2Free field theory

III Quantum Field Theory



2.5 The Heisenberg picture
We have tried to make our theory Lorentz invariant, but it’s currently in a terrible
state. Indeed, the time evolution happens in the states, and the operators depend
on states only. Fortunately, from IID Principles of Quantum Mechanics, we
know that there is a way to encode the time evolution in the operators instead,
via the Heisenberg picture.
Recall that the equation of motion is given by the Schr¨odinger equation
i
d
dt
|ψ(t)i = H |ψ(t)i.
We can write the solution formally as
|ψ(t)i = e
iHt
|ψ(0)i.
Thus we can think of
e
iHt
as the “time evolution operator” that sends
|ψ(0)i
forward in time by t.
Now if we are given an initial state
|ψ(0)i
, and we want to know what an
operator
O
S
does to it at time
t
, we can first apply
e
iHt
to let
|ψ(0)i
evolve to
time
t
, then apply the operator
O
S
, then pull the result back to time 0. So in
total, we obtain the operator
O
H
(t) = e
iHt
O
S
e
iHt
.
To evaluate this expression, it is often convenient to note the following result:
Proposition. Let A and B be operators. Then
e
A
Be
A
= B + [A, B] +
1
2!
[A, [A, B]] +
1
3!
[A, [A, [A, B]]] + ··· .
In particular, if [A, B] = cB for some constant c, then we have
e
A
Be
A
= e
c
B.
Proof. For λ a real variable, note that
d
dλ
(e
λA
Be
λA
) = lim
ε0
e
(λ+ε)A
Be
(λ+ε)A
e
λA
Be
λA
ε
= lim
ε0
e
λA
e
εA
Be
εA
B
ε
e
λA
= lim
ε0
e
λA
(1 + εA)B(1 εA) B + o(ε)
ε
e
λA
= lim
ε0
e
λA
(ε(AB BA) + o(ε))
ε
e
λA
= e
λA
[A, B]e
λA
.
So by induction, we have
d
n
dλ
n
(e
λA
Be
λA
) = e
λA
[A, [A, ···[A, B] ···]]e
λA
.
Evaluating these at λ = 0, we obtain a power series representation
e
λA
Be
λA
= B + λ[A, B] +
λ
2
2
[A, [A, B]] + ··· .
Putting λ = 1 then gives the desired result.
In the Heisenberg picture, one can readily directly verify that the commutation
relations for our operators
φ
(
x, t
) and
π
(
x, t
) now become equal time commutation
relations
[φ(x, t), φ(y, t)] = [π(x, t), π(y, t)] = 0, [φ(x, t), π(y, t)] =
3
(x y).
For an arbitrary operator O
H
, we can compute
dO
H
dt
=
d
dt
(e
iHt
O
S
e
iHt
)
= iHe
iHt
O
S
e
iHt
+ e
iHt
O
S
(iHe
iHt
)
= i[H, O
H
],
where we used the fact that the Hamiltonian commutes with any function of
itself. This gives the time evolution for the operators.
Recall that in quantum mechanics, when we transformed into the Heisenberg
picture, a miracle occurred, and the equations of motion of the operators became
the usual classical equations of motion. This will happen again. For
O
S
=
φ
(
x
),
we have
˙
φ(x, t) = i[H, φ(x, t)]
= i
Z
d
3
y
1
2
[π
2
(y, t) + (
y
φ(y, t))
2
+ m
2
φ
2
(y, t), φ(x, t)]
= i
Z
d
3
y
1
2
[π
2
(y, t), φ(x, t)]
= i
Z
d
3
y
1
2
π(y, t)[π(y, t), φ(x, t)] + [π(y, t), φ(x, t)]π(y, t)
= i
Z
d
3
y π(y, t)(
3
(x y))
= π(x, t).
Similarly, we have
˙π(x, t) = i[H, π(x, t)]
=
i
2
Z
d
3
y
[(
y
φ(y, t))
2
, π(x, t)] + m
2
[φ
2
(y, t), π(x, t)]
Now notice that
y
does not interact with π(x, t). So this becomes
=
Z
d
3
y
(
y
φ(y, t) ·
y
δ
3
(x y)) m
2
φ(y, t)δ
3
(x y)
Integrating the first term by parts, we obtain
=
Z
d
3
y
(
2
y
φ(y, t))δ
3
(x y) m
2
φ(y, t)δ
3
(x y)
=
2
φ(x, t) m
2
φ(x, t).
Finally, noting that ˙π =
¨
φ and rearranging, we obtain
µ
µ
φ(x, t) + m
2
φ(x, t) = 0.
This is just the usual Klein–Gordon equation!
It is interesting to note that this final equation of motion is Lorentz invariant,
but our original
dφ
dt
= i[H, φ]
is not. Indeed,
dφ
dt
itself singles out time, and to define
H
we also need to single
out a preferred time direction. However, this statement tells us essentially that
H
generates time evolution, which is true in any frame, for the appropriate
definition of H and t in that frame.
What happens to the creation and annihilation operators? We use our
previous magic formula to find that
e
iHt
a
p
e
iHt
= e
iE
p
t
a
p
e
iHt
a
p
e
iHt
= e
iE
p
t
a
p
.
These extra factors of
e
±iE
p
t
might seem really annoying to carry around, but
they are not! They merge perfectly into the rest of our formulas relativistically.
Indeed, we have
φ(x) φ(x, t) =
Z
d
3
p
(2π)
3
1
p
2E
p
a
p
e
iE
p
t
e
ip·x
+ a
p
e
iE
p
t
e
ip·x
=
Z
d
3
p
(2π)
3
1
p
2E
p
a
p
e
ip·x
+ a
p
e
ip·x
,
where now
p · x
is the inner product of the 4-vectors
p
and
x
! If we use our
relativistically normalized
a
(p) =
p
2E
p
a
p
instead, then the equation becomes
φ(x) =
Z
d
3
p
(2π)
3
1
2E
p
a(p)e
ip·x
+ a(p)
e
ip·x
,
which is made up of Lorentz invariant quantities only!
Unfortunately, one should note that we are not yet completely Lorentz-
invariant, as our commutation relations are equal time commutation relations.
However, this is the best we can do, at least in this course.
Causality
Since we are doing relativistic things, it is important to ask ourselves if our
theory is causal. If two points
x, y
in spacetime are space-like separated, then a
measurement of the field at
x
should not affect the measurement of the field at
y
. So measuring
φ
(
x
) after
φ
(
y
) should give the same thing as measuring
φ
(
y
)
after φ(x). So we must have [φ(x), φ(y)] = 0.
Definition
(Causal theory)
.
A theory is causal if for any space-like separated
points x, y, and any two fields φ, ψ, we have
[φ(x), ψ(y)] = 0.
Does our theory satisfy causality? For convenience, we will write
∆(x y) = [φ(x), φ(y)].
We now use the familiar trick that our
φ
is Lorentz-invariant, so we can pick
a convenient frame to do the computations. Suppose
x
and
y
are space-like
separated. Then there is some frame where they are of the form
x = (x, t), y = (y, t)
for the same t. Then our equal time commutation relations tell us that
∆(x y) = [φ(x), φ(y)] =
3
(x y) = 0.
So our theory is indeed causal!
What happens for time-like separation? We can derive the more general
formula
∆(x y) = [φ(x), φ(y)]
=
Z
d
3
p d
3
q
(2π)
6
1
p
4E
p
E
q
[a
p
, a
q
]e
i(p·xq·y)
+ [a
p
, a
q
]e
i(p·x+q·y)
+[a
p
, a
q
]e
i(p·xq·y)
+ [a
p
, a
q
]e
i(p·x+q·y)
=
Z
d
3
p d
3
q
(2π)
6
1
p
4E
p
E
q
(2π)
3
δ
3
(p q)
e
i(p·x+q·y)
e
i(p·xq·y)
=
Z
d
3
p
(2π)
3
1
2E
p
(e
ip·(yx)
e
ip·(yx)
).
This doesn’t vanish for time-like separation, which we may wlog be of the form
x = (x, 0), y = (x, t), since we have
[φ(x, 0), φ(x, t)] e
imt
e
imt
.