Physics 5040
Spring 2009
Problem Set 3 Solutions
1. The Lagrangian density is
L =
i~
~2
∇ψ · ∇ψ ∗ + V ψψ ∗ − (ψ ∗ ψ̇ − ψ ψ̇ ∗ )
2m
2
and the Euler-Lagrange equations are
∂L
∂L
= 0.
− ∂µ
∂ψr
∂(∂µ ψr )
In this case, r = 1, 2 refers to ψ and ψ ∗ . Then
∂L
i~
= V ψ ∗ + ψ̇ ∗
∂ψ
2
and
∂µ
∂L
∂(∂µ ψ)
=−
(since ∇2 = ∇ · ∇ =
∂L
+ ∂i
= ∂t
∂(∂i ψ)
∂ ψ̇
i~
~2 i ∗
= ∂t − ψ ∗ + ∂i −
∂ψ
2
2m
∂L
~2 2 ∗
i~ ∗
ψ̇ +
∇ ψ
2
2m
∂i ∂i = −∂i ∂ i ) so the Euler-Lagrange equation yields
~2 2
∂ψ ∗
−
∇ + V ψ ∗ = −i~
.
2m
∂t
P
Similarly,
∂L
i~
= V ψ − ψ̇
∂ψ ∗
2
and
∂µ
∂L
∂(∂µ ψ ∗ )
= ∂t
∂L
∂ ψ̇ ∗
= ∂t
i~
~2 i
ψ + ∂i −
∂ψ
2
2m
=
+ ∂i
~2 2
i~
ψ̇ +
∇ ψ
2
2m
1
∂L
∂(∂i ψ ∗ )
which yields
~2 2
∂ψ
−
∇ + V ψ = i~
.
2m
∂t
2. (a) We have
L =
and hence
1
1
(∂µ ϕ)(∂ µ ϕ) − m2 ϕ2 = g µν (∂µ ϕ)(∂ν ϕ) − m2 ϕ2
2
2
π(x) =
∂L
= ∂0 ϕ(x) = ϕ̇(x) .
∂ ϕ̇(x)
(b) From the definition
1
H = π ϕ̇ − L = ϕ̇2 − [ϕ̇2 − ∇ϕ · ∇ϕ − m2 ϕ2 ]
2
=
1 2
[ϕ̇ + ∇ϕ · ∇ϕ + m2 ϕ2 ]
2
and hence
1
d3 x [ϕ̇2 + ∇ϕ · ∇ϕ + m2 ϕ2 ] .
2
√
(c) We simply do the integral over k0 . Using ωk = + k2 + m2 we have
Z
Z ∞
d3 k
d4 k
2
2
2πδ(k − m )θ(k0 ) =
dk0 δ(k02 − k2 − m2 )
4
3
(2π)
(2π)
k0
0
Z ∞
3
d k
dk0 δ(k02 − ωk2 )
=
(2π)3 0
Z ∞
d3 k
=
dk0 δ[(k0 − ωk )(k0 + ωk )]
(2π)3 0
Z ∞
d3 k
1
=
δ(k0 − ωk )
dk0
(2π)3 0
2ωk
H=
Z
=
d3 k
.
(2π)3 2ωk
(d) We have
f (p) =
Z
d3 q δ(p − q)f (q)
=
Z
d3 q
(2π)3 2ωq δ(p − q)f (q) .
(2π)3 2ωq
But from part (c) we know that d3 q/[(2π)3 2ωq ] is Lorentz invariant, so if
the number f (p) is to be the same in any Lorentz frame (i.e., f (p) = f (p′ )
2
where p and p′ refer to the same momentum as seen in two different
Lorentz frames), it must also be that the quantity (2π)3 2ωq δ(p − q) is a
Lorentz invariant.
(e) Using the definition of fk (x) we simply compute:
Z
←
→
d3 x fk∗ (x)i∂0 fk′ (x)
=i
′
′
d3 x
[eikx ∂0 e−ik x − ∂0 eikx e−ik x ]
(2π)3 [2ωk 2ωk′ ]1/2
Z
′
d3 x
[−iωk′ − iωk ]ei(k−k )x
3
1/2
(2π) [2ωk 2ωk′ ]
Z
ωk′ + ωk i(ωk −ωk′ )t
d3 x −i(k−k′ )·x
=
e
e
1/2
(2π)3
[2ωk 2ωk′ ]
=i
=
Z
ωk′ + ωk i(ωk −ωk′ )t
e
δ(k − k′ )
[2ωk 2ωk′ ]1/2
= δ(k − k′ )
since k = k′ implies that ωk =
√
k2 + m2 = ωk′ .
Similarly,
Z
←
→
d3 x fk∗ (x)i∂0 fk∗′ (x)
=i
Z
′
′
d3 x
[eikx ∂0 eik x − ∂0 eikx eik x ]
1/2
′
]
2ω
k
k
(2π)3 [2ω
′
d3 x
[iωk′ − iωk ]ei(k+k )x
3
1/2
′
(2π) [2ωk 2ωk ]
Z
ωk′ − ωk i(ωk +ωk′ )t
d3 x −i(k+k′ )·x
=−
e
e
1/2
(2π)3
[2ωk 2ωk′ ]
=i
Z
=−
ωk′ − ωk i(ωk +ωk′ )t
e
δ(k + k′ )
[2ωk 2ωk′ ]1/2
=0
where we still have k = −k′ implies that ωk = ωk′ .
In exactly the same manner, we also have
Z
←
→
d3 x fk (x)i∂0 fk′ (x)
3
=i
Z
′
′
d3 x
[e−ikx ∂0 e−ik x − ∂0 e−ikx e−ik x ]
1/2
′
]
2ω
k
k
(2π)3 [2ω
′
d3 x
[−iωk′ + iωk ]e−i(k+k )x
1/2
3
′
]
(2π) [2ωk 2ωk
Z
ωk′ − ωk −i(ωk +ωk′ )t
d3 x i(k+k′ )·x
=
e
e
(2π)3
[2ωk 2ωk′ ]1/2
=i
=
Z
ωk′ − ωk −i(ωk +ωk′ )t
e
δ(k + k′ )
[2ωk 2ωk′ ]1/2
= 0.
(f) Using the expansion
ϕ(x) =
Z
d3 k
[fk (x)ak + fk∗ (x)a†k ]
[(2π)3 2ωk ]1/2
together with the results of part (e) we have
Z
←
→
d3 x fk∗′ (x)i∂0 ϕ(x)
=
Z
d3 k
[(2π)3 2ωk ]1/2
=
Z
d3 k
δ(k′ − k)ak
[(2π)3 2ωk ]1/2
=
Z
←
→
←
→
d3 x [fk∗′ (x)i∂0 fk (x)ak + fk∗′ (x)i∂0 fk∗ (x)a†k ]
ak ′
[(2π)3 2ωk′ ]1/2
so that
ak ′ =
Z
←
→
d3 x [(2π)3 2ωk′ ]1/2 fk∗′ (x)i∂0 ϕ(x) .
Similarly
Z
←
→
d3 x ϕ(x)i∂0 fk′ (x) =
=
Z
d3 k
[(2π)3 2ωk ]1/2
=
Z
d3 k
a† δ(k − k′ )
[(2π)3 2ωk ]1/2 k
=
a†k′
[(2π)3 2ωk′ ]1/2
Z
←
→
←
→
d3 x [ak fk (x)i∂0 fk′ (x) + a†k fk∗ (x)i∂0 fk′ (x)]
4
and therefore also
a†k′ =
Z
←
→
d3 x [(2π)3 2ωk′ ]1/2 ϕ(x)i∂0 fk′ (x) .
(g) Note that ak and a†k are independent of time, and hence both may be
evaluated at the same time. Note also that the symbol ∂0 refers to two
different things in the commutator below—in one term it is really ∂/∂x0
and in the other term it is ∂/∂y 0 . Then using π(x) = ϕ̇(x) in the ETCR’s
together with the results of part (e) we have
Z
←
→
←
→
[ak , a†k′ ] = d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2 [fk∗ (x)i∂0 ϕ(x), ϕ(y)i∂0 fk′ (y)]
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
=
Z
d3 x (2π)3 (2ωk 2ωk′ )1/2 {fk∗ (x)i∂0 fk′ (x) − i∂0 fk∗ (x)fk′ (x)}
=
Z
←
→
d3 x (2π)3 (2ωk 2ωk′ )1/2 {fk∗ (x)i∂0 fk′ (x)}
=
=
=
[fk∗ (x)iϕ̇(x) − i∂0 fk∗ (x)ϕ(x), ϕ(y)i∂0 fk′ (y) − iϕ̇(y)fk′ (y)]
fk∗ (x)i∂0 fk′ (y)[iϕ̇(x), ϕ(y)] + i∂0 fk∗ (x)fk′ (y)[ϕ(x), iϕ̇(y)]
fk∗ (x)i∂0 fk′ (y)i(−i)δ(x − y) + i∂0 fk∗ (x)fk′ (y)i(i)δ(x − y)
= (2π)3 (2ωk 2ωk′ )1/2 δ(k − k′ )
= (2π)3 2ωk δ(k − k′ ) .
And similarly we evaluate
Z
←
→
←
→
[ak , ak′ ] = d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2 [fk∗ (x)i∂0 ϕ(x), fk∗′ (y)i∂0 ϕ(y)]
=
=
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
[fk∗ (x)iϕ̇(x) − i∂0 fk∗ (x)ϕ(x), fk∗′ (y)iϕ̇(y) − i∂0 fk∗′ (y)ϕ(y)]
fk∗ (x)(−i)∂0 fk∗′ (y)[iϕ̇(x), ϕ(y)] − i∂0 fk∗ (x)fk∗′ (y)[ϕ(x), iϕ̇(y)]
5
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
=
Z
d3 x (2π)3 (2ωk 2ωk′ )1/2 {−fk∗ (x)i∂0 fk∗′ (x) + i∂0 fk∗ (x)fk∗′ (x)}
=
Z
←
→
d3 x (2π)3 (2ωk 2ωk′ )1/2 {−fk∗ (x)i∂0 fk∗′ (x)}
=
fk∗ (x)(−i)∂0 fk∗′ (y)i(−i)δ(x − y) − i∂0 fk∗ (x)fk∗′ (y)i(i)δ(x − y)
=0
and obviously also
Z
←
→
←
→
[a†k , a†k′ ] = d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2 [ϕ(x)i∂0 fk (x), ϕ(y)i∂0 fk′ (y)]
=
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
=
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
=
Z
d3 x d3 y (2π)3 (2ωk 2ωk′ )1/2
=
Z
d3 x (2π)3 (2ωk 2ωk′ )1/2 {i∂0 fk (x)fk′ (x) − fk (x)i∂0 fk′ (x)}
=
Z
←
→
d3 x (2π)3 (2ωk 2ωk′ )1/2 {−fk (x)i∂0 fk′ (x)}
[ϕ(x)i∂0 fk (x) − iϕ̇(x)fk (x), ϕ(y)i∂0 fk′ (y) − iϕ̇(y)fk′ (y)]
i∂0 fk (x)fk′ (y)[ϕ(x), −iϕ̇(y)] − fk (x)i∂0 fk′ (y)[iϕ̇(x), ϕ(y)]
i∂0 fk (x)fk′ (y)(−i)iδ(x − y) − fk (x)i∂0 fk′ (y)i(−i)δ(x − y)
= 0.
(h) Another straight-forward computation now using part (g):
Z
′
d3 k
d3 k ′
+
−
[ak , a†k′ ]e−ikx eik y
[ϕ (x), ϕ (y)] =
(2π)3 2ωk (2π)3 2ωk′
Z
′
d3 k
d3 k ′
=
(2π)3 2ωk δ(k − k′ )e−ikx eik y
3
3
′
(2π) 2ωk (2π) 2ωk
Z
d3 k
e−ik(x−y)
=
(2π)3 2ωk
= i∆+ (x − y) .
6
3. (a) By direct calculation we have, for example,
0 1
0 −i
1
0
σ1 σ2 =
=i
= iσ3 = iε123 σ3 .
1 0
i
0
0 −1
And in general we find that (for i 6= j)
σi σj = iεijk σk
which also implies that for i 6= j we have σj σi = −σi σj . Therefore, for
i 6= j it follows that
σi σj + σj σi = 0 .
Also by direct calculation we can verify that
σi 2 = I .
Combining these last two results we have
σi σj + σj σi = 2Iδij .
For example,
−i
0
+
0
i
i
0
−i 0
=
+
0 −i
0 i
σ1 σ2 + σ2 σ1 =
0 1
1 0
0
i
−i
0
0
1
1
0
= 0.
Then adding the equations
σi σj − σj σi = 2iεijk σk
and
σi σj + σj σi = 2Iδij
it is easy to see that
2σi σj = 2Iδij + 2iεijk σk
or simply
σi σj = Iδij + iεijk σk .
(b) Using the last result from part (a) we have (where there is no difference
between upper and lower indices since we are in R3 )
(a · σ)(b · σ) = ai σi bj σj = ai bj (Iδij + iεijk σk )
= ai bi I + iai bj εijk σk = (a · b)I + i(a × b)k σk
= (a · b)I + i(a × b) · σ .
7
(c) Expanding the exponential we have
2
3
θ·σ
1
θ·σ
θ·σ
1
+ (−i)3
+ ···
+ (−i)2
2
2!
2
3!
2
2
3
4
1 θ·σ
i θ·σ
1 θ·σ
θ·σ
−
+
+
+ ··· .
=I −i
2
2!
2
3!
2
4!
2
e−iθ·σ/2 = I + (−i)
From part (b) we see that
(â · σ)2 = (â · â)I = I
(â · σ)3 = â · σ
(â · σ)4 = I .
Therefore, letting a · σ = θ · σ/2 = (θ/2)(θ̂ · σ) we obtain
e
−iθ·σ/2
2
3
4
θ
1 θ
1 θ
1 θ
= I − i(θ̂ · σ) −
I + i(θ̂ · σ)
+
I + ···
2 2! 2
3! 2
4! 2
2
4
3
1 θ
1 θ
1 θ
θ
=I 1−
+
+ · · · − i(θ̂ · σ) −
+ ···
2! 2
4! 2
2 3! 2
= I cos
θ
θ
− i(θ̂ · σ) sin .
2
2
Using the explicit form of each σi we have
θ̂ · σ = θ̂1 σ1 + θ̂2 σ2 + θ̂3 σ3
# "
"
# "
0 −iθ̂2
0 θ̂1
θ̂3
+
=
+
θ̂1 0
iθ̂2
0
0
#
"
θ̂1 − iθ̂2
θ̂3
=
θ̂1 + iθ̂2
−θ̂3
#
"
θ̂−
θ̂3
=
θ̂+ −θ̂3
0
−θ̂3
#
and therefore
e−iθ·σ/2 = I cos
=
"
θ
θ
− i(θ̂ · σ) 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
8
#
.
4. (a) We have the Lagrangian density
L = ∂µ ϕ† ∂ µ ϕ − m2 ϕ† ϕ = g µν ∂µ ϕ† ∂ν ϕ − m2 ϕ† ϕ
and the Euler-Lagrange equations
∂L
∂L
=0
− ∂µ
∂ϕr
∂(∂µ ϕr )
where r = 1, 2 corresponds to ϕ and ϕ† . Then
∂L
= −m2 ϕ†
∂ϕ
and
∂L
= ∂ µ ϕ†
∂(∂µ ϕ)
so that the equation of motion for ϕ† is
(∂µ ∂ µ + m2 )ϕ† = 0 .
Similarly
∂L
∂L
= −m2 ϕ
and
= ∂ µϕ
∂ϕ†
∂(∂µ ϕ† )
and hence the equation of motion for ϕ is
(∂µ ∂ µ + m2 )ϕ = 0 .
Therefore the above Lagrangian describes two distinct free Klein-Gordon
particles, each of mass m. Since the canonical momentum is defined by
πr (x) =
∂L
∂ ϕ̇r (x)
we easily have
π(x) =
∂L
= ϕ̇† (x)
∂ ϕ̇
π † (x) =
∂L
= ϕ̇(x) .
∂ ϕ˙†
(b) We have
ϕ′ (x) = e−iΛ ϕ(x) ≈ ϕ(x) − iΛϕ(x)
ϕ′† (x) = eiΛ ϕ† (x) ≈ ϕ† (x) + iΛϕ† (x)
and hence
δϕ(x) = −iΛϕ(x)
and
δϕ† (x) = iΛϕ† (x) .
Under this global phase change we have δxα = 0 so that
jµ =
∂L
∂L
∆ϕr − T µ α δxα =
δϕr
∂ϕr,µ
∂ϕr,µ
= (∂ µ ϕ† )δϕ + (∂ µ ϕ)δϕ† = −iΛ[(∂ µ ϕ† )ϕ − (∂ µ ϕ)ϕ† ]
←
→
= iΛ(ϕ† ∂ µ ϕ) .
9
In particular,
j 0 = iΛ(ϕ† ϕ̇ − ϕ̇† ϕ)
so we may take the conserved charge to be
Z
Q = i d3 x (ϕ† ϕ̇ − ϕ̇† ϕ) .
Note that this is independent of time.
(c) We compute using [ab, c] = a[b, c] + [a, c]b and the ETCR’s:
Z
[Q, ϕ(x)] = i d3 y [ϕ† (y)ϕ̇(y), ϕ(x)] − [ϕ̇† (y)ϕ(y), ϕ(x)]
= −i
Z
d3 y [ϕ̇† (y), ϕ(x)]ϕ(y)
= −i
Z
d3 y (−i)δ(x − y)ϕ(y)
= −ϕ(x) .
Similarly
[Q, ϕ (x)] = i
Z
=i
Z
d3 y ϕ† (y)[ϕ̇(y), ϕ† (x)]
=i
Z
d3 y ϕ† (y)(−i)δ(x − y)
†
d3 y [ϕ† (y)ϕ̇(y), ϕ† (x)] − [ϕ̇† (y)ϕ(y), ϕ† (x)]
= ϕ† (x) .
(d) Use the expression for Q derived in part (b) and the Fourier expansions of
ϕ and ϕ† given in the statement of the problem. Breaking up the integral
for Q into parts we first have
Z
Z
d3 x d3 k d3 k ′ d3 x ϕ† ϕ̇ =
bk fk (x) + a†k fk∗ (x)
1/2
3
(2π) [2ωk 2ωk′ ]
× (iωk′ ) − ak′ fk′ (x) + b†k′ fk∗′ (x)
Z
d3 x d3 k d3 k ′
=
(iωk′ ) − bk ak′ fk fk′ + bk b†k′ fk fk∗′
1/2
3
(2π) [2ωk 2ωk′ ]
− a†k ak′ fk∗ fk′ + a†k b†k′ fk∗ fk∗′ .
10
Noting that all the x-dependence in this equation lies in the fk terms,
consider one of them:
Z
Z
Z
′
−i(ωk +ωk′ )t i(k+k′ )·x
e−i(k+k )x
e
3
3
3 e
d x fk fk ′ = d x
= d x
3
1/2
3
′
(2π) [2ωk 2ωk ]
(2π) [2ωk 2ωk′ ]1/2
=
e−2iωk t
e−i(ωk +ωk′ )t
δ(k + k′ ) .
δ(k + k′ ) =
1/2
2ωk
[2ωk 2ωk′ ]
Similarly, it is easy to see that
Z
e2iωk t
δ(k + k′ )
d3 x fk∗ fk∗′ =
2ωk
and
Z
3
d
x fk fk∗′
=
Z
d3 x fk∗ fk′ =
1
δ(k − k′ ) .
2ωk
Using these results we now have
Z
Z
iωk′
d3 k d3 k ′
d3 x ϕ† ϕ̇ =
1/2
3
2ωk
(2π) [2ωk 2ωk′ ]
−2iωk t
δ(k − k′ ) + bk b†k′ δ(k − k′ )
× − b k ak ′ e
=
Z
− a†k ak′ δ(k − k′ ) + a†k b†k′ e2iωk t δ(k + k′ )
i
d3 k
− bk ak e−2iωk t + bk b†k − a†k ak + a†k b†k e2iωk t
3
(2π) 2ωk 2
Now for the second part of Q:
Z
Z
d3 x d3 k d3 k ′
d3 x ϕ̇† ϕ =
(iωk ) − bk fk (x) + a†k fk∗ (x)
3
1/2
(2π) [2ωk 2ωk′ ]
× ak′ fk′ (x) + b†k′ fk∗′ (x)
Z
d3 x d3 k d3 k ′
=
(iωk ) − bk ak′ fk fk′ − bk b†k′ fk fk∗′
3
1/2
(2π) [2ωk 2ωk′ ]
+ a†k ak′ fk∗ fk′ + a†k b†k′ fk∗ fk∗′
=
Z
Finally we have
i
d3 k
− bk ak e−2iωk t − bk b†k + a†k ak + a†k b†k e2iωk t .
(2π)3 2ωk 2
Q=i
Z
d3 x (ϕ† ϕ̇ − ϕ̇† ϕ)
11
=i
=
Z
Z
d3 k
i
2bk b†k − 2a†k ak
3
(2π) 2ωk 2
d3 k †
a ak − b†k bk − (2π)3 2ωk δ(0) .
(2π)3 2ωk k
12