In a quantum harmonic oscillator of characteristic angular frequency ω, place a state that is offset from the bottom of the potential by some displacement x 0 as ψ ψ ( x ) = ( m Ω Ω π π ℏ ℏ ) 1 / 4 exp ⁡ ⁡ ( − − m Ω Ω ( x − − x 0 ) 2 2 ℏ ℏ ) , where Ω describes the width of the initial state but need not be the same as ω. Through integration over the propagator, we can solve for the full time-dependent solution. After many cancelations, the probability densities reduce to | Ψ Ψ ( x , t ) | 2 ∼ ∼ N ( x 0 cos ⁡ ⁡ ( ω ω t ) , ℏ ℏ 2 m Ω Ω ( cos 2 ⁡ ⁡ ( ω ω t ) + Ω Ω 2 ω ω 2 sin 2 ⁡ ⁡ ( ω ω t ) ) ) | Φ Φ ( p , t ) | 2 ∼ ∼ N ( − − m x 0 ω ω sin ⁡ ⁡ ( ω ω t ) , ℏ ℏ m Ω Ω 2 ( cos 2 ⁡ ⁡ ( ω ω t ) + ω ω 2 Ω Ω 2 sin 2 ⁡ ⁡ ( ω ω t ) ) ) , where we have used the notation N ( μ μ , σ σ 2 ) to denote a normal distribution of mean μ and variance σ. Copying the variances above and applying trigonometric identities, we can write the product of the standard deviations as σ σ x σ σ p = ℏ ℏ 2 ( cos 2 ⁡ ⁡ ( ω ω t ) + Ω Ω 2 ω ω 2 sin 2 ⁡ ⁡ ( ω ω t ) ) ( cos 2 ⁡ ⁡ ( ω ω t ) + ω ω 2 Ω Ω 2 sin 2 ⁡ ⁡ ( ω ω t ) ) = ℏ ℏ 4 3 + 1 2 ( Ω Ω 2 ω ω 2 + ω ω 2 Ω Ω 2 ) − − ( 1 2 ( Ω Ω 2 ω ω 2 + ω ω 2 Ω Ω 2 ) − − 1 ) cos ⁡ ⁡ ( 4 ω ω t )