillator in the n = 0 state is 1ho/2 by substituting the wave function wo = Ae First, calculate dvo/dx, using A, x, and a. dyo/dx = Second, calculate dvo/dx?, using A, x, and a. dyo/dx2 = Third, calculate a?x?wo-dayo/dx?, using A, x, and a. a3x240 - dạyo/dx? Fourth, calculate (a?x240-d2vo/dx2)/yo, using A, X, and a. (22x200-2vo/dx?)/- 1 Finally, calculate E = [(a3x240 - 02yo/dx2)/49]h?/(2m), using A, X, a, m, w, and h. E =

Answer 1

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Yo= A e-ax/ - ci s Cotelo (x) dx = 5 Are -ax? da =) AZ 1 A2 = 1/2 (a)/2 a Then, drola a2 Do, imo se, A- men t Now, We know H Xed) E %10) Thus, -h2 2m d2Colx] + vens Yola) = to leas =-axA e 2 A (-a [2x] e-98/2 d2%(n) = -9 A e-9X2/2 - 2x (-22)A e da d146 = (22x31-aA) e-92²/2 = (62x²a) Yolz) Then, ? [?x2_2] C (m) + mw2x2 4.636) =E4069) [a-a2x²] Yo (2) +12m mw2x2 Yo Cx) dn -ar22 am 72 2 am Elorno [ a- m2w 2 2 2740 (m) + m2w2X? Yo (n)= 2 mE Yolm) Ee t² 7² a lola) = 2mE Young t2 so, compaining we get E-a ħ² h2] = to ti am Menu prooved). = mo am