https://www.awa.tohoku.ac.jp/~suekane/kougi/19_soryuushi-I/

-

Plain wave Solution of Dirac Equation

Apply

€

p0γ0 − pxγ1 − pyγ2 − pzγ 3( )−m[ ]w = 0

€

p0 −m − ! p ⋅ ! σ ( )! p ⋅ ! σ ( ) −p0 −m

%

& '

(

) *

uv%

& ' (

) * = 0

€

i γ0∂∂t

+γ1∂∂x

+γ2∂∂y

+γ 3∂∂z

$

% &

'

( ) −m

+

, -

.

/ 0 ψ x( ) = 0

è

Matrix form

191029 2素粒子I

ψ x( ) = wexp −i p0t −!p!x( )"# $%=

uv

&

'(

)

*+exp −i p0t −

!p!x( )"# $%

191029 3

€

p0 = ± ! p 2 + m2 ≡±E

€

p0 −m( )u − ! p ⋅ ! σ v = 0 "" 1( )! p ⋅ ! σ u − p0 + m( )v = 0 "" 2( )

% & '

€

1( ) ⇒ u =! p ⋅ ! σ v

p0 −m

€

! p ⋅ ! σ ( ) ! p ⋅ ! σ ( )vp0 −m

− p0 + m( )v = 0

€

p02 − ! p 2 −m2( )v = 0

(2)

n=0

€

p02 − ! p 2 −m2 = 0

v=0 u=0 ψ=0

素粒子I

ç ()

€

ψ+ x( ) =1

E + mE + m( )u! p ⋅ ! σ ( )u

%

& '

(

) * exp i ! p ! x −Et( )[ ]For :

€

p0 = E

For :

€

p0 = −E

€

ψ− x( ) =1

E + m

! p ⋅ ! σ ( )vE + m( )v

&

' (

)

* + exp −i ! p ! x −Et( )[ ]

191029 4素粒子I

191029 5

€

ψ ! p x( ) =1

E + mE + m( )u! p ⋅ ! σ ( )u

%

& '

(

) * e−ipx +

! p ⋅ ! σ ( )vE + m( )v

%

& '

(

) * eipx

,

- .

/

0 1

€

! p = 0 è

€

ψ ! p =0 x( ) =u0#

$ % &

' ( e−imt +

0v#

$ % &

' ( eimt

*

+ ,

-

. /

:

素粒子I

€

! p = 0

€

ψ x( ) = u1

1000

#

$

% % % %

&

'

( ( ( (

e−imt +u2

0100

#

$

% % % %

&

'

( ( ( (

e−imt + v1

0010

#

$

% % % %

&

'

( ( ( (

eimt + v2

0001

#

$

% % % %

&

'

( ( ( (

eimt

€

i γ0∂∂t

+γ1∂∂x

+γ2∂∂y

+γ 3∂∂z

$

% &

'

( ) −m

+

, -

.

/ 0 ψ x( ) = 0

6191029 素粒子I

?

€

exp imt[ ] = exp −i −m( )t[ ]

t

x

e-, E<0t

x

e-, E>0€

exp imt[ ] = exp −im −t( )[ ]

7191029 素粒子I

191029 8

( )

x=vtt

xt

x

t x

t

x

x=(-v)(-t)

素粒子I

t

x

e-, E<0

e+, E>0

t

x

e+, E>0

191029 素粒子I 9

?

0

(e+)

OK.

t

x

e-

e+,

191029 素粒子I 10

How to show particle movement

x x

t

xx

t

x

t

x

AntiparticleParticle travelling backward in time(OK to write either way)

physical move diagram

191029 11

E<0

素粒子I

How to show interaction

Collision: A+B => A+B

x

y

A B

physical move

x

t

AB

diagram

x

t

AB

x

t

A

Collision: A+B => A+B E>0 E<0

B

E>0

191029 12素粒子I

Antiparticlesname charge 1st 2nd 3rd EM W S

Antilepton +1

0

antiquark -2/3

+1/3€

e+

€

µ+

€

τ+

€

ν e

€

ν µ

€

ν τ

€

u

€

c

€

t

€

d

€

s

€

b

=> Particle with opposite intrinsic quantum numbers (such as charge)=> However, mass and energy are positive=> Annihilate with particle producing gauge boson and energy=> Can be created from vacuum with particle

if enough energy is supplied.

e- e+

13191029 素粒子I 13

191029 14

*

素粒子I

!"ug

p0

191029 素粒子I 15

Hydrogen Atom Levels

-20

-15

-10

-5

0

JP

0+ 1+ 0- 1- 2-

1S

2S 1P

ΔE=21cm

2P

Hydrogen Atom Levels

-20

-15

-10

-5

0

JP

0+ 1+ 0- 1- 2-

1S

2S 1P

ΔE=21cm

2P

-2.5

-.5

-7.5

-20

1/2 ( 1/2 )

191029 素粒子I 16

Hydrogen Atom Levels

-20

-15

-10

-5

0

JP

0+ 1+ 0- 1- 2-

1S

2S 1P

ΔE=21cm

2P

-2.5

-.5

-7.5

-20

0.01%

(n=0, l=0) S=0 S=1( )

S=1 S=0

uu,dd,ss

ρ, ω, φ

Generation of non-flavor neutral mesons

€

e+ + e− →γ * J =1( )→ q + q meson

ECM(GeV)€

R =σ e+e− → qq ( )

σ e+e− →γ * → qq ( )

191029 17素粒子I

my phD thesis

# ̅# %!%

191029 素粒子I 18

Positronium Energy Levels

cc quark system Energy level

ψeenlm x[ ] = Rnl r[ ]Ylm θ,φ[ ]e−iEnt

ψcc x[ ] = R 'nl r[ ]Ylm θ,φ[ ]e−iEnlt

10%0.01%

There are small energy gaps

191029l=0 l =l

(1)

(2) e+-e--

Hydrogen Atom Levels

-20

-15

-10

-5

0

JP

0+ 1+ 0- 1- 2-

1S

2S 1P

ΔE=21cm

2P

素粒子I 19

2

191029 20素粒子I

1/2

⇑ ⇓è

t

ψS t[ ] =α t[ ] ⇑ +β t[ ] ⇓ =α t[ ]β t[ ]

⎛

⎝⎜⎜

⎞

⎠⎟⎟

191029 素粒子I 21

&' = 0 11 0 , &+ = 0 −-

- 0 , &. = 1 00 −1

&⃗ = &'0⃗' + &+0⃗+ + &.0⃗.

= 0 0⃗'0⃗' 0 + 0 −-0⃗+

-0⃗+ 0 + 0⃗. 00 −0⃗.

= 0⃗. 0⃗' − -0⃗+0⃗' + -0⃗+ −0⃗. = 0⃗. 0⃗2

0⃗3 −0⃗.4 5 &⃗ = 4'&' + 4+&+ + 4.&.

= 4. 4' − -4+4' + -4+ −4. = 4. 42

43 −4.

2

€

! B = Bx ,By,Bz( )

191029 22素粒子I

6⃗=68̂

6⃗=68̂E=6⃗4=648̂

8̂:

8̂ &⃗ = 0. 0' − 0+0' + 0+ −0.

E H HI=64&⃗

€

i˙ α ˙ β

$

% & '

( ) = µ

Bz B−B+ −Bz

$

% &

'

( ) α

β

$

% & '

( )

€

ψ t( ) =α t( )⇑ +β t( )⇓ =α

β

'

( ) *

+ ,

€

i ˙ ψ = HIψ

191029 23素粒子I

2

€

! B = Bx ,By,Bz( )

6⃗=68̂

8̂ &⃗

191029 素粒子I 24

€

˙ α ˙ β

$

% & '

( ) = −iµ

Bz 00 −Bz

$

% &

'

( ) α

β

$

% & '

( )

€

! B = 0,0,Bz( )z ,

€

˙ α = −iµBzα˙ β = +iµBzβ

% & '

€

α t( ) =α 0( )exp −iµBzt[ ]β t( ) = β 0( )exp iµBzt[ ]

% & '

€

ψ t( ) =α 0( )e−iµBz t ⇑ +β 0( )eiµBz t ⇓

€

! B

€

! s

191029 素粒子I 25

€

ψ t( ) =α 0( )e−iµBz t ⇑ +β 0( )eiµBz t ⇓

b(0)=0, a(0)=0

€

ψ+ t( ) = e−iµBz t ⇑ , ψ− t( ) = eiµBz t ⇓

( )

ψ t[ ] = A x[ ]exp −iEt[ ]

× A t.

€

˙ α ˙ β

$

% & '

( ) = −iµBx

0 11 0$

% &

'

( ) α

β

$

% & '

( )

€

! B = Bx ,0,0( )z

€

˙ α = −iµBxβ˙ β = −iµBxα

% & '

191029 26素粒子I

€

! B

€

! s

€

⇑

€

⇓µBx

€

˙ α + ˙ β = −iµBx α +β( )˙ α − ˙ β = +iµBx α −β( )

% & '

€

α t( ) +β t( ) =C+ exp −iµBxt[ ]α t( )−β t( ) =C− exp +iµBxt[ ]

% & '

€

α t( ) =12C+e

−iµBx t +C−eiµBx t( )

β t( ) =12C+e

−iµBx t −C−eiµBx t( )

%

& '

( '

€

ψ t( ) =C+e

−iµBx t +C−eiµBx t

2⇑ +

C+e−iµBx t −C−e

iµBx t

2⇓

€

ψ+ t( ) =⇑ + ⇓2

e−iµBx t , ψ− t( ) =⇑ − ⇓2

eiµBx t€

=C+

2⇑ + ⇓2

e−iµBx t +C−

2⇑ − ⇓2

eiµBx t

, C_=0 C+=0

€

! B

€

! s

191029 素粒子I 27

€

! B = Bx ,0,0( )

€

! B

€

! s

€

ψ 0( ) = ⇑

€

P⇑ t( ) = cos2 µBxt, P⇓ t( ) = sin2 µBxtt

P

€

⇑

€

⇓

Spin oscillates due to perpendicular magnetic field with w=µB

Probabilities to be spin-up and -down are

t=0

€

C+ =C−

€

ψ t( ) =e−iµBx t + eiµBx t

2⇑ +

e−iµBx t − eiµBx t

2⇓

= cos µBxt( )⇑ + i sin µBxt( )⇓

191029 素粒子I 28

m1 m2

V

r

S1 S2

(1)Magnetic moment of particle 1 makesmagnetic field at particle 2.

(2)magnetic moment of particle 2 feels magnetic field and energy level changes

€

E = ! µ 2 ⋅! B = ! µ 2 ⋅ κ

! µ 1( ) =κ ! µ 1 ⋅! µ 2( )

quantize the spin and get Hamiltonian

€

HI = µ1µ2κ! σ 1 ⋅! σ 2( )

€

! B =κ ! µ 1

spin-spin interaction (EM case)

€

i dψdt

= A ! σ 1 ⋅! σ 2( )ψ (A≡µ1µ2k)

p eg

191029 29素粒子I

m1 m2

V

r

S1 S2

Basic states are

spin-spin interaction (EM case)

€

ψ = s1, s2

€

! σ 1 ⋅! σ 2( )↑↑ =

! σ ↑( ) ⋅ ! σ ↑( ) = ↑↑

! σ 1 ⋅! σ 2( )↑↓ =

! σ ↑( ) ⋅ ! σ ↓( ) = ↑ ! e z + ↓ ! e +( ) ↑ ! e − − ↓

! e z( ) = −↑↓ + 2↓↑! σ 1 ⋅! σ 2( )↓↑ ="= 2↑↓ − ↓↑

! σ 1 ⋅! σ 2( )↓↓ ="= ↓↓

'

(

) )

*

) )

Spin part wave function

€

↑↓ , ↓↑ , ↑↑ , ↓↓

Then

€

ψ t( ) =C↑↑ t( )↑↑ +C↓↓ t( )↑↑ +C↑↓ t( )↑↓ +C↓↑ t( )↓↑

€

! σ 1 ⋅! σ 2( )ψ

191029 素粒子I 30

191029 31

€

i dψdt

= A ! σ 1 ⋅! σ 2( )ψ

€

ddt

C↑↑

C↓↓

C↑↓

C↓↑

$

%

& & & &

'

(

) ) ) )

= −iA

1 0 0 00 1 0 00 0 −1 20 0 2 −1

$

%

& & & &

'

(

) ) ) )

C↑↑

C↓↓

C↑↓

C↓↑

$

%

& & & &

'

(

) ) ) )

€

ψ t( ) =C↑↑ t( )↑↑ +C↓↓ t( )↑↑ +C↑↓ t( )↑↓ +C↓↑ t( )↓↑

Equation of motion:

p p

素粒子I