Quantum Field Theory Path Integral from Lattice Limit

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 (ψ (\vec{x}), \partial_{a} ψ (\vec{x}) \forall (a \in [D]))

In the definition of the path integral it will be discreetized to

\int 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} : = \lim_{Δ x \to 0} \lim_{N \to \infty} \int (\prod_{\vec{n}} d ϕ_{\vec{n}}) e^{\frac{i}{ℏ} \sum_{\vec{n}} Δ x ℒ (ϕ_{\vec{n}}, \frac{ϕ_{\vec{n} + 𝟏_{k}} - ϕ_{\vec{n}}}{Δ x} \forall (k \in [D]))}

where $𝟏_{k}$ is a basis the vector at index $k$

Transfer matrix

Choose some arbitrary direction and split it off from the others, so $\vec{n} = ⟨ n_{t}, {\vec{n}}_{w} ⟩$ (usually the "time" direction so the other dimensions represent the spacial dimension)

= \lim_{Δ x \to 0} \lim_{N \to \infty} \int (\prod_{\vec{n}} d ϕ_{\vec{n}}) e^{\frac{i}{ℏ} \sum_{n_{t}} \sum_{{\vec{n}}_{w}} Δ x ℒ (ϕ_{\vec{n}}, \frac{ϕ_{\vec{n} + 𝟏_{k}} - ϕ_{\vec{n}}}{Δ x} \forall (k \in [D]))}

An exponent of sums becomes a produc of exponent

= \lim_{Δ x \to 0} \lim_{N \to \infty} \int (\prod_{\vec{n}} d ϕ_{\vec{n}}) \prod_{n_{t}} e^{\frac{i}{ℏ} \sum_{{\vec{n}}_{w}} Δ x ℒ (ϕ_{\vec{n}}, \frac{ϕ_{\vec{n} + 𝟏_{k}} - ϕ_{\vec{n}}}{Δ x} \forall (k \in [D]))}

= \lim_{Δ x \to 0} \lim_{N \to \infty} \int \prod_{n_{t}} (\prod_{{\vec{n}}_{w}} d ϕ_{\vec{n}}) e^{\frac{i}{ℏ} \sum_{{\vec{n}}_{w}} Δ x ℒ (ϕ_{\vec{n}}, \frac{ϕ_{\vec{n} + 𝟏_{k}} - ϕ_{\vec{n}}}{Δ x} \forall (k \in [D]))}

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

𝐓 (ϕ_{w, a}, ϕ_{w, b}) = e^{\frac{i}{ℏ} \sum_{{\vec{n}}_{w}} Δ x ℒ (ϕ_{w, a, {\vec{n}}_{w}}, \frac{ϕ_{w, b, {\vec{n}}_{w}} - ϕ_{w, a, {\vec{n}}_{w}}}{Δ x}, \frac{ϕ_{w, a, {\vec{n}}_{w} + 𝟏_{k}} - ϕ_{w, a, {\vec{n}}_{w}}}{Δ x} \forall (k \in w))}

for $ϕ_{w} \in ℝ^{(N^{D - 1})}$ . 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 just now equal to

\lim_{Δ x \to 0} \lim_{N \to \infty} \int \prod_{n_{t}} (\prod_{{\vec{n}}_{w}} d ϕ_{\vec{n}}) 𝐓 (ϕ_{n_{t}, {\vec{n}}_{w}}, ϕ_{n_{t} + 1, {\vec{n}}_{w}})

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

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

so the path integral can now be written as

\lim_{Δ x \to 0} \lim_{N \to \infty} \int \prod_{n_{t}} (\prod_{{\vec{n}}_{w}} d ϕ_{\vec{n}}) ⟨ ϕ_{n_{t} + 1, {\vec{n}}_{w}} | 𝐓 | ϕ_{n_{t}, {\vec{n}}_{w}} ⟩

Move the integrals around

= \lim_{Δ x \to 0} \lim_{N \to \infty} \int (\prod_{{\vec{n}}_{w}} d ϕ_{N - 1, {\vec{n}}_{w}}) (\prod_{{\vec{n}}_{w}} d ϕ_{0, {\vec{n}}_{w}}) ⟨ ϕ_{N - 1, {\vec{n}}_{w}} | 𝐓 \int (\prod_{{\vec{n}}_{w}} d ϕ_{N - 2, {\vec{n}}_{w}}) | ϕ_{N - 2, {\vec{n}}_{w}} ⟩ ⟨ ϕ_{N - 2, {\vec{n}}_{w}} | 𝐓 \int (\prod_{{\vec{n}}_{w}} d ϕ_{N - 3, {\vec{n}}_{w}}) | ϕ_{N - 3, {\vec{n}}_{w}} ⟩ \dots ⟨ ϕ_{2, {\vec{n}}_{w}} | 𝐓 \int (\prod_{{\vec{n}}_{w}} d ϕ_{N - 3, {\vec{n}}_{w}}) | ϕ_{1, {\vec{n}}_{w}} ⟩ ⟨ ϕ_{1, {\vec{n}}_{w}} | 𝐓 | ϕ_{0, {\vec{n}}_{w}} ⟩

Because it forms an orthonormal basis

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

the path integral simplifies to

\int 𝒟 ϕ e^{\frac{i}{ℏ} S [ϕ]} = \lim_{Δ x \to 0} \lim_{N \to \infty} \int (\prod_{{\vec{n}}_{w}} d ϕ_{N - 1, {\vec{n}}_{w}}) (\prod_{{\vec{n}}_{w}} d ϕ_{0, {\vec{n}}_{w}}) ⟨ ϕ_{N - 1, {\vec{n}}_{w}} | 𝐓^{N} | ϕ_{0, {\vec{n}}_{w}} ⟩

If we fix the endpoints

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

Or more conciely

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