2. Variational Principle. The energy of a system with wave function ψ is given by where...

Variation method problem Let trial wave fun Psi (x) = A e^-alpha x^2/2 - (1) given H = p^2/2m + lambda x^2 = -h^2/2m partial differential^2/partial differential x^2 + lambda x^4 Hence ground state Energy < E> = integral Psi* H Psi dx/integral Psi* Psi dx < E> = A^2 (- h^2/2m) integral^+infinity_-infinity e^-alpha x^2/2 partial differential^2/partial differential x^2 (e^-alpha x^2/2)/A^2 integral^+infinity_-infinity e^-2 alpha x^2 dx + A^2 integral^+infinity_-infinity lambda x^4 e^-alpha x^2 dx/A^2 integral^+infinity_-infinity e^-alpha x^2 dx use integral integral^infinity_a x^n e^-alpha x^2 dx = squareroot h + 1/2/2 (squareroot infinity n + 1/2) then we will get < E> = h^2 alpha/2 (2m) + lambda times squareroot 5/2/(alpha)^5/2/squareroot 1/2/(alpha)^1/2 < E> = h^2 alpha/4m + lambda times 3/2 middot 1/2 middot squareroot pi times alpha^1/2/(alpha)^5/2 squareroot pi < E> = h^2 alpha/4m + 3 lambda/4 alpha^2 - (2) For minimum Elegy d < t>/d alpha = 0 => h^2/4m - 2 times 3 lambda/4 alpha^2 = 0