• Nucleons in nucleus behave not like particles, moving with momenta p but like waves with de Broglie wavelength λB~h/p: • Because p ~250 MeV/c < E =939 MeV it’s behaviour is described by non-relativistic Quantum Mechanics (NR QM).• 1900 M. Planck – blackbody radiation: emission of el-magn waves by the heated bodies: light is not only a wave but also a particle (γ-particle) carried a portion of energy E=hν• 1905 A. Einstein – photoelectric effect • 1924 De Broglie - particles have also a wave origine - λB~h/p (based on Einstein photoeffect and Compton scattering)• 1927 Thompson , Davisson and Germer experiment showed the diffraction behaiviour of the electrons with wavelength λB~h/p

M Born - the state of Quantum System is described by the wave function Ψ(x,t) (or probability amplitude) which connected with probability P to find the system in volume dV as dP = | Ψ|2 dV where | Ψ|2 is probability density P= ∫V dp = ∫V | Ψ|2 dV – total propability with normalization to unity:

∫V →∞| Ψ|2 dV=1 The wave function Ψ(x,t) is the state vector similar to the r(t) in classical mechanics

Superposition principle: the difference between bullets passed through two slits and electrons scattered on two small slits is that one bullet always fly through only one slit wih probability P1 or P2

and in the case of two opened slits the detector will detect P12=P1+P2

while electron behaves as a wave so the probability that two electrons will pass though two slits is the same as in the case of waves with interferomety term P12=| ψ12 |2=|ψ1+ψ2|2 ≠ P1+P2=|ψ1|2+ |ψ2|2

i.e. In the case of microworld the superposition principle is valid not for probabilities but for probability amplitudes: ψ12 =ψ1+ψ2 here P1 =|ψ1|2 (P2=|ψ2|2 ) is the probability to find electron in the detector if only 1st (2nd) slit is open. It means that each electron feels the existance of two slits, and the wave of one electron from 1st slit is interfering with the wave of the same electron of the other slit. Superposition principle is valid not only for 2 but for many states ψ =ψ1+ψ2 +ψ3 +…+ψN

and the particle can be described by the electro-magnetic wave packat with group velocity vgr=p/m=vpart

=> 1925 E Schroedinger QM : particle is not classical : its momenta and position cannot be simultaneously measured exactly but with some accuracy : 1927 Heisenberg uncertainty principle : for position x and momentum p : ΔxΔpx< ћ/2 , or for energy E and time t ΔEΔt < ћ/2 (E=hν) where ћ=h/2π – Planck’s constant or for angular momentum L and its projection on z lz and on xy Δφ : ΔφΔlz < ћ/2

B.R. Martin. Nuclear and Particle Physics. Appendix A. Some results in Quantum mechanics.

A.1. Barrier penetration.

QM: Schrodinger equation• The Schrodinger equation plays the role of Newton's laws and conservation of energy

in classical mechanics - i.e., it predicts the future behavior of a dynamic system. It is a wave equation in terms of the wavefunction ψ(x,p,t) which predicts analytically and precisely the probability P= |ψ(x,p,t) |2 of the event.

• The kinetic (T) and potential (V) energies are transformed into the Hamiltonian which acts upon the wave function to generate the evolution of the wavefunction in time and space.

T + V = E

• The Schrodinger equation gives the quantized energies of the system and gives the form of the wavefunction so that other properties may be calculated.

• Operators in Quantum Mechanics • Associated with each measurable parameter in a physical system is a quantum mechanical operator. Such

operators arise because in quantum mechanics you are describing nature with waves (the wavefunction) rather than with discrete particles whose motion and dymamics can be described with the deterministic equations of Newtonian physics. Part of the development of quantum mechanics is the establishment of the operators associated with the parameters needed to describe the system. Some of those operators are listed below.

• It is part of the basic structure of quantum mechanics that functions of position are unchanged in the Schrodinger equation, while momenta take the form of spatial derivatives. The Hamiltonian operator contains both time and space derivatives.

One dimensional Schrodinger equation2

22

Nonrelativistic QM equation for spinless particle-wave packet is simply the energy conservation law:

( ) - energy conservation2

p̂=-i ( )

( ) - time dep2

tot

pV x E

m

x y z

i Vt m

22

/

endent Schrodinger eq.

( ) - if V=V(x) - time independent Schrodinger eq2

( , ) ( ) - solution of Schr. eq. for time independent case conditions on wavefunction :

iEt

H V Em

x t x e

and must be contineous across the any boundary:

lim ( ) ( ) 0 - should always be fulfilled

lim 0 - can be violated in the casx a x a

d

dxa a

d d

dx dx

2

1

e of discontineous V(x)

If we know ( , ) then we know

P= *( , ) ( , ) - probability to find particle (wave package) between (x1,x2)

* 1 - normalization condition: the total probability s

x

x

x t

x t x t dx

dx

x x x

hould be 1

* ( ) - physical meaning of the observable f(x) is statistical average over all possible states

ˆ ˆ<p > = *p *(-i ) -i * for momentum operator p

For stationary

f f x dx

dx dx dxx x

/xcase, V(x), time dependent part is cancelling and <p > doesnot depend on t.

*j= ( * ) - particle current density

2

iEte

mi x x

Free-particle in one dimension2

2 2

2

ikx -ikx 2

Stationary Schrodinger eq. for with H= and looks like 2

and has solution:2(x)=A' sin(kx) + B' cos(kx) or e

free particle V

quivalently

(x)=A e + B e , where k 2 /

(x)=0 p

H Em

dE

m dx

mE

2

i(kx- t) -i(kx+ t)

i(kx- t)

-i(kx+ t)

(x)=A e + B e - time-dependent solution with E=

. A e - a wave traveling to x=+

II. B e - a wave traveling to x=-

Boundary conditions are not valid in this cas

I

2 2

e because both

cos kx and sin diverge at x= .

In contrast we put at x=- a source (accelerator) which emits particles

with the rate I particles per second with momentum p= k in positive directi

kx

on x=+ .

Then B=0 because there is no particles travelling in opposite direction to x=- .

The particle current should coinside with initial current of the emitted particles:

*j= ( * ) (

2 2mi x x mi

-i(kx- t) i(kx- t) i(kx- t) -i(kx

i(kx-

- t)

2

)

2

t

A* e ( )A e A e ( )A* e )

2 | | | | , so 2

A= / .

The plane wave solution for free

(x,t)=

particle is

with and . / e E= p= km

ik ik

kj A ik A I

mi m

mI

I

k

k

Infinite potential cell2

free particle in box

, at x>a and x<0 with H= +V(x) V(x)=

0, at

Stationary Schrodinger eq. for

and

looks like

0

2

pH E

x am

2 2

2

at 0<x<a and has solution:2(x)=A sin(kx) + B cos(kx) with continuity at x=0 and x=a :

(0)=0 =>B=0,

( )=0 =>Asin(k )=0,A

dE

m dx

a L

2 2 2 22

2

2 2

0 then

sin(ka)=0 with solutions at kL= n where n=1,2,3 ... and k 2 /

- here energy E is quantized, only certain values are permitted

A is found from normalization condition fo

2 2

r

nn

kE n

m mL

mE

0

2 2 2 2

0

2 2

0 0

2

nx wave function (x)=A sin( )

* 1

nx| | sin 1. Using cos2 1 2sin x what gives sin (1 cos 2 ) / 2 one gets

1 n 1 n1 | | (1 cos 2 ) | | ( cos 2 )

2 2

1 n n = | | ( cos 2 2

2 2

a

L

a a

a

dx

A dx x x xa

A x dx A a xdxa a

aA a xd x

n a a

2

2

0

22 2

00

22

2

| |)

2

| | 1= a , because cos sin | sin(2 ) sin(0) 0 at n=1,2,3...

2

A= 2 / .

The plane wave solution for free particle in the in

2 (x,t

finite box is

)= sin E = for and 2

n=

a

n

n n

n

Aa

Axdx x n

a

nxn

a a ma

1,2,3....

V(x)= V(x)=

V(x)=0

0 a x

2 22

2

2 (x,t)= sin E =

2 for and n=1,2,3... for a=L n n

nxn

L L mL

Potential Barier, E>V0 and (tunnelling effect) E<V0

0V

0E >V

0

0, at x<0

V( x)= V , at a≥x≥0

0, at x>0

0V

0 a x 0 a xI II III IIIIII

0E <V

2

0

22

0

2

1 2 2

Stationary Schrodinger eq. for step potential

wit

p +V( x) ψ=Eψ V

2m

0, at x<0h

- ∇ h

is writen as

+V( x) ψ=Eψ V( x)= V , at a≥x≥02m

0, at x>0

ψ'' +k( x) ψ=0

ψ( 0)=ψ( 0) , ψ

( a)=

3

1 2 2 3

1 2

2 0 02

2 0 02

3 2

ψ( a)

ψ'( 0)=ψ'( 0) , ψ'( a)=ψ'( a)

2mk = , x<0

h

2mk( x)= ( E-V ), E >V

h k( x)= , 0<x<

at

ifwhe a

2miκ( x)=i ( V -E), E <V

h

2mk = , x>a

h∂ρ

div j+ = ∇j=0 -c

re at

i

ontinuity eq∂t

f

at

1

2

ik x

. ρ( x,t)=|ψ( x,t)| -propability density,

h j( x,t)= ψ*( ∇ψ)-( ∇ψ*) ψ - flux

2mi I region x<

where

For the the solution is the sum of

incoming particle( free particle

0, V( x)=0

A solution)

and

e

wave

��

1

1 1

-ik x

ik x -ik x1 1 1

2 211 1 1 1 2

2 21

Be

ψ( x)=A e + B e =2mE

hk j = |A | -|B

reflected from the barrier :

with h k

.

| = j -

jm

0

22

0

22 2 2 2 2

For the

V( x)=V >E or

equations are

II region at 0<x<a,

for V( x)=V <E

ψ '' +k ( x) ψ =0

ψ '' - κ ( x ) ψ or =0

2 2 2 2ik x -ik x κ x -κ x2 2 2 2

2 22 22 2 2 2

ψ( x)=A e + B e ψ( x)=Ce + De

h

solutions are or

currents are or

Finally in t

k hκ j = |A | -|B | j =2 Im C *D

m mIII rehe it is onlygion( x>a) outg oing

3ik x3

233

23 1

particles

with k k = 2mE /hψ( x)=Fe =

hkj = |F|

m

1 2 2 3

1 2 2 3

5 A,B,C,D,F can 4 boundary conditions,

one - A

coefficients be defined from

it means that for example, is defined from somewhere , .

ψ( 0)=ψ( 0) , ψ( a)=ψ( a)

ψ'( 0)=ψ'( 0) , ψ'( a)=ψ'( a)

else

1 1 2 2 1 1

A+ B= C+ D ( 1) A+ B= C+ D

Syst

ik A- ikB=iκ C+-iκ D ( 2)

em

of

linear equations

ik A- ikB2 2 1 2 2 1

2 2 1 2 1

1

2

2 2iκ a -iκ a ik a -κ a κ a ik a

-κ a κ a ik a -κ a κ a ik a2

-

2 1 2 2

i a

1

k1 2

1 2 2

=-κ C+κ D Ce + De = Fe ( 3) Ce + De = Fe iκ Ce +-iκ De =ikFe ( 4) -κ C

2Ae (

e +κ De =ikFe

F =kκ )

2kκ coshκ

2 21

2 21 2 2

2 21 2

2 2 2 22 1 2 2

x -x

x -x

2

2

02

2 0

a - i( k -κ ) si

sinhx=( e -e )/2

coshx=( e +e )/2

|F|T= = transmission coefficient for barrier V >E.

|A|

At

nhκ a

16k κ4κ k +( k +κ ) sinh κ a

κ a≪1, T ≈ 1,( κ = 2m/h( V -E))

κ

where

2

2 2-2κ a1 2

2 2 22

21

T becomes very sensitive to the

E of the particle due to the second exponential term.It is especially important for life time of the

4k κT

nuce

a

i

= e ≪1,( k +κ

in the cas

≫1

e

,

of

)

α-de

the first term is due to the reflection losses at two boundaries x=0 and x=a,the decrea

cay or tunnelin

sing exponent d

g of α-particle through nuclear pote

escribes the amplitude decay within

ncial.

the b

1/2

2 0

2exp 2 ( )

x mT dx V x Eat x

2-2 κ

arrier.the first term is slowly varying with E and is usually neglected.

The last result can be used for the case of arbitarary smooth potential

e

V( x)

– –

WKB approximationWentzel Kramers Brillouin

2 22 2 20 2

2

κ ->-ik , sin and takes

For the c plac

h κ a->sin k a at k a=πn T=1 - resonance transm

ase V <Eissie

on.

1

31 1

ik x -ik x ik x

1 1 1ψ( x)=A e +B e ψ( x)=Fe

2 22 2x x

2

ik -ikψ( x)=A e +B e

1

31 1

ik x -ik x ik x

1 1 1ψ( x)=A e +B e ψ( x)=Fe

2 2x x

2

-κ +κψ( x)=Ce +De

A.2.Density of the states and Three dimensional boxIn the case of the well with infinite potential barrierthe solution of the Schrodinger eq.

are the standing waves vanishing at the ends of the well

corresponding to discrete energy levels always > 0because of uncertainty principle:Δp˜1/L -> Δ(E=p2/2m)˜1/L2 .For 3 dimensional well the energy orand the wavefunction are where nx>0, ny>0, nz>0

The number of the states is proportional to the number of lattice points dinstanced from each other by (π/L): (L/π)3

Or in the Fermi sphere it will be 1/8 of the sphere (because nxyz>0)n(k0)=1/8 ּ4/3πk0

3 ּ(L/π)3=V/(2π)3ּ4/3 πk03 - for all k<k0 while for k0<k<k0+dk0:

dn(k0)=V/(2π)3ּ4πk02dk0, dn(p)=V/(2πћ)3ּ4πp2dp, dn(E)=V/(2π)3ּ4πk0

2dk0 where ρ(k0)=dn(k0)/dk0=V/(2π)3ּ4πk0

2 is density of states for k0<k<k0+dk0

ρ(p)=dn(p)/dp=V/(2πћ)3ּ4πp2 is density of states for momenta p<p<p+dp:k=ћpρ(E)=ρ(p)dp(E)/dE=V/(2πћ)3ּ4πp2/v =V/(2πћ)3ּ4πmp - for E<E<E+dE, E=p2/2m,p=mvρ(E)=V/(2πћ)3ּp2/vdΩ for solid angle Ω, d3p=p2dpdΩ ¨ dE=p/mdp=vdp For particles with spin spin multiplicity factor should be included: for spin=1/2 according to the Pauli principle such factor is 2.

A.3 Perturbation theory and the Second Golden Rule In perturbation theory the Hamiltonian at any time t may be written as H(t)=H0+V(t), where H0 is unperturbed Hamiltonian and V(t) is small.

The solution for eigenfunctions of H starts by expanding in the terms of the complete set of energy eigenfunctions |un> of H0 : H0 |un> =En |un>

|ψ(t)>=Σcn(t) |un> exp{-iEnt/ћ}, where En are the corresponding energies.

If |ψ(t)> is normalized to unity <ψ(t)|ψ(t)>=1 then |cn(t)|2 is probability that

at time t the system will be in the state |un>. Substituting it in Schrodinger eq :

iћ ∂cf(t)/∂t=ΣVfn(t) cn(t) exp{-iωfnt}, where matrix element

Vfn(t)= <uf|V(t)|un> and angular frequency ωfn=(Ef-En)/ ћ.

Initial state of the system |ui> at t=0 then cn(0)=δni and ci(t)=ci(0)+1/iћ∫0tVii(t’)dt’

Final state of the system f≠i cf(t)= 1/iћ∫0tVfi(t’)ei ω

fi t’dt’ general, for V(t)

For const V =0 and then V0 cf(t)= Vfi/(ћ ωfi )(1-ei ω

fi t)

Probability for transition from state i to state f : Pfi=|cf (t)|2 =4|Vfi|2/ћ2 [sin2(1/2 ωfi

t)/ω2fi

]. At large t it is valid only if

Ћ | ωfi |= |Ef-En|<2πћ/t (uncertainty principle) then [sin2(1/2 ωfi t)/ω2

fi ]->1/2 πћtδ(Ef-Ei).

Pfi =2πt/ћ |Vfi|2δ(Ef-Ei) – probability

dPfi /dt= 2π/ћ |Vfi|2δ(Ef-Ei) – transition probability per unit time – valid for discrete final states.

dTfi /dt= ∫ dPfi /dt ρ(Ef)dEf = 2π/ћ [|Vfi|2 ρ(Ef)]Ef=Ei– transition probability per unit time – for continious spectra

Dirac Delta Function

Appendix C 1911 Rutherford scattering

In Ernest Rutherford's laboratory, Hans Geiger and Ernest Marsden (a 20 yr old undergraduate student) carried out experiments to study the scattering of alpha particles by thin metal foils. In 1909 they observed that alpha particles from radioactive decays occasionally scatter at angles greater than 90°, which is physically impossible unless they are scattering off something more massive than themselves. This led Rutherford to deduce that the positive charge in an atom is concentrated into a small compact nucleus. During the period 1911-1913 in a table-top apparatus, they bombarded the foils with high energy alpha particles and observed the number of scattered alpha particles as a function of angle. Based on the Thomson model of the atom, all of the alpha particles should have been found within a small fraction of a degree from the beam, but Geiger and Marsden found a few scattered alphas at angles over 140 degrees from the beam. Rutherford's remark "It was quite the most incredible event that ever happened to me in my life. It was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The scattering data was consistent with a small positive nucleus which repelled the incoming positively charged alpha particles. Rutherford worked out a detailed formula for the scattering (Rutherford formula), which matched the Geiger-Marsden data to high precision.Only if cos θ <0 or θ >90o

Appendix C 2. Rutherford scattering CM versus QMClassical mechanics : Rutherford's model enables us to derive a formula for the angular distribution of scattered a-particles. Basic Assumptions: • a) Scattering is due to Coulomb interaction between a-particle and positively charged atomic nucleus. • b) Target is thin enough to consider only single scattering (and no shadowing) • c) The nucleus is massive and fixed. • This simplifies the calculation, but it could be avoided by working in the centre of mass frame. • d) Scattering is elastic

Finite potential cell2

0

free particle in box

, at x>a and x<0 with H= +V(x) V(x)=

0, at

Stationary Schrodinger eq. for

and

looks like

0

2

VpH E

x am

2 2

2

( ) at 0<x<a and has solution:2(x)=A sin(kx) + B cos(kx) with continuity at x=0 and x=a :

(0)=0 =>B=0,

( )=0 =>Asin(k

dV x E

m dx

a

2 2

2 2 2 22

2

)=0,A 0 then

sin(ka)=0 with solutions at kL= n where n=1,2,3 ... and k 2 /

- here energy E is quantized, only certain values are permitted

A is found from normalization condi

2

t

2n n

L

mE

kE n

m mL

0

2 2 2 2

0

2 2

0 0

2

nxion for wave function (x)=A sin( )

* 1

nx| | sin 1. Using cos2 1 2sin x what gives sin (1 cos 2 ) / 2 one gets

1 n 1 n1 | | (1 cos 2 ) | | ( cos 2 )

2 2

1 n = | | ( cos 2

2 2

a

L

a a

a

dx

A dx x x xa

A x dx A a xdxa a

aA a

n a

2

0

22 2

00

2 22

2

n | |2 )

2

| | 1= a , because cos sin | sin(2 ) sin(0) 0 at n=1,2,3...

2

A= 2 / .

The plane wave solution for free particle in the infinite box

2 (x,t)=

is

sin E =for 2

a

n n

n n

Axd x a

a

Axdx x n

nxn

a

a a ma

and n=1,2,3....

0 0( ) ( ) 0 ( ) V x V V x V x V

0 a x

I II III

Discrete energy for the particle in the boxand continious energy for free particle

1 2 2 3

1 2 2 3

5 A,B,C,D,F can 4 boundary conditions,

one - A

coefficients be defined from

it means that for example, is defined from somewhere , .

ψ( 0)=ψ( 0) , ψ( a)=ψ( a)

ψ'( 0)=ψ'( 0) , ψ'( a)=ψ'( a)

else

1 1 2 2

A+ B= C+ D ( 1) - B+ C+ D + 0*F = A

System of linear equ

ik A- ikB=-κ C

ations

+κ D2 2 1 2 2 1

2 2

2 1 2 1-κ a κ a ik a -κ a κ a ik a

-κ a κ a2 2

( 2) + B-κ /ikC+ κ /ikD + 0*F = A Ce + De = Fe ( 3) e C+ e D - e F = 0-κ Ce +κ De 1 2 2 1

2

ik a -κ a κ a ik a1 1 2

2 1 2

κ a2

1

1

=ikFe ( 4) -e C+ e D-ik /κ e F = 0

- B+ C+ ( 1+2) ( 3)*( 1-κ /ik )*e :

D + 0*F= A

( 1-κ /ik ) C+( 1+κ /ik ) D = 2 2 1

2 2

2

1

2 1

2κ a κ a ik a2 1 2 1 2 1

2κ a κ a ik a2 1 2 1 2 1

κ a2

κ a ik a2

2

1

1

2A ( 1-κ /ik ) C+( 1-κ /ik ) e D- ( 1-κ / ik ) e e F = 0

{ ( 1+κ /ik )-( 1-κ /ik ) e } D+( 1-κ /ik ) e e F =2( 1+2)-( 3) *( 1-κ /ik )

A ( 5)

2e D-( 1+ik /κ ) e

*e :

( 3)+( 4)/κ :

F =

2 2 1 2 1

2 2 2 1

2κ a -κ a ik a κ a ik a2 1 2 1 1 2 2 1

κ a -κ a κ a -ik a2 1 2 1 1 2 2 1

0=>

1{ ( 1+κ /ik )-( 1-κ /ik ) e }( 1+ik /κ ) e e F+( 1-κ /ik ) e e F =2A ( 6)

2{ ( 1+κ /ik ) e -( 1-κ /ik ) e } ( 1+ik /κ ) F+2( 1-κ /ik ) e F =4Ae ( 6)

F=

2 1-κ a ik a1 2

1D= ( 1+ik / κ ) e e F

2

1

2 2 2

1

2 2

1

2

-ik a

-κ a κ a κ a2 1 2 1 1 2 2 1

-ik a

-κ a κ a2 1 1 2 2 1 2 1 1 2

-ik a

-κ a2 1 1 2 2 1 1

4Ae=

{ ( 1+κ /ik ) e -( 1-κ /ik ) e } ( 1+ik /κ )+2( 1-κ /ik ) e4Ae

= =( 1+κ /ik ) ( 1+ik /κ ) e +{ 2( 1-κ /ik )-( 1-κ /ik ) ( 1+ik /κ ) } e

4Ae =

( 2+κ /ik +ik /κ ) e +( 1-κ /ik ) ( 1-ik /κ 2

1

2 2

1

2 2

1

κ a2

-ik a

-κ a κ a2 2 2 21 2 2 1 1 2 1 2 2 1 1 2

-ik a1 2

-κ a κ a2 2 2 21 2 2 1 1 2 2 1

-ik a1 2

2 21 2 2 1 2

=) e

4Ae = =

( 2ikκ +κ -k ) e /( ikκ )+{ 2ikκ -κ +k } e /( ikκ )4Ae ( ikκ )

= =(

2Ae ( kκ )2

2ikκ +κ

kκ cos

-k ) e +{ 2ikκ -κ +k } e

hκ

a - i( k -κ ) s

=inhκ

2 2 4 =

2 2

x -x x -x

21 2 1 2

2 2 2 2 21 2 2 1 2 2 1 2 2 1 2 2

2 21 2

κ a κ a2 2 2 22 1

2

1 1 2

sinhx=( e -e )/2,coshx=( e +e )/2

2kκ 2kκ|F|T= = *

|A| 2kκ coshκ a - i( k -κ ) sinhκ a 2kκ coshκ a + i( k -κ ) sinhκ a 4k κ

4κ k ( e +e +2)/4+( k -

2k +κ

a

κ

where

2 24

2 2 2 24 4

2 24 4

2 21

= =

=

T

2 2

2 2 2 2

2 2

κ a κ a2

2 21 2

κ a κ a κ a κ a2 2 2 22 1 1 1 2 2

2 21 2

κ a κ a2 2 2 22 1

2 21 2

2

1 1 2

2 2 22 1

2

2 2

) ( e +e -2)/44k κ

4κ k ( e +e -2+4)/4+( k -2k κ +κ ) ( e +e -2)/44k κ

44k κ

4κ k +(

κ k 4/4+( k +2k κ +κ ) ( e

k +κ ) sin

+e -2)/

h κ

4

0transmission coefficient for barrier V .Ea