If you solve the Time Independent Schrodinger equation for the Harmonic Oscillator, that is [tex] -\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} + \frac{1}{2}kx^2 \Psi = E \Psi [/tex] The quantization of energy comes from the boundary conditions (ie, [itex] \Psi = 0 [/itex] when [itex] x= \infty [/itex] or [itex] x = -\infty [/itex]). The permitted energy levels will be [tex] E_n = (n+\frac{1}{2}) \hbar \omega [/tex] So the lowest Energy is not E=0.
I could give a hand-wave argument. We have E=1/2mv^2+1/2kx^2. If E=0 both x and v are zero, which contradicts Heisenberg.