Wolfram Alpha:

```The Quantum Harmonic Oscillator
-------------------------------

Hooke's Law:  F = -kx

Newton's Law:  ma = -kx

∴ x = -(m/k)a or,

x = -(m/k)d2x/dt2

A solution to this equation is:

x = Asin(ωt) where ω = √(k/m)

Now, the PE of the spring, V, is given by:

V = (1/2)kx2

= (1/2)mω2x2

Schrodinger's Wave Equation becomes:

(-h2/2m)d2ψ/dx2 + (1/2)mω2x2ψ = Eψ

With solution:

ψ = Cexp(-αx2/2)

Substituting back in the wave equation gives:

(-h2/2m)[-α + α2x2]ψ + (1/2)mω2x2ψ = Eψ

∴ (h2α/2m)ψ - (h2/2m)α2x2ψ + (1/2)mω2x2ψ = E

We require E -> 0 as x -> ∞ for normalization of ψ.
The second and third term will dominate the first
term at large x.  Therefore:

(-h2/2m)α2x2ψ + (1/2)mω2x2ψ = 0

(-h2/2m)α2x2ψ = -(1/2)mω2x2ψ

(h2/2m)α2 = (1/2)mω2

∴ α = mω/h

Substituting this back into the first term gives:

(h2α/2m) => hω/2

Therefore, the lowest energy, E0 = hω/2.

Derivation From Uncertainty Principle, ΔxΔp = h/2
-------------------------------------------------

E = p2/2m + (1/2)mω2x2

= (Δp)2/2m + (1/2)mω2(Δx)2

= h2/8m(Δx)2 + (1/2)mω2(Δx)2

Differentiate and set = 0 to minimize:

-h2/4m(Δx)3 + mω2Δx = 0

∴ Δx = √(h/2mω)

Substituting this back into the original equation
gives:

E0 = h2/8m(Δx)2 + (1/2)mω2(Δx)2

= hω/4 + hω/4

= hω/2
```