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1 Particle Physics

Dr M.A. Thomson

´ Ñµ ¼

µ- µ-

e- e+ e- e+ µ+ µ+ µ- µ-

e- e+ e- e+ µ+ µ+

Part II, Lent Term 2004 HANDOUT V

Dr M.A. Thomson Lent 2004 2

Spin, Helicity and the

Dirac Equation

★ Upto this point we have taken a hands-off approach to “spin”.

★ Scattering cross sections calculated for spin-less particles

★ To understand the WEAK interaction need to understand SPIN

★ Need a relativistic theory of quantum mechanics that includes spin

★ µ The DIRAC EQUATION

Dr M.A. Thomson Lent 2004 3 ★ SPIN complicates things.... e- µ- The process γ + e+ µ represents the sum over all possible spin states µ- µ- µ- µ-

e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ-

e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ-

e- e+ e- e+ e- e+ e- e+ µ+ µ+ µ+ µ+ µ- µ- µ- µ-

e- e+ e- e+ e- e+ e- e+

µ+ µ+ µ+ µ+

Å Å

¾ ¾ ¾

Å Å Å

since ORTHOGONAL SPIN states.

½

¾

¾ Å ´ µ

Cross-section : sum over all spin assignments, averaged over initial spin states.

Dr M.A. Thomson Lent 2004 4 The Klein-Gordon Equation Revisited

Schrodinger¨ Equation for a free particle can be

written as

½

¾

Ö

Ø ¾Ñ

Derivatives : 1st order in time and 2nd order in

space coordinates µ not Lorentz invariant

From Special Relativity:

¾ ¾ ¾

Ô · Ñ

from Quantum Mechanics:

Ô Ö

Ø

Combine to give the Klein-Gordon Equation:

¾

¾ ¾

´Ö Ñ µ

¾ Ø Second order in both space and time derivatives - by construction Lorentz invariant. ★ Negative energy solutions

µ anti-particles ★ BUT negative energy solutions also give

negative particle densities !?

£

¼

Try another approach.....

Dr M.A. Thomson Lent 2004 5 Weyl Equations

(The massless version of the Dirac Equation)

Klein-Gordon Eqn. for massless particles:

¾

¾

´ Ö µ ¼

¾

Ø

¾

¾

´ Ô µ ¼

Try to factorize 2nd Order KG equation ➔

equation linear in Ö AND :

Ø

Öµ´ · Öµ ¼ ´

Ø Ø with as yet undetermined constants

Gives the two decoupled WEYL Equations

´ · · µ ·

Ü Ý Þ

Ü Ý Þ Ø

· · µ ´

Ý Þ Ü

Ü Ý Þ Ø both linear in space and time derivatives. BUT must satisfy the Klein-Gordon Equation

i.e. in operator form

´ Ôµ´ · Ôµ ¼

must satisfy

¾

¾

´ Ô µ ¼

Dr M.A. Thomson Lent 2004 6

Weyl equations give:

¾

´ Ô Ô Ô Ô Ô Ô

Ü Ü Ý Ý Þ Þ

Ü Ü Ý Ý Þ Þ

Ô Ô Ô Ô

Ü Ý Ý Ü

Ü Ý Ý Ü

Ô Ô Ô Ô

Ý Þ Þ Ý

Ý Þ Þ Ý

Ô Ô Ô Ô µ ¼

Þ Ü Ü Þ

Þ Ü Ü Þ

Therefore in order to recover the KG equation:

¾

´ Ô Ô Ô Ô Ô Ô µ ¼

Ü Ü Ý Ý Þ Þ

require:

¾ ¾ ¾

½

Ü Ý Þ

´ · µ ¼ Ø

Ü Ý Ý Ü

µ

Ü Þ , Ý , ANTI-COMMUTE

The simplest choice for are the Pauli spin matrices

½ ¼ ¼ ½

¼ -

Þ Ý Ü

¼ ½ ¼ ½ ¼ -

Hence solutions to the Klein-Gordon equation

´ Ôµ´ · Ôµ ¼

are given by the Weyl Equations:

Ôµ ¼ ´

´ · Ôµ ¼

¾ ¢ ¾

Since are matrices, need 2 component

wave-functions - WEYL SPINORS.

½

´Ø Ô Öµ

Æ

¾ The wave-function is forced to have a new degree of freedom - the spin of the fermion.

Dr M.A. Thomson Lent 2004 7

Consider the FIRST Weyl Equation

´ Ôµ ¼

´ · · · µ ¼

Ü Ý Þ

Ø Ü Ý Þ

For a plane wave solution:

½

´Ø Ô Öµ

Æ

¾

the ﬁrst Weyl Equation gives

´ Ô Ô Ô µ ¼

Ü Ü Ý Ý Þ Þ

´ Ô µ ¼

where

Ô Ô · Ô · Ô

Ü Ü Ý Ý Þ Þ

Ô Ô Ô

Þ Ü Ý

Ô · Ô Ô

Ü Ý Þ

Hence for the ﬁrst WEYL equation, the SPINOR solutions of

Ô µ ¼

´ are given the coupled equations:

µ

Ô · ´Ô Ô µ

Þ ½ Ü Ý ¾ ½

µ

´Ô · Ô µ Ô

Ü Ý ½ Þ ¾ ¾

Dr M.A. Thomson Lent 2004

8

★ Ô Þ Ô Ô

Choose along axis, i.e. Þ :

´ Ôµ ¼

½

´ · Ôµ ¼

¾

½ ¼

There are two solutions · and .

¼ ½

½

★ ·Ô

The ﬁrst solution, · , requires , i.e. ¼

a positive energy (particle) solution. Similarly, the second

¼

Ô

, requires , i.e. a negative energy ½ (anti-particle) solution.

★ Back to the FIRST WEYL equation

´ Ôµ ¼

Ô

´

·½ ¼

Ô

Ô

Ô ½ ¼

★ The solutions of the WEYL equations are Eigenstates of

the HELICITY operator.

Ô

À

Ô

½ with Eigenvalues ·½ and respectively. σ p

HELICITY is the projection of a particle’s SPIN onto its ﬂight direction. (Recall WEYL equations applicable for massless particles) Dr M.A. Thomson Lent 2004 9

★ Interpret the two solutions of the FIRST WEYL equation

·½

as a RIGHT-HANDED À particle and a

½ LEFT-HANDED anti-particle À . E > 0, H = +1 E < 0, H = -1

RH particle LH anti-particle

★ The SECOND WEYL equation:

´ · Ôµ ¼

has LEFT-HANDED particle and RIGHT-HANDED anti-particle solutions. E < 0, H = +1 E > 0, H = -1

RH anti-particle LH particle

SUMMARY:

★ By factorizing the Klein-Gordon equation into a form

linear in the derivatives µ force particles to have a non-commuting degree of freedom, SPIN !

★ Still obtain anti-particle solutions

★ Probability densities always positive

★ ‘Natural’ states are the Helicity Eigenstates

★ Weyl Equations are the ultra-relativistic (massless) limit of the Dirac Equation

Dr M.A. Thomson Lent 2004 10 Spin in the Fundamental Interactions

★ The ELECTROMAGNETIC, STRONG, and WEAK interactions are all mediated by VECTOR (spin-1) ﬁelds. In the massless limit, the fundamental fermion states are eigenstates of the helicity operator. HANDEDNESS ★ (CHIRALITY) plays a central roleˆ in the interactions between the ﬁeld bosons and the fermions; the only allowed couplings are: LH particle to LH particle RH particle to RH particle LH anti-particle to LH anti-particle RH anti-particle to RH anti-particle LH particle to RH anti-particle

RH particle to LH anti-particle

· ·

EXAMPLE Of the 16 possibilities ONLY the following SPIN combinations contribute to the cross-section µ- µ-

e- e+ e- e+ µ+ µ+ µ- µ-

e- e+ e- e+

µ+ µ+

¾

Å

All other SPIN combinations give zero

Dr M.A. Thomson Lent 2004 11 Solutions of the Weyl Equations

Consider the general case of a particle with travelling at an Þ

angle with respect to the -axis

Ô Ô Ó×

Þ

Ô Ô ×Ò

Ü

Ôµ ¼

WEYL 1 ´

´ Ôµ

µ

Ô · Ô

Þ ½ Ü ¾ ½

µ

Ô Ô

Ü ½ Þ ¾ ¾

·Ô

For the positive energy solution :

µ

Ó× · ×Ò

½ ¾ ½

µ

×Ò Ó×

½ ¾ ¾

×Ò

µ

½ ¾

´½ Ó× µ

Therefore

¾×Ò Ó×

½

¾ ¾

¾

¾

´½ Ó× · ×Ò µ

¾

¾ ¾

Ó×

½

¾

×Ò

¾

¾

¾ ¾

· ½ ½

Normalizing such that ¾ gives:

Ó×

½

¾

×Ò

¾ ¾

Dr M.A. Thomson Lent 2004 12 ★ So the positive energy solution to the ﬁrst Weyl equation

(RH particle) gives

+ Ó×

´Ø Ô Öµ

¾

Æ ÊÀ ÖÑÓÒ

ÊÀ

+ ×Ò ¾

This is still a Eigenvalue of the helicity operator with

À ·½µ ´ i.e. a RH particle but now referred to an external axis. The positive energy solution to the SECOND Weyl equation

(LH particle) gives

- ×Ò

´Ø Ô Öµ

¾

Æ ÄÀ ÖÑÓÒ

ÄÀ

+ Ó× ¾

Not much more than the Quantum Mechanical rotation

properties of spin- ½ . ¾ ★ The spin part of a RH particle/anti-particle wave-function

can be written

+ Ó×

¾

´ µ · ×Ò Ó×

Ê

¾ ¾

+ ×Ò ¾

★ Similarly for a LH particle/anti-particle

- ×Ò

¾

· Ó× ´ µ ×Ò

Ä

¾ ¾

+ Ó×

¾

Ô´ µ

For particles/anti-particles with momentum , i.e. an

·

angle to the z-axis:

· Ó× ´ · µ ×Ò

Ê

¾ ¾

×Ò ´ · µ Ó×

Ä

¾ ¾

Dr M.A. Thomson Lent 2004

13

· ·

Application to

µ- µ-

e- e+ e- e+ ★ Four helicity combinations µ+ µ+ - - contribute to the cross-section. µ µ e- e+ e- e+ µ+ µ+

★ Consider the ﬁrst diagram µ-

· ·

- +

Ä Ê Ä

Ê e e

With the direction deﬁn- µ+

ing the Þ axis:

★ The spin parts of the wave-functions :

´¼µ ´ µ

· Ä

= Ê =

Ê Ä

´ µ ´ · µ

·

Ä

= Ê

Ä Ê

¾

¾

Ó× Ó× µ ×Ò ×Ò ´

=- - + -

¾ ¾ ¾ ¾

★ Giving the contribution the matrix element:

¾

¾ ¾

Å

· ·

½

¾

Ê Ä Ê Ä

Õ

¾

¾

Ó×

¾

Õ

¾

½

¾

´½ · Ó× µ

Õ ¾

Dr M.A. Thomson Lent 2004 14 ★ Perform same calculation for the four allowed helicity combinations µ- - + α 2 θ e e M1 cos ( /2) µ+ → 1 θ 2 (1+cos ) µ- - + α 2 θ e e M2 sin ( /2) µ+ → 1 θ 2 (1-cos ) µ- - + α 2 θ e e M3 sin ( /2) µ+ → 1 θ 2 (1-cos ) µ- - + α 2 θ e e M4 cos ( /2) µ+ → 1 θ 2 (1+cos ) ★ For unpolarized electron/positron beams : each of the above process contributes equally. Therefore SUM over all matrix elements and AVERAGE over initial spin states. Giving the total Matrix Element (remember spin-states are

orthogonal so sum squared matrix elements) :

¨ ©

½

¾ ¾ ¾ ¾ ¾

Å Å · Å · Å · Å

½ ¾ ¿

½

½ ½

¾ ¾ ¾

´½ · Ó× µ · ´½ Ó× µ · Å

Õ

½ ½

¾ ¾

´½ Ó× µ · ´½ · Ó× µ

¾

¾

´½ · Ó× µ Å

Õ ★ Nothing more than the QM properties of a SPIN-1 particle

decaying to two SPIN- ½ particles ¾

Dr M.A. Thomson Lent 2004 15 + f e µ f p2 γ θ √α √α - + Qf e e µ 1 p - 1 q2

e f f

¾ ¾

´Õ ×µ Electron/Positron beams along Þ -axis

Using the spin-averaged matrix element

¾

¾

´½ · Ó× µ Å

Õ

¾

´«µ

¾

¾

Å ´½ · Ó× µ

¾

×

¾

¾

¾ Å

¿

ª ´¾ µ

¾

´«µ × ½

¾

¾ ´½ · Ó× µ

¾ ¿

× ´¾ µ

¾ ¾

« É

¾

´½ · Ó× µ

×

·

ÕÕ

for .

Ó×

The angle is determined from

the measured directions of the

Ó× jets. is plotted since it is not possible to uniquely identify which jet corresponds to the quark and which corresponds to the anti-

quark. The curve shows the ex-

¾

· Ó× µ pected ´½ distribution.

QUARKS are SPIN- ½ ¾

Dr M.A. Thomson Lent 2004 16

(yet again)

·

Total cross section for

ª

ª

¾ ¾

¾

« É

¾

´½ · Ó× µ×Ò

×

¼ ¼

¾ ¾

·½

« É

¾

´½ · Ý µÝ ´Ý Ó× µ

¾×

½

¾ ¾

« É

¿×

¾

«

· ·

´ µ

¿×

· ·

´ µ

· for collider data at centre-of-mass ener- gies 8-36 GeV

★ This is the complete lowest order calculation of the

· ·

cross-section (in the limit of massless fermions). Dr M.A. Thomson Lent 2004 17 The Dirac Equation

NON-EXAMINABLE

WEYL Equations describe massless SPIN- ½ particles. But ¾ all known fermions are MASSIVE. Again start from the KG

equation.

¾

¾ ¾

´Ö Ñ µ

¾

Ø

¾

¾

¾

À ´Ô · Ñ µ

Write down equation LINEAR in space and time derivatives

À ´« Ô · ¬ Ñµ

and require it to be a solution of the KG equation:

À ´« Ô · « Ô · « Ô · ¬ Ñµ

Ü Ý Þ

Ü Ý Þ

¾

¾

¾

À « Ô ·

Ü

·´« « · « « µÔ Ô ·

Ü Ý Ý Ü

Ü Ý

·´« ¬ · ¬« µÔ Ñ ·

Ü Ü

Ü

¾ ¾

·¬ Ñ

For this to satisfy Klein-Gordon equation:

¾

¾ ¾ ¾

¾

À ´Ô · Ô · Ô · Ñ µ

Ü Ý Þ

require

¾ ¾

« ¬ ½ ´ Ü Ý Þ µ

« « · « « ¼ ´ µ

« ¬ · ¬« ¼

Now require 4 anti-commuting matrices.

Dr M.A. Thomson Lent 2004 18

The Dirac Equation:

À ´« Ô · ¬ Ñµ

Can be written in a slightly different form

´ « Ö · ¬ Ñµ

Ø

¬ ´ ¬«Ö · Ñµ

Ø

¾

¬ ¬«Ö ¬ Ñµ ¼

´ + - =

Ø

¾

½ ¾ ¿ ¼

¬ Ñµ ¼

´ + + + - =

Ø Ü Ý Þ

´¬ ¬«µ with

Giving

´ Ñµ ¼

with

¾

¼

´ µ ½

¾ ¾ ¾

½ ¾ ¿

´ µ ´ µ ´ µ ½

µ ¼ ´ µ ´

Identify the as matrices which must satisfy the anti-commutation relations above. The Pauli spin matrices provide only 3 anti-commuting matrices and the lowest

dimension matrices satisfying these requirements are

¢ ¾ ¢ ¾ . The -matrices are closely related to the Pauli spin matrices.

Dr M.A. Thomson Lent 2004

19

¼ ½

½ ¼ ¼ ¼

¼ ½ ¼ ¼ ½ ¼

¼

¼ ¼ ½ ¼ ½

- ¼ -

¼ ¼ ½

¼ -

½ ¼

¼ ¼ ¼ ½

¼ ¼ ¼ ½ ¼

½

Ü

¼ ½ ¼ ¼ ¼

- - Ü

¼ ¼ ¼

- ½

¼ ½

¼ ¼

¼ -

¼

¼ ¼ ¼

Ý

¾

¼

¼ ¼ ¼

- Ý

¼ ¼ ¼

-

¼ ½

¼ ¼ ½ ¼

¼ ¼ ¼ ½ ¼ ¿

- Þ

½ ¼ ¼ ¼ ¼

- - Þ

¼ ½ ¼ ¼ ½

also deﬁne ¼

¼ ¼ ½ ¼

¼ ½ ¼ ¼ ¼ ½

¼ ½ ¾ ¿

½ ¼ ½ ¼ ¼ ¼

¼ ½ ¼ ¼

Solutions to the Dirac Equation are written as

four-component Dirac SPINORs

½ ¼

½

¾

¿

NOTE: this is not the only possible representation

of the matrices - just the most commonly used Dr M.A. Thomson Lent 2004 20

Rest Frame Solutions of the Dirac Equation

´ Ñµ ¼

Dirac Equation:

¼ ½ ¾ ¿

´ Ñµ ¼

+ + + -

Ø Ü Ý Þ

Ô ¼

Consider a particle at REST: Ü , etc.

Ü

Dirac Equation becomes:

¼

´ Ñµ ¼

Ø

½ ¼

¼ ½ ½ ¼

" Ø

½ ¼ ¼ ¼

½

½

" Ø ¼ ½ ¼ ¼

¾ ¾

Ñ

½ ¼ ¼ ¼

" Ø

-

¿

¿

¼ ¼ ¼ ½

-

" Ø

½ ¾

Ñ Ñ

½ ¾

Ø Ø

·Ñ

Giving two orthogonal solutions:

¼ ½ ¼ ½

½ ¼

¼ ½

ÑØ ÑØ

Ù ´Øµ Ù ´Øµ

½ ¾

¼ ¼

¼ ¼

i.e. positive energy spin-up and spin-down PARTICLES

The two other equations

¿

Ñ Ñ

¿

Ø Ø

Ñ

give two orthogonal solutions:

¼ ½ ¼ ½

¼ ¼

¼ ¼

·ÑØ ·ÑØ

Ù ´Øµ Ù ´Øµ

¿

¼ ½

½ ¼

i.e. ve energy spin-up and spin-down ANTI-PARTICLES

Dr M.A. Thomson Lent 2004 21 The DIRAC equation ★ gives PARTICLE/ANTI-PARTICLE solutions ★ requires the particles/anti-particles to have an additional degree of freedom (SPIN) ! ★ in the massless limit, the DIRAC equation reduces to the two uncoupled WEYL equations ★ In general the Dirac Equation gives FOUR simultaneous equations for the components of the SPINOR.

e.g. more general solutions

¼ ½ ½ ¼

¼

½

½

¼

Ô

´Ô Ô µ

Þ

Ü Ý

Ù Æ Ù Æ

½ ¾

´ ·Ñµ

´ ·Ñµ

´Ô ·Ô µ

Ô

Ü Ý

Þ

´ ·Ñµ

´ ·Ñµ

Ô

¾ ¾

´Ù Ù µ Ô · Ñ ¾

For ½

¼ ½ ½ ¼

Ô

´Ô Ô µ

Þ

Ü Ý

´ Ñµ

´ Ñµ

´Ô ·Ô µ

Ô

Ü Ý

Þ

Ù Æ Ù Æ

¿

´ Ñµ ´ Ñµ

½ ¼

¼ ½

Ô

¾ ¾

´Ù Ù µ Ô · Ñ For ¿ The DIRAC equation lives in the realm of PART III Particle Physics.

Dr M.A. Thomson Lent 2004 22 Lorentz Structure of Interactions

NON-EXAMINABLE e- e- e Electron Current

γ 1 Propagator q2

ppe Proton Current

Matrix element Å factorises into 3 terms :

Å Ù Ù

Electron Current

$

¢ Photon Propagator

¾

Õ

¢ Ù Ù Ô Ô Proton Current

★ Fermions are 4-component SPINORS.

¢ ★ µ interaction enters as matrices. ★ Lorentz invariance allows only ﬁve possible forms for

the interaction: SCALAR ÙÙ, PSEUDO-SCALAR

Ù Ù Ù Ù Ù

Ù , VECTOR , AXIAL-VECTOR ,

Ù TENSOR Ù ★ Electro-magnetic and Strong forces are VECTOR interactions - which determines the HELICITY structure.

Treats helicity states symmetrically µ PARITY CONSERVATION

★ The WEAK interaction has a different form: (V-A) i.e.

´½ µ . Projects out a single helicity

combination : µ PARITY VIOLATION The WEAK interaction is the subject of next lecture

Dr M.A. Thomson Lent 2004

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