Solution to Homework Set #1, Problem #2. J. Schindler

Physics 217
Quantum Field Theory I
Fall 2010
Solution to Homework Set #1, Problem #2.
J. Schindler
2. Lorentz Invariance
p
R∞
(a) Show that −∞ dk 0 δ(k 2 − m2 )θ(k 0 ) = 2ω1k , where ωk ≡ ~k 2 + m2 .
(b) Show that the integration measure d4 k is Lorentz invariant.
R d3 k
is Lorentz invariant.
(c) Show that 2ω
k
Solutions
(a) From calculus we have
Z ∞
X g(x0 )
dx δ(f (x))g(x) =
|f 0 (x0 )|
−∞
x
(1)
0
where x0 are the zeros of f (x).
2
Our integral has this form with f (k 0 ) = k 0 − ~k 2 − m2 and f 0 (k 0 ) = 2k 0 ,
and zeros x0 = ±ωk . So to apply the above identity we write
Z ∞
Z ∞
2
0
2
2
0
dk δ(k − m )θ(k ) =
dk 0 δ(k 0 − ~k 2 − m2 )θ(k 0 )
(2)
−∞
−∞
θ(k 0 ) θ(k 0 ) +
(3)
=
2|k 0 | ωk 2|k 0 | −ωk
θ(ωk ) θ(−ωk )
+
2ωk
2ωk
1
=
2ωk
=
(4)
(5)
since θ(ωk ) = 1 and θ(−ωk ) = 0. (b) Under an arbitrary coordinate transformation xi → x0j , we have from
calculus that the measure transforms as
0 ∂xi n
n 0
d x = det
d x
(6)
∂xj h 0i
∂x
where ∂xji is the Jacobian matrix of partial derivatives of the transformation. This property reflects the fact that the volume of a parallelepiped
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Physics 217
Quantum Field Theory I
Fall 2010
relates to the determinant of its defining vectors, and gives the factor r2 sin θ
in integrals in 3d spherical coordinates.
If the coordinate transformation is a Lorentz transformation Λ, we have
0
0
k µ = Λµ ν k ν
and thus
(7)
0
∂k µ
0
= Λµ ν
ν
∂k
and
0
∂k µ
det
∂k ν
(8)
= det Λ.
(9)
But we know by definition that ΛT gΛ = g, which implies
||ΛT gΛ|| = ||ΛT || ||g|| ||Λ|| = ||g|| ||Λ||2 = ||g||
(10)
||Λ|| = ±1
(11)
where ||M || ≡ det(M ), and using ||M T || = ||M ||. Moreover, if Λ is proper
orthochronus, ||Λ|| = 1 (but this is irrelevant due to the absolute magnitude
sign above).
Finally, we can write the answer:
µ0 ∂k 4
4 0
d k = |det Λ| d4 k = d4 k,
(12)
d k = det
∂k ν which demonstrates that d4 k = d4 k 0 is Lorentz invariant. (c) Integrating over all momenta.
(i) Stating the problem:
We want to show that
d3 k
(13)
2ωk
is Lorentz invariant. To do so, we must specify what fourvectors we
are integrating over. In fact, for this integral to be invariant we must
integrate over the fourvectors of some constant magnitude m. Denoting
Z
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Physics 217
Quantum Field Theory I
Fall 2010
the domain of integration by Σ = {k µ | k 2 = m2 , k 0 > 0}, the correct
statement of the problem says
Z 3 0
Z 3
d k 0 µ0
dk
µ
ϕ(k ) =
ϕ (k )
(14)
2ωk
2ωk0
Σ0
Σ
under a Lorentz transformation k 0 = Λk.
The domain Σ specifies at which vectors the integrand ϕ(k µ ) is to be
evaluated at. In the expression above, this means that
Z 3
Z 3
dk
dk
µ
ϕ(k ) =
ϕ(ωk , k i ).
(15)
2ωk
2ωk
Σ
(ii) Proof using parts (a) and (b):
Consider, for any scalar field ϕ(k µ ) = ϕ(k 0 , k i ), the integral
Z
d4 k δ(k 2 − m2 )θ(k 0 )ϕ(k µ ).
(16)
R4
Under the Lorentz transformation k 0 = Λk, we already showed that the
measure d4 k = d4 k 0 is invariant. The scalar field θ(k 0 )ϕ(k µ ) is automatically covariant, although its expression changes under the transformation. And the argument (k 2 − m2 ) of the delta function is obviously
invariant so we will treat δ(k 2 − m2 ) as invariant as well. To write out
the full transformation, we have
Z
k → k 0 = Λk
(17)
Z
0
0
d4 k δ(k 2 − m2 )θ(k 0 )ϕ(k µ ) → d4 k 0 δ(k 02 − m2 )θ(Λ0 µ0 k µ )ϕ0 (k µ ).
R4
R4
(18)
R
But, by evaluating the dk 0 as above, the left-hand side yields
Z
Z 3
dk
4
2
2
0
µ
ϕ(ωk , k i ).
d k δ(k − m )θ(k )ϕ(k ) =
2ωk
R4
R3
3
(19)
Physics 217
Quantum Field Theory I
Meanwhile, the right-hand side reads
Z
0
0
d4 k 0 δ(k 02 − m2 )θ(Λ0 µ0 k µ )ϕ0 (k µ ).
Fall 2010
(20)
R4
0
Now inspect the quantity δ(k 02 − m2 )θ(Λ0 µ0 k µ ). This is zero except on
the hyperbola k 02 = m2 . But this hyperbola is timelike separated from
the origin. Thus for any (orthochronus) Lorentz transformation, the
0
conditions k 02 = m2 and k 0 > 0 imply that k 0 > 0 (drawing it helps to
0
0
see this is true). Therefore, δ(k 02 − m2 )θ(Λ0 µ0 k µ ) = δ(k 02 − m2 )θ(k 0 ).
So the right-hand integral reduces to
Z
Z 3 0
dk 0
0
4 0
02
2
00
0 µ0
d k δ(k − m )θ(k )ϕ (k ) =
ϕ (ωk0 , k i )
(21)
0
2ωk
R4
R3
Combining these results, we have shown that
k → k 0 = Λk
Z 3
Z 3 0
dk
dk 0
0
i
ϕ(ωk , k ) →
ϕ (ωk0 , k i )
0
2ωk
2ωk
R3
(22)
(23)
R3
0
Note that the prime in ϕ0 (k µ ) denotes that the expression of ϕ may
have changed due to the change of labelling of points. In particular,
0
0
ϕ0 (k µ ) = ϕ(k µ (k µ )). This shows that the integral is covariant (as it
must be), and the expression of the integral is invariant if and only
if both ϕ and the limits of integration are invariant. In practice this
means that all boundaries of integration must be infinity, and ϕ = ϕ(k 2 )
only.
(iii) Note: Using the exact same argument, we can easily prove that for any
f (k 2 )
Z
d3 k
∂f (24)
0
f (k2 )=0
∂k
is Lorentz invariant. This implies that the importance of our result is
not just its invariance, but also the fact that the integral runs over Σ,
the set of all possible momenta for a particle of fixed mass.
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Physics 217
Quantum Field Theory I
Fall 2010
(iv) Note: The weaker claim, which omits ϕ(k µ ), is meaningless, since
Z 3
dk
1=∞
(25)
2ωk
R3
and finite limits of integration break the invariance (although not the
covariance).
(v) Note about the equivalent rotational problem:
The equivalent rotational problem would show that
Z
dx dy
p
2
a − x2 − y 2
is invariant under rotations, using the integral
Z
d3 x δ(~x2 − a2 ).
(26)
(27)
R3
However, if you perform this integral, you don’t get the area of the
sphere. In fact, the result doesn’t even have units of area.
This tells us that our invariant integral measure does not have the right
area element for integrating over its hyperbola, so we shouldn’t throw
it around willy-nilly for integrating arbitrary functions. However if for
some other reason we come upon an integral of that form, it’s great,
since we know that it is Lorentz invariant.
(vi) Note: Even though its argument is invariant, the assumption that
δ(k 2 − m2 ) → δ(k 02 − m2 ) needs to be proved to make the preceding proof rigorous. Since δ is a distribution, not a function, the normal
guarantee of covariance doesn’t apply. The simple transformation rule
postulated is not obvious for an arbitrary (non-Lorentz) transformation,
even if norm-preserving. Thus it is not trivial in the Lorentz case.
(vii) Alternate proof by brute force:
Another tactic is to prove this by brute force. Once we know the constraint equation defining Σ, this is possible. We can directly write down
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Physics 217
Quantum Field Theory I
Fall 2010
the 3d coordinate transformation ~k → k~0 . The Lorentz transformation
gives us the system
0
0
i0 = 1, 2, 3
k i = Λi µ k µ ,
and we can use the constraint equation
q
0
k = ~k 2 + m2
(28)
(29)
to eliminate k 0 from the system above, leaving the nonlinear system of
transformation equations
0
0
k i = f i (k i ).
(30)
To find the transformation, we then simply evaluate the Jacobian determinant
i0 ∂f
J = | det(
)|
(31)
∂k i
If
ωk
(32)
J=
ωk 0
then the result is proved. This proof is more satisfying because it relies
on only local quantities, but on the other hand it takes a lot of algebra.
(viii) Note: Counterexample
Note, for example, that the following integrals (as well as all others not
over Σ), which have a similar form to the one above, are NOT Lorentz
invariant:
Z
d3 k
→ Not L.I.
(33)
2ωk
k0 =const
Z
d3 k
→ Not L.I.
(34)
2ωk
k0 =(k1 )2 −k2
This again points to the role of Σ in determining what integral measure
we want. Integrating over arbitrary surfaces, even spacelike, is not good
enough.
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