Functional Method
We solve Problem 9.2 in Peskin–Schroeder.
(a)
We would like to evaluate the partition function
Z = tr[e−βH ]
(1)
using the path integral. First of all, the trace of an arbitrary operator O over
the Hilbert space can be taken in any basis, e.g., energy eigenstates |ni or
position eigenstates |xi,
X
hn|O|ni
trO =
=
Zn Z X
hn|xidxhx|O|yidyhy|ni
n
Z Z
=
dxdyhx|O|yi
X
hy|nihn|xi
n
Z Z
=
dxdyhx|O|yihy|xi
Z Z
dxdyhx|O|yiδ(x − y)
=
Z
=
dxhx|O|xi.
(2)
The next step is to divide up the Boltzmann factor e−βH into many many
small pieces,
Z
−βH
tr[e
] = dxhx|e−βH |xi
Z
=
dx
N
−1
Y
hx|e−βH/N |xN −1 idxN −1 hxN −1 |e−βH/N |xN −2 idxN −2 hxN −2 | · · ·
i=1
· · · |x1 idx1 hx1 |e−βH/N |xi
(3)
1
2
p
For the Hamiltonian H = 2m
+ V (x), we compute the matrix element
−H
hx|e
|yi for small = β/N . We find
2 /2m
hx|e−H |yi = hx|e−p
2
e−V (x) e−O( ) |yi
(4)
because of the Baker–Campbell–Hausdorff formula. We ignore O(2 ) piece
in the exponent. Then we insert the complete set of momentum eigenstates,
Z
2
2
−H
hx|e
|yi =
hx|e−p /2m |pidphp|e−V (x) e−O( ) |yi
Z
1
1
2
e−p /2m eipx/~ √
e−V (y) e−ipy/~
=
dp √
2π~
2π~
r
1
2πm −m(x−y)2 /2~2 −V (y)
=
e
e
(5)
2π~
Therefore, using the notation x0 = xN = x,
Z Y
N
dx
−βH
√ i e−SE /~ ,
tr[e
]=
(6)
2π~2 i=1
where
SE
= ~
N X
m(xi − xi−1 )2
i=1
N X
2~2
+ V (xi )
~β
m(xi − xi−1 )2
=
+ V (xi )
N i=1
2(~β/N )2
#
I ~β " 2
m dx
+ V (x) dτ.
→
2 dτ
0
(7)
At the last step, the limit N → ∞ was taken. This is nothing but the
Euclidean action, namely the action after the Wick rotation t = −iτ . The
path x(τ ) satisfies the periodic boundary condition x(~β) = x(0).
(b)
The specified expansion (~ = 1 below),1
1 X
x= √
xn e2πinτ /β
β n
1
I use τ instead of t to distinguish real and imaginary times.
2
(8)
implies that x−n = x∗n . In particular, x0 is real. The Euclidean action is
I β X 1 2πin 2πim
2 1
SE =
dτ
+ω
xn e2πinτ /β xm e2πimτ /β
2
β
β
β
0
n,m
∞ 1 2 2 X (2πn)2
2
=
ω x0 +
(9)
+ ω |xn |2 .
2
2
β
n=1
In the second line, we used m = −n from the τ integral. Therefore, the
path integral is Gaussian and is given by the product of eigenvalues up to an
overall β-dependent (but ω-independent) factor,
" ∞ #−1
Y (2πn)2
+ ω2
Z ∝ ω
2
β
n=1
"
#−1 −1
∞ βω Y
β 2ω2
βω
1+
.
(10)
∝
= sinh
2 n=1
(2πn)2
2
Here, we used the infinite product representation of sinh z. Up to a βdependent (but ω-independent) factor, it is2
Z ∝
βω/2
−1
e−βω/2
e
− e−βω/2
.
=
1 − e−βω
(11)
This precisely matches the known expression of the partition function for a
harmonic oscillator,
Z=
∞
X
e−βω(n+1/2) =
n=0
e−βω/2
.
1 − e−βω
(12)
As a prepration to the next problem, it is instructive to rewrite the Euclidean action as
I ~β
1
SE =
dτ x(τ ) −∂τ2 + ω 2 x(τ ),
(13)
2
0
and hence
Z = [det(−∂τ2 + ω 2 )]−1/2 =
2
e−βω/2
.
1 − e−βω
(14)
One can of course work out the overall factor carefully with a lot more work. See
http://hitoshi.berkeley.edu/221A/pathintegral.pdf
3
(c)
The partition function is given in terms of the path integral
Z
Z = Dφ(~x, τ )e−SE ,
(15)
where the Euclidean action is (~ = c = 1)
Z
I β
I β
i Z
1h 2
1
2
2 2
~
dτ
SE = d~x
dτ φ(−∂ 2 + m2 )φ.
φ̇ + (∇φ) + m φ = d~x
2
2
0
0
(16)
Here, φ̇ = dφ/dτ . The path integral therefore yields
~ 2 + m2 )]−1/2 .
Z = [det(−∂ 2 + m2 )]−1/2 = [det(−∂τ2 − ∇
~ = i~p,
This determinant is a product of many momentum modes ∇
Y
Z=
[det(−∂τ2 + p~2 + m2 )]−1/2 .
(17)
(18)
p
~
Using the result from the part (b), the determinant is
Z=
Y e−βω(~p)/2
,
1 − e−βω(~p)
(19)
p
~
p
where ω(~p) = p~2 + m2 . This expression is indeed the partition function for
a relativistic boson of mass m.
(d)
We are given the Euclidean action
I β h
i
SE =
dτ ψ̄ ψ̇ + ω ψ̄ψ ,
(20)
0
Because of the anti-periodic boundary condition ψ(β) = −ψ(0), we expand
it in Fourier series
1
1
1 X
1 X
ψ(τ ) = √
ψn e2πi(n+ 2 )τ /β ,
ψ̄(τ ) = √
ψ̄n e−2πi(n+ 2 )τ /β . (21)
β n
β n
4
Then the Euclidean action is
I β
1
1X
2πi
1
−2πi(n+ 12 )τ /β
dτ
SE =
ψ̄n e
m+
+ ω ψm e2πi(m+ 2 )τ /β
β n,m
β
2
0
X
1
2πi
n+
+ ω ψn .
(22)
=
ψ̄n
β
2
n
Therefore the path integral over Grasmannian variables gives
Z Y
Y 2πi 1
−SE
dψn dψ̄n e
=
Z=
n+
+ω
β
2
n
n
This infinite product can be rewritten as
∞ Y
1
2πi
1
2πi
n+
+ω −
n+
+ω
Z =
β
2
β
2
n=0
" #
2
∞
2
Y
2π
1
=
+ ω2 .
n+
β
2
n=0
It is useful to recall the infinite product representation of cosh,
∞ Y
x2
cosh x =
1+ 2
.
π (n + 21 )2
n=0
(23)
(24)
(25)
We find
Z =
2 ∞ Y
2π
n=0
β
1
n+
2
2 "
1+
βω
2π(n + 21 )
2 #
βω
∝ eβω/2 + e−βω/2 .
(26)
2
On the other hand, the canonical quantization of the same Lagrangian
in the real time is based on the anti-commutation relations {ψ, ψ̄} = 1 and
{ψ, ψ} = {ψ̄, ψ̄} = 0 with the Hamiltonian H = ω 21 (ψ̄ψ −ψ ψ̄) = ω ψ̄ψ − 21 .
The ground state is annihilated by the annihilation operator ψ|0i = 0 while
the excited state is |1i = ψ̄|0i. Because of the anticommutation {ψ̄, ψ̄} =
2ψ̄ 2 = 0, one cannot occupy the sate with more than once, ψ̄|1i = ψ̄ 2 |0i = 0.
The energy eigenvalues are
1
ω
H = ω ψ̄ψ −
=± .
(27)
2
2
∝ cosh
5
Therefore, the partition function is
Z = eβω/2 + e−βω/2 = 2 cosh
βω
,
2
(28)
which agrees with the path integral up to an overall factor.
(e)
The Euclidean path integral including the gauge fixing as in Eq. (9.56)
(Peskin–Schroeder) is
Z
R 4 1 2
1
2
Z = DAµ e− d x( 4 Fµν − 2ξ (∂µ Aµ ) ) (det ∂ 2 )
(29)
up to an overall constant. In Feynman gauge (ξ = 1), the calculation is
extremely simple.
The Euclidean action can be rewritten as
Z
1
1 2
2
4
F − (∂µ Aµ )
SE =
dx
4 µν 2
Z
1
=
d4 x [∂µ Aν (∂µ Aν − ∂ν Aµ ) + (∂ν Aν )(∂µ Aµ )]
2
Z
1
=
d4 x −Aν (∂ 2 Aν − ∂ν (∂µ Aµ )) − Aν ∂ν (∂µ Aµ )
2
Z
1
= −
d4 xAν ∂ 2 Aν .
(30)
2
Therefore, it is nothing but a collection of four scalar fields (µ = 0, 1, 2, 3),
and its integral over Aµ yields [(det ∂ 2 )−1/2 ]4 = (det ∂ 2 )−2 . On the other
hand, the Faddeev–Popov determinant (det ∂ 2 ) cancels one of the powers
and hence
Z = (det ∂ 2 )−1 .
(31)
This is square of the determinant of a massless boson, correctly account for
two polarization states of the photon for the partition function.
6