In this problem you will derive the ground-state energy of the harmonic oscillator using the precise form of the uncertainty principle, Δx Δp ≥ h/2, where Δx and Δp are defined to be the standard deviations (Δx)² = [(x – x_av)²]_av and (Δp)² = [(p – p_av)²]_av (see Equation 18-31). Proceed as follows:

1. Write the total classical energy in terms of the position x and momentum p using

U(x) = ½ mω²x² and K = p²/2m.

2. Use the result of Equation 18–35 to write (Δx)² = [(x – x_av)² ]_av = (x²) – x²_av and (Δp)²

= [(p - p_av)²]_av = (p²)_av – p²_av

3. Use the symmetry of the potential energy function to argue that x_av and p_av must be zero, so that (Δx)² = (x)_av and (Δp)² = (p²)_av.

4. Assume that Δp = h/2Δx to eliminate (p²)_av from the average energy E_av = (p²)_av/2m + ½mω²(x²)_av and write E_av as E_av = h₂/8mZ + ½ mω²Z, where Z = (x²)_av.

5. Set dE/dZ = 0 to find the value of Z for which E is a minimum.

6. Show that the minimum energy is given by (E_av)_min = + ½hω.

Transcribed Image Text:

36-31 y(x, y, z) = A sin k,x sin k,y sin k,z h? 0?y(x,x2) h? a2w(x1,x2) 2m +Uy(x1,x2) = Ey(x1,X2) 36-35 ôx} ôx3 2m