Physics 5040
Spring 2009
Problem Set 3
1. Consider the Lagrangian density
L =
~2
i~
∇ψ · ∇ψ ∗ + V ψψ ∗ − (ψ ∗ ψ̇ − ψ ψ̇ ∗ )
2m
2
where ψ and ψ ∗ are to be treated as independent variables. Show that this
leads to the time-dependent Schrödinger equation for ψ and it’s complex conjugate.
2. We have seen that the Euler-Lagrange equations applied to the Lagrangian
density
1
L = (∂µ ϕ)(∂ µ ϕ) − m2 ϕ2
2
leads to the Klein-Gordon equation ( + m2 )ϕ = 0.
(a) Recall that the Lagrangian in classical point particle mechanics is of the
form L = T − V = mẋ2 /2 − V (x). Observing that
∂L
= mẋi = pi
∂ ẋi
we define the canonical momentum in general by
pi =
∂L
.
∂ ẋi
In a (relativistic) field theory, we similarly define the field momentum by
π(x) =
∂L
.
∂ ϕ̇(x)
What is the canonical momentum of the K-G field?
(b) The Hamiltonian density is defined by H = π ϕ̇ − L . What is the Hamiltonian H of the field ϕ?
±ikx
(c) The K-G equation has plane
where k 2 = k02 − k2 =
√ wave solutions e
2
2
2
m . If we define ωk = + k + m , then we may write a general real
solution as a linear superpostion of plane waves in the form
Z
d4 k
ϕ(x) =
(ak e−ikx + a†k eikx )2πδ(k 2 − m2 )θ(k0 )
(2π)4
1
where the step function θ(k0 ) is required to avoid double counting in the
R
integral d4 k over all space, and ak , a†k are hermitian conjugate expansion coefficients. This form is really telling you that in the exponentials
we have k0 = ωk .
Consider the factor
d4 k
2πδ(k 2 − m2 )θ(k0 ) .
(2π)4
Since Lorentz transformations are orthogonal (i.e., ΛT gΛ = g so that
det Λ = +1 for proper transformations), it follows that d4 k is unchanged
under a Lorentz transformation (i.e., the Jacobian of the transformation
is just 1). It is also obvious that k 2 − m2 is unchanged under a Lorentz
transformation since both k 2 and m2 are Lorentz scalars. Furthermore,
a proper orthochronous Lorentz transformation can’t change the sign of
k0 . This is easiest to see from the basic equation ∆t′ = γ(∆t − β∆x).
Since β < 1 and ∆x < ∆t, we must have ∆t′ > 0 if and only if ∆t > 0.
Therefore the above factor is manifestly Lorentz invariant. Show that
Z
d3 k
d4 k
=
2πδ(k 2 − m2 )θ(k0 ) .
4
(2π)3 2ωk
k0 (2π)
Because of this, the term
d3 k
(2π)3 2ωk
is referred to as the invariant volume element. Note that the numerical factors aren’t necessary to maintain invariance.
R
(d) By considering the integral f (p) = d3 q δ(p − q)f (q) (where p2 = q 2 =
m2 ), argue that
(2π)3 2ωp δ(p − q)
is also Lorentz invariant. (This is called the invariant delta function.)
(e) We may now write
Z
d3 k
ϕ(x) =
(ak e−ikx + a†k eikx ) .
(2π)3 2ωk
If we let
fk (x) =
e−ikx
[(2π)3 2ωk ]1/2
then we have
ϕ(x) =
Z
d3 k
[fk (x)ak + fk∗ (x)a†k ] .
[(2π)3 2ωk ]1/2
2
Define an “inner product” on functions f (x), g(x) by
Z
←
→
hf (x), g(x)i := d3 x f ∗ (x)i∂0 g(x)
:=
Z
d3 x f ∗ (x)[i∂0 g(x)] − [i∂0 f ∗ (x)]g(x)
where ∂0 = ∂/∂x0 = ∂/∂t. Show that
Z
←
→
d3 x fk∗ (x)i∂0 fk′ (x) = δ(k − k′ )
while
Z
3
d
←
→
x fk∗ (x)i∂0 fk∗′ (x)
=
Z
←
→
d3 x fk (x)i∂0 fk′ (x) = 0 .
In other words, the functions fk (x) form an orthonormal set with respect
to the above inner product.
(f) Show that
ak =
Z
←
→
d3 x [(2π)3 2ωk ]1/2 fk∗ (x)i∂0 ϕ(x)
a†k
Z
←
→
d3 x [(2π)3 2ωk ]1/2 ϕ(x)i∂0 fk (x) .
and
=
(g) So far all of this has applied to a classical field. We go over to a quantum
field by analogy with ordinary quantum mechanics where we have (with
~ = 1) the operator commutation relations [xi , pj ] = iδij . Thus we require
the equal time canonical commutation relations
[ϕ(x, t), π(y, t)] = iδ(x − y)
and
[ϕ(x, t), ϕ(y, t)] = [π(x, t), π(y, t)] = 0 .
Show that these relations imply
[ak , a†k′ ] = (2π)3 2ωk δ(k − k′ )
and
[ak , ak′ ] = [a†k , a†k′ ] = 0 .
This shows that the coefficients ak and a†k become the annihilation and
creation operators of a harmonic oscillator. Thus the field ϕ(x) acts to
annihilate and create quanta with 4-momentum k µ .
3
(h) Let us now write
ϕ(x) =
Z
d3 k
(ak e−ikx + a†k eikx )
(2π)3 2ωk
:= ϕ+ (x) + ϕ− (x)
where
ϕ (x) =
Z
d3 k
ak e−ikx
(2π)3 2ωk
ϕ− (x) =
Z
d3 k
a† eikx .
(2π)3 2ωk k
+
and
are referred to as the positive and negative frequency parts of ϕ(x). Show
that
Z
d3 k
+
−
e−ik(x−y) := i∆+ (x − y)
[ϕ (x), ϕ (y)] =
(2π)3 2ωk
where ∆+ (x − y) is the same function defined in Exercise 15 of Problem
Set 1. This shows how the Green’s function arises in a natural way in
quantum field theory.
3. We have seen that the rotation operator for a vector in R3 can be written in
the form
R(θ) = eθ·J .
Now let’s take a look at how the spatial rotation operator is defined in quantum mechanics. If we rotate a vector x in R3 , then we obtain a new vector
x′ = R(θ)x where R(θ) is the matrix that represents the rotation. In two
dimensions this is
′ x
cos θ − sin θ
x
=
.
y′
sin θ
cos θ
y
If we have a scalar wavefunction ψ(x), then under rotation we obtain a new
wavefunction ψR (x′ ), where ψ(x) = ψR (x′ ) = ψR (R(θ)x). (See the figure
below. This is for an active transformation.)
ψR (x′ )
x′
ψ(x)
θ
x
Alternatively, we can write
ψR (x) = ψ(R−1 (θ)x).
4
Since R is an orthogonal transformation (it preserves the length of x) we know
that R−1 (θ) = RT (θ), and in the case where θ ≪ 1 we then have
x + θy
R−1 (θ)x =
.
−θx + y
Expanding ψ(R−1 (θ)x) with these values for x and y we have
ψR (x) = ψ(x + θy, y − θx) = ψ(x) − θ[x∂y − y∂x ]ψ(x)
or, using pi = −i∂i this is
ψR (x) = ψ(x) − iθ[xpy − ypx ]ψ(x) = [1 − iθLz ]ψ(x).
For finite θ we exponentiate this to write ψR (x) = e−iθLz ψ(x), and in the
case of an arbitrary angle θ in R3 this becomes
ψR (x) = e−iθ·L ψ(x) .
This was for spatial rotations. In general, for a particle with total angular
momentum J (where [Ji , Jj ] = iεijk Jk ) we define the rotation operator to be
R(θ) = e−iθ·J .
In the particular case of a spin one-half particle we have J = S = σ/2 where
the Pauli matrices σ are defined by
0 1
0 −i
1
0
σ1 =
σ2 =
σ3 =
1 0
i
0
0 −1
and obey the commutation relations
[σi , σj ] = 2iεijk σk .
(a) Show that the Pauli matrices obey the relations
σi σj = iεijk σk
for i 6= j
and
[σi , σj ]+ := σi σj + σj σi = 2Iδij
and hence that we also have the very useful result
σi σj = Iδij + iεijk σk .
(b) Show that for any vectors a and b we have
(a · σ)(b · σ) = (a · b)I + i(a × b) · σ .
5
(c) Show that the rotation operator for spin one-half particles is given by
e−iθ·σ/2 = I cos
=
"
θ
θ
− i(σ · θ̂) sin
2
2
cos θ/2 − iθ̂3 sin θ/2
−iθ̂− sin θ/2
−iθ̂+ sin θ/2
cos θ/2 + iθ̂3 sin θ/2
#
where θ̂± = θ̂1 ± iθ̂2 .
4. In Problem 2 you looked at some properties of the real Klein-Gordon field
which describes neutral scalar particles (i.e., spinless neutral particles). In
this problem you will look at some properties of the complex Klein-Gordon
field which, as you will see, describes charged spinless particles. So, consider
the Lagrangian density
L = ∂µ ϕ† ∂ µ ϕ − m2 ϕ† ϕ
where I have written ϕ† rather than ϕ∗ because we will be treating the fields ϕ
as quantum mechanical operators rather than simply complex classical fields.
(a) Treating ϕ and ϕ† as independent fields, what are the corresponding equations of motion for this Lagrangian density? What are the canonical momenta π and π † ?
(b) Now observe that this Lagrangian density is invariant under the global
phase transformation
ϕ(x) → ϕ′ (x) = e−iΛ ϕ(x)
and
†
ϕ† (x) → ϕ′ (x) = eiΛ ϕ† (x)
where Λ ≪ 1 is a constant. Then we know that there exists a conserved
current j µ with ∂µ j µ = 0 and where
jµ =
∂L
∆ϕr − T µ α δxα
∂ϕr,µ
with
∆ϕr (x) = δϕr (x) + ∂µ ϕr (x)δxµ
and
T µα =
∂L
ϕr,α − δαµ L .
∂ϕr,µ
What is the conserved charge
Q=
Z
j 0 (x) d3 x ?
Note that any overall factor of the constant Λ can be dropped from Q
since d(ΛQ)/dt = 0 implies that dQ/dt = 0 also.
6
(c) Following exactly the same approach as we did for the real K-G field, we
expand the complex fields as
Z
d3 k
ϕ(x) =
ak fk (x) + b†k fk∗ (x)
[(2π)3 2ωk ]1/2
and
†
ϕ (x) =
Z
d3 k
† ∗
b
f
(x)
+
a
f
(x)
k
k
k
k
[(2π)3 2ωk ]1/2
where we write fk∗ (x) for the complex conjugate of the ordinary function fk (x), and a†k for the “complex conjugate” (i.e., the adjoint) of the
quantum mechanical operator ak . We have labelled the expansion coefficients of the positive and negative frequency parts by ak (resp. bk ) and
b†k (resp. a†k ) because the fields are complex so there is no requirement
that ak and b†k be complex conjugates of each other. We have added the
dagger on bk because with hindsight we know that it behaves as a creation
operator. We now impose the canonical equal time commutation relations
(ETCR’s) on the fields:
[ϕ(x, t), π(y, t)] = [ϕ(x, t), ϕ̇† (y, t)] = iδ(x − y)
and
[ϕ† (x, t), π † (y, t)] = [ϕ† (x, t), ϕ̇(y, t)] = iδ(x − y)
and where all other commutators vanish.
Show that
[Q, ϕ(x)] = −ϕ(x)
and
[Q, ϕ† (x)] = +ϕ† (x) .
Hint : Recall the commutator identity [ab, c] = a[b, c] + [a, c]b.
To help understand what these mean, let |qi be an eigenstate of Q, so
that Q|qi = q|qi. Then Qϕ|qi = ϕ(Q − 1)|qi = (q − 1)ϕ|qi and hence ϕ
decreases the eigenvalue of |qi by one. Similarly, Qϕ† |qi = ϕ† (Q + 1)|qi =
(q + 1)ϕ† |qi and therefore ϕ† increases the eigenvalue of |qi by one.
(d) Solving for the coefficients ak , a†k , bk , b†k as we did in the real case, we can
then use the ETCR’s to finally show that (after a lot of algebra which
wouldn’t hurt you to work out for yourself)
[ak , a†k′ ] = [bk , b†k′ ] = (2π)2 2ωk δ(k − k′ )
and
[ak , bk′ ] = [ak , b†k ] = 0 .
7
Thus we see that both a†k and b†k create harmonic oscillator quanta.
Show that
Q=
Z
d3 k †
a ak − b†k bk − (2π)3 2ωk δ(0) .
(2π)3 2ωk k
Dropping the infinite constant (2π)3 2ωk δ(0), we see that Q represents the
total number of a quanta minus the total number of b quanta, and hence
Q is essentially just the total charge operator.
8