Path Integral from Lattice Limit

Classical Lagrangian

Start with the definition of action $S$ across some path $\vec{x} (t)$ where $ℒ (\vec{x}, \dot{\vec{x}})$ is the Lagrangian density.

S (\vec{x}) = \int_{t_{a}}^{t_{b}} ℒ (\vec{x} (t), \frac{d \vec{x} (t)}{d t}) d t

From the principle of least action the path should be a minimum, so the variational derivative should be zero given an perturbation $δ \vec{x} (t)$ that preserves the same start and end points so $δ \vec{x} (t_{a}) = 0, δ \vec{x} (t_{b}) = 0$

\lim_{ϵ \to 0} \frac{S (\vec{x} + ϵ δ \vec{x}) - S (\vec{x})}{ϵ} = 0

\lim_{ϵ \to 0} \frac{1}{ϵ} \int_{t_{a}}^{t_{b}} ℒ (\vec{x} (t) + ϵ δ \vec{x}, \frac{d \vec{x} (t)}{d t} + ϵ \frac{d δ \vec{x}}{d t}) - ℒ (\vec{x} (t), \frac{d \vec{x} (t)}{d t}) d t = 0

Assume we can swap the derivative and integral.

\int_{t_{a}}^{t_{b}} \lim_{ϵ \to 0} \frac{ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t} + ϵ \frac{d δ \vec{x} (t)}{d t}) - ℒ (\vec{x} (t), \frac{d \vec{x} (t)}{d t})}{ϵ} d t = 0

Create a dummy value.

\int_{t_{a}}^{t_{b}} \lim_{ϵ \to 0} \frac{ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t} + ϵ \frac{d δ \vec{x} (t)}{d t}) - ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t}) + ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t}) - ℒ (\vec{x} (t), \frac{d \vec{x} (t)}{d t})}{ϵ} d t = 0

\int_{t_{a}}^{t_{b}} \lim_{ϵ \to 0} \frac{ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t} + ϵ \frac{d δ \vec{x} (t)}{d t}) - ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t})}{ϵ} + \frac{ℒ (\vec{x} (t) + ϵ δ \vec{x} (t), \frac{d \vec{x} (t)}{d t}) - ℒ (\vec{x} (t), \frac{d \vec{x} (t)}{d t})}{ϵ} d t = 0

See these are just definitions for partial derivaitves (gradients).

\int_{t_{a}}^{t_{b}} \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \dot{\vec{x}}} \cdot \frac{d δ \vec{x} (t)}{d t} + \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \vec{x}} \cdot δ \vec{x} (t) d t = 0

Integrate by parts on left side.

\frac{\partial ℒ (\vec{x} (t_{b}), \frac{\partial \vec{x} (t_{b})}{d t})}{\partial \dot{\vec{x}}} \cdot δ \vec{x} (t_{b}) - \frac{\partial ℒ (\vec{x} (t_{a}), \frac{\partial \vec{x} (t_{a})}{d t})}{d \dot{\vec{x}}} \cdot δ \vec{x} (t_{a}) + \int_{t_{a}}^{t_{b}} - (\frac{d}{d t} \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \dot{\vec{x}}}) \cdot δ \vec{x} (t) + \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \vec{x}} \cdot δ \vec{x} (t) d t = 0

Remember $δ \vec{x} (t_{a}), δ \vec{x} (t_{b}) = 0$ .

\int_{t_{a}}^{t_{b}} (\frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \vec{x}} - (\frac{d}{d t} \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{d \dot{\vec{x}}})) \cdot δ \vec{x} (t) d t = 0

Since this must be true for arbitrary $δ \vec{x}$

\frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{\partial \vec{x}} - (\frac{d}{d t} \frac{\partial ℒ (\vec{x} (t), \frac{\partial \vec{x} (t)}{d t})}{d \dot{\vec{x}}}) = \vec{0}

which is the Euler Lagrange equation. Next we define a Hamiltonian $ℋ (\vec{x}, \vec{q}) = \dot{\vec{x}} \cdot \vec{q} - ℒ (\vec{x}, \dot{\vec{x}})$ where $\vec{q} = \frac{\partial ℒ (\vec{x}, \dot{\vec{x}})}{\partial \dot{\vec{x}}}$ , usually interpreted as momentum.

\frac{\partial ℋ (\vec{x}, \vec{q})}{\partial \vec{x}} = - \frac{ℒ (\vec{x}, \dot{\vec{x}})}{\vec{x}} = - \frac{d}{d t} \frac{\partial ℒ (\vec{x}, \dot{\vec{x}})}{\dot{\vec{x}}} = - \dot{\vec{q}}

So we get Hamiltons equations.

\dot{\vec{x}} = \frac{\partial ℋ (\vec{x}, \vec{q})}{\partial \vec{q}}

\dot{\vec{q}} = - \frac{\partial ℋ (\vec{x}, \vec{q})}{\partial \vec{x}}

Phase Space Path Integral

We postulate that there exists an operator $\hat{H}$ that

⟨ x_{2} | e^{- Δ t \frac{i}{ℏ} \hat{H}} | x_{1} ⟩ \propto \int \frac{d p}{τ ℏ} e^{Δ t \frac{i}{ℏ} (p \frac{x_{2} - x_{1}}{Δ t} - H (\frac{x_{1} + x_{2}}{2}, p))} + O (Δ t^{2})

So that it becomes exact in the limit that $Δ t$ approaches zero. Start with a Taylor series expansion on both sides

⟨ x_{2} | (1 - Δ t \frac{i}{ℏ} \hat{H} + O (Δ t^{2})) | x_{1} ⟩ \propto \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} (1 - Δ t \frac{i}{ℏ} H (\frac{x_{1} + x_{2}}{2}, p) + O (Δ t^{2}))

And multiply out

δ (x_{2} - x_{1}) - ⟨ x_{2} | Δ t \frac{i}{ℏ} \hat{H} | x_{1} ⟩ + O (Δ t^{2}) \propto \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} (1 - Δ t \frac{i}{ℏ} H (\frac{x_{1} + x_{2}}{2}, p) + O (Δ t^{2}))

split the integral

δ (x_{2} - x_{1}) - ⟨ x_{2} | Δ t \frac{i}{ℏ} \hat{H} | x_{1} ⟩ + O (Δ t^{2}) \propto \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} - \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} Δ t \frac{i}{ℏ} H (\frac{x_{1} + x_{2}}{2}, p) + O (Δ t^{2})

And simplify

δ (x_{2} - x_{1}) - ⟨ x_{2} | Δ t \frac{i}{ℏ} \hat{H} | x_{1} ⟩ + O (Δ t^{2}) \propto δ (x_{2} - x_{1}) - \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} Δ t \frac{i}{ℏ} H (\frac{x_{1} + x_{2}}{2}, p) + O (Δ t^{2})

Then we can see that

- ⟨ x_{2} | Δ t \frac{i}{ℏ} \hat{H} | x_{1} ⟩ \propto - \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} Δ t \frac{i}{ℏ} H (\frac{x_{1} + x_{2}}{2}, p)

And canceling common terms

⟨ x_{2} | \hat{H} | x_{1} ⟩ \propto \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} H (\frac{x_{1} + x_{2}}{2}, p)

So in position basis the Hamiltonian operator is defined as

⟨ x_{2} | \hat{H} | x_{1} ⟩ \propto \int \frac{d p}{τ ℏ} e^{\frac{i}{ℏ} p (x_{2} - x_{1})} H (\frac{x_{1} + x_{2}}{2}, p)

Defining Lattice

In $D$ dimensional spacetime we define each point of a $N \times \dots \times N$ lattice to be indexed by elements of the set

\vec{n} \in [N]^{D}

where $[N]$ is the set of natural numbers smaller than $N$

[N] = {x \in ℕ | x < N}

Create a field configuration assigning a real number to each point on the lattice (although this can be trivialy generalized to spinors, vectors, tensors etc). It can be represented by a $N^{D}$ dimensional vector space over the real numbers

([N]^{D} \to ℝ) ≅ ℝ^{(N^{D})}

A field configuration gives a real number for each point in the lattice

ϕ \in ℝ^{(N^{D})}

And the field configuration can be indexed by a point on the lattice

ϕ_{\vec{n}} \in ℝ

Defining Path Integral

So given a classical action defined by it's Lagrangian density

S [ϕ] = \int_{ℝ^{D}} d \vec{x} ℒ (ϕ (\vec{x}), \partial_{0} ϕ (\vec{x}), \partial_{1} ϕ (\vec{x}), \dots, \partial_{D - 1} ϕ (\vec{x}))

We can write it like

S [ϕ] = \int_{ℝ^{D}} d \vec{x} ℒ (ϕ (\vec{x}), \nabla ϕ (\vec{x}))

Where $(\nabla ϕ (\vec{x}))_{k} = \partial_{k} ϕ (\vec{x})$ .

To discreetize it to a lattice we define $\forall (k \in [D]) . L_{k} \in ℝ$ to be the length of the lattice in each direction. Make a discrete version of the action. Defined a discreete derivative where the spacing of the lattice points in each direction are $Δ k = \frac{L_{k}}{N}$ and a basis vector $𝟏_{k}$ in each direction so

(\nabla ϕ_{\vec{n}})_{k} = \frac{ϕ_{\vec{n} + 𝟏_{k}} - ϕ_{\vec{n}}}{Δ k}

Then

S^{'} [ϕ] = \sum_{\vec{n} \in [N]^{D}} (\prod_{k \in [D]} Δ k) ℒ (\frac{ϕ_{\vec{n}} + ϕ_{\vec{n + 𝟏_{t}}}}{2}, \nabla ϕ_{\vec{n}})

When the indices go off the edge of the lattice it just loops around, called periodic boundary conditions. Here $t$ is a chosen direction (usually time) The path integral is now defined as below. (You can also take the limit as the $L$ 's go to infinity so the lattice covers the whole universe).

\int 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} : = \lim_{N \to \infty} \int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \exp (\frac{i}{ℏ} S^{'} [ϕ])

Transfer matrix

So the path integral is defined as the limit of

\int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \exp (\frac{i}{ℏ} \sum_{\vec{n} \in [N]^{D}} (\prod_{k \in [D]} Δ k) ℒ (\frac{ϕ_{\vec{n}} + ϕ_{\vec{n + 𝟏_{t}}}}{2}, \nabla ϕ_{\vec{n}}))

Split the "time" direction off from the others, so $\vec{n} = ⟨ t, \vec{w} ⟩$ .

\int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \exp (\frac{i}{ℏ} \sum_{t \in [N]} \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D]} Δ k) ℒ (\frac{ϕ_{\vec{n}} + ϕ_{\vec{n + 𝟏_{t}}}}{2}, \nabla ϕ_{\vec{n}}))

An exponent of sums becomes a product of exponent

\int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \prod_{t \in [N]} \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D]} Δ k) ℒ (\frac{ϕ_{\vec{n}} + ϕ_{\vec{n + 𝟏_{t}}}}{2}, \nabla ϕ_{\vec{n}}))

Define a transfer matrix $𝐓 \in (ℝ^{(N^{D - 1})} \to ℝ) \times (ℝ^{(N^{D - 1})} \to ℝ) \to ℂ$ . Define each element of the matrix to be

𝐓 (ω_{a}, ω_{b}) = \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D]} Δ k) ℒ (\frac{ω_{a, \vec{w}} + ω_{b, \vec{w}}}{2}, \frac{ω_{b, \vec{w}} - ω_{a, \vec{w}}}{Δ t}, \frac{ω_{a, \vec{w} + 𝟏_{k}} - ω_{a, \vec{w}}}{Δ k} \forall (k \in [D] / t)))

for $ω \in ℝ^{(N^{D - 1})}$ (intuitively 3d time slices of 4d spacetime). So the whole matrix is defined by one-hot vectors (more generally any orthonormal basis). So the time derivative is now the difference between the two, while the spacial derivatives are the same. Back to the integral its now just equal to

\int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \prod_{t \in [N]} 𝐓 (ϕ_{t}, ϕ_{t + 1})

The elements of the matrix can also be extracted with one-hot vectors where $| ω ⟩ \in ℝ^{(N^{D - 1})} \to ℝ$ is a one-hot vector with a single 1 at $ω$

𝐓 (ω_{a}, ω_{b}) = ⟨ ω_{b} | 𝐓 | ω_{a} ⟩

so the path integral can now be written as

\int (\prod_{\vec{n} \in [N]^{D}} d ϕ_{\vec{n}}) \prod_{t \in [N]} ⟨ ϕ_{t + 1} | 𝐓 | ϕ_{t} ⟩

Move the integrals around

= \int (\prod_{\vec{w}} d ϕ_{N - 1, \vec{w}}) (\prod_{\vec{w}} d ϕ_{0, \vec{w}}) ⟨ ϕ_{N - 1, \vec{w}} | 𝐓 \int (\prod_{\vec{w}} d ϕ_{N - 2, \vec{w}}) | ϕ_{N - 2, \vec{w}} ⟩ ⟨ ϕ_{N - 2, \vec{w}} | 𝐓 \int (\prod_{\vec{w}} d ϕ_{N - 3, \vec{w}}) | ϕ_{N - 3, \vec{w}} ⟩ \dots ⟨ ϕ_{2, \vec{w}} | 𝐓 \int (\prod_{\vec{w}} d ϕ_{1, \vec{w}}) | ϕ_{1, \vec{w}} ⟩ ⟨ ϕ_{1, \vec{w}} | 𝐓 | ϕ_{0, \vec{w}} ⟩

Because it forms an orthonormal basis so all the middle terms disappear.

\int (\prod_{\vec{w}} d ϕ_{t, \vec{w}}) | ϕ_{t, \vec{w}} ⟩ ⟨ ϕ_{t, \vec{w}} | = I

the path integral simplifies to

\int 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} = \lim_{N \to \infty} \int (\prod_{\vec{w}} d ϕ_{N, \vec{w}}) (\prod_{\vec{w}} d ϕ_{0, \vec{w}}) ⟨ ϕ_{N, \vec{w}} | 𝐓^{N} | ϕ_{0, \vec{w}} ⟩

But remember we assumed that it was periodic as if it was in a repeating $L_{0} \times \dots \times L_{N} - 1$ box so $ϕ_{0, \vec{w}} = ϕ_{N - 1, \vec{w}}$ and the integral is actually just over one of them.

= \lim_{N \to \infty} \int (\prod_{\vec{w}} d ϕ_{0, \vec{w}}) ⟨ ϕ_{0, \vec{w}} | 𝐓^{N} | ϕ_{0, \vec{w}} ⟩ = Tr (𝐓^{N})

Instead we can fix the endpoints in time to some field configuration. Here $L_{t} = t_{1} - t_{0}$ and the time direction would be non periodic.

\int_{ϕ_{t_{0}}}^{ϕ_{t_{1}}} 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} = \lim_{N \to \infty} ⟨ ϕ_{t_{1}, {\vec{n}}_{w}} | 𝐓^{N} | ϕ_{t_{0}, {\vec{n}}_{w}} ⟩

Or more concicely below.

\int_{ω_{0}}^{ω_{1}} 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} = \lim_{τ \to \infty} \lim_{N \to \infty} ⟨ ω_{1} | 𝐓^{N} | ω_{0} ⟩

Most of quantum field theory can be derived from this.

Legendre transform

Momentum Eigenstates

We now define another orthogonal basis. Using the previous orthonormal basis of elements like $| ω ⟩$ . Each element of the new basis $| π ⟩$ is defined

⟨ ω | π ⟩ = \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{\vec{w}})

And you can prove they're orthogonal by

⟨ π_{a} | π_{b} ⟩ = \int 𝒟 ω ⟨ π_{a} | ω ⟩ ⟨ ω | π_{b} ⟩

= \int 𝒟 ω \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{a, \vec{w}}) \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{b, \vec{w}})

= \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} (π_{b, \vec{w}} - π_{a, \vec{w}}))

If we expand the $𝒟 ω$ notation

= \int (\prod_{\vec{w} \in [N]^{D - 1}} d ω_{\vec{w}}) \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} (π_{b, \vec{w}} - π_{a, \vec{w}}))

We see that these are really just a bunch of (unnormalized) Dirac delta functions.

= (\frac{τ ℏ}{\prod_{k \in [D] / t} δ_{k}})^{(N^{D - 1})} \prod_{\vec{w} \in [N]^{D - 1}} δ (π_{b, \vec{w}} - π_{a, \vec{w}})

Because they're unnormalized we define the measure $𝒟 π$ to be

𝒟 π = \prod_{\vec{w} \in [N]^{D - 1}} (d π_{\vec{w}} \frac{\prod_{k \in [D] / t} Δ_{k}}{τ ℏ})

So that

\int 𝒟 π | π ⟩ ⟨ π | = I

We will mess around with this

⟨ ω_{b} | T | ω_{a} ⟩

Insert the momentum basis identity

= \int 𝒟 π ⟨ ω_{b} | π ⟩ ⟨ π | T | ω_{a} ⟩

And from construction of the momentum basis

= \int 𝒟 π \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) ⟨ π | T | ω_{a} ⟩

and then insert another identity, using position basis this time

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) ⟨ π | ω ⟩ ⟨ ω | T | ω_{a} ⟩

And again from the construction of the momentum basis, and the exponent is negatived since its conjugated.

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{\vec{w}}) ⟨ ω | T | ω_{a} ⟩

And then from the definition of the transfer matrix

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{\vec{w}}) \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D]} Δ_{k}) ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots))

And

\prod_{k \in [D]} Δ_{k} = Δ t \prod_{k \in [D] / t} Δ_{k}

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{\vec{w}} π_{\vec{w}}) \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots))

Rewrite as $ω_{\vec{w}} = ω_{a, \vec{w}} + (ω_{\vec{w}} - ω_{a, \vec{w}})$

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} (ω_{a, \vec{w}} + (ω_{\vec{w}} - ω_{a, \vec{w}})) π_{\vec{w}}) \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots))

Then just move it over

= \int 𝒟 π \int 𝒟 ω \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{a, \vec{w}} π_{\vec{w}}) \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots) - \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}} π_{\vec{w}}))

Move the terms that don't depend in front

= \int 𝒟 π \exp (\frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{b, \vec{w}} π_{\vec{w}}) \exp (- \frac{i}{ℏ} (\prod_{k \in [D] / t} Δ_{k}) \sum_{\vec{w} \in [N]^{D - 1}} ω_{a, \vec{w}} π_{\vec{w}}) \int 𝒟 ω \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots) - \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}} π_{\vec{w}}))

So now we have

= \int 𝒟 π ⟨ ω_{b} | π ⟩ ⟨ π | ω_{a} ⟩ \int 𝒟 ω \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots) - \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}} π_{\vec{w}}))

(TODO make this more rigorous) In the limit that $N \to \infty$ so that $Δ t \to 0$ the integral on the right will converge to just the stationary points where the derivative with respect to the components of $ω$ are zero. If the derivative isn't zero then the "swirling" of the exponent gets infinitely fast and cancells out. So the derivative of the exponent is

\frac{\partial}{\partial ω_{\vec{w}}} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}}, \dots) - \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ_{t}} π_{\vec{w}})

= \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (\frac{\partial ℒ}{\partial ω_{\vec{w}}} \frac{1}{2} + \frac{\partial ℒ}{\partial {\dot{ω}}_{\vec{w}}} \frac{1}{Δ t} - \frac{1}{Δ_{t}} π_{\vec{w}})

= \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D] / t} Δ_{k}) (Δ t \frac{\partial ℒ}{\partial ω_{\vec{w}}} \frac{1}{2} + \frac{\partial ℒ}{\partial {\dot{ω}}_{\vec{w}}} - π_{\vec{w}})

And in the limit the term goes to zero

= \sum_{\vec{w} \in [N]^{D - 1}} (\prod_{k \in [D] / t} Δ_{k}) (\frac{δ ℒ}{δ {\dot{ω}}_{\vec{w}}} - π_{\vec{w}})

So for the derivative to be zero, this restricts $ω$ to the field configurations where

\frac{δ ℒ (\frac{ω_{\vec{w}} + ω_{a, \vec{w}}}{2}, \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ t}, \dots)}{δ {\dot{ω}}_{\vec{w}}} - π_{\vec{w}} = 0

Let $ω_{\vec{w}}^{'} = \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{2}$ and ${\dot{ω}}_{\vec{w}}^{'} = \frac{ω_{\vec{w}} - ω_{a, \vec{w}}}{Δ t}$ in the limit as delta t is small. So then

π_{\vec{w}} = \frac{δ ℒ (ω_{\vec{w}}^{'}, {\dot{ω}}_{\vec{w}}^{'}, \dots)}{δ {\dot{ω}}_{\vec{w}}}

If an inverse function exists (this assumption is adding an extra requirement to the Lagrangian) so that

{\dot{ω}}_{\vec{w}}^{'} = f (ω_{\vec{w}}^{'}, π_{\vec{w}})

And now the "velocity" can be written as a function of the "position" and "momentum". The whole equality now simplifies to

⟨ ω_{b} | T | ω_{a} ⟩ = 𝒩 \int 𝒟 π ⟨ ω_{b} | π ⟩ ⟨ π | ω_{a} ⟩ \int 𝒟 ω \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℒ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, f (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, π_{\vec{w}}), \dots) - f (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, π_{\vec{w}}) π_{\vec{w}}))

(TODO get an expression for it) Here $𝒩$ is some constant depending on how flat the stationary point is.

Now we define a Hamiltonian density defined by the Legedre transform of the Lagrangian density

ℋ (ω_{\vec{w}}, π_{\vec{w}}) = ℒ (ω_{\vec{w}}, f (ω_{\vec{w}}, π_{\vec{w}}), \dots) - f (ω_{\vec{w}}, π_{\vec{w}}) π_{\vec{w}}

So now its

𝒩 \int 𝒟 π ⟨ ω_{b} | π ⟩ ⟨ π | ω_{a} ⟩ \int 𝒟 ω \exp (\frac{i}{ℏ} \sum_{\vec{w} \in [N]^{D - 1}} (Δ_{t} \prod_{k \in [D] / t} Δ_{k}) (ℋ (\frac{ω_{a, \vec{w}} + ω_{\vec{w}}}{2}, π_{\vec{w}})))