     # Quantum Mechanics Lecture 8: Relativistic ... Quantum Mechanics Lecture 8: Relativistic Quantum Mechanics

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• Quantum Mechanics Lecture 8: Relativistic Quantum Mechanics

Contents:

1. Notation and units.

2. Relativistic kinematics.

3. The Schrödinger equation.

4. Relativistic wave equation (Klein-Fock-Gordon equation); antiparticles.

5. Electromagnetic interaction of spin-0 particles.

5.1 Perturbation theory; Fermi’s Golden Rule.

5.2 Elastic scattering of spin-0 particles.

6. Calculation of the cross section; phase space; Mandelstam variables.

7. The Dirac equation.

7.1 Derivation.

7.2 Continuity equation.

7.3 Covariant form of the Dirac equation.

7.4 Properties of the γ matrices.

7.6 Plane wave solutions.

7.7 Antiparticles.

7.8 Completeness relations.

7.9 Helicity.

7.10 Bilinear covariants.

8. Electrodynamics of spin-1/2 particles.

8.1 Transition matrix.

8.2 Trace theorems.

8.3 Completion of the calculation of the electron tensor.

8.4 Cross section of elastic eµ scattering in the CMS.

8.5 Electron-positron annihilation into muon pairs.

8.6 Electron-muon elastic scattering in the LAB frame.

Figures 1a and 1b.

1

• 1. Notation and units.

We denote the components of the space-time 4-vector x by x0 = ct, x1 = x, x2 = y, x3 = z or, in compact form,

x = (x0, x1, x2, x3) contravariant components

The dual vector has the following covariant components:

x0 = ct, x1 = −x, x2 = −y, x3 = −z.

The invariant square of x is:

x2 ≡ x·x = xµxµ = (x0)2 − (x1)2 − (x2)2 − (x3)2 = invariant

where summation over the repeated index µ, one upper and one lower, is implied (Einstein summation convention). We define the metric tensor gµν:

gµν = diag(1,−1,−1,−1)

hence xµ = gµνxν , µ = 0, 1, 2, 3

this is called the raising of the subscripts. The dual metric tensor gµν is given by

gµν = diag(1,−1,−1,−1)

hence xµ = gµνx

ν , µ = 0, 1, 2, 3

(lowering of the superscripts).

Lorentz boost with boost velocity v in x direction:

x′ = γ(x− vt), y′ = y, z′ = z, (1) ct′ = γ(ct− vx/c) (2)

where γ = 1/ √

1 − v2/c2. To get the Inverse transformation change the sign of v and exchange the primed and unprimed coordinates.

The energy-momentum 4-vector p of a particle of mass m is

p = (E/c, ~p)

and its Lorentz invariant square is

p2 = pµpµ = (E/c) 2 − ~p 2 = (mc)2 = invariant

We get the rest energy E0 by setting the momentum equal to nought, hence:

E0 ≡ E|~p=0 = mc2

2

• The relativistic kinetic energy is defined by

T = E −mc2

and we can easily check that in the nonrelativistic limit |~p| � mc we get the well known formula T = p2/2m.

Energy and momentum expressed in terms of the particle velocity v are

E = γmc2, ~p = γm~v, hence ~v = c ~pc

E .

Units: We use natural units, defined by c = 1 and h̄ = 1. To carry out the conversion to GeV units use the conversion factors

h̄c = 197.327 MeV fm, (h̄c)2 = 0.3894 GeV2 mbarn

where 1 fm = 10−15 m (femto-meter or Fermi), and 1 mbarn = 10−31 m2 (milli-barn). For numerical estimates use the following approximate values:

h̄c = 200MeV fm = 200 eV nm, (3)

(h̄c)2 = 0.4 GeV2 mbarn (4)

Typical particle masses (approximate values):

Particle symbol Mass in MeV/c2

Electron e− 0.5 Muon µ− 106 Charged pion π± 140 Neutral pion π0 135 Charged kaon K± 494 Neutral kaon K0 498 Proton p 938 Neutron n 940

2. Relativistic kinematics.

For a collision of particles a and b, which gives rise to the creation of particles c, d, . . ., we write the reaction equation

a+ b→ c + d+ . . . Particles a and b are the incident particles, particles c, d, etc. are the outgoing particles. The four-momenta of particles a and b will be denoted by p1 and p2, respectively, and those of the outgoing particles by p3, p4 etc. Their masses are given by p

2 i = m

2 i .

The kinematics of collision processes is described in various reference frames of which we discuss here the laboratory frame (LAB) and the centre-of mass frame (CMS) in some detail.

LAB frame: particle a is the incident particle, particle b is the target particle. Their four- momenta are

p1 = (Elab, 0, 0, plab), p2 = (m2, 0, 0, 0) (5)

3

• The square of the total four-momentum of the system is

s = (p1 + p2) 2 = m21 +m

2 2 + 2m2Elab (6)

Solving for plab we get

plab = √

[s− (m1 −m2)2][s− (m1 +m2)2]/2m2 (7)

CMS frame: by definition of the CMS, the incident particles have 3-momenta of equal magnitude and op- posite direction:

pcms1 = (E ∗ 1 , 0, 0, p

∗), pcms2 = (E ∗ 2 , 0, 0,−p∗). (8)

hence s = (E∗1 + E

∗ 2)

2. (9)

Thus √ s is the total CMS energy of the system. Solving for p∗ and comparing with Eq. (7) we

get

p∗ = plab m2√ s

(10)

3. The Schrödinger equation

Fundamental for quantum mechanics is the concept of particle-wave duality. Formally particle-wave duality is expressed by the Einstein-de Broglie relations:

E = h̄ω, ~p = h̄~k, (11)

which relate the particle characteristics E (energy) and ~p (momentum) to the wave character-

istics ω (frequency) and ~k (wave vector). Here h̄ = h/2π, where h is Planck’s constant. Implied in the statement of particle-wave duality is also that the particle must be describable

in terms of a wave function. Moreover, we demand that the wave function contains the complete information on the state of motion of the particle at time t. This implies that the wave function must satisfy an evolution equation in time. This is the Schrödinger equation

ih̄ ∂ψ(x, t)

∂t = Ĥψ(x, t) (12)

where Ĥ is the Hamiltonian operator or Hamiltonian.

4. Relativistic wave equation (Klein-Fock-Gordon equation); antiparticles.

To construct a relativistic wave equation we shall use the relations that provide the transition from classical mechanics to quantum mechanics:

E → ih̄ ∂ ∂t , (13)

px → −ih̄ ∂

∂x , py → −ih̄

∂y , pz → −ih̄

∂z ,

4

• together with the relativistic energy-momentum relation

E2 = (pc)2 +m2c4

which immediately gives the Klein-Fock-Gordon equation:

−h̄2∂2t φ(x) = [

−(h̄c)2 (

∂2x + ∂ 2 y + ∂

2 z

)

+m2c4 ]

φ(x) (14)

where we have used the shorthand form of the differential operators:

∂t ≡ ∂

∂t , ∂x ≡

∂x , ∂y ≡

∂y , ∂z ≡

∂z .

We can rewrite the KFG equation in a relativistically symmetric form which clearly exhibits its relativistic invariance. To do this we define the four-dimensional generalization of the momentum operator:

p̂ = (

p̂0, p̂1, p̂2, p̂3 )

= (

ih̄∂0, ih̄∂1, ih̄∂2, ih̄∂3 )

=

(

i h̄

c ∂t,−ih̄∂x,−ih̄∂y,−ih̄∂z

)

and hence the four-dimensional generalization of the Laplacian operator:

∂µ∂ µ =

1

c2 ∂2t −∇2

With these definitions the KFG equation takes on the manifestly invariant form

[

∂µ∂ µ +

(

mc h̄

)2 ]

φ(x) = 0 (15)

provided that the wave function φ(x) is a Lorentz scalar or pseudoscalar. This latter point appears here as a mathematical requirement, but it should be stressed that

it has an important physical implication: we can apply the KFG equation only to particles which are described by scalar or pseudoscalar wave functions. Such particles do exist, for instance pions and kaons, which are pseudoscalar particles, but electrons are not scalar particles and their wave functions are correspondingly not scalars.

Note that mc/h̄ is the inverse of the Compton wavelength of the particle of mass m.

To derive a continuity equation we write down the KFG equation for the complex conjugate wave function φ∗(x), multiply it by φ(x), multiply Eq. (14) by φ∗(x) and subtract the resulting equations. As a result we get

∂tρ+ ∇·~ = 0 (16)

where

ρ(x) = i (φ∗∂tφ− φ∂tφ∗) , ~ = −i (φ∗∇φ− φ∇φ∗)

Important is that ρ(x) can be negative as well as positive, so it is not a probability density.

In the particular case of a plane wave, φ(x) = N exp(i(~p·~r − Et)/h̄), we have

ρ = 2E|N |2, ~ = 2~p|N |2

5

• where E = ±

(pc)2 +m2c4

Thus the negative values of ρ are related to negative energies. These negative energy solutions cannot be dropped as unphysical because the remaining positive energy wave functions would no longer form a complete set. Therefore a physical explanation for these states is needed. This was given independently by Feynman and by Stückelberg. As before, in the nonrelativistic case, we can get the equivalent integral form of the continuity equation:

d

dt

V ρ(~r, t)d3r = −

S ~·d~S (17)

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