Density Matrix for Harmonic Oscillator

in which the Hamiltonian takes the formH = ħω )(− ∂22 ∂X 2 + X 2by defining f = (ħω/2)β we get the nicer looking differential equationwith the initial condition− ∂∂f ρ(X , X ′ , f ) =ρ(X , X ′ , f = 0) = δ(x − x ′ ) = δ((18)(− ∂2∂X 2 + X 2 )ρ(X , X ′ , f ) (19)√ħmω (X − X ′ )) =√ mωħ δ(X − X ′ ) (20)where I used the identity δ(g (y)) = |g ′ (y 0 )| −1 δ(y − y 0 ) where y 0 is a root of f (y).We solve this differential equation similar to how we solved the Fokker-planck equation: by making an ansatz and then comparing coefficients in powers of X .Ansatz: ρ = e −(a(f )X 2 +b(f )X +c(f )) (21)Note that X ′ only enters differential equation as a parameter ⇒ a,b,c may dependon X ′ . The left hand side of Eq. (19) can be writtenLHS: − ∂ρ ( ∂a∂f = ∂f X 2 + ∂b∂f X + ∂c )ρ (22)∂fand the right hand sideRHS:)(− ∂2∂X 2 + X 2 ρ = ∂ [− ∂ ]∂X ∂X ρ + X 2 ρ= ∂ [ ](2aX + b)ρ + X 2 ρ∂X= [ 2a − (2aX + b) 2 + X 2] ρ= [ 2a − 4a 2 X 2 − 4abX − b 2 + X 2] ρPutting it all together we have[ ∂a∂f X 2 + ∂b∂f X + ∂c ]ρ = [ 2a − 4a 2 X 2 − 4abX − b 2 + X 2] ρ (24)∂fCancelling ρ on both sides and comparing powers of X gives us three equations:X 2 :X 1 :X 0 :∂a∂f = 1 − 4a2 ⇒ a = 1 2 coth2f∂b∂f = −4ab ⇒ b = Asinh2f∂c∂f = 2a − b2 ⇒ c = 1 2 ln[ sinh2f ] + A24(23)2 coth2f − ln[B] (25)