f t i dt 1 2 m y 2 Going back to the path integral we now have x f t f x i t i

# F t i dt 1 2 m y 2 going back to the path integral we

• Notes
• 72

This preview shows page 47 - 50 out of 72 pages.

f t i dt 1 2 m ˙ y 2 Going back to the path integral, we now have: x f , t f | x i , t i ⟩ = C exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! Z N - 1 Y k = 1 d 3 y k exp ˆ i m 2 N - 1 X k = 1 ˙ y 2 k ε ! = C exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! Z N - 1 Y k = 1 d 3 y k exp ˆ i m 2 N - 1 X k = 1 ( y k + 1 - y k ) 2 ε ! where we have defined the discretised version of ˙ y : ˙ y k = y k + 1 - y k ε How do we do this integral? • Feynman and Hibbs: guess the answer • Weinberg: do each Gaussian integral one at a time until you can find a recursion formula • We’re going to use matrices x f , t f | x i , t i ⟩ = C exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! Z N - 1 Y k = 1 d 3 y k exp - i m 2 ε K i j y i y j | {z } = y T K y where we have defined the ( N - 1) × ( N - 1) matrix K i j = 2 - 1 0 0 - 1 2 - 1 0 0 - 1 2 . . . . . . - 1 0 0 - 1 2 47
which is tridiagonal, non-degenerate and symmetric so can therefore be diagonalised by an orthogonal matrix O to give D = O T KO . The entries of D will be the eigenvalues of K . Chang- ing variables to z = O T y , such that y T K y = z T O T KOz : x f , t f | x i , t i ⟩ = C exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! Z N - 1 Y l = 1 d 3 z l (Jacobian)exp - im 2 ε z T Dz where the Jacobian is | det O | = 1. Doing the integral and using z T Dz = z 2 k λ k : Z N - 1 Y l = 1 d 3 z l exp - im 2 ε z T Dz = exp - 3 i π 4 ( N - 1) N - 1 Y k = 1 πε m λ k 3 2 = exp - 3 i π 4 ( N - 1) πε m · 3( N - 1) 2 ˆ N - 1 Y k = 1 λ k ! - 3 2 = exp - 3 i π 4 ( N - 1) πε m · 3( N - 1) 2 (det D ) - 3 2 = exp - 3 i π 4 ( N - 1) πε m · 3( N - 1) 2 (det K ) - 3 2 Eigenvalues of K are λ k = 2 1 - cos k π N · for k = 1,..., N 1 . A standard result is: N - 1 Y k = 1 x 2 - 2 x cos k π N + 1 = 1 + x 2 + x 4 + ... + x 2( N - 1) = 1 - x 2 N 1 - x 2 Expanding about x for small δ : 1 - ( x + δ ) 2 N 1 - ( x + δ ) 2 = 1 - x 2 N - 2 Nx δ 1 - x 2 - 2 x δ so, for x = 1 we find that Q N - 1 k = 1 λ k = N . Putting this into the path integral: x f , t f | x i , t i ⟩ = C exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! exp - 3 i π 4 ( N - 1) πε m · 3( N - 1) 2 N - 3 2 Substituting back in for C , we find some of the terms cancel nicely to give x f , t f | x i , t i ⟩ = exp ˆ i m 2 ( x f - x i ) 2 t f - t i ! m 2 π ( t f - t i ) 3 2 which is the same solution found by solving the Schrödinger equation. If V ( x ) 6= 0, then the path integral takes a similar form, e.g. if V ( x ) = 1 2 m ω 2 x 2 , then this changes the eigenvalues of the K matrix. 48
6.2 Path integrals in field theory In field theory, there are an infinite number of generalised coordinates, φ ( x ), and the Hamil- tonian is of the form H R d 3 x ( π 2 + V ( φ ) ) . The action now becomes I [ φ ] = Z d 4 x - 1 2 a φ∂ a φ - V ( φ ) where V ( φ ) = 1 2 m 2 φ 2 + .... In 4d, the only possible interaction is 1 4! λφ 4 . Introducing a source term into the action, this gives: I [ φ ] = Z d 4 x - 1 2 a φ∂ a φ - V ( φ ) + J ( x ) φ Classically, - φ - V 0 ( φ ) = - J ( x ). Let’s look at the case where V ( φ ) = 1 2 m 2 φ 2 : Z = Z D φ exp i Z d 4 x - 1 2 ( ∂φ ) 2 - 1 2 m 2 φ 2 + J φ = Z Y x d φ ( x ( t )) exp i Z d 4 x - 1 2 ( ∂φ ) 2 - 1 2 m 2 φ 2 + J φ = Z Y x , t d φ exp i Z d 4 x - 1 2 ( ∂φ ) 2 - 1 2 m 2 φ 2 + J φ = Z Y x , t d φ exp i Z d 4 x - 1 2 φ A φ + J φ where A = - + m 2 = Z Y x , t d φ exp i Z d 4 x - 1 2 ( φ - J A - 1 ) A

#### You've reached the end of your free preview.

Want to read all 72 pages?