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

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