Question

For a simple harmonic oscillator, show that the expectation value of x, defined as <x>mn = ∫ Ψ*m (x) Ψ n(x) dx is √1/2a^2 for the n=0 and m=1states.Use the

result ∞∫0 x^1/2 exp ^-x dx= √π 

Answer 1

The wave function is initially in the ground state of the oscillator with classical frequency !. The wave function is $Ψ_0 (x) = (\frac{\omega m}{\pi h})^2e^{- \frac{\omega m}{\pi h}x^2}$

The spring constant changes in real time, while the wav function does not. As a result, the wave function stays unchanged immediately following the change in spring constant. It is, however, no longer an eigenfunction of the hamiltonian operator. Any function, on the other hand, may be written as a linear combination of the new eigenfunctions, n0, and we may write

$a_n = \int^{\infin}_{-\infin} Ψ_n' (x)Ψ_0 (x) dx$

$a_n = \int^{\infin}_{-\infin} (\frac{\omega' m}{\pi h})^2e^{- \frac{\omega' m}{\pi h}x^2} (\frac{\omega m}{\pi h})^2e^{- \frac{\omega m}{\pi h}x^2}$

$<x>mn = ∫ Ψ*m (x) Ψ n(x) dx = \sqrt{\frac{1}{2a^2}}$