3Interacting fields

III Quantum Field Theory



3.2 Interaction picture
How do we actually study interacting fields? We can imagine that the Hamil-
tonian is split into two parts, one of which is the “free field” part
H
0
, and the
other part is the “interaction” part
H
int
. For example, in the
φ
4
theory, we can
split it up as
H =
Z
d
3
x
1
2
(π
2
+ (φ)
2
+ m
2
φ
2
)
| {z }
H
0
+
λ
4!
φ
4
|{z}
H
int
.
The idea is that the interaction picture mixes the Schr¨odinger picture and the
Heisenberg picture, where the simple time evolutions remain in the operators,
whereas the complicated interactions live in the states. This is a very useful
trick if we want to deal with small interactions.
To see how this actually works, we revisit what the Schr¨odinger picture and
the Heisenberg picture are.
In the Schr¨odinger picture, we have a state
|ψ(t)i
that evolves in time, and
operators
O
S
that are fixed. In anticipation of what we are going to do next, we
are going to say that the operators are a function of time. It’s just that they are
constant functions:
d
dt
O
S
(t) = 0.
The states then evolve according to the Schr¨odinger equation
i
d
dt
|ψ(t)i
S
= H |ψ(t)i
S
.
Since
H
is constant in time, this allows us to write the solution to this equation
as
|ψ(t)i
S
= e
iHt
|ψ(0)i
S
.
Here
e
iHt
is the time evolution operator, which is a unitary operator. More
generally, we have
|ψ(t)i
S
= e
iH(tt
0
)
|ψ(0)i
S
= U
S
(t, t
0
) |ψ(t
0
)i
S
,
where we define
U
S
(t, t
0
) = e
iH(tt
0
)
.
To obtain information about the system at time
t
, we look at what happens
when we apply O
S
(t) |ψ(t)i
S
.
In the Heisenberg picture, we define a new set of operators, relating to the
Schr¨odinger picture by
O
H
(t) = e
iHt
O
S
(t)e
iHt
|ψ(t)i
H
= e
iHt
|ψ(t)i
S
.
In this picture, the states
|ψ(t)i
H
are actually constant in time, and are just
equal to |ψ(0)i
S
all the time. The equations of motion are then
d
dt
O
H
= i[H, O
H
]
d
dt
|ψ(t)i
H
= 0.
To know about the system at time t, we apply
O
H
(t) |ψ(t)i
H
= e
iHt
O
S
(t) |ψ(t)i
S
.
In this picture, the time evolution operator is just the identity map
U
H
(t, t
0
) = id .
In the interaction picture, we do a mixture of both. We only shift the “free”
part of the Hamiltonian to the operators. We define
O
I
(t) = e
iH
0
t
O
S
(t)e
iH
0
t
|ψ(t)i
I
= e
iH
0
t
|ψ(t)i
S
.
In this picture, both the operators and the states change in time!
Why is this a good idea? When we studied the Heisenberg picture of the free
theory, we have already figured out what conjugating by
e
iH
0
t
does to operators,
and it was really simple! In fact, it made things look nicer. On the other hand,
time evolution of the state is now just generated by the interaction part of the
Hamiltonian, rather than the whole thing, and this is much simpler (or at least
shorter) to deal with:
Proposition. In the interaction picture, the equations of motion are
d
dt
O
I
= i[H
0
, O
I
]
d
dt
|ψ(t)i
I
= H
I
|ψ(t)i
I
,
where H
I
is defined by
H
I
= (H
int
)
I
= e
iH
0
t
(H
int
)
S
e
iH
0
t
.
Proof.
The first part we’ve essentially done before, but we will write it out again
for completeness.
d
dt
(e
iH
0
t
O
S
e
iH
0
t
) = lim
ε0
e
iH
0
(t+ε)
O
S
e
iH
0
(t+ε)
e
iH
0
t
O
S
e
iH
0
t
ε
= lim
ε0
e
iH
0
t
e
iH
0
ε
O
S
e
iH
0
ε
O
S
ε
e
iH
0
t
= lim
ε0
e
iH
0
t
(1 + iH
0
ε)O
S
(1 iH
0
ε) O
S
+ o(ε)
ε
e
iH
0
t
= lim
ε0
e
iH
0
t
(H
0
O
S
O
S
H
0
) + o(ε)
ε
e
iH
0
t
= e
iH
0
t
i[H
0
, O
S
]e
iH
0
t
= i[H
0
, O
I
].
For the second part, we start with Schr¨odinger’s equation
i
d |ψ(t)i
S
dt
= H
S
|ψ(t)i
S
.
Putting in our definitions, we have
i
d
dt
(e
iH
0
t
|ψ(t)i
I
) = (H
0
+ H
int
)
S
e
iH
0
t
|ψ(t)i
I
.
Using the product rule and chain rule, we obtain
H
0
e
iH
0
t
|ψ(t)i
I
+ ie
iH
0
t
d
dt
|ψ(t)i
I
= (H
0
+ H
int
)
S
e
iH
0
t
|ψ(t)i
I
.
Rearranging gives us
i
d |ψ(t)i
I
dt
= e
iH
0
t
(H
int
)
S
e
iH
0
t
|ψ(t)i
I
.
To obtain information about the state at time
t
, we look at what happens
when we apply
O
I
(t) |ψ(t)i
I
= e
iH
0
t
O
S
(t) |ψ(t)i
S
.
Now what is the time evolution operator
U
(
t, t
0
)? In the case of the Schr¨odinger
picture, we had the equation of motion
i
d
dt
|ψ(t)i
S
= H |ψ(t)i
S
.
So we could immediately write down
|ψ(t)i = e
iH(tt
0
)
|ψ(t
0
)i.
In this case, we cannot do that, because
H
I
is now a function of time. The next
best guess would be
U
I
(t, t
0
) = exp
i
Z
t
t
0
H
I
(t
0
) dt
0
.
For this to be right, it has to satisfy
i
d
dt
U
I
(t, t
0
) = H
I
(t)U
I
(t, t
0
).
We can try to compute the derivative of this. If we take the series expansion of
this, then the second-order term would be
(i)
2
2
Z
t
t
0
H
I
(t
0
) dt
0
2
.
For the equation of motion to hold, the derivative should evaluate to
iH
I
times
the first order term, i.e.
iH
I
(i)
Z
t
t
0
H
I
(t
0
) dt
0
.
We can compute its derivative to be
(i)
2
2
H
I
(t)
Z
t
t
0
H
I
(t
0
) dt
0
+
(i)
2
2
Z
t
t
0
H
I
(t
0
) dt
0
H
I
(t).
This is not equal to what we wanted, because
H
I
(
t
) and
H
I
(
t
0
) do not commute
in general.
Yet, this is a good guess, and to get the real answer, we just need to make a
slight tweak:
Proposition
(Dyson’s formula)
.
The solution to the Schr¨odinger equation in
the interaction picture is given by
U(t, t
0
) = T exp
i
Z
t
t
0
H
I
(t
0
)dt
0
,
where
T
stands for time ordering: operators evaluated at earlier times appear
to the right of operators evaluated at leter times when we write out the power
series. More explicitly,
T {O
1
(t
1
)O
2
(t
2
)} =
(
O
1
(t
1
)O
2
(t
2
) t
1
> t
2
O
2
(t
2
)O
1
(t
1
) t
2
> t
1
.
We do not specify what happens when
t
1
=
t
2
, but it doesn’t matter in our case
since the operators are then equal.
Thus, we have
U(t, t
0
) = 1 i
Z
t
t
0
dt
0
H
I
(t
0
) +
(i)
2
2
Z
t
t
0
dt
0
Z
t
t
0
dt
00
H
I
(t
00
)H
I
(t
0
)
+
Z
t
t
0
dt
0
Z
t
0
t
0
dt
00
H
I
(t
0
)H
I
(t
00
)
)
+ ··· .
We now notice that
Z
t
t
0
dt
0
Z
t
t
0
dt
00
H
I
(t
00
)H
I
(t
0
) =
Z
t
t
0
dt
00
Z
t
00
t
0
dt
0
H
I
(t
00
)H
I
(t
0
),
since we are integrating over all
t
0
t
0
t
00
t
on both sides. Swapping
t
0
and
t
00
shows that the two second-order terms are indeed the same, so we have
U(t, t
0
) = 1 i
Z
t
t
0
dt
0
H
I
(t
0
) + (i)
2
Z
t
t
0
dt
0
Z
t
0
t
0
dt
00
H
I
(t
00
)H
I
(t
0
) + ··· .
Proof.
This from the last presentation above, using the fundamental theorem of
calculus.
In theory, this time-ordered exponential is very difficult to compute. However,
we have, at the beginning, made the assumption that the interactions
H
I
are
small. So often we can just consider the first one or two terms.
Note that when we evaluate the time integral of the Hamiltonian, a convenient
thing happens. Recall that our Hamiltonian is the integral of some operators
over all space. Now that we are also integrating it over time, what we are doing
is essentially integrating over spacetime a rather relativistic thing to do!
Some computations
Now let’s try to do some computations. By doing so, we will very soon realize
that our theory is very annoying to work with, and then we will later come up
with something more pleasant.
We consider the scalar Yukawa theory. This has one complex scalar field and
one real scalar field, with an interaction Hamiltonian of
H
I
= gψ
ψφ.
Here the
ψ, φ
are the “Heseinberg” versions where we conjugated by
e
iH
0
t
, so we
have
e
ip·x
rather than
e
ip·x
in the integral. We will pretend that the
φ
particles
are mesons, and that the ψ-particles are nucleons.
We now want to ask ourselves the following question suppose we start
with a meson, and we let it evolve over time. After infinite time, what is the
probability that we end up with a ψ
¯
ψ pair, where
¯
ψ is the anti-ψ particle?
φ
ψ
¯
ψ
We denote our initial state and “target” final states by
|ii =
p
2E
p
a
p
|0i
|fi =
p
4E
q
1
E
q
2
b
q
1
c
q
2
|0i.
What we now want to find is the quantity
hf|U(, −∞) |ii.
Definition (S-matrix). The S-matrix is defined as
S = U(, −∞).
Here
S
stands for “scattering”. So we often just write our desired quantity
as
hf|S |ii
instead. When we evaluate this thing, we will end up with some
numerical expression, which will very obviously not be a probability (it contains
a delta function). We will come back to interpreting this quantity later. For
the moment, we will just try to come up with ways to compute it in a sensible
manner.
Now let’s look at that expression term by term in the Taylor expansion of
U
.
The first term is the identity 1, and we have a contribution of
hf|1 |ii = hf|ii = 0
since distinct eigenstates are orthogonal. In the next term, we are integrating
over
ψ
ψφ
. If we expand this out, we will get loads of terms, each of which is a
product of some
b, c, a
(or their conjugates). But the only term that contributes
would be the term
c
q
2
b
q
1
a
p
, so that we exactly destroy the
a
p
particle and
create the
b
q
1
and
c
q
2
particles. The other terms will end up transforming
|ii
to
something that is not
|fi
(or perhaps it would kill
|ii
completely), and then the
inner product with |fi will give 0.
Before we go on to discuss what the higher order terms do, we first actually
compute the contribution of the first order term.
We want to compute
hf|S |ii = hf|1 i
Z
dt H
I
(t) + ···|ii
We know the
1
contributes nothing, and we drop the higher order terms to get
i hf|
Z
dt H
I
(t) |ii
= ig hf|
Z
d
4
x ψ
(x)ψ(x)φ(x) |ii
= ig
Z
d
4
x hf|ψ
(x)ψ(x)φ(x) |ii.
We now note that the
|ii
interacts with the
φ
terms only, and the
hf|
interacts
with the
ψ
terms only, so we can evaluate
φ
(
x
)
|ii
and
ψ
(
x
)
ψ
(
x
)
|fi
separately.
Moreover, any leftover a
, b
, c
terms in the expressions will end up killing the
other object, as the a
’s commute with the b
, c
, and h0|a
= (a |0i)
= 0.
We can expand φ in terms of the creation and annihilation operators a and
a
to get
φ(x) |ii =
Z
d
3
q
(2π)
3
p
2E
p
p
2E
q
(a
q
e
iq·x
+ a
q
e
iq·x
)a
p
|0i
We notice that nothing happens to the
a
q
terms and they will end up becoming
0 by killing h0|. Swapping a
q
and a
p
and inserting the commutator, we obtain
Z
d
3
q
(2π)
3
p
2E
p
p
2E
q
([a
q
, a
p
] + a
p
a
q
)e
iq·x
|0i
=
Z
d
3
q
(2π)
3
p
2E
p
p
2E
q
(2π)
3
δ
3
(p q)e
iq·x
|0i
= e
ip·x
|0i.
We will use this to describe the creation and destruction of mesons.
We can similarly look at
ψ
(x)ψ(x) |fi
=
Z
d
3
k
1
d
3
k
2
(2π)
6
1
p
4E
k
1
E
k
2
(b
k
1
e
ik
1
·x
+ c
k
1
e
ik
1
·x
)(b
k
2
e
ik
2
·x
+ c
k
2
e
ik
2
·x
) |fi
As before, the only interesting term for us is when we have both
c
k
1
and
b
k
2
present to “kill” b
q
1
c
q
2
in |f i. So we look at
Z
d
3
k
1
d
3
k
2
(2π)
6
p
4E
q
1
E
q
2
p
4E
k
1
E
k
2
c
k
1
b
k
2
b
q
1
c
q
2
e
i(k
1
+k
2
)·x
|0i
=
Z
d
3
k
1
d
3
k
2
(2π)
6
(2π)
6
δ
3
(q
1
k
2
)δ
3
(q
2
k
1
)e
i(k
1
+k
2
)·x
|0i
= e
i(q
1
+q
2
)·x
|0i.
So we have
hf|S |ii ig
Z
d
4
x (e
i(q
1
+q
2
)·x
|0i)
(e
ip·x
|0i)
= ig
Z
d
4
x e
i(q
1
+q
2
p)·x
= ig(2π)
4
δ
4
(q
1
+ q
2
p).
If we look carefully at what we have done, what allowed us to go from the initial
φ
state
|ii
=
p
2E
p
a
p
|0i
to the final
ψ
¯
ψ
state
|fi
=
p
4E
q
1
E
q
2
b
q
1
c
q
2
|0i
was
the presence of the term
c
q
1
b
q
2
a
p
in the
S
matrix (what we appeared to have
used in the calculation was
c
q
1
b
q
2
instead of their conjugates, but that is because
we were working with the conjugate ψ
ψ |fi rather than hf|ψ
ψ).
If we expand the
S
-matrix to, say, second order, we will, for example, find
a term that looks like (
c
b
a
)(
cba
). This corresponds to destroying a
ψ
¯
ψ
pair
to produce a
φ
, then destroying that to recover another
ψ
¯
ψ
pair, which is an
interaction of the form
ψ
¯
ψ
ψ
¯
ψ
φ