Assignment I
Solutions
• 1.1 (a) It is possible to understand the strong force between a neutron and a proton as being effectively mediated by a boson called the pion, whose mass is 139 MeV. Use this to estimate the range of the strong force.
(b) Similarly the weak force is mediated by theW-boson, whose mass is about 80 GeV. What range do you estimate this force has?
Solution
(a) If the mediating particle is massive, then it cannot have any energy less than its rest mass, and so by the uncertainty principle, the time for exchange of the virtual particle is
t' ¯h
mc2
and so its range will be
r≈ct' ¯h
mc
If the mediating particle is the pion, then
r' ¯h
mπc = 197
139fm'10
−13cm
which is about the size of a nucleus.
(b) Applying these results for the weak interactions, we have
r' ¯h
mWc
= 197
80,000fm'2.5×10
−16 cm
• 1.3 An experimentalist wants to probe distances ofd≤10−20cm. How much collision energy must the machine be able to produce? How does this compare with the maximum energy of the LHC? If the size of the machine scales with the energy, how large would this machine have to be?
Solution The momentum needed must not be smaller than that given by the uncertainty principle, which is
p≥ ¯h
d =
197 MeV fm
10−20 cm '200×10
7 MeV = 2000 TeV
where d∼10−20 cm = 10−7 fm. By comparison, the energy of the LHC is about 7 TeV, which is between 1 and 2 orders of magnitude smaller. The size of this machine would have to be 2000/7 times the size of the LHC (which is 27 km), or about 7,700 km in circumference.
• 2.2 Suppose we write the velocityv of a particle asv=ctanhη, where η is a parameter called the rapidity.
(b) Consider a succession of two boosts, both in thexdirection, with velocitiesv1 andv2. What is the value of the rapidity parameter for the combination of these transformations in terms of the rapidity parametersη1
andη2 for each?
Solution (a) Sincev=ctanhη we haveβ =v/c= tanhη and
γ= p 1
1−v2/c2 =
1
p
1−tanh2η
= coshη
Hence the Lorentz transformation is
Λµν =
γ −βγ 0 0
−βγ γ 0 0
0 0 1 0
0 0 0 1
=
coshη −sinhη 0 0 −sinhη coshη 0 0
0 0 1 0
0 0 0 1
(b) For two boosts we havev1=ctanhη1andv2=ctanhη2, yielding
Λµ1ν =
coshη1 −sinhη1 0 0 −sinhη1 coshη1 0 0
0 0 1 0
0 0 0 1
Λµ2ν=
coshη2 −sinhη2 0 0 −sinhη2 coshη2 0 0
0 0 1 0
0 0 0 1
so the combined boost is Λµ3ν = Λµ1αΛα
2ν which gives
Λµ3ν =
coshη1 −sinhη1 0 0 −sinhη1 coshη1 0 0
0 0 1 0
0 0 0 1
coshη2 −sinhη2 0 0 −sinhη2 coshη2 0 0
0 0 1 0
0 0 0 1
=
coshη1coshη2+ sinhη1sinhη2 −sinhη1coshη2−sinhη2coshη1 0 0 −sinhη1coshη2−sinhη2coshη1 coshη1coshη2+ sinhη1sinhη2 0 0
0 0 1 0
0 0 0 1
=
cosh(η1+η2) −sinh(η1+η2) 0 0
−sinh(η1+η2) cosh(η1+η2) 0 0
0 0 1 0
0 0 0 1
and so the new rapidity parameter isη3= (η1+η2). Note that this means that
v3=ctanhη3=ctanh(η1+η2) =c
tanhη1+ tanhη2
1−tanhη1tanhη2 =
(v1+v2)
1−v1v2/c2
which is the velocity addition formula.
(a) Find an expression for the angle that the lepton is emitted relative to the original direction of motion.
(b) Suppose a pion of speedv emits a muon at angleθ, and another pion at speedv0 emits an electron at the same angle, each having emitted the antineutrinos at right angles to the direction of motion. How much larger or smaller isv compared tov0?
SolutionFor simplicity, let’s setc= 1 for this problem.
(a) Momentum conservation sets pµ π =p
µ
¯
ν` +p
µ
m`. Isolating the lepton 4-momentum in terms of the pion and
antineutrino 4-momenta yields
m2` =m2π+mν¯2`−2EπE¯ν`
after squaring both sides since the spatial momenta of the pion and neutrino are orthogonal. Looking at the 0th and spatial components of the momenta, we find that
Eπ =E¯ν` +Em` pπ=pm`cosθ
where the second of these equations also follows because the spatial momenta of the pion and neutrino are orthogonal. Inserting both of these into the first equation gives
m2` =m2π+m2ν−2Eπ Eπ−
r
p2
π cos2θ+m
2
`
!
and solving for cosθyields, usingEπ =γmπ=mπ/ √
1−v2 andp2
π=E
2
π−m
2
π,
cosθ=r 2v
(1−v2)21−m2`−m2ν`¯
m2
π
2
+ 4v2−(1−v2)m2ν`¯
m2
π
This clearly vanishes for smallv, indicating that the antineutrino and the lepton must come out back-to-back (i.e. θ = π2) when the pion is at rest, and approaches unity as v → 1, indicating that the lepton must be emitted almost collinear with the pion for high speeds.
(b) In this case we must equation the two expressions for cosθ, at differentv’s andm`’s. This gives
2v
s
(1−v2)2
1−m2µ−m2νµ¯
m2
π
2
+ 4
v2−(1−v2)m
2 ¯
νµ
m2
π
= 2v
0
r
(1−v02)21−m2e−m2¯νe
m2
π
2
+ 4v02−(1−v02)m2νe¯
m2
π
which upon squaring becomes a quadratic equation inv2 (or alternativelyv02). Since antineutrino masses are small, they can be neglected, in which case the solution is
v2 = 1
2v02(M2−m2
e)4
(M2−m2µ)2(1 +v04) + 2(m2µ−m2e)(2M2−m2µ−m2e)v02
−(1−v02)(M2−m2µ)
q
(M2−m2
µ)2(1 +v04) + 2v02(M4−4m2eM2+ 2m2µM2−m4µ+ 2m4e)
• 2.11 ParticleA collides into particleB, which is at rest, and three or more particles are produced as a result:
A+B −→C1+C2+· · ·+CN.
Find the threshold energy for this reaction to take place in terms of the rest masses of the particles.
SolutionMomentum conservation gives
pµA+pµB = N
X
I=1 pµI
⇒ ~pA=
PN
I=1~pI
EA+mB=P N I=1EI
)
in the rest frame ofB
from which it is tempting to infer that A must have at least as much energy as the sum of the rest masses of the final state particles minus the rest mass of B. However this isn’t right because we must also conserve momentum, meaning that at least one particle in the final state must have an energy greater than its rest energy. The problem becomes that of how this momentum is distributed over the final state particles; we want this done in such a way thatEA is minimized as a function of these momenta.
Consider first 2 particles in the final state. the above equations become
pA = p1+p2
~
p⊥1 = −~p⊥2
EA+mB =
q
p2
1+|~p⊥1|2+m21+
q
p2
2+|~p⊥2|2+m22
where~p⊥I refers to the components of momenta transverse to the direction of A andpI refers to the component of momenta that is parallel/antiparallel to A. It is clear from the above that we must set~p⊥I = 0 to minimize
EA because the last equation is an increasing function of the |~p⊥I|. Inserting this into the above equations yields
EA=
q
p2
1+m21+
r
(
q
E2
A−m
2
A−p1)2+m22−mB
for the threshold energy. Note thatEAis now fully determined in terms of the rest masses of the particles and the final momentump1.
For more than two particles in the final state the logic is similar. The equations become
pA = N
X
I=1 pI
N
X
I=1 ~
p⊥I = 0
EA+mB = N
X
J=1
q
p2
Since all of the transverse momenta add to zero, and since the last equation indicates thatEAis an increasing function of these momenta, we must set~p⊥J = 0 for every J. Now if we combine the first equation with the last we get
EA= N−1
X
J=1
q
p2
J+m2J+
v u u t(p−
N−1
X
J=1
pJ)2+m2N −mB
upon eliminating pN−1 in terms of the other momenta, whereEA2 =p2A+m2A =p2+m2A, settingpA=pfor convenience.
We can regardEA (and thereforep) as a function of (p1, p2, . . . , pN−1). The threshold energy must be greater
than the minimum ofEA. We can find this minimum by setting ∂EA
∂pJ = 0 (or alternatively
∂p
∂pJ = 0) and solving
for thepJ. Taking the pJ-derivative of both sides of the above equation gives
0 = pJ
EJ +p−
PN−1
J=1 pJ
EN
= pJ
EJ − pN
EN
where EJ =
p
p2
J+m
2
J forJ = 1, . . . , N, and the last equality follows from conservation of the momenta in the direction ofp(pA=p=PNI=1pI). We see that all the EpJ
J are equal to each other. Furthermore, since the
transverse momenta have been set to zero, we havepJ =γJmJvJ and EJ =γJmJ where γJ = 1/
p
1−v2
J. Hence pJ
EJ = vJ and we see that all the final velocities must be equal for EA to be minimized. Momentum
conservation gives
p= N
X
J=1
γJmJvJ =vγ N
X
J=1 mJ
withγ= 1/√1−v2. Solving for vyields
v= q p
p2+ (PN
J=1mJ)2 ⇒γ=
q
p2+ (PN
J=1mJ)2
PN
J=1mJ
Finally energy conservation gives
EA= N
X
J=1
EJ−mB =γ N
X
J=1
mJ−mB =
v u u tp2+ (
N
X
J=1
mJ)2−m
B =
v u u tEA2 + (
N
X
J=1
mJ)2−m2A−mB
and solving forEA yields
EA= (PN
J=1mJ)2−m2A−m2B 2mB
for the threshold energy.
Here’s another way to do it that is more elegant, and makes use of Lorentz transformations. Since the square of any 4-momentum is the same in any reference frame, we have
(pA+pB)2= (p0A+p
0
B)
2
where the prime refers to the CM system and unprimed quantities are in the lab system. This equation becomes
m2A+m2B+ 2EAmB= (EA0 +E
0
B)
The right-hand side is the total energy of the system in the lab frame. Energy conservation implies that
EA0 +EB0 = N
X
J=1 EJ0 =
N
X
J=1
q
|~p0
J|2+m
2
J ≥ N
X
J=1 mJ
where the last inequality follows because it is possible for all spatial momenta to be zero in the CMs and still satisfy momentum conservation. Hence
m2A+m2B+ 2EAmB≥( N
X
J=1 mJ)2
or
EA= (PN
J=1mJ) 2−m2
A−m
2
B 2mB
for the threshold energy.
• 3.4 An orthogonal matrix is defined by the relationRTR= 1, whereR is an N×N matrix, where ‘T’ refers to the transpose, RT
ij=Rji
(a) Show that the set of all orthogonalN×N matrices forms a group.
(b) Show that the set of all orthogonalN×N matrices of determinant 1 forms a group.
Solution
Multiplication of matrices is always associative, so property (4) is automatically satisfied. We therefore need to check only the first three properties in each case.
(a) To check closure consider the productR1R2, whereRT1R1= 1 andR2TR2= 1. Closure implies thatR1R2
must be orthogonal, ie . (R1R2)T(R1R2) = 1. Let’s check this:
(R1R2)T(R1R2) = RT2RT1
(R1R2) =RT2 RT1R1
R2=RT2R2= 1
where the 3rd equality follows from matrix associativity. It is clear that an identity exists, and it is also clear that the inverse of any element is its transpose, so all four properties of a group are satisfied.
(b) Since the above relation, along with inverse and identity, also hold for matrices with unit determinant, the only property we need to check is closure insofar as we must ensure not only that (R1R2)T(R1R2) = 1 but also that det (R1R2) = 1 whenever det (R1) = 1 and det (R2) = 1 . This is easy to check:
det (R1R2) = det (R1) det (R2) = 1×1 = 1
Note that the set of orthogonal matrices with negative unit determinant does NOT form a group, because if det (R1) =−1 and det (R2) =−1 then
• 3.6 A symplectic matrix is defined by the relation STκS =κ, where S is an 2N×2N matrix, where κis a matrix of the form
κ=
0 IN −IN 0
whereIN is anN×N identity matrix. Show that the set of all symplecticN×N matrices forms a group.
Solution
Once again we note that multiplication of matrices is always associative, so property (4) is automatically satisfied. We therefore need to check only the first three properties in each case. To check closure consider the product S1S2, where S1TκS1 =κ and ST2κS2 = κ. Closure implies that S1S2 must be symplectic, ie .
(S1S2)
T
κ(S1S2) =κ. Let’s check this:
(S1S2)
T
κ(S1S2) = S2TS
T
1
κ(S1S2) =S2T S
T
1κS1
S2=S2TκS2=κ
where the 3rd equality follows from matrix associativity. It is clear that an identity exists since ITκI =κ and the product of any matrix with the 2N×2N matrix I is itself. The inverse is a bit more subtle. Since
κ2=−I, we have
−κSTκS=I
which shows thatS−1=−κSTκ. Hence all four properties of a group are satisfied.
• 3.8 (a) Find the general form of a Lorentz transformation for an arbitrary velocity~vas shown in the diagram below.
(b) From your answer in part (a), work out the general form for velocity addition for two velocities~v and~u.
(c) Under the velocity addition formula in part (b), does the set of velocities form a group? Why or why not?
(a) The coordinate associated with the axis parallel to the direction of motion is the only one that undergoes a non-trivial transformation. Hence
~
x0k=γ(~xk−βct~ ) ~
x0⊥ =~x⊥
t0 =γ(t−β~•~x c )
whereβ~ =~v/c γ=q 1
1−|~β|2
~
x=~x⊥+β~•~x |β~|2
~
β=~x⊥+~xk
and so if we add the two equations for the components of~xwe get~x0=x~0⊥+~x0k which yields
~
x0 = ~x+ (γ−1)
~ β•~x
~ β
2 β~−γ ~βct
t0 = γ(t−β~•~x
c )
(b) The velocity in the primed frame is
~u0= ∆~x
0
∆t0 =
∆~x+ (γ−1)β~•∆~x |~β|2
~
β−γ ~βc∆t
γ(∆t−β~•∆~x c )
=
~
u+ (γ−1)~β•~u |β~|2
~ β−γ ~βc
γ(1−~β•~u c )
or, taking~β=~v/c,
~ u0=
~
u−γ~v+ (γ−1)~v•~u
|~v|2~v
γ(1−~v•~u c2 )
Now set~v→ −~v so that we are adding velocities in the unprimed frame:
~ u⊕~v=
~u+γ~v+ (γ−1)~v•~u
|~v|2~v
γ(1 +~vc•2~u)
We can simplify this using|~v|2=c2 1−γ−2; this gives
~ u⊕~v=
~u+γ~v1 + γγ+1~vc•2~u
γ(1 +~vc•2~u)
(A) Closure – we must have|~u|< cfor all velocities:
|~u⊕~v|2 =
~
u+γ~v1 + γγ+1~vc•2~u
γ(1 +~vc•2~u)
2 =
|~u|2+ 2γ~v•~u1 + γγ+1~vc•2~u
+γ2|~v|21 + γγ+1~vc•2~u
2
γ2(1 +~v•~u c2 )2
=
|~u|2+ 2γ~v•u~1 + γγ+1~v•~u c2
+ γ2−1
c21 + 2γ γ+1
~ v•~u
c2 +
γ2
(γ+1)2
~ v•~u
c2
2
γ2(1 +~v•~u c2 )2
= |~u|
2
−c2+c2γ2 1 +~v•~u c2
2
γ2(1 +~v•~u c2 )2
= c2 1−(c
2− |~u|2
)(c2− |~v|2
) (c2+~v•~u)2
!
< c2 So closure is OK
(B) Check identity and inverse
~
u⊕0 = ~u So identity is OK
~u⊕(−~u) =
~u−γu~u
1− γu
γu+1 |~u|2
c2
γu(1−|~uc2|2)
= γu~u
1−γu
1− γu
γu+ 1
γ2
u−1
γ2
u
= 0 So inverse is OK
(C) Check associativity, which means that~u⊕(~v⊕w~) = (~u⊕~v)⊕w~
~
u⊕(~v⊕w~) = ~u⊕
~v+γww~
1 + γw
γw+1
~ v•w~
c2
γw(1 +~vc•2w~)
=
~ u+γw~v
1 + γw
γw+1
~v•~u c2
γvγw(1 +~vc•2~u)(1 +
~ v•w~
c2 )
+
~ u+γww~
1 + γw
γw+1
~ u•w~
c2
γw(1 + ~uc•2w~)
1 + γw
γw+1
~ v•w~
c2
(1 +~v•w~ c2 )
However
(~u⊕~v)⊕w~ =
~ u+γv~v
1 + γv
γv+1
~ v•~u
c2
γv(1 +~vc•2~u)
⊕w~
=
~ u+γww~
1 + γw
γw+1
~ u•w~
c2
γvγw(1 +~vc•2~u)(1 +
~ u•w~
c2 )
+
~v+γww~
1 + γw
γw+1
~v•w~ c2
γw(1 +~vc•2w~)
1 + γv
γv+1
~ v•~u
c2
(1 +~vc•2~u)
6= ~u⊕(~v⊕w~) So the ⊕ operation is NOT associative!
• 3.9 For matricesAandB, show
exp [−A]BexpA= exp(−adA)B where adAB≡[A,B]
Solution Use induction. Suppose
adNAB= N X r=0 N r
(−1)rAN−rBAr
Then
adNA+1B = N X r=0 N r
(−1)rad
A AN−rBAr
= N X r=0 N r
(−1)rAN−r(AB−BA)Ar
= N X r=0 N r
(−1)r AN+1−rBAr
−AN−rBA1+r
= AN+1B+ N+1 X r=1 (−1)r N r
AN+1−rBAr−
N r−1
(−1)AN+1−rBAr
= AN+1B+ N+1
X
r=1
N+ 1
r
(−1)rAN+1−rBAr= N+1
X
r=0
N+ 1
r
(−1)rAN+1−rBAr
Next consider the function
f(`) = (exp [−`A]Bexp [`A])
We have
f(`) =
∞
X
N=0 `N
N!
dNf
d`N
`=0
where
dNf
d`N `=0 = " d d` N
(exp [−`A]Bexp [`A])
# `=0 = N X r=0 N r
dexp [−`A]
d`
N−r B
dexp [`A]
d` r! `=0 = N X r=0 N r
(−1)N−rAN−rBAr
=adN(−A)B
with the last line following from the equations above. Hence
exp [−A]Bexp [A] =f(1) =
∞
X
N=0
1
N!
dNf
d`N `=0 = ∞ X N=0 1
N!ad N
• 4.1 Suppose an operator F has real expectation values, ie. hψ|F|ψi is real for any wavefunctionψ. Show thatF is Hermitian.
Solution
hψ|F|ψi=hψ|F|ψi∗=
ψF† ψ
Since the above relation holds for any wavefunctionψ, then F=F†.
• 4.2 Consider the following action in classical mechanics
S=
Z
dt
"
1 2m
d~x dt
2
−V(~x)
#
(a) Suppose we want to make a transformation that rescales the coordinates by a constant factor of σ, i.e.
~
x0=σ ~x. How must the time rescale in order that the action remains invariant if the potentialV(~x) = 0?
(b) Under what circumstances is the action invariant under this transformation ifV(~x)6= 0? Find the general form of the potential.
(c) Find the Noether current associated with this transformation and show that it is conserved when the equations of motion are satisfied.
Solution
(a) Note that
~
x0 =σ ~x⇒~x0(t0) =σ ~x(t)
and so
S[~x0] =
Z
dt0
"
1 2m
d~x0(t0)
dt0
2#
=
Z
dtdt 0
dt
"
1 2m
σd~x(t)
dt dt dt0
2#
=
Z
dt
"
1 2m
d~x(t)
dt
2 dt dt0σ
2
#
which means that we must havet=σ−2t0, or
(b) If the potential is nonzero, we obtain
S[~x0] =
Z
dt0
"
1 2m
d~x0(t0)
dt0
2
−V(~x0(t0))
#
=
Z
σ2dt
"
1 2m
σd~x(t)
dt dt dt0
2
−V(σ~x(t))
#
=
Z
σ2dt
"
1 2m
σd~x(t)
dt
1
σ2
2
−V(σ~x)
#
=
Z
dt
"
1 2m
d~x dt
2
−σ2V(σ~x)
#
and so the action is invariant provided
σ2V(σ~x) =V(~x) (1) Note that this also implies that
V(~x) = V0 |~x|2 whereV0 is a constant.
(c) Sinceσ= 1 is the identity transformation, we have
d~x0(t) dσ
σ=1
= d
dσ
σ ~x(t/σ2)
σ=1
= ~x(t/σ2)−σ
2t
σ3
d~x t/σ2
d(t/σ2)
σ=1
= ~x(t)−2td~x dt dt0
dσ
σ=1
= 2t
and so we must findG such thatδσL = dtdG. First, note that we can expand the relationσ
2V(σ~x) =V(~x)
aboutσ= 1 :
This implies
δσL =
d dσ
1 2m
d~x0 dt
2
−V(~x0)
! σ=1
= md~x dt ·
d dt
d~x0 dσ
σ=1
− d~x
0
dσ
σ=1
·∇~V
= md~x dt ·
d dt
~x(t)−2td~x dt
−
~
x(t)−2td~x dt
·∇~V
= −m
d~x dt
2
+ 2V −2td~x dt ·
md 2~x dt2 −∇~V
= −2L−2tdL dt
= −2d
dt(tL)
and so we see thatG=−2tL. Hence
J =
~ p·d~x
0(t)
dα −G
σ=1
=~p·
~
x−2td~x dt
+ 2tL
= 2t
L−~p·d~x
dt
+~p·~x
= −2tH+~p·~x
whereH is the Hamiltonian. To check thatJis conserved we note that the equations of motion are
d~p dt =−
~
∇V
and so we compute
dJ
dt = −2H−2t dH
dt + d~p
dt ·~x+p~· d~x
dt
= −2H−~x·∇~V +~p·d~x
dt
using the eqs. of motion
= −2
~p
·p~
2m +V
−~x·∇~V +p~· ~p
m
= −(2V +~x·∇~V)
= 0
where dHdt = 0, which can be shown directly.
• 4.5 Show that the operatorsP˜·P˜ andP˜·L˜commute with all elements of the algebra in question #4.4.
Solution From #4.4 we have
and so
h
Pi,P˜·P˜
i
= Pk[Pi, Pk] + [Pi, Pk]Pk = 0
h
Li,P˜·P˜
i
= Pk[Li, Pk] + [Li, Pk]Pk =−i¯hPkiklPl−i¯hiklPlPk=−ih¯(ikl+ilk)PkPl= 0
h
Pi,P˜·˜L
i
= Pk[Pi, Lk] + [Pi, Pk]Lk =i¯hPkiklPl= 0
h
Li,P˜·˜L
i
