topics in standard model - universiteit leidenboyarsky/media/lecture1.pdf · quantum...

Topics in Standard Model

Alexey BoyarskyAutumn 2013

Course material

Books used in this course:

Bjorken-Drell: James D. Bjorken, Sidney David Drell Relativisticquantum mechanics, Vol. 1, McGraw-Hill, 1964

Aitchison-Hey: I.J.R. Aitchison, A.J.G. Hey, Gauge Theories inParticle Physics, Vol. 1: From Relativistic Quantum Mechanics toQED, 3rd edition

Cheng-Li: T.-P. Cheng and L.-F. Li, Gauge theory of elementaryparticle physics. Clarendon Press, Oxford, 1982.

Landau-Lifshitz-vol2: L.D. Landau, E.M. Lifshitz (1975). TheClassical Theory of Fields. Vol. 2 (4th ed.). Butterworth-Heinemann.

Landau-Lifshitz-vol3: L.D. Landau, E.M. Lifshitz (1977). Quantum

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 1

Course material

Mechanics: Non-Relativistic Theory. Vol. 3 (3rd ed.). PergamonPress.

Landau-Lifshitz-vol4: V.B. Berestetskii, E.M. Lifshitz, L.P. Pitaevskii(1982). Quantum Electrodynamics. Vol. 4 (2nd ed.). Butterworth-Heinemann


Maxwell’s equations

¥ In 1860s Maxwell wrote a set of field equations describingelectromagnetism

div ~E = ρ (1)

div ~B = 0 (2)

curl ~E = −1c

∂ ~B

∂t(3)

curl ~B =1c

∂ ~E

∂t+ ~ (4)

we will always work in vacuum, so that ε = µ = 1 and as a result ~D = ~E and ~B = ~H! The

CGS system of units is used, so that ε0 = µ0 = 1 and dimensions of ~E and ~B are the same!



¥ Check that as a consequence of Maxwell’s equations (1)–(4)

1. One can introduce functions A0 and ~A such that

~B = curl ~A

~E =− 1

c

∂ ~A

∂t− ~∇A0

(5)

and Eqs. (2)–(3) are satisfied identically2. Eqs. (5) define A0 and ~A up to an arbitrary scalar function λ(t, ~x) such that

changing~A→ ~A+ ~∇λ

A0 → A0 −1

c

∂λ

∂t

(6)

does not change the l.h.s. of Eq. (5)3. Derive equations on ~A and ~A0 based on Eqs. (1) and (4)4. Show that current should be conserved ∂ρ

∂t + div~ = 0

5. Show that in a space without sources ~E and ~B obey wave equation:

1

c2∂2 ~E

∂t2−∆~E = 0 (7)



¥ Maxwell has predicted that in the empty space electric andmagnetic fields can exist – electromagnetic waves.

¥ Equation (7) means that the electromagnetic wave propagatesthrough the space with the speed of light c.

¥ Hertz had discovered radio-waves and confirmed this prediction

¥ Maxwell’s discovery meant that there is no “action-at-a-distance”but rather two electric (magnetic) bodies interact via emitting andreceiving electromagnetic waves

¥ Maxwell’s equations possess Lorentz symmetry .1 To see this,define 4-vector Aµ = (A0, ~A) (based on (5)) and define the fieldstrength

Fµν ≡ ∂Aµ∂xν

− ∂Aν∂xµ

(8)1To remind yourself about 4-vectors, Lorentz transformations and Lorentz covariant formulation of

electrodynamics, see Landau & Lifshitz, vol. 2, §§4–7, §§16–17, §23



where xµ = (ct, ~x).

¥ Find the relation between Fµν and ~E, ~B so that (8) is equivalent to (5)

¥ The Maxwell’s equations take the form

∂Fµν

∂xµ= jν (9)

where jµ = (ρ,~) – 4-vector of electromagnetic current and Fµν =ηµαηνβFαβ where ηαβ is the Minkowski metric

¥ Lorentz symmetry is a rotation in the 4-dimensional space (givenby the 4×4 antisymmetric matrix Λµν) so that any vector translatesas

Xµ = ΛµνXν (10)

and field strength Fµν translates as

Fµν = ΛµαΛνβF

αβ (11)



¥ Show that (9) is equivalent to (1) and (4)

¥ Show that Eq. (9) with jµ = 0 can be rewritten as 4 wave equations for 4components of Aµ, supplemented by the Lorentz gauge condition ∂µAµ = 0


Photons

¥ To explain some experiments (black body radiation; photoelectriceffect) one was led to assume that the electromagnetic radiation isquantized and that a “quantum of light” – photon – exists (Einstein

1905)

¥ This has been confirmed by Compton in 1923 (Comptonscattering)

`

λ′ − λ =h

mec(1− cos θ)

– X-rays, scattering of atomicelectrons change wave length

– Naturally explained by theelastic scattering of two particles(electron and photon)

– Wave would not change itsfrequency (Thomson scattering)


Quantum Mechanics

¥ The atomic physics led to the creation of Quantum mechanics

¥ The quantum system is described by Schrodinger equation:

i~∂ψ(x, t)∂t

= Hψ(x, t) (12)

where the operator H is called Hamiltonian

¥ The operator Hamiltonian is built based on correspondenceprinciple: if a classical system is described by the HamiltonianH(x, p) then the quantum system is described by H(x,−i~∂x). Forexample for a particle of mass m in external field we get

Classical H(x, p) =p2

2m+ V (x) ⇒ Quantum H = − ~

2

2m∂2

∂x2+ V (x)

(13)


Solution of Schodinger equation

¥ Recall some important facts about the Schrodinger equation:2

– Consider

0 =

Zd

3x

»ψ∗“i~∂ψ

∂t−Hψ

”− ψ

“−i~∂ψ

∗

∂t−Hψ

∗”–

⇔ ∂

∂t

Zd

3x|ψ(x)|2 = 0

(14)

– |ψ(x)|2 is interpreted as probability density and∫d3x|ψ(x)|2 =

const as the full probability– ~J = i~

2m(ψ∗∇ψ − ψ∇ψ∗) is interpreted as a probability densitycurrent so that

∂|ψ|2∂t

+ div ~J = 0 (15)

Check this relation– Uncertainty principle ∆x∆p ∼ ~

2To remind yourself about this material, see e.g. Landau & Lifshitz, Vol. 3, §19


Solution of Schodinger equation

– Solution of the Schodinger equation for the hydrogen atom waspredicting the energy spectrum

En = −mee4

2~2n2, n = 1, 2, . . . (16)

for each n the degeneracy (the number of states) of the energylevel is

∑n−1l=0 (2l + 1) = n2.

This latter fact contradicted to the observations as the lowest energy level ofHydrogen contained two electrons


Pauli Hamiltonian

¥ In order to explain the emission spectra of elements Pauliintroduced a spin (“a two-valued quantum degree of freedom” ).Its interpretation was not known. Pauli postulated Pauli (or Pauli-Schrodinger) equation

HPauli =1

2m

[~σ · (−i~~∇− e ~A

)]2

+ eA01 (17)

where Pauli matrices ~σ = (σx, σy, σz) are defined as

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)(18)

and 1 is 2 × 2 identity matrix. Pauli Hamiltonian acts on the 2-component wave-function

i~∂

∂t

(ψ↑ψ↓

)= HPauli

(ψ↑ψ↓

)(19)


Pauli matrices

Show that

– Pauli matrices are Hermitian– For each Pauli matrix σ2

i = 1– For any two Pauli matrices

{σi, σj} ≡ σiσj + σjσi = 2δij1 i, j = 1, 2, 3 (20)

– σxσy = iσz and in general

[σi, σj] = 2iεijkσk (21)

– As a consequence of (20) and (21) one finds

(~σ · ~X)(~σ · ~Y ) = ~X · ~Y 1 + i~σ · ( ~X × ~Y ) (22)

– Use the identity (22) to rewrite the Pauli Hamiltonian as

HPauli =1

2m(−i~~∇− e ~A)

2 − ~e2m

~σ · ~B + eA01 (23)


Pauli matrices

– Compute the matrix U = ei2ασz (as the Taylor expansion of the exponent) and

show that matrix U is unitary. Try to repeat this for arbitrary ei2~α·~σ

– Show that any 2× 2 unitary matrix U with detU = 1 can be represented as

U = exp“ i2(n0 + ~n · σ)

”; n

20 + ~n

2= 1 (24)


Quantum mechanics and relativity

Attempts to blend quantum mechanics and special relativity led to theproblem of infinities.3

¥ Electromagnetic energy of uniformly charged ball of radius reshould be smaller than the total mass of electron (mec

2)?

mec2 >

e2

re=⇒ re >

e2

mc2= 3× 10−13 cm (25)

– larger than the size of atomic nucleus (Hydrogen rH ∼ 1.75 ×10−13 m)

¥ Pauli’s idea of spin meant that electron possesses magneticmoment: µe = e~

2mec. The energy of magnetic sphere with radius

re would be µ2er3e

which would exceed mec2 for r ∼ 1

me

(e~c4

)2/3(even

larger distances than (25!)3See an interesting exposition of the historical perspectives at

http://people.bu.edu/gorelik/cGh_Bronstein_UFN-200510_Engl.htm, Sec. 4


Relativistic Schodinger equation 4

¥ Free Schrodinger equation is not Lorentz invariant (indeed, it isobtained from E = p2

2m dispersion relation)

¥ Replace r.h.s. by relativistic dispersion relation E =√p2c2 +m2c4

Relativistic Schrodinger equation ? i~∂ψ

∂t=

√−~2~∇2 +m2c4 ψ (26)

¥ How to make sense of square root? Try to take square of Eq. (26)?

− ~2∂2ψ

∂t2=

(−~2c2∇2 +m2c4

)ψ ⇔

(¤ +

(mc~

)2)ψ = 0 (27)

This is Klein-Gordon equation. Here ¤ = 1c2∂2

∂t2− ~∇2 is the

D’Alembert operator. Show that ¤ is Lorentz invariant

3The presentation of this topic follows Bjorken & Drell, Chap. 1, Sec. 1.1–1.3


Problems with Klein-Gordon equation

¥ By taking square of (26) we have included negative energy states(with E = −

√p2c2 +m2c4)

¥ Probabilistic interpretation is gone? Indeed, let us checkprobability density |ψ|2:

∂

∂t

∫d3x |ψ(x)|2 6= 0 (28)

¥ Repeating trick similar to Eq. (14) we find that the following currentis conserved as a consequence of Klein-Gordon equation:

Jµ = ψ∗∂ψ

∂xµ− ψ

∂ψ∗

∂xµ(29)

that is1c

∂J0

∂t+ div ~J = 0 (30)


Problems with Klein-Gordon equation

¥ Can we use J0 as the probability density? No!

J0 = ψ∗∂ψ

∂t− ψ

∂ψ∗

∂t= |ψ|2∂(argψ)

∂t< 0 (31)

for any negative energy solution


Dirac equation 5

¥ Dirac (1928) had suggested that one needs linear in time evolution.Indeed, if i~∂tψ = Hψ than the probability density of the form (28)is conserved for any Hermitian Hamiltonian (c.f. (14)).

¥ Dirac proposed the form

i~∂ψ

∂t=

[−i~c

(αx

∂

∂x+ αy

∂

∂y+ αz

∂

∂z

)+ βmc2

]

︸︷︷︸Dirac Hamiltonian

ψ (32)

¥ Take square of Dirac Hamiltonian. We should arrive to the Klein-Gordon equation. This means that for i, j = 1, 2, 3

αiαj + αjαi = 2δij (33)

αiβ + βαi = 0 (34)4The presentation of this topic similar to that of Bjorken & Drell, Chap. 1, Sec. 1.1–1.3


Dirac equation 7

and additionally β2 = 1

¥ ⇒Both αi nor β are not ordinary numbers (they do not commute).

¥ Eq. (33) looks very much like Eq. (20). But guess that αi = σi iswrong there are only 3 Pauli matrices and we need 4

¥ Another guess:6

αi =(

0 σiσi 0

); β =

(1 00 −1

)(35)

Show that properties (33)–(34) are satisfied. Find different representation ofαi, β

6Show that αi, β are even-dimensional matrices


Properties of Dirac gamma-matrices 8

¥ Define 4 γ-matrices (or Dirac gamma-matrices):

γ0 = β; γi = βαi (36)

¥ Define 4-vector γµ = (γ0, γ1, γ2, γ3)

¥ Dirac equation can be rewritten as

i~γµ∂ψ

∂xµ= 0 (37)

– explicitly Lorentz-covariant

¥ Properties of γ-matrices:

γµγν + γνγµ = 2ηµν1 (38)7See Bjorken & Drell, Chap. 2, Sec. 2.1–2.2


Properties of Dirac gamma-matrices 9

¥ Defineσµν =

i

2[γµ, γν] (39)

¥ Show that [σµν, σλρ] commute as operators of Lorentz rotation in3 + 1


Standard Model: great success of particle physics


Standard Model


Standard Model

¥ The Standard Model describes all the confirmed data obtainedusing particle accelerators and has enabled many successfultheoretical predictions.

¥ It has been tested with amazing accuracy, and its calculablequantum corrections play an essential role. Its only missing featureis a particle, called the Higgs boson, whose coupling to the otherparticles is believed to generate their masses.


Standard Model and particle astrophysics

The SM has also been successful in explaining phenomena in theEarly Universe and certain astrophysical systems.

¥ It plays an essential role in the epoch of Big Bang nucleosynthesis– observed abundances of elements are sensitive to details of theStandard Model.

¥ Energy production of main-sequence stars is controlled by nuclearphysics and weak β decay.

¥ Weak neutral-current processes are important for neutrino emissionfrom the collapsing core and ejection of outer layers of thesupernova progenitor.

¥ Propagation of cosmic rays through the Universe.


End of Lecture 1

End of Lecture I


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