2QFT in one dimension (i.e. QM)

III Advanced Quantum Field Theory



2.1 Quantum mechanics
In
d
= 1, there are two possible connected, compact manifolds
M
we can
have M = S
1
or M = I = [0, 1].
M = I
φ
We will mostly be considering the case where
M
=
I
. In this case, we need to
specify boundary conditions on the field in the path integral, and we will see that
this corresponds to providing start and end points to compute matrix elements.
We let
t
[0
,
1] be the worldline coordinate parametrizing the field, and we
write our field as
x : I N
for some Riemannian manifold (
N, g
), which we call the target manifold. If
U N
has coordinates
x
a
with
a
= 1
, ··· , n
=
dim
(
N
), then we usually write
x
a
(t) for the coordinates of x(t). The standard choice of action is
S[x] =
Z
I
1
2
g( ˙x, ˙x) + V (x)
dt,
where ˙x is as usual the time derivative, and V (x) is some potential term.
We call this theory a non-linear
σ
-model. This is called
σ
-model because
when people first wrote this theory down, they used
σ
for the name of the field,
and the name stuck. This is non-linear because the term
g
(
˙x, ˙x
) can be very
complicated when the metric
g
if it is not flat. Note that +
V
(
x
) is the correct
sign for a Euclidean worldline.
Classically, we look for the extrema of
S
[
x
] (for fixed end points), and the
solutions are given by the solutions to
d
2
x
a
dt
2
+ Γ
a
bc
˙x
b
˙x
c
= g
ab
V
x
b
.
The left-hand side is just the familiar geodesic equation we know from general
relativity, and the right hand term corresponds to some non-gravitational force.
In the case of zero-dimensional quantum field theory, we just wrote down
the partition function
Z =
Z
e
S
,
and assumed it was an interesting thing to calculate. There wasn’t really a
better option because it is difficult to give a physical interpretation to a zero-
dimensional quantum field theory. In this case, we will see that path integrals
naturally arise when we try to do quantum mechanics.
To do quantum mechanics, we first pick a Hilbert space
H
. We usually take
it as
H = L
2
(N, dµ),
the space of square-integrable functions on N.
To describe dynamics, we pick a Hamiltonian operator
H
:
H H
, with the
usual choice being
H =
1
2
+ V,
where the Laplacian is given by
∆ =
1
g
a
(
gg
ab
b
).
As usually, the
g refers to the square root of the determinant of g.
We will work in the Heisenberg picture. Under this set up, the amplitude for
a particle to travel from x N to y N in time T is given by the heat kernel
K
T
(y, x) = hy|e
HT
|xi.
Note that we have
e
HT
instead of
e
iHT
because we are working in a Euclidean
world. Strictly speaking, this doesn’t really make sense, because
|xi
and
|yi
are not genuine elements of
H
=
L
2
(
N,
d
µ
). They are
δ
-functions, which aren’t
functions. There are some ways to fix this problem. One way is to see that the
above suggests K
T
satisfies
t
K
t
(y, x) + HK
t
(y, x) = 0,
where we view
x
as a fixed parameter, and
K
t
is a function of
y
, so that
H
can
act on K
t
. The boundary condition is
lim
t0
K
t
(y, x) = δ(y x),
and this uniquely specifies
K
t
. So we can define
K
t
to be the unique solution to
this problem.
We can reconnect this back to the usual quantum mechanics by replacing
the Euclidean time
T
with
it
, and the above equation gives us the Schr¨odinger
equation
i
K
t
t
(y, x) = HK
t
(y, x).
We first focus on the case where
V
= 0. In the case where (
N, g
) = (
R
n
, δ
), we
know from, say, IB Methods, that the solution is given by
K
t
(y, x) =
1
(2πt)
n/2
exp
|x y|
2
2t
.
For an arbitrary Riemannian manifold (
N, g
), it is in general very hard to write
down a closed-form expression for
K
t
. However, we can find the asymptotic form
lim
t0
K
t
(y, x)
a(x)
(2πt)
n/2
exp
d(y, x)
2
2t
,
where
d
(
x, y
) is the geodesic distance between
x
and
y
, and
a
(
x
) is some invariant
of our manifold built from (integrals of) polynomials of the Riemann curvature
that isn’t too important.
Here comes the magic. We notice that since
I =
Z
d
n
z |zihz|
is the identity operator, we can write
K
t
1
+t
2
(y, x) = hy|e
T H
|xi
=
Z
d
n
z hy|e
t
1
H
|zihz|e
t
2
H
|xi
=
Z
d
n
z K
t
2
(y, z)K
t
1
(z, x).
For flat space, this is just reduces to the formula for convolution of Gaussians.
This is the concatenation property of the heat kernel.
Using this many times, we can break up our previous heat kernel by setting
t =
T
N
for some large N N. Then we have
K
T
(y, x) =
Z
N1
Y
i=1
d
n
x
i
K
t
(x
i
, x
i1
),
where we conveniently set x
0
= x and x
N
= y.
The purpose of introducing these
t
is that we can now use the asymptotic
form of K
t
(y, t). We can now say
hy
1
|e
HT
|y
0
i
= lim
N→∞
1
2πt
nN/2
Z
N1
Y
i=1
d
n
x
i
a(x
i
) exp
t
2
d(x
i+1
, x
i
)
t
2
!
.
This looks more-or-less like a path integral! We now dubiously introduce the
path integral measure
Dx
?
lim
N→∞
1
2πt
nN/2
N1
Y
i=1
d
n
x
i
a(x
i
),
and also assume our map
x
(
t
) is at least once-continuously differentiable, so that
lim
N→∞
N1
Y
i=1
exp
t
2
d(x
i+1
, x
i
)
t
2
!
?
exp
1
2
Z
dt g( ˙x, ˙x)
= exp(S[x]).
Assuming these things actually make sense (we’ll later figure out they don’t),
we can write
hy
1
|e
HT
|y
0
i =
Z
C
T
[y
1
,y
0
]
Dx e
S[x]
,
where
C
T
[
y
1
, y
0
] is the space of “all” maps
I N
such that
x
(0) =
y
0
and
x(1) = y
1
.
Now consider an arbitrary V 6= 0. Then we can write
H = H
0
+ V (x),
where H
0
is the free Hamiltonian. Then we note that for small t, we have
e
Ht
= e
H
0
t
E
V (x)t
+ o(t).
Thus, for small t, we have
K
t
(y, x) = hy|e
Ht
|xi
hy|e
H
0
t
e
V (x)t
|xi
a(x)
(2πt)
n/2
exp
1
2
d(y, x)
t
2
+ V (x)
!
t
!
.
Then repeating the above derivations will again give us
hy
1
|e
HT
|y
0
i =
Z
C
T
[y
1
,y
0
]
Dx e
S[x]
.
Before we move on to express other things in terms of path integrals, and
then realize our assumptions are all wrong, we make a small comment on the
phenomena we see here.
Notice that the states
|y
0
i H
and
hy
1
| H
we used to evaluate our
propagator here arise as boundary conditions on the map
x
. This is a general
phenomenon. The co-dimension-1 subspaces (i.e. subspaces of
M
of one dimen-
sion lower than
M
) are associated to states in our Hilbert space
H
. Indeed, when
we did quantum field theory via canonical quantization, the states corresponded
to the state of the universe at some fixed time, which is a co-dimension 1 slice of
spacetime.
The partition function
We can naturally interpret a lot of the things we meet in quantum mechanics via
path integrals. In statistical physics, we often called the quantity
Tr
H
(
e
HT
)
the partition function. Here we can compute it as
Tr
H
(e
HT
) =
Z
d
n
y hy|e
HT
|yi.
Using our path integral formulation, we can write this as
Tr
H
(e
HT
) =
Z
d
n
y
Z
C
I
[y,y]
Dx e
S
=
Z
C
S
1
Dx e
S
,
where we integrate over all circles. This is the partition function
Z
(
S
1
,
(
N, g, V
))
of our theory. If one is worried about convergence issues, then we would have some
problems in this definition. If we work in flat space, then
K
T
(
y, y
) =
hy|e
HT
|yi
is independent of
y
. So when we integrate over all
y
, the answer diverges (as
long as
K
T
(
y, y
) is non-zero). However, we would get a finite result if we had a
compact manifold instead.
Correlation functions
More interestingly, we can do correlation functions. We will begin by considering
the simplest choice local operators.
Definition
(Local operator)
.
A local operator
O
(
t
) is one which depends on
the values of the fields and finitely many derivatives just at one point t M.
We further restrict to local operators that do not depend on derivatives.
These are given by functions
O
:
N R
, and then by pullback we obtain an
operator O(x(t)).
Suppose the corresponding quantum operator is
ˆ
O
=
O
(
ˆx
), characterized by
the property
ˆ
O|xi = O(x) |xi.
If we want to evaluate this at time t, then we would compute
hy
1
|
ˆ
O(t) |y
0
i = hy
1
|e
H(T t)
ˆ
Oe
Ht
|y
0
i.
But, inserting a complete set of states, this is equal to
Z
d
n
x hy
1
|e
H(T t)
|xihx|
ˆ
Oe
Ht
|yi
=
Z
d
n
x O(x) hy
1
|e
H(T t)
|xihx|e
Ht
|yi.
Simplifying notation a bit, we can write
hy
1
|
ˆ
O(t) |y
0
i =
Z
d
n
x O(x(t))
Z
C
[T,t]
[y
1
,x
t
]
Dx e
S[x]
Z
C
[t,0]
[x
t
,y
0
]
Dx e
S[x]
.
But this is just the same as
Z
C
[T,0]
[y
1
,y
0
]
Dx O(x(t))e
S[x]
.
More generally, suppose we have a sequence of operators
O
n
, ··· , O
1
we want
to evaluate at times
T > t
n
> t
n1
> ··· > t
1
>
0, then by the same argument,
we find
hy
1
|e
HT
ˆ
O
n
(t
n
) ···
ˆ
O
1
(t
1
) |y
0
i =
Z
C
[0,T ]
[y
0
,y
1
]
Dx O
n
(x(t
n
)) ···O(x(t
1
))e
S[x]
.
Note that it is crucial that the operators are ordered in this way, as one would
see if they actually try to prove this. Indeed, the
ˆ
O
i
are operators, but the
objects in the path integral, i.e. the
O
(
x
(
t
i
)) are just functions. Multiplication
of operators does not commute, but multiplication of functions does. In general,
if {t
i
} (0, T ) are a collection of times, then we have
Z
Dx
n
Y
i=1
O
i
(x(t
i
))e
S[x]
= hy
1
|e
HT
T
n
Y
i=1
ˆ
O
i
(t
i
) |y
0
i,
where T denotes the time ordering operator. For example, for n = 2, we have
T [
ˆ
O
1
(t
1
)
ˆ
O
2
(t
2
)] = Θ(t
2
t
1
)
ˆ
O
2
(t
2
)
ˆ
O
1
(t
1
) + Θ(t
1
t
2
)
ˆ
O
1
(t
1
)
ˆ
O
2
(t
2
),
where Θ is the step function.
It is interesting to note that in this path integral formulation, we see that
we get non-trivial correlation between operators at different times only because
of the kinetic (derivative) term in the action. Indeed, for a free theory, the
discretized version of the path integral looked like
S
kin
[x] =
X
i
1
2
x
i+1
x
i
t
2
t,
Now if we don’t have this term, and the action is purely potential:
S
pot
[x] =
X
i
V (x
i
),
then the discretized path integral would have factorized into a product of integrals
at these sampling points
x
i
. It would follow that for any operators
{O
i
}
that
depend only on position, we have
*
Y
i
O
i
(x(t
i
))
+
=
Y
i
hO
i
(x(t
i
))i,
and this is incredibly boring. When we work with higher dimensional universes,
the corresponding result shows that if there are no derivative terms in the
potential, then events at one position in spacetime have nothing to do with
events at any other position.
We have already seen this phenomena when we did quantum field theory
with perturbation theory the interactions between different times and places
are given by propagators, and these propagators arise from the kinetic terms in
the Lagrangian.
Derivative terms
We now move on to consider more general functions of the field and its derivative.
Consider operators
O
i
(
x, ˙x, ···
). We might expect that the value of this operator
is related to path integrals of the form
Z
Dx O
1
(x, ˙x, ···)|
t
1
O
2
(x, ˙x, ···)|
t
2
e
S[x]
But this can’t be right. We were told that one of the most important properties
of quantum mechanics is that operators do not commute. In particular, for
p
i
= ˙x
i
, we had the renowned commutator relation
[ˆx
i
, ˆp
j
] = δ
i
j
.
But in this path integral formulation, we feed in functions to the path integral,
and it knows nothing about how we order
x
and
˙x
in the operators
O
i
. So what
can we do?
The answer to this is really really important. The answer is that path
integrals don’t work.
The path integral measure
Recall that to express our correlation functions as path integrals, we had to take
the limits
Dx
?
= lim
N→∞
1
(2πt)
nN/2
N1
Y
i=1
d
n
x
i
a(x
i
),
and also
S[x]
?
= lim
N→∞
N1
X
n=1
1
2
x
n+1
x
n
t
2
t.
Do these actually make sense?
What we are trying to do with these expressions is that we are trying to
regularize our path integral, i.e. find a finite-dimensional approximation of the
path integral. For quantum field theory in higher dimensions, this is essentially
a lattice regularization.
Before we move on and try to see if this makes sense, we look at another way
of regularizing our path integral. To do so, we decompose our field into Fourier
modes:
x
a
(t) =
X
kZ
x
a
k
e
2πikt/T
,
and then we can obtain a regularized form of the action as
S
N
[x] =
N
X
k=N
1
2
k
2
x
a
k
x
a
k
.
Under this decomposition, we can take the regularized path integral measure to
be
D
N
x =
Y
|k|≤N
d
n
x
k
.
This is analogous to high-energy cutoff regularization. Now the natural question
to ask is do the limits
lim
N→∞
Z
D
N
x, lim
N→∞
S
N
[x]
exist?
The answer is, again: NO! This is in fact not a problem of us not being able
to produce limits well. It is a general fact of life that we cannot have a Lebesgue
measure on an infinite dimensional inner product space (i.e. vector space with
an inner product).
Recall the following definition:
Definition
(Lebesgue measure)
.
A Lebesgue measure d
µ
on an inner product
space V obeys the following properties
For all non-empty open subsets U R
D
, we have
vol(U) =
Z
U
dµ > 0.
If U
0
is obtained by translating U , then
vol(U
0
) = vol(U).
Every
x V
is contained in at least one open neighbourhood
U
x
with
finite volume.
note that we don’t really need the inner product structure in this definition.
We just need it to tell us what the word “open” means.
We now prove that there cannot be any Lebesgue measure on an infinite
dimensional inner product space.
We first consider the case of a finite-dimensional inner product space. Any
such inner product space is isomorphic to
R
D
for some
D
. Write
C
(
L
) for an
open hypercube of side length
L
. By translation in variance, the volume of any
two such hypercubes would be the same.
Now we note that given any such hypercube, we can cut it up into 2
D
hypercubes of side length L/2:
Then since
C
(
L
) contains 2
D
disjoint copies of
C
(
L/
2) (note that it is not exactly
the union of them, since we are missing some boundary points), we know that
vol(C(L))
2
D
X
i=1
vol(C(L/2)) = 2
D
vol(C(L/2)),
Now in the case of an infinite dimensional vector space,
C
(
L
) will contain
infinitely many copies of
C
(
L/
2). So since
vol
(
C
(
L/
2)) must be non-zero, as it
is open, we know
vol
(
C
(
L
)) must be infinite, and this is true for any
L
. Since
any open set must contain some open hypercube, it follows that all open sets
have infinite measure, and we are dead.
Theorem.
There are no Lebesgue measures on an infinite dimensional inner
product space.
This means whenever we do path integrals, we need to understand that we
are not actually doing an integral in the usual sense, but we are just using a
shorthand for the limit of the discretized integral
lim
N→∞
1
2πt
nN/2
Z
N1
Y
i=1
d
n
x
i
exp
1
2
|x
i+1
x
i
|
t
2
t
!
.
as a whole. In particular, we cannot expect the familiar properties of integrals
to always hold for our path integrals.
If we just forget about this problem and start to do path integrals, then we
would essentially be writing down nonsense. We can follow perfectly logical steps
and prove things, but the output will still be nonsense. Then we would have
to try to invent some new nonsense to make sense of the nonsense. This was,
in fact, how renormalization was invented! But as we will see, that is not what
renormalization really is about.
Note that we showed that the measure D
x
doesn’t exist, but what we really
need wasn’t Dx. What we really needed was
Z
Dx e
S[x]
.
This is no longer translation invariant, so it is conceivable that it exists. Indeed,
in the case of a 1D quantum field theory, it does, and is known as the Wiener
measure.
In higher dimensions, we are less certain. We know it doesn’t exist for QED,
and we believe it does not exist for the standard model. However, we believe
that it does exist for Yang–Mills theory in four dimensions.
Non-commutativity in QM
Now we know that the path integral measure doesn’t exist, and this will solve
our problem with non-commutativity. Indeed, as we analyze the discretization
of the path integral, the fact that
[ˆx, ˆp] 6= 0
will fall out naturally.
Again, consider a free theory, and pick times
T > t
+
> t > t
> 0.
We will consider
Z
Dx x(t) ˙x(t
)e
S
= hy
1
|e
H(T t)
ˆxe
H(tt
)
ˆpe
Ht
|y
0
i,
Z
Dx x(t) ˙x(t
+
)e
S
= hy
1
|e
H(tt
+
)
ˆpe
H(t
+
t)
ˆxe
Ht
|y
0
i.
As we take the limit t
±
t, the difference of the right hand sides becomes
hy
1
|e
H(tt)
[ˆx, ˆp]e
Ht
|y
0
i = hy
1
|e
HT
|y
0
i 6= 0.
On the other hand, in the continuum path integral, the limit seems to give the
same expression in both cases, and the difference vanishes, naively. The problem
is that we need to regularize. We cannot just bring two operators together in
time and expect it to behave well. We saw that the path integral was sensitive to
the time-ordering of the operators, so we expect something “discrete” to happen
when the times cross. This is just like in perturbation theory, when we bring two
events together in time, we have to worry that the propagators become singular.
Normally, we would have something like
x(t) ˙x(t
) x(t) ˙x(t
+
) = x
t
x
t
x
t
δt
δt
x
t
x
t
+
+δt
x
t
+
δt
.
In the regularized integral, we can keep increasing
t
and decreasing
t
+
, until
we get to the point
x
t
x
t
x
tt
t
x
t
x
t+∆t
x
t
t
.
Now that
t
±
have hit
t
, we need to look carefully what happens to the individual
heat kernels. In general, we stop taking the limit as soon as any part of the
discretized derivative touches
x
t
. The part of the integral that depends on
x
t
looks like
Z
d
n
x
t
K
t
(x
t+∆t
, x
t
)x
t
x
t
x
tt
t
x
t+∆t
x
t
t
K
t
(x
t
, x
tt
).
Using the fact that
K
t
(x
t
, x
tt
) exp
(x
t
x
tt
)
2
2∆t
,
we can write the integral as
Z
d
n
x
t
x
t
x
t
K
t
(x
t+∆t
, x
t
)K
t
(x
t
, x
tt
)
,
Now integrating by parts, we get that this is equal to
Z
d
n
x
t
K
t
(x
t+∆t
, x
t
)K
t
(x
t
, x
tt
) = K
2∆t
(x
t+∆t
, x
tt
).
So we get the same as in the operator approach.