In d -dimensional free space, corresponding to V ( r ) = constant on R d , this equation can be solved using Fourier transforms. If the particle is initialized at t = 0 with position r ′ and momentum p ′ , corresponding to initial condition f ( r , p , 0 ) = δ δ ( r − − r ′ ) δ δ ( p − − p ′ ) , then the solution is f ( r , p , t ) = 1 ( 2 π π σ σ X σ σ P 1 − − β β 2 ) d × × exp ⁡ ⁡ [ − − 1 2 ( 1 − − β β 2 ) ( | r − − μ μ X | 2 σ σ X 2 + | p − − μ μ P | 2 σ σ P 2 − − 2 β β ( r − − μ μ X ) ⋅ ⋅ ( p − − μ μ P ) σ σ X σ σ P ) ] where σ σ X 2 = k B T m ξ ξ 2 [ 1 + 2 ξ ξ t − − ( 2 − − e − − ξ ξ t ) 2 ] ; σ σ P 2 = m k B T ( 1 − − e − − 2 ξ ξ t ) β β = k B T ξ ξ σ σ X σ σ P ( 1 − − e − − ξ ξ t ) 2 μ μ X = r ′ + ( m ξ ξ ) − − 1 ( 1 − − e − − ξ ξ t ) p ′ ; μ μ P = p ′ e − − ξ ξ t . In three spatial dimensions, the mean squared displacement is ⟨ ⟨ r ( t ) 2 ⟩ ⟩ = ∫ ∫ f ( r , p , t ) r 2 d r d p = μ μ X 2 + 3 σ σ X 2