Particle physics course of the University of Oxford (Physics Part C major option, C4)

An incoming fermion contributes a Dirac spinor

u_{s} (p)

and an outgoing fermion contributes its Dirac adjoint

\overset{u}{ˉ}_{s} (p) = u_{s}^{†} (p) γ^{0}

. The label

s

specifies a spin state;

p

is the on-shell four-momentum. Incoming and outgoing antifermions instead contribute

\overset{v}{ˉ}_{s} (p)

and

v_{s} (p)

The exchanged boson is a virtual photon, represented in the Feynman rules by the photon propagator

- i g_{μν} / (q^{2} + i ϵ)

in Feynman gauge. Its momentum transfer is

q = p_{1} - p_{3} = p_{4} - p_{2}

; unlike an external real photon, it need not satisfy

q^{2} = 0

. Each Feynman vertex contracts a gamma matrix with this propagator and contributes

- i e γ^{μ}

in a conventional QED sign convention. The coupling constant

e

sets the interaction strength, with fine-structure constant

α = e^{2} / (4 π)

. One may reverse the sign of the photon field and both vertex signs without changing the result. The gamma matrices obey

{γ^{μ}, γ^{ν}} = 2 g^{μν} I

, with Minkowski metric

g = diag (1, - 1, - 1, - 1)

Solution

There is one tree-level Feynman diagram: the distinguishable electron and muon exchange a photon in the

t

channel. There is no identical-particle exchange diagram.

Figure 1.
Electron-muon scattering by one-photon exchange
. **Details**.

Multiplying the external Dirac spinors, the two Feynman vertices and the photon propagator gives

i M = [\overset{u}{ˉ}_{3} (- i e γ^{μ}) u_{1}] \frac{- i g _{μν}}{q ^{2} + i ϵ} [\overset{u}{ˉ}_{4} (- i e γ^{ν}) u_{2}], M = \frac{e ^{2}}{q ^{2}} [\overset{u}{ˉ}_{3} γ^{μ} u_{1}] [\overset{u}{ˉ}_{4} γ_{μ} u_{2}] .

The overall phase of this scattering amplitude is convention dependent. Four external spin labels with two choices each give

2^{4} = 16

possible configurations before taking a limit. For

E ≫ m, M

, the vector QED interaction preserves fermion chirality, which becomes helicity for each external particle. Thus the surviving configurations, written as

(h_{e}, h_{μ})_{in} \to (h_{e}, h_{μ})_{out}

, are

LL \to LL, L R \to L R, R L \to R L, RR \to RR .

Here

L, R

specify helicity along each particle's own momentum. The other configurations require at least one mass-suppressed flip and vanish in the strictly massless limit.

Solution

Define the Mandelstam variables for the scattering labels by

s = (p_{1} + p_{2})^{2}

t = (p_{1} - p_{3})^{2} = q^{2}

and

u = (p_{1} - p_{4})^{2}

. Dropping the electron and muon masses gives

p_{1} \cdot p_{2} = p_{3} \cdot p_{4} = \frac{s}{2}, p_{1} \cdot p_{4} = p_{3} \cdot p_{2} = - \frac{u}{2} .

These identities follow by expanding the corresponding squared four-momenta, using

p_{i}^{2} = 0

and four-momentum conservation. Substitution into the supplied spin-averaged matrix element yields

\overline{∣ M ∣^{2}} = \frac{8 e ^{4}}{t ^{2}} (\frac{s ^{2}}{4} + \frac{u ^{2}}{4}) = 2 e^{4} \frac{s ^{2} + u ^{2}}{t ^{2}} .

The

1/4

initial spin average is already included in the given expression; no further averaging factor belongs here.

Solution

The intermediate photon carries spin

J = 1

. In the massless limit, a back-to-back fermion-antifermion pair coupled to its vector current has opposite helicities, so the initial and final projections along their respective axes are

m, m^{'} = \pm 1

. A rotation between these axes contributes the Wigner D-matrix elements

d_{m^{'} m}^{1} (θ)

. The relevant elements have magnitudes

(1 + cos θ) /2

and

(1 - cos θ) /2

. Summing the four squared helicity amplitudes gives the angular factor

2 (\frac{1 + c o s θ}{2})^{2} + 2 (\frac{1 - c o s θ}{2})^{2} = 1 + cos^{2} θ .

The common coupling and normalization factors recover the result in d. The

m = 0

terms do not contribute: the massless external vector currents eliminate the corresponding helicity configurations.

Solution

The favoured weak charged current transition is

\overset{c}{ˉ} \to \overset{s}{ˉ} W^{-}

followed by

W^{-} \to d \overset{u}{ˉ}

. The spectator quark

u

combines with

\overset{u}{ˉ}

to form the

u \overset{u}{ˉ} / 2

component of the neutral pion, while

d \overset{s}{ˉ}

forms

K^{0}

. The product of CKM matrix factors is

V_{cs} V_{u d}^{*}

. The doubly Cabibbo suppressed transition is

\overset{c}{ˉ} \to \overset{ˉ}{d} W^{-}

followed by

W^{-} \to s \overset{u}{ˉ}

, with factors

V_{c d} V_{u s}^{*}

; the final

s \overset{ˉ}{d}

is the neutral antikaon. Both are colour-suppressed internal-emission Feynman diagrams, with the same spectator routing.

Figure 1.
Favoured and doubly suppressed neutral-charm decay diagrams
. **Details**.

In the two-generation approximation to the CKM matrix,

V_{u d} = V_{cs} = cos θ_{C}

V_{u s} = sin θ_{C}

and

V_{c d} = - sin θ_{C}

. For equal magnitudes of the hadronic matrix elements,

\frac{∣ A ( D ˉ ^{0} \to K ˉ ^{0} π ^{0} ) ∣}{∣ A ( D ˉ ^{0} \to K ^{0} π ^{0} ) ∣} = tan^{2} θ_{C}, \frac{Γ _{suppressed}}{Γ _{favoured}} ≃ tan^{4} θ_{C} .

The rate ratio is about

2.8 \times 1 0^{- 3}

with the supplied CKM matrix magnitudes. The minus sign of a particular product of CKM matrix elements must be combined with the phases of the hadronic matrix elements and kaon basis states; it is not by itself the observable interference sign.

b

Solution

Choose the exchange-positive CP convention for neutral kaon mixing,

CP ∣ K^{0} ⟩ = ∣ \overset{ˉ}{K}^{0} ⟩

. Neglecting CP violation, the kaon CP eigenstates can then be written

∣ K_{S} ⟩ = \frac{∣ K ^{0} ⟩ + ∣ K ˉ ^{0} ⟩}{2}, ∣ K_{L} ⟩ = \frac{∣ K ^{0} ⟩ - ∣ K ˉ ^{0} ⟩}{2} .

Writing

A = A (\overset{ˉ}{D}^{0} \to K^{0} π^{0})

and

B = A (\overset{ˉ}{D}^{0} \to \overset{ˉ}{K}^{0} π^{0})

, projection on these coherent final states gives

A_{S} = \frac{A + B}{2} = \frac{A}{2} (1 + B / A), A_{L} = \frac{A - B}{2} = \frac{A}{2} (1 - B / A) .

This is the convention implicit in the printed sum. Changing the sign convention for the neutral antikaon changes both its coefficient in the kaon CP eigenstates and

B

; it cannot change either physical decay rate. The extremely small

K_{S}

–

K_{L}

mass difference can be ignored in the two-body phase space.

c

Solution

Put

r = B / A

. Equal two-body phase space gives

Γ_{S, L} \propto ∣ A ∣^{2} ∣1 \pm r ∣^{2} /2

, so the decay-rate asymmetry is exactly

R = \frac{∣1 + r ∣ ^{2} - ∣1 - r ∣ ^{2}}{∣1 + r ∣ ^{2} + ∣1 - r ∣ ^{2}} = \frac{2 Re r}{1 + ∣ r ∣ ^{2}} .

The stated magnitude assumption fixes

∣ r ∣ = tan^{2} θ_{C}

. Retaining the relative strong phase

δ

in the convention of b therefore gives

R = \frac{2 tan ^{2} θ _{C} cos δ}{1 + tan ^{4} θ _{C}} = 2 tan^{2} θ_{C} cos δ + O (tan^{6} θ_{C}) .

A magnitude assumption alone does not determine $R$ . For example,

r = i tan^{2} θ_{C}

has the stipulated magnitude but gives

R = 0

, whereas a real positive

r

gives a positive asymmetry. With the additional aligned-amplitude assumption

δ = 0

, conventionally used for this idealised extraction, the intended leading result is

R ≃ 2 tan^{2} θ_{C}

. The sign of

r

here refers to the complete amplitude after choosing the kaon phase convention, not to a bare CKM matrix product. A flavour-symmetry relation can supply further information, but is stronger than the printed magnitude assumption.

d

Solution

The long-lived, electrically neutral kaon often traverses the particle detector without decaying into a reconstructable charged final state. Identifying it through a hadronic interaction or missing four-momentum is much less efficient than reconstructing the short-lived kaon through its nearby two-pion decay vertex. The efficiencies must be applied to the observed counts before taking their asymmetry:

N_{S} = \frac{1107}{0.93} = 1190.32, N_{L} = \frac{49}{0.05} = 980, R = \frac{N _{S} - N _{L}}{N _{S} + N _{L}} = 0.09691.

Using the leading aligned-amplitude approximation from c,

θ_{C} = arctan R /2 = 0.21667 rad = 12.4 1^{\circ} .

The exact aligned formula instead gives

12.4 3^{\circ}

, an immaterial correction here. Treating the observed counts as independent Poisson random variables and the efficiencies as known, propagation of uncertainty gives

σ_{S} = \frac{1107}{0.93} = 35.78, σ_{L} = \frac{49}{0.05} = 140,

σ_{R}^{2} = [\frac{2 N _{L}}{( N _{S} + N _{L} ) ^{2}}]^{2} σ_{S}^{2} + [\frac{2 N _{S}}{( N _{S} + N _{L} ) ^{2}}]^{2} σ_{L}^{2}, σ_{R} = 0.0723.

(4)

The separate contributions to

σ_{R}

are

0.0149

and

0.0708

; the small

K_{L}

sample dominates. Linear propagation of uncertainty yields

σ (θ_{C}) ≃ 4.4 9^{\circ}

, although the nonlinear extraction and physical boundary make a symmetric confidence interval a rough summary. Unknown efficiency uncertainties, especially that of

K_{L}

, would be additional systematic contributions.

The supplied CKM matrix gives

sin θ_{C} ≃ 0.224

, hence

θ_{C} ≃ 12.9 4^{\circ}

and

R ≃ 2 (0.224/0.975)^{2} = 0.1056

. The asymmetry difference is only about

0.12

statistical standard deviations, so the measurement is compatible. Without the phase assumption, these data measure

tan^{2} θ_{C} cos δ

to leading order and do not uniquely measure the Cabibbo angle.

3

a

Solution

The

Δ^{++}

Delta baryon has electric charge

+ 2

and valence quark content

uuu

. In its ground state the spatial wavefunction has orbital angular momentum

L = 0

and is symmetric. Its spin is

3/2

; the maximal projection state

u_{↑} u_{↑} u_{↑}

is symmetric in spin, and its quark flavour wavefunction is symmetric as well. Three identical fermions require an antisymmetric total wavefunction by the Pauli exclusion principle. A further internal degree of freedom must therefore supply the antisymmetry.

With three color charges, that factor is the normalized colour singlet

∣ C ⟩ = \frac{∣ r g b ⟩ + ∣ g b r ⟩ + ∣ b r g ⟩ - ∣ r b g ⟩ - ∣ g r b ⟩ - ∣ b g r ⟩}{6} .

It changes sign under exchange of any two quarks. The remaining symmetric factors then give the required antisymmetric total state. This argument motivates color charge independently of the already present quark flavour and spin labels.

b

Solution

For quarks in an

N_{c}

-dimensional internal space

V

, complete antisymmetrization of three labels corresponds to a Young diagram with one column of three boxes, the partition

(1, 1, 1)

. Its representation is the exterior power

⋀^{3} V

and has dimension

dim ⋀^{3} V = (3 N _{c}) .

For

N_{c} = 2

, the column is too tall:

⋀^{3} C^{2} = 0

. With only two labels at least two of the three coincide, and the Young tableau column antisymmetrizer annihilates the state. For

N_{c} = 3

, the filled column

r, g, b

spans a one-dimensional determinant representation. It is a colour singlet of SU(3) because every group element has determinant one.

Equivalently, the tensor product of group representations decomposes as

3 \otimes 3 \otimes 3 = 1 0_{[3]} \oplus 8_{[2, 1]} \oplus 8_{[2, 1]} \oplus 1_{[1, 1, 1]} .

The two mixed-symmetry terms account for

16

dimensions, so the total is

10 + 8 + 8 + 1 = 27

. Antisymmetry alone establishes at least three internal states. Exactly three follows in the usual special unitary group model when the observed three-quark baryon must be a colour singlet: for

N_{c} > 3

the antisymmetric third exterior power has dimension greater than one and is not a singlet. This extra assumption should be made explicit.

c

Solution

Away from narrow resonances and quark thresholds, the QED annihilation cross section into each accessible quark flavour is its squared fractional electric charge times the muon-pair cross section, multiplied by the number

N_{c}

of distinguishable color charges. Thus the inclusive hadronic ratio is

R_{had} = \frac{σ ( e ^{+} e ^{-} \to hadrons )}{σ ( e ^{+} e ^{-} \to μ ^{+} μ ^{-} )} ≃ N_{c} \sum_{f} Q_{f}^{2} .

With

N_{c} = 3

, the continuum predictions are

R_{u d s} = 2

R_{u d sc} = 10/3

and

R_{u d sc b} = 11/3

as successive charm quark and bottom quark channels become accessible. The measured normalization and steps support three color charges. At leading order these hadronic events form two back-to-back jets from the quark and antiquark; quantitative comparisons include QCD corrections and avoid the nonperturbative threshold region.

The extra resolved jet is explained by

e^{+} e^{-} \to γ^{*} \to q \overset{q}{ˉ} g

, with gluon radiation from either the quark or antiquark. One possible color charge routing is initial

q_{r} \overset{q}{ˉ}_{\overset{r}{ˉ}}

followed by

q_{r} \to q_{b} + g_{r \overset{ˉ}{b}}

; the final

q_{b}

\overset{q}{ˉ}_{\overset{r}{ˉ}}

and

g_{r \overset{ˉ}{b}}

combine to a net colour singlet.

Figure 1.
Colour flow in a three-jet process and in gluon-fusion Higgs production
. **Details**.

The three-jet cross section contains one additional strong interaction vertex in the amplitude, so its ratio to the two-jet rate is of order

α_{s} = g_{s}^{2} / (4 π)

times a resolution-dependent kinematic coefficient. A resolved fraction of order

15%

is consistent with perturbative QCD and a strong coupling constant much larger than the fine-structure constant. It is not an equality

α_{s} = 0.15

: the jet definition and cuts are needed. Photon radiation has a much smaller electromagnetic rate. The large hadronic third-jet rate supports gluon radiation; identifying the radiated object specifically as a colour-octet gluon also uses the QCD vertex and colour-flow description, not merely the existence of three visible jets.

d

Solution

Represent the color charges by

r = (1, 0, 0)^{T}

g = (0, 1, 0)^{T}

and

b = (0, 0, 1)^{T}

. The Gell-Mann matrices satisfy

λ^{4} r = b, λ^{5} r = ib, \frac{λ ^{4} - i λ ^{5}}{2} = ∣ b ⟩ ⟨ r ∣.

The physical quark-gluon Feynman vertex contains

g_{s} T^{a}

with

T^{a} = λ^{a} /2

; the off-diagonal combination therefore allows

r \to b

. Color charge conservation assigns the emitted gluon the colour-flow label

r \overset{ˉ}{b}

The claimed impossibility of $r \to r$ is false. A direct counterexample from the printed matrices is

λ^{3} r = r, λ^{8} r = \frac{r}{3} .

Thus both diagonal gluons have nonzero

r \to r

vertex elements,

(T^{3})_{rr} = 1/2

and

(T^{8})_{rr} = 1/ (23)

. They correspond to the traceless combinations

\frac{r r ˉ - g g ˉ}{2}, \frac{r r ˉ + g g ˉ - 2 b b ˉ}{6} .

A Gell-Mann matrix is a generator rather than a finite rotation, but exponentiating a diagonal generator also leaves the red basis direction fixed up to a phase. The actual excluded object is the identity generator, proportional to the colour singlet

(r \overset{r}{ˉ} + g \overset{g}{ˉ} + b \overset{ˉ}{b}) / 3

. Adding it would extend SU(3) to unitary group and introduce a ninth singlet vector boson. A colour-preserving vertex among the existing eight gluons does not require that extra boson and does not make the ordinary diagonal gluons colour singlets.

e

Solution

In gluon-fusion Higgs production, two gluons couple to a closed top quark loop, and the Higgs boson couples to that massive quark through its Yukawa interaction. The right panel of the diagram in c shows one explicit colour-flow component: internal

t_{r} \to t_{b}

at one gluon insertion, then

t_{b} \to t_{r}

at the other; the colour-neutral Higgs boson insertion preserves the internal color charge. Include both loop orderings. The color charge factor is

tr (T^{a} T^{b}) = \frac{1}{2} δ^{ab},

so the two incoming gluons are projected onto a colour singlet. The Yukawa coupling is

y_{t} = 2 m_{t} / v

, and the

h \overset{ˉ}{t} t

vertex has strength

m_{t} / v

. No direct tree-level

h gg

mass coupling is needed. Integrating out a sufficiently heavy top quark generates an effective interaction proportional to

(h / v) G_{μν}^{a} G^{a μν}

, which is allowed by gauge invariance even though the gluon is massless.

There is also no kinematic obstacle: two massless particles with nonparallel momenta can have nonzero total invariant mass. In particular, head-on gluons have

\overset{s}{^} = 4 E_{1} E_{2}

, providing the energy needed for the Higgs boson at the LHC.

4

a

i

Solution

The Lorentz force is

d p / d t = q v \times B

. Because it is perpendicular to the velocity, it changes the direction of momentum without changing its magnitude. For a charged particle in a uniform magnetic field, the radius of curvature is

R = p / (∣ q ∣ B)

. Over the short track, the bending angle is

ϕ ≃ L / R

. Equivalently, integrate the force for a time

L / v

while neglecting the small rotation in its direction. The magnitude of the acquired momentum component perpendicular to the incident direction is

∣Δ p_{⊥} ∣ = p sin ϕ ≃ pϕ = ∣ q ∣ B L .

Its direction follows

q v \times B

; a signed component can therefore be written

qB L

after fixing an oriented bending axis. Chord length and arc length agree at this order in

L / R

ii

Solution

The sagitta is the maximum distance between the circular track and its chord. The perpendicular bisector of a chord of length

L

gives a right triangle with hypotenuse

R

and sides

L /2

and

R - s

. Hence

s = R - R^{2} - L^{2} /4 = \frac{L ^{2}}{8 R} + O (L^{4} / R^{3}) ≃ \frac{∣ q ∣ B L ^{2}}{8 p} .

At fixed electric charge, magnetic field and lever arm,

p = ∣ q ∣ B L^{2} / (8 s)

, so differentiation and propagation of uncertainty give

\frac{σ ( p )}{p} = \frac{σ ( s )}{s} = \frac{8 p σ ( s )}{∣ q ∣ B L ^{2}} .

Thus a fixed spatial measurement uncertainty produces a relative momentum uncertainty growing linearly with

p

. If momentum is expressed in

GeV / c

B

in tesla,

L, s

in metres and

z = q / e

, restore the usual conversion

p = 0.2998∣ z ∣ BR

b

i

Solution

For azimuthally symmetric Rutherford scattering, integrate over the solid angle,

d Ω = 2 π sin θ d θ

. Substituting

u = sin (θ /2)

gives

sin θ d θ = 4 u d u

, so

σ = \frac{2 π K}{p ^{2}} \int_{θ_{m i n}}^{θ_{m a x}} \frac{sin θ}{sin ^{4} ( θ /2 )} d θ = \frac{4 π K}{p ^{2}} [csc^{2} \frac{θ _{m i n}}{2} - csc^{2} \frac{θ _{m a x}}{2}] .

The small lower cutoff dominates when

θ_{m a x} ≫ θ_{m i n}

and

θ_{m i n} ≪ 1

. Using

sin (θ_{m i n} /2) ≃ θ_{m i n} /2

, the leading total cross section is

σ ≃ \frac{16 π K}{p ^{2} θ _{m i n}^{2}} .

A nonzero cutoff is essential: the unscreened forward cross section otherwise diverges.

ii

Solution

Let

ℓ = ln (θ_{m a x} / θ_{m i n})

. The supplied second moment gives the single-scatter root mean square angle

θ_{rms, 1} = ⟨ θ^{2} ⟩ ≃ 2 θ_{m i n} ℓ .

For independent scatters with uniform azimuth, the small angular-deflection vectors have zero mean. Their cross terms average to zero, so their variances add. After

N

scatters,

RMS (θ) = N ⟨ θ^{2} ⟩ ≃ θ_{m i n} 2 N ℓ .

This is the root mean square magnitude of the two-dimensional angular deflection. Its projection into one specified plane has a variance smaller by a factor of two, hence a root mean square smaller by

2

iii

Solution

For a thickness

x

with number density

ρ_{n}

of scattering centres, the mean count is

N = ρ_{n} x σ

. Using the total cross section in i and the variance addition in ii,

RMS (θ)^{2} ≃ ρ_{n} x \frac{16 π K}{p ^{2} θ _{m i n}^{2}} 2 θ_{m i n}^{2} ℓ = \frac{32 π K ρ _{n} x ℓ}{p ^{2}} .

Thus

RMS (θ) ≃ 32 π K ρ_{n} ln (θ_{m a x} / θ_{m i n}) \frac{x}{p} .

The stated assumptions make the prefactor independent of

x

and approximately independent of

p

. The lower-cutoff factors cancel except inside the slowly varying logarithm. Treating

N

as a Poisson random variable rather than a fixed count gives the same result after replacing

N

by its mean.

iv

Solution

The magnetic bending angle is

ϕ ≃ ∣ q ∣ B L / p

. For a momentum estimate based on this angle, propagation of uncertainty gives

σ (p) / p = σ (ϕ) / ϕ

. Comparing the total Multiple Coulomb scattering angle with the bend therefore gives the commonly intended characteristic estimate

(\frac{σ ( p )}{p})_{ms} \sim \frac{p RMS ( θ )}{∣ q ∣ B L} = \frac{32 π K ρ _{n} x ln ( θ _{m a x} / θ _{m i n} )}{∣ q ∣ B L} .

Its independence of

p

is the key result for relativistic particles: the scattering angle and magnetic bend both scale as

1/ p

A precise standard deviation requires specifying how tracks are measured. If

ϕ

is obtained from independently measured entrance and exit directions and all the material lies between them, the bending-plane projected root mean square is

RMS (θ) / 2

. For that estimator the equality is

(\frac{σ ( p )}{p})_{ms, angle} = \frac{16 π K ρ _{n} x l n ( θ _{m a x} / θ _{m i n} )}{∣ q ∣ B L} .

A three-point sagitta fit weights scatter positions differently again. The paper does not specify these measurement details, so it fixes the scaling and characteristic estimate, but not one universal geometric coefficient for all track fits.

c

Solution

At high momentum, the fixed spatial measurement uncertainty dominates because ii grows as

p

. At lower, still relativistic momentum, Multiple Coulomb scattering dominates the relative uncertainty, which approaches the approximately constant term in iv. Below the relativistic regime the additional velocity dependence of scattering must also be included.

Three complementary improvements are: increase

B

and the tracking lever arm

L

to enlarge the magnetic bend; improve hit-position resolution and alignment to reduce

σ (s)

; and reduce passive material and detector-layer thickness to reduce

x

and hence Multiple Coulomb scattering. For independent contributions one commonly writes

\frac{σ ( p )}{p} ≃ (A p)^{2} + C^{2}, A = \frac{8 σ ( s )}{∣ q ∣ B L ^{2}},

where

C

is the scattering term for the specified material distribution and track estimator.

5

a

Solution

The covariant derivative

D_{μ}

differentiates a field while preserving its transformation law under the local electroweak gauge symmetry

S U (2)_{L} \times U (1)_{Y}

. Here

\partial_{μ}

is the ordinary spacetime derivative and

μ

a Lorentz index. The gauge fields

W_{μ}^{j}

j = 1, 2, 3

, are the three SU(2) fields;

B_{μ}

is the U(1) field. The corresponding dimensionless coupling constant is

g

for SU(2) and

g^{'}

for U(1).

The

τ_{j}

are weak isospin generators in the representation of the differentiated field, with

τ_{j} = σ_{j} /2

on a doublet and

τ_{j} = 0

on an SU(2) singlet. The repeated index

j

is summed. The generator

Y /2

supplies the field's weak hypercharge in the convention

Q = τ_{3} + Y /2

, where

Q

is its fractional electric charge. The factor

- i

follows the convention of Hermitian generators. The two connection terms compensate the derivatives of spacetime-dependent gauge transformations.

b

Solution

Set

G = g^{2} + g^{'2}

and

s_{W} = sin θ_{W}

. First expand the charged term using the ladder operators:

τ^{+} W^{+} + τ^{-} W^{-} = \frac{( τ _{1} + i τ _{2} ) ( W ^{1} - i W ^{2} ) + ( τ _{1} - i τ _{2} ) ( W ^{1} + i W ^{2} )}{2} = 2 (τ_{1} W^{1} + τ_{2} W^{2}) .

Substituting the neutral gauge fields as well, the covariant derivative becomes

D_{μ} = \partial_{μ} - i g (τ_{1} W_{μ}^{1} + τ_{2} W_{μ}^{2}) - i [g (τ_{3} - Q s_{W}^{2}) + \frac{e g ^{'}}{G} Q] W_{μ}^{3} - i [- g^{'} (τ_{3} - Q s_{W}^{2}) + \frac{e g}{G} Q] B_{μ} .

Matching

W^{3}

g τ_{3}

requires

g s_{W}^{2} = \frac{e g ^{'}}{G} .

Matching

B

g^{'} (Q - τ_{3}) = g^{'} Y /2

requires

g^{'} s_{W}^{2} + \frac{e g}{G} = g^{'} .

(4)

For

g, g^{'} > 0

, elimination yields

sin^{2} θ_{W} = \frac{g ^{'2}}{g ^{2} + g ^{'2}}, tan θ_{W} = \frac{g ^{'}}{g}, θ_{W} = arctan \frac{g ^{'}}{g}, e = \frac{g g ^{'}}{g ^{2} + g ^{'2}} .

(5)

Equivalently,

e = g sin θ_{W} = g^{'} cos θ_{W}

and

1/ e^{2} = 1/ g^{2} + 1/ g^{'2}

. The conventional Weinberg angle lies in the first quadrant.

c

Solution

Write

f = (v + h) / 2

. The Pauli matrices give the weak isospin operators

τ^{+} = (0010), τ^{-} = (0100), τ_{3} = \frac{1}{2} (10 0 - 1) .

For the Higgs doublet

Φ = (0, f)^{T}

with weak hypercharge

Y = 1

τ^{+} Φ = (f 0), τ^{-} Φ = 0, τ_{3} Φ = (0 - f /2), Q Φ = (τ_{3} + \frac{1}{2} I) Φ = 0.

The Higgs field direction, including its vacuum expectation value, is electrically neutral. Consequently the electromagnetic U(1) leaves the vacuum invariant: electroweak symmetry breaking preserves electromagnetism and the photon remains massless. The result concerns this neutral vacuum direction; it does not say that both components of the full Higgs doublet have zero electric charge.

d

Solution

Using c and omitting

\partial_{μ} Φ

, the covariant derivative is

D_{μ} Φ = (- i g f W_{μ}^{+} / 2 i G f Z_{μ} /2), G = g^{2} + g^{'2} .

The two components are orthogonal, so their squared magnitudes add;

(W_{μ}^{+})^{†} = W_{μ}^{-}

. Thus the Higgs field kinetic contribution to the Lagrangian density is

L = (v^{2} + 2 v h + h^{2}) (\frac{g ^{2}}{4} W_{μ}^{+} W^{- μ} + \frac{G ^{2}}{8} Z_{μ} Z^{μ}) .

The two

v^{2}

terms are gauge boson mass terms. Comparing with

m_{W}^{2} W^{+} W^{-} + \frac{1}{2} m_{Z}^{2} Z^{2}

gives

m_{W} = gv /2, m_{Z} = G v /2

and

m_{W} / m_{Z} = cos θ_{W}

. The factor

1/2

for the real Z boson field is essential.

The two

2 v h

terms are the cubic Higgs boson coupling to W bosons and Higgs boson coupling to Z bosons. They produce the three-leg

h W^{+} W^{-}

and

h ZZ

Feynman vertices, allowing Higgs decay to two W bosons, Higgs decay to two Z bosons when kinematically available, and the corresponding production mechanisms with virtual gauge bosons. The two

h^{2}

terms are quartic

hh W^{+} W^{-}

and

hh ZZ

interactions, including vector-boson scattering into a Higgs boson pair. Written in terms of the masses,

L \supset m_{W}^{2} W^{+} W^{-} + \frac{1}{2} m_{Z}^{2} Z^{2} + \frac{2 m _{W}^{2}}{v} h W^{+} W^{-} + \frac{m _{Z}^{2}}{v} h Z^{2} + \frac{m _{W}^{2}}{v ^{2}} h^{2} W^{+} W^{-} + \frac{m _{Z}^{2}}{2 v ^{2}} h^{2} Z^{2} .

The corresponding Feynman vertex factors are

2 i m_{V}^{2} g_{μν} / v

for one

h

and

2 i m_{V}^{2} g_{μν} / v^{2}

for two

h

, for

V = W, Z

, after differentiating with respect to identical fields. These six contributions contain no photon mass or direct

h AA

interaction because

Q Φ = 0

Figure 1.
Mass insertions and cubic and quartic Higgs-gauge vertices
. **Details**.

6

a

Solution

The lepton mixing matrix is unitary, so each row has unit norm. Explicitly its electron row obeys

∣ U_{e 1} ∣^{2} + ∣ U_{e 2} ∣^{2} + ∣ U_{e 3} ∣^{2} = c_{13}^{2} (c_{12}^{2} + s_{12}^{2}) + s_{13}^{2} = 1,

where

c_{ij} = cos θ_{ij}

and

s_{ij} = sin θ_{ij}

. This is also the normalization needed for unit electron neutrino survival at

L = 0

Use

Re (1 - e^{i x}) = 1 - cos x = 2 sin^{2} (x /2)

and define

Δ_{ij} = (ϕ_{i} - ϕ_{j}) /2 = Δ m_{ij}^{2} L / (4 E)

in natural units. The exact vacuum neutrino oscillation result becomes

P_{ee} = 1 - 4 \sum_{i < j} ∣ U_{e i} ∣^{2} ∣ U_{e j} ∣^{2} sin^{2} Δ_{ij} .

The given mass-squared differences satisfy

∣Δ m_{12}^{2} /Δ m_{23}^{2} ∣ ≃ 0.027 ≪ 1

, so the two fast oscillation frequencies are close:

Δ_{13} = Δ_{23} + Δ_{12}

. Approximating their phases as equal combines their coefficients into

4 s_{13}^{2} c_{13}^{2} (c_{12}^{2} + s_{12}^{2}) = sin^{2} 2 θ_{13},

while

4 c_{12}^{2} s_{12}^{2} c_{13}^{4} = c_{13}^{4} sin^{2} 2 θ_{12}

. Hence

P_{ee} ≃ 1 - cos^{4} θ_{13} sin^{2} 2 θ_{12} sin^{2} Δ_{12} - sin^{2} 2 θ_{13} sin^{2} Δ_{23} .

(4)

The complex CP phase cancels from every

∣ U_{e i} ∣^{2}

. The phase-replacement approximation is uniformly reliable for

∣ Δ_{12} ∣ ≪ 1

, especially at short reactor baselines; closeness of the mass-squared differences alone does not guarantee close sine values at arbitrarily large

L / E

. At medium baselines the distinct fast frequencies can beat, so the exact three-frequency expression is preferable if their detailed pattern matters.

b

Solution

Restoring units in the oscillation phase gives

Δ_{ij} = Δ m_{ij}^{2} L / (4 E ℏ c)

. With

ℏ c = 1.97327 \times 1 0^{- 7} eV m

, one kilometre is

5.06773 \times 1 0^{9} eV^{- 1}

. Thus

Δ_{ij} = \frac{5.06773 \times 1 0 ^{9}}{4 \times 1 0 ^{6}} \frac{Δ m _{ij}^{2} [ eV ^{2} ] L [ km ]}{E [ MeV ]}, k_{practical} = 1.26693 \times 1 0^{3} .

Here

k_{practical}

denotes the coefficient in

Δ = k_{practical} Δ m^{2} [eV^{2}] L [km] / E [MeV]

. The familiar

1.267

applies when

E

is in GeV, not MeV. Strictly, the dimensionless definition

Δ = 2 k (ϕ_{i} - ϕ_{j})

keeps

k = 1/4

once the

ϕ_{j}

already include the unit conversion. The intended numerical question concerns the practical coefficient multiplying the numerical values of

Δ m^{2} L / E

c

Solution

Using the two-frequency approximation in a and

L

in kilometres, the survival probability is

P_{ee} (L) ≃ 1 - 0.71374 sin^{2} (0.0337849 L) - 0.0954915 sin^{2} (1.26693 L) .

The slow coefficient is

cos^{4} (9^{\circ}) sin^{2} (6 0^{\circ}) = 0.71374

; the fast one is

sin^{2} (1 8^{\circ}) = 0.09549

. Since

sin^{2} x

has period

π

, the two individual oscillation periods are

T_{12} = \frac{π E}{k _{practical} ∣Δ m _{12}^{2} ∣} = 92.99 km, T_{23} = \frac{π E}{k _{practical} ∣Δ m _{23}^{2} ∣} = 2.480 km .

The first broad trough is near

46.5

km and the second near

139.5

km; the slow component returns to its crest near

93

km. Small, rapid ripples of depth about

0.0955

sit on the broad envelope. Averaging over unresolved fast oscillations replaces their squared sine by

1/2

, giving

⟨ P_{ee} ⟩ ≃ 0.95225 - 0.71374 sin^{2} (π L /92.99)

Figure 1.
Reactor survival probability and the limits of merging the fast frequencies
. **Details**.

There is no single short period for the combined curve. If the supplied ratio

0.003/0.00008 = 75/2

is taken as exact, its mathematical common period is

2 T_{12} = 75 T_{23} ≃ 186

km, outside the plotted range; the approximate physical data do not warrant interpreting that commensurability as exact.

The figure also shows the unmerged expression

P_{ee}^{exact} = 1 - 4 c_{13}^{4} c_{12}^{2} s_{12}^{2} sin^{2} Δ_{12} - 4 c_{13}^{2} s_{13}^{2} (c_{12}^{2} sin^{2} Δ_{13} + s_{12}^{2} sin^{2} Δ_{23}) .

For that comparison take

Δ m_{23}^{2} = 0.003

and enforce the identity

Δ m_{13}^{2} = Δ m_{23}^{2} + Δ m_{12}^{2} = 0.00308 eV^{2}

, with a consistent sign convention. This illustrates the limitation identified in a: the simplified envelope is useful, but its individual fast ripples need not match the three-frequency result over 150 km. The supplied

θ_{23}

has no effect on vacuum electron-flavour disappearance.

d

i

Solution

For a rate measurement of the small mixing angle

θ_{13}

, choose the first fast disappearance maximum,

Δ_{23} = \frac{π}{2}, L_{opt} ≃ \frac{π E}{2 k _{practical} ∣Δ m _{23}^{2} ∣} = 1.24 km (E = 3 MeV) .

A 47 km baseline instead lies near the slow oscillation maximum associated with

θ_{12}

. Its much lower flux, approximately falling as

1/ L^{2}

, and the many fast oscillations washed out by a reactor's broad energy spectrum make it a poor choice for this simple

θ_{13}

design. A medium-baseline spectral experiment can still constrain neutrino oscillation parameters if it has enough events and resolution; 47 km is not intrinsically useless, but it is not the optimal short-baseline rate experiment. In a,

θ_{13}

also affects the slow envelope through

cos^{4} θ_{13}

, so the sensitivity there is not identically zero.

ii

Solution

Matched near and far particle detectors are a good design. A near particle detector, close enough that the fast oscillation phase is small, measures the unoscillated reactor spectrum and normalization. A far particle detector near the first fast maximum measures the disappearance. Their appropriately geometry-corrected ratio cancels much of the uncertain reactor flux and common cross section and detection efficiency. Relative calibration, different backgrounds, multiple reactor distances and any residual near-site oscillation still need treatment. The near site should retain enough shielding to suppress cosmic backgrounds.^[ref]

iii

Solution

A fission reactor supplies predominantly electron antineutrinos, with typical energies of a few MeV. Even if they oscillate to muon antineutrinos, producing a real muon through a charged current requires roughly the muon rest energy, over

100

MeV for a free nucleon target. For example,

E_{th} (\overset{ν}{ˉ}_{μ} p \to μ^{+} n) = \frac{( m _{n} + m _{μ} ) ^{2} - m _{p}^{2}}{2 m _{p}} ≃ 113 MeV .

Thus a muon spectrometer cannot observe this appearance channel from a reactor. Measure electron antineutrino disappearance instead, most commonly through inverse beta decay,

\overset{ν}{ˉ}_{e} p \to e^{+} n

, using the prompt Positron and delayed neutron signal.

iv

Solution

The radiation length

X_{0}

is the characteristic length for high-energy electron energy loss by bremsstrahlung; smaller

X_{0}

means more radiative loss for the same traversed thickness. To limit that loss, minimize

x / X_{0}

, choosing less material or a larger

X_{0}

, rather than minimizing

X_{0}

itself. At reactor Positron energies of a few MeV, ionization and subsequent annihilation matter more than high-energy bremsstrahlung. A sufficiently large scintillation counter measures the deposited energy, so containment of the Positron and annihilation photons, light yield and calibration are the relevant design issues.^[ref]

v

Solution

A Positron signal is correct for the reactor's antineutrinos, through inverse beta decay,

\overset{ν}{ˉ}_{e} + p \to e^{+} + n, E_{\overset{ν}{ˉ}, th} ≃ 1.806 MeV .

A true electron neutrino charged current interaction instead produces an electron, for example

ν_{e} + n \to e^{-} + p

. This corrects the source's loose use of “neutrino” for the reactor beam.

After slowing, the Positron annihilates with an electron, generally producing two

511

keV photons. The prompt visible energy includes both the Positron kinetic energy and annihilation energy; neglecting the small neutron recoil,

E_{prompt} ≃ T_{e^{+}} + 2 m_{e} ≃ E_{\overset{ν}{ˉ}} - 0.782 MeV .

A delayed neutron capture supplies a powerful coincidence tag. Detecting only two annihilation photons discards useful prompt energy and does not uniquely identify the charged current event.

vi

Solution

The relevant resolution condition is on the oscillation phase, not solely on which mixing angle is being measured. Since

Δ \propto L / E

σ_{Δ} ≃ ∣Δ∣ \frac{σ _{E}}{E} .

At the same medium baseline, the fast

θ_{13}

-dependent ripples have phases about

37.5

times larger than the slow

θ_{12}

component. Resolving those ripples therefore requires a much better fractional energy resolution. At 47 km,

∣ Δ_{23} ∣ ≃ 59.5

; avoiding substantial phase smearing requires a resolution appreciably better than

1/59.5 ≃ 1.7%

However, a short-baseline

θ_{13}

experiment near 1.24 km and a

θ_{12}

experiment near 46.5 km each operate near their relevant first maximum,

Δ ≃ π /2

. Their intrinsic fractional phase-resolution requirements are then comparable. The smaller

θ_{13}

disappearance amplitude demands careful relative calibration and background control, but it does not by itself impose a universally finer energy resolution. The proposal is conditional on baseline and analysis, rather than a general requirement.

vii

Solution

In vacuum the electron-flavour survival probability depends only on

∣ U_{e i} ∣^{2}

and the mass-squared differences. Complex conjugation of the lepton mixing matrix therefore leaves it unchanged:

P (ν_{e} \to ν_{e}) = P (\overset{ν}{ˉ}_{e} \to \overset{ν}{ˉ}_{e}), \frac{\partial P _{ee}}{\partial δ} = 0.

Thus a reactor disappearance measurement cannot search for the intrinsic CP violation phase in the standard three-neutrino model. Sensitivity to that phase calls for suitable appearance channels and, usually, comparison of neutrino and antineutrino beams. Matter can produce differences between propagation of neutrinos and antineutrinos; these are not by themselves intrinsic CP violation, and are small for the proposed short reactor baseline.

2021 (A11047W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The source is on bundle pages 34–41. Its cover gives Trinity Term 2021 despite the attachment's 2020 filename. These solutions use natural units

ℏ = c = 1

unless a conversion is written explicitly.

1

a

Solution

In a relativistic two-body decay, four-momentum conservation in the parent's rest frame requires

p_{1}^{*} = - p_{2}^{*}

and

M = p^{* 2} + m_{1}^{2} + p^{* 2} + m_{2}^{2} .

For fixed daughter masses, this fixes a unique

p^{*}

, independent of direction. In the Lorentz-invariant phase space integral, the Dirac delta distribution enforcing conservation of energy therefore performs the radial momentum integration. Only the solid angle remains to be integrated; the resulting

p^{*}

can be taken outside that integral, even if the matrix element depends strongly on direction.

For three or more final particles, four-momentum conservation generally leaves continuously variable daughter energies and momenta, so the decay rate includes integrations over them, or equivalently over independent invariant masses in a Dalitz plot. Integrating over a variable daughter invariant mass is also necessary when an unstable daughter is treated with a finite decay width rather than a fixed mass.

b

Solution

Useful examples with no net quark flavor are the eta c meson

η_{c} = c \overset{c}{ˉ}

and the eta b meson

η_{b} = b \overset{ˉ}{b}

. Examples with nonzero quark flavor are the kaon

K^{+} = u \overset{s}{ˉ}

and the D meson

D^{+} = c \overset{ˉ}{d}

. These are pseudoscalar mesons: in the lowest quark model configuration the constituent spins combine to

S = 0

, the relative orbital angular momentum is

L = 0

, and the total angular momentum and parity are

J^{P} = 0^{-}

. For the self-conjugate examples, meson parity and charge conjugation also gives

C = (- 1)^{L + S} = + 1

. Electrical neutrality alone would not imply zero net quark flavor; for example

K^{0} = d \overset{s}{ˉ}

has strangeness.

Restoring reduced Planck constant, decay width and mean lifetime satisfy

Γ τ = ℏ

. Thus the representative numbers correspond to

τ_{20 MeV} = \frac{6.582 \times 1 0 ^{- 22} MeV s}{20 MeV} = 3.29 \times 1 0^{- 23} s, Γ_{1 ps} = 6.582 \times 1 0^{- 10} MeV .

The mean lifetimes differ by at least

3.0 \times 1 0^{10}

. Hidden-heavy-flavor quarkonium can undergo quark-antiquark annihilation into gluons, followed by hadronization. A

0^{-+}

color-singlet state can annihilate into two gluons. A lightest open-flavor meson cannot simply annihilate through the strong interaction, which conserves net quark flavor. With no lighter accessible final state carrying that flavor, it must undergo a weak interaction changing a constituent's flavor. The small weak interaction amplitude, CKM matrix factors and available Lorentz-invariant phase space produce a much smaller decay width.

Qualification. A decay width of order tens of MeV is not a universal property of flavor-neutral pseudoscalar mesons. The neutral pion has

Γ ≃ 7.8 eV

and decays predominantly by the electromagnetic interaction, a concrete counterexample. Its mass lies below the lightest allowed hadronic final state. The examples above illustrate the intended annihilation-versus-open-flavor comparison; the numerical scale depends on thresholds and selection rules.^[ref]

c

Solution

Assume the charmonium state annihilates through a virtual photon, with lepton universality and negligible higher-order corrections. For a vector meson of mass

M

, the partial decay width into a charged lepton pair contains both the relativistic two-body decay factor

β_{ℓ}

and a vector-current spin factor:

Γ_{ℓℓ} = C (1 + \frac{2 m _{ℓ}^{2}}{M ^{2}}) 1 - \frac{4 m _{ℓ}^{2}}{M ^{2}}, β_{ℓ} = 1 - 4 m_{ℓ}^{2} / M^{2} .

^[ref] For example, contracting the vector polarization sum with the Dirac trace for the two final fermions gives a factor proportional to

M^{2} + 2 m_{ℓ}^{2}

, explaining why retaining only

β_{ℓ}

is insufficient. Using

m_{τ} = 1.777 GeV

m_{μ} = 0.1057 GeV

and

M = 3.686 GeV

gives

\frac{Γ _{ττ}}{Γ _{μμ}} ≃ 0.388, B_{ττ} ≃ 0.008 \times 0.388 = 3.10 \times 1 0^{- 3} = 0.31%.

The supplied total decay width also gives

Γ_{μμ} = 2.352 keV

and

Γ_{ττ} ≃ 0.91 keV

. It cancels from the branching fraction ratio.

The decisive change at

3.770 GeV

is crossing the open-charm D meson-pair threshold, approximately

3.73 GeV

. The strong interaction can now create a light quark-antiquark pair and rearrange

c \overset{c}{ˉ}

into

D \overset{ˉ}{D}

, without annihilating the heavy pair. This increases the total decay width from hundreds of keV to tens of MeV. The leptonic decay partial decay width is additionally smaller because

ψ (3770)

is predominantly a

1^{3} D_{1}

state, whereas

ψ (3686)

is predominantly

2^{3} S_{1}

; the short-distance annihilation overlap is suppressed for nonzero orbital angular momentum. A roughly hundredfold increase in total decay width together with an order-ten reduction in leptonic partial decay width explains the order-thousand reduction in branching fraction. Attributing all three orders of magnitude to the threshold alone would miss the second effect.^[ref]

d

Solution

At an electron-positron collider tuned to the charmonium resonance, select the mixed leptonic tau lepton decays

τ^{-} \to e^{-} \overset{ν}{ˉ}_{e} ν_{τ}, τ^{+} \to μ^{+} ν_{μ} \overset{ν}{ˉ}_{τ},

and their charge conjugates. Require exactly two well-reconstructed oppositely charged tracks, one identified electron and one identified muon, little extra visible activity, visible energy below the beam total, and significant missing momentum pointing into the instrumented region. The multiple neutrinos prevent full reconstruction of each tau lepton, but the known initial four-momentum, lepton spectra and missing energy support a statistical signal extraction. Off-resonance data constrain nonresonant electron-positron annihilation into a tau lepton pair; that continuum is present at both beam energies.^[ref]

A resonant muon pair is one possible detector background: one muon can be misidentified as an electron, while radiation escaping along the beam or a badly measured track produces apparent missing energy. Particle-identification requirements and rejection of almost back-to-back tracks near the beam momentum suppress it. This background needs detector mismeasurement; an ideal measurement would distinguish it.

Above the open-charm threshold,

ψ (3770) \to D \overset{ˉ}{D}

supplies an additional resonant background. Semileptonic decays of the two D mesons can give an electron, a muon and neutrinos; unreconstructed hadrons let some events pass the selection. This on-shell D meson-pair background is kinematically forbidden at

ψ (3686)

e

Solution

The three structures are the radiative cascades

ψ (2 S) \to γ_{l} χ_{c J}, χ_{c J} \to γ_{h} J / ψ, J = 0, 1, 2.

In the quark model, the J/psi meson and

ψ (2 S)

are respectively

1^{3} S_{1}

and

2^{3} S_{1}

charmonium states. Thus

L = 0

S = 1

J = 1

, and meson parity and charge conjugation gives

J^{PC} = 1^{--}

. The intermediate chi c mesons are

1^{3} P_{J}

L = 1

S = 1

, and addition of angular momentum permits

J = 0, 1, 2

. Consequently

χ_{c 0} : 0^{++}, χ_{c 1} : 1^{++}, χ_{c 2} : 2^{++} .

These are electric dipole transitions, changing

L

by one and preserving

S

to leading order. A singlet

1^{1} P_{1}

state has

J^{PC} = 1^{+-}

and is not another member of these spin-triplet cascades.

The primary photon energy is

(M_{2 S}^{2} - M_{χ_{J}}^{2}) / (2 M_{2 S})

, about

261

171

and

128 MeV

for

J = 0, 1, 2

. The secondary photon remains more energetic even after its boost into the parent rest frame. Hence

m^{2} (J / ψ γ_{h}) = M_{χ_{J}}^{2}

: with this quantity on the vertical axis, each resonance gives a horizontal band. A finite decay width and detector measurement resolution broaden each band; photon ordering removes the mirrored duplication of an unordered Dalitz plot.

The faint band beside the negative-slope diagonal boundary is

ψ (2 S) \to J / ψ π^{0}

, followed by

π^{0} \to γγ

. The neutral pion is predominantly

(u \overset{u}{ˉ} - d \overset{ˉ}{d}) / 2

, a

^{1} S_{0}

pseudoscalar meson with

J^{PC} = 0^{-+}

. This production transition violates isospin, accounting for its small rate. To locate the band, set

X = m^{2} (J / ψ γ_{l})

and

Y = m^{2} (J / ψ γ_{h})

. Four-momentum conservation gives

X + Y = M_{2 S}^{2} + M_{J / ψ}^{2} - m_{γγ}^{2} .

For the neutral pion,

m_{γγ}^{2} = m_{π^{0}}^{2}

is small, putting the line just inside the

m_{γγ}^{2} = 0

boundary. The allowed

J / ψ η

channel, with the eta meson also

0^{-+}

, produces a parallel line further inside, at

m_{γγ}^{2} = m_{η}^{2}

. The positive-slope edge

X = Y

instead comes from ordering the two photon energies; proximity to that edge alone would not identify the two-photon state.

2

a

Solution

Use the Minkowski metric

(+, -, -, -)

and choose the incoming electron along

+ z

, with the scattered electron in the

x z

plane. In natural units, neglecting the electron mass, the four-momenta are

p_{1} = (E, 0, 0, E), p_{2} = (m_{p}, 0, 0, 0), p_{3} = (E^{'}, E^{'} sin θ, 0, E^{'} cos θ) .

The momentum transfer entering the proton is therefore

q = p_{1} - p_{3} = (E - E^{'}, - E^{'} sin θ, 0, E - E^{'} cos θ) .

The outgoing hadronic system carries

p_{X} = p_{2} + q

. The energy transfer is

ν = E - E^{'}

, while

q

need not point along the original beam.

Figure 1.
Fixed-target deep inelastic scattering: laboratory directions and a schematic detector
. **Details**.

b

Solution

With the four-momenta in a,

q^{2} = - 2 E E^{'} (1 - cos θ)

p_{2} \cdot q = m_{p} (E - E^{'})

, and

p_{2} \cdot p_{1} = m_{p} E

. Thus the measured beam energy

E

, scattered electron energy

E^{'}

and scattering angle

θ

determine

Q^{2} = 4 E E^{'} sin^{2} \frac{θ}{2}, y = 1 - \frac{E ^{'}}{E}, x = \frac{2 E E ^{'} sin ^{2} ( θ /2 )}{m _{p} ( E - E ^{'} )} .

Here

Q^{2}

measures the spacelike momentum transfer,

y

is the fractional energy transfer, and the Bjorken scaling variable

x

is the struck parton's longitudinal momentum fraction in the leading collinear description. The hadronic invariant mass satisfies

W^{2} = (p_{2} + q)^{2} = m_{p}^{2} + Q^{2} (1/ x - 1)

; large

Q^{2}

and

W^{2}

select deep inelastic scattering.

A beam spectrometer measures

E

before a thin stationary hydrogen target. Tracking layers on both sides of a magnetic spectrometer measure the outgoing electron direction, charge and curvature; an electromagnetic calorimeter measures its energy. Matching the track to its particle shower, with downstream muon rejection where useful, separates the scattered electron from hadronic tracks. A downstream hadronic detector measures the hadronic activity and supports event-quality checks. The sketch in a shows these separate measurements.

The scattered electron provides a clean, single-particle reconstruction. Reconstructing the entire hadronic system requires summing many charged and neutral fragments, correcting unequal particle calorimeter response and losses through beam holes, and sometimes accounting for escaping neutrinos. Thus the leptonic definition is experimentally convenient. It is not a fundamental requirement of Lorentz invariance: exact reconstruction of

p_{X}

gives the same

q = p_{X} - p_{2}

, and hadronic or combined reconstructions can be preferable in some kinematic regions.

c

Solution

At leading order in QED, the electron exchanges one spacelike virtual photon with a charged parton inside the proton.

Figure 1.
One-photon-exchange Feynman diagram for inclusive electron-proton deep inelastic scattering
. **Details**.

The two Feynman vertices contribute

e

and

e e_{q}

, while the photon propagator contributes

1/ q^{2} = - 1/ Q^{2}

. Hence the scattering amplitude scales as

e^{2} e_{q} / Q^{2}

and its squared absolute value as

e^{4} e_{q}^{2} / Q^{4}

. Since the fine-structure constant is

α = e^{2} / (4 π)

, the electromagnetic coupling dependence is

α^{2} / Q^{4}

. The numerical

4 π

in the stated differential scattering cross-section additionally uses the flux, Lorentz-invariant phase space, azimuthal integration and the convention defining the deep-inelastic structure function; it cannot be deduced from the propagator alone.

For each measured

x, Q^{2}

bin form the reduced quantity

F_{2} (x, Q^{2}) = \frac{x Q ^{4}}{4 π α ^{2} [ 1 - y + y ^{2} /2 ]} \frac{d ^{2} σ}{d x d Q ^{2}} .