Research Progress
Srivatsan
February 22, 2014
1 Lorentz transformations
a) The denition of
(4)
(p) is:
(4)
(p) =
1
(2)
4
_
d
4
xe
ipx
. (1)
Here, k x = x
.
We could equally well choose to integrate with respect to x
(4)
(p) =
1
(2)
4
_
d
4
xe
ip x
=
1
(2)
4
_
d
4
xe
ip
=
1
(2)
4
_
d
4
xe
i px
=
(4)
( p).
(2)
In the rst line, we used the fact that the integration measure is invariant,
d
4
x = d
4
x. In the second line we dened p
= p
.
b) We want to show that:
(3)
(
) =
(3)
(
) (3)
Integrating against an arbitrary function, we have:
F(
) =
_
d
3
k
(3)
(
)F(
k) =
_
d
3
k
(3)
(
)F(
k)
=
_
d
3
k
(3)
(
)F(
1
k)
=
_
d
3
p
(3)
( p
)F( p)
=
_
d
3
k
(3)
(
)F(
k).
(4)
1
As F was arbitrary this implies:
(3)
(
) =
(3)
(
). (5)
Note, in arriving at the last line, we used
d
3
k
k
=
d
3
k
k
, which we show in
part (d). Here,
k
i
.
c) From part (b) we have:
_
a
k
, a
_
= (2)
3
(3)
(
) =
k
(2)
3
(3)
(
)
=
k
_
a
k
, a
_
.
(6)
We can then guess that:
a
k
=
k
a
k
a
k
=
k
a
k
.
(7)
d)
We want to show:
_
d
3
k
(2)
3
1
2
k
f(k) =
_
d
3
k
(2)
3
1
2
k
f(
k) (8)
for any proper, orthocronos Lorentz transformation. In order to show this,
rst note that:
_
d
3
k
(2)
3
1
2
k
f(k) =
_
d
4
k
(2)
4
(2)(k
2
+ m
2
)(k
0
)f(k). (9)
One can derive (??) by performing the k
0
integral. Note that, as, k
2
+m
2
=
(k
0
)
2
+
k
2
+ m
2
, changing variables from k
0
to u = (k
0
)
2
, changes the
measure as dk
0
=
1
2k
0
du. Though the delta function k
0
=
_
k
2
+ m
2
has
two solutions, the theta function (k
0
) picks out only one.
2
With this relation we have:
_
d
3
k
(2)
3
1
2
k
f(k) =
_
d
4
k
(2)
4
(2)(k
2
+ m
2
)(k
0
)f(k)
=
_
d
4
k
(2)
4
(2)(
k
2
+ m
2
)(
k
0
)f(
k)
=
_
d
4
k
(2)
4
(2)(k
2
+ m
2
)(k
0
)f(
k) =
_
d
3
k
(2)
3
1
2
k
f(
k).
(10)
In getting to the last line, we used (
k
0
) = (k
0
). This holds, as we are only
considering the connected part of the Lorentz group, which does not change
the sign of k
0
.
2 Problem with relativistic quantum mechanics
a)
Beginning with the Schrodinger equation, i
t
=
1
2m
2
, and =
||
2
, we have:
t
=
t
=
i
2m
_
_
=
i
2m
_
=
J
(11)
b) Proceeding as before, but now with
2
t
=
2
m
2
and =
i
2m
(
t
t
) .
t
=
i
2m
_
2
t
2
t
_
=
i
2m
_
_
=
J
(12)
c)
The expression here for , =
i
2m
(
t
t
) =
1
m
Im(
t
), is
not positive denite, and so cannot be interpreted as a probability density.
3
3 Commutation relations of annihilation and cre-
ation operators
a) The eld and momentum at time t can be written in terms of the eld
and momentum at time 0 as: (x, t) = e
iHt
(x, 0)e
iHt
and (x, t) =
e
iHt
(x, 0)e
iHt
. With this the equal time commutation relations can be
written as
_
(x, t), (x
, t)
= e
iHt
_
(x, 0), (x
, 0)
e
iHt
_
(x, t), (x
, t)
= e
iHt
_
(x, 0), (x
, 0)
e
iHt
_
(x, t), (x
, t)
= e
iHt
_
(x, 0), (x
, 0)
e
iHt
.
(13)
As, e
iHt
0e
iHt
= 0, and e
iHt
i
3
(x x
)e
iHt
= i
3
(x x
), it is enough to
impose the commutation relations at t = 0.
_
(x, 0), (x
, 0)
= 0
_
(x, t), (x
, t)
= 0
_
(x, 0), (x
, 0)
= 0
_
(x, t), (x
, t)
= 0
_
(x, 0), (x
, 0)
= i
(3)
(x x
)
_
(x, t), (x
, t)
= i
(3)
(x x
)
(14)
b)
(x, t) =
_
d
3
k
(2)
3
1
_
2
k
_
a
k
e
ikx
+ a
k
e
ikx
_
(x, t) =
_
d
3
k
(2)
3
i
_
k
2
_
a
k
e
ikx
a
k
e
ikx
_
(15)
Transforming expression (??), we have:
(
k) =
1
_
2
k
_
a
k
+ a
k
_
(
k) = i
_
k
2
_
a
k
a
k
_
.
(16)
With this,
a
k
=
_
k
2
(
k) + i
1
2
k
(
k)
a
k
=
_
k
2
(
k) i
1
2
k
(
k).
(17)
4
It is also convenient to have the expressions:
(
k) =
1
2
k
_
a
k
+ a
k
_
(
k) = i
_
k
2
_
a
k
a
k
_
.
(18)
c) Form the commutation relations, (??), we also have the momentum space,
equal time, relations:
_
(
k), (
)
_
= 0
_
(
k), (
)
_
= 0
_
(
k), (
)
_
= i(2)
3
(3)
(
k +
).
(19)
With these it is strait forward to calculate the commutators of creation and
annihilation operators.
_
a
k
, a
=
i
2
__
(
k), (
)
_
+
_
(
k), (
)
__
= 0
_
a
k
, a
= 0
_
a
k
, a
_
=
i
2
__
(
k), (
)
_
_
(
k), (
)
__
= (2)
3
(3)
(
).
(20)
Here we have used the fact that [(
k), (
)] = [(
k), (
)] = 0.
4 Expressing Nother charges in terms of creation
and annihilation operators
In analogy to the last problem set we have:
H =
1
2
_
d
3
x
_
2
+
2
+ m
2
2
_
P
i
=
_
d
3
x
_
_
M
=
_
d
3
x
_
x
T
0
x
T
0
_
(21)
5
Note, the convention for P
i
diers by a sign from that on the last problem
set. a) Beginning with H, we can rewrite (??) in momentum space.
H =
1
2
_
d
3
k
(2)
3
_
(
k, t)(
k, t) + (
k
2
+ m
2
. .
k
)(
k, t)(
k, t)
_
.
(22)
Using expression (??), or equivalently, (??), we can write H in terms of a
and a
.
H =
_
d
3
k
(2)
3
k
2
_
a
k
a
k
+ a
k
a
k
_
=
_
d
3
k
(2)
3
k
a
k
a
k
..
N
k
+
_
d
3
k
(2)
3
(2)
3
(3)
(0)
. .
C
H
=
_
d
3
k
(2)
3
k
N
k
+ C
H
.
(23)
b)
Again it is easiest to work in momentum space.
P
i
= i
_
d
3
k
(2)
3
(
k, t)(
k, t)k
i
=
1
2
_
d
3
k
(2)
3
_
a
k
a
k
__
a
k
+ a
k
_
k
i
=
1
2
_
d
3
k
(2)
3
_
a
k
a
k
a
k
a
k
+ a
k
a
k
a
k
a
k
+
_
a
k
, a
k
__
k
i
=
_
d
3
k
(2)
3
k
i
a
k
a
k
..
N
1
2
_
d
3
k
(2)
3
(2)
3
(3)
(0)k
i
. .
C
P
i
=
_
d
3
k
(2)
3
k
i
N
k
C
P
i .
(24)
Here we used the fact that expressions such as,
_
d
3
ka
k
a
k
k
i
= 0.
c)
To write M
in terms of creation and annihilation operators, it is useful
to write down the expression for the stress-energy tensor.
T
1
2
+ m
2
2
_
. (25)
6
For M
we need
T
0
=
1
2
0
_
2
+
+ m
2
2
_
=
_
H(x),
P(x)
_
.
(26)
With this we have:
M
0i
=
_
d
3
x
_
tP
i
(x) x
i
H(x)
_
= tP
i
_
d
3
k
(2)
3
k
i
(
k
N
k
),
M
ij
=
_
d
3
x
_
x
i
P
j
(x) x
j
P
i
(x)
_
=
_
d
3
k
(2)
3
_
k
j
k
i
k
i
k
j
_
N
k
.
(27)
Here we have dropped the normal ordering constants.
5 Second quantization
a) Varying the action L = i
t
1
2m
V (x)
with respect
to
and yields:
i
t
+
1
2m
2
V (x) = 0
i
t
+
1
2m
V (x)
= 0.
(28)
This is indeed the Schrodinger equation.
b) The conjugate momentum to is given as:
=
L
= i
. (29)
The Hamiltonian is:
H =
_
d
3
x[
t
L] =
_
d
3
x
_
1
2m
+ V (x)
_
(30)
7
c) The canonical commutation relations are:
_
(x, t),
(x
, t)
_
= i
(3)
(x x
). (31)
This can also be written,
_
(x, t),
(x
, t)
_
=
(3)
(x x
). (32)
d) Writing,
(x, t) =
n
a
n
e
iEnt
u
n
(x),
(x, t) =
n
a
n
e
iEnt
u
n
(x). It is
enough to consider the commutator, (??), at t = 0.
_
(x, 0),
(x
, 0)
_
=
m,n
_
a
m
a
n
a
n
a
m
_
u
m
(x)u
n
(x
) =
(3)
(x x
)
(33)
Multiplying by u
m
(x), u
n
(x
), integrating d
3
x, d
3
x
and dropping the primes
on the indices, we have:
_
a
m
, a
n
_
=
_
d
3
xd
3
x
u
m
(x)u
n
(x
)
(3)
(x x
) =
mn
. (34)
Similarly, [
(x, 0),
(x
, 0)] = 0 implies [ a
m
, a
n
] = [ a
m
, a
n
] = 0.
e)
Plugging the expression for
into (??) we have:
H =
_
d
3
x
(
1
2m
2
+ V (x))
=
_
d
3
x
m,n
a
m
a
n
e
i(EnEm)t
u
m
(
1
2m
2
+ V (x))u
n
=
_
d
3
x
m,n
a
m
a
n
e
i(EnEm)t
E
n
u
m
u
n
=
n
a
n
a
n
. .
Nn
E
n
.
(35)
In going from the second line to the third line, we used the fact that u
n
satises the time independent Schrodinger equation, (
1
2m
2
+ V (x))u
n
=
E
n
u
n
. In getting the last line we used the orthonormality of u
n
.
f)
8
A generic state can be written as an arbitrary combination of creation
operators acting on the vacuum, with each creation operator aloud to act
an arbitrary number of times.
| |{m
i
} = (a
1
)
m
1
(a
2
)
m
2
. . . |0. (36)
Here, the {m
i
} range from 0 up to . Note, do not confuse this notation
with that used in part (h).
g)
A physical interpretation of the states |{m
i
} is that this state has m
i
indistinguishable particles in the energy level i. The reason for this inter-
pretation comes from looking at the Hamiltonian. The energy level for the
state is just E
tot
=
i
E
i
m
i
. This is precisely the energy for many particles
interacting with an external potential with m
i
particles in energy level i.
h) From the picture above the wave function would just be the two
particle wave function for one particle in state n
1
and another in state n
2
.
n
1
n
2
(x
1
, x
2
, t) =
1
2
e
i(En
1
+En
2
)t
(u
n
1
(x
1
)u
n
2
(x
2
) + u
n
2
(x
1
)u
n
1
(x
2
)) .
(37)
We can also get this expression from the second quantized eld.
1
2
0|(x
1
, t)(x
2
, t)|n
1
, n
2
=
1
2
0|
m,n
e
i(Em+En)t
u
m
(x
1
)u
n
(x
2
) a
m
a
n
a
n
1
a
n
2
|0
=
1
2
e
i(En
1
+En
2
)t
(u
n
1
(x
1
)u
n
2
(x
2
) + u
n
2
(x
1
)u
n
1
(x
2
)) .
(38)
9
