Seperating Free Energy into Potential and Kinetic Terms

e^{- \frac{F (Q, P)}{k_{B} T}} = \int_{V} d \vec{q} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \prod_{i} δ (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \frac{\partial H}{\partial p_{j}}) e^{- \frac{H (\vec{p}, \vec{q})}{k_{B} T}}

Then if the Hamiltonian is in the quadratic form

H (\vec{q}, \vec{p}) = \frac{1}{2} p^{⊤} A (q) p + U (q) = \frac{1}{2} \sum_{k} \sum_{l} 𝐆_{k l} (q) p_{k} p_{l} + U (q)

and

\frac{\partial H (\vec{q}, \vec{p})}{\partial p_{j}} = \sum_{k} S_{j k} (q) p_{k}

then

e^{- \frac{F (Q, P)}{k_{B} T}} = \int_{V} d \vec{q} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \prod_{i} δ (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k}) e^{- \frac{\frac{1}{2} \sum_{k} \sum_{l} G_{k l} (q) p_{k} p_{l} + U (q)}{k_{B} T}}

= \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \prod_{i} δ (\frac{\partial F}{\partial P_{i}} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k}) e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l})}

And assume that the free energy is also quadratic or something

e^{- β (P^{⊤} R (Q) P + V (Q))} = \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \prod_{i} δ (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k}) e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l})}

substitute in the fourier definition for the dirac delta

= \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \prod_{i} \frac{1}{τ} \int_{- \infty}^{\infty} d k e^{i k (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l})}

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} \prod_{i} e^{i k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l})}

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i \sum_{i} k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})} e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l})}

combine the exponents

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- β (\frac{1}{2} \sum_{k} \sum_{l} S_{k l} (q) p_{k} p_{l}) + i \sum_{i} k_{i} (\sum_{l} R_{i l} P_{l} - \sum_{j} \frac{\partial f_{i}}{\partial q_{j}} \sum_{k} S_{j k} (q) p_{k})}

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{- \infty} d \vec{p} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- β \frac{1}{2} p^{⊤} 𝐒 p + i k^{⊤} (𝐑 P - 𝐀 𝐒 p)}

swap the order of integration

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} \int_{- \infty}^{- \infty} d \vec{p} e^{- β \frac{1}{2} p^{⊤} 𝐒 p + i k^{⊤} (𝐑 P - 𝐀 𝐒 p)}

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} \int_{- \infty}^{- \infty} d \vec{p} e^{- β \frac{1}{2} p^{⊤} 𝐒 p - i k^{⊤} 𝐀 𝐒 p}

Do complete the squares on the exponent

- β \frac{1}{2} p^{⊤} 𝐒 p - i k^{⊤} 𝐀 𝐒 p

= - \frac{1}{2} p^{⊤} (β 𝐒) p + (- i 𝐒 𝐀^{⊤} k)^{⊤} p

And assume its positive definite symmetric and invertable then do complete the squares

= - \frac{1}{2} (p - i (\frac{1}{β} 𝐀^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐀^{⊤} k)) - \frac{1}{2} \frac{1}{β} (𝐀^{⊤} k)^{⊤} 𝐒 (𝐀^{⊤} k)

Then substitute it back in to get

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} \int_{- \infty}^{- \infty} d \vec{p} e^{- \frac{1}{2} (p - i (\frac{1}{β} 𝐀^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐀^{⊤} k)) - \frac{1}{2} \frac{1}{β} (𝐀^{⊤} k)^{⊤} 𝐒 (𝐀^{⊤} k)}

stuff

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} e^{- \frac{1}{2} \frac{1}{β} (𝐀^{⊤} k)^{⊤} 𝐒 (𝐀^{⊤} k)} \int_{- \infty}^{- \infty} d \vec{p} e^{- \frac{1}{2} (p - i (\frac{1}{β} 𝐀^{⊤} k))^{⊤} (β 𝐒) (p - i (\frac{1}{β} 𝐀^{⊤} k))}

for an arbitrary multidimensional gaussian and symmetric positive definite matrix $A$ .

\int d \vec{x} e^{(x - a)^{⊤} 𝐌 (x - a)} = \sqrt{\frac{π^{n}}{det 𝐌}}

You can show it works even iff the offset $a$ is an complex vector

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{i k^{⊤} 𝐑 P} e^{- \frac{1}{2} \frac{1}{β} (𝐀^{⊤} k)^{⊤} 𝐒 (𝐀^{⊤} k)} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

now the expression is independent of the fine grain momentum. Do complete the squares again.

i k^{⊤} 𝐑 P - \frac{1}{2} \frac{1}{β} (𝐀^{⊤} k)^{⊤} 𝐒 (𝐀^{⊤} k)

rearrange a bit

= - \frac{1}{2} \frac{1}{β} k^{⊤} (𝐀 𝐒 𝐀^{⊤}) k + i (𝐑 P)^{⊤} k

complete the squares does

= - \frac{1}{2} \frac{1}{β} (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐀 𝐒 𝐀^{⊤}) (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P) - β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P

and note this requires that $𝐀 𝐒 𝐀^{⊤}$ is invertable symmetric positive definite. Substitute it back in.

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- \frac{1}{2} \frac{1}{β} (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐀 𝐒 𝐀^{⊤}) (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P) - β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) e^{- β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P} \int_{- \infty}^{\infty} \dots \int_{- \infty}^{\infty} d k_{1} \dots d k_{N} e^{- \frac{1}{2} \frac{1}{β} (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P)^{⊤} (𝐀 𝐒 𝐀^{⊤}) (k + i β (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P)} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

Then do the same thing again with the gaussian

= \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) e^{- β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{β^{N} τ^{N}}{det (𝐀 𝐒 𝐀^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

This gives a solution for seperating the free energy

e^{- β (P^{⊤} R (Q) P + V (Q))} = \frac{1}{τ^{N}} \int_{V} d \vec{q} e^{- β U (q)} \prod_{i} δ (Q_{i} - f_{i} (\vec{q})) e^{- β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P} \sqrt{\frac{β^{N} τ^{N}}{det (𝐀 𝐒 𝐀^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}}

This gives a final expression of

\exp (- β (P^{⊤} 𝐑 (Q) P + V (Q))) = \frac{1}{τ^{N}} \int_{V} d \vec{q} \exp (- β U (q) + \ln (\sqrt{\frac{β^{N} τ^{N}}{det (𝐀 𝐒 𝐀^{⊤})}} \sqrt{\frac{\frac{1}{β^{n}} τ^{n}}{det 𝐒}})) \exp (- β P 𝐑^{T} (𝐀 𝐒 𝐀^{⊤})^{- 1} 𝐑 P) \prod_{i} δ (Q_{i} - f_{i} (\vec{q}))