Download M = m 1 + m 2

States, operators and matrices
Starting with the most basic form
of the Schrödinger equation, and
the wave function ():

 jm  jm

   c j ,m jm
  Aei kx t   Be i kx t 
j ,m
The state of a quantum mechanical
system is . This state can be expanded
to a vector and the system can be more
complicated (e.g. with spin, etc.):
   jm

Mˆ jm   jm jm
Note: to extract the expectation
value of a property, one needs
to sum over the expanded state:

 2  2
i

t
2m x 2

Dirac notation
 Mˆ 

c
j ,m
2
j ,m
*
jm

M jm   c 2j ,m jm
j ,m
jm Mˆ jm   c 2j ,m jm
j ,m
j ,m
Combining operators
Operator T, transforms the original state.
M  jm   jm jm
Here, a hermitian matrix M was introduced,
to extract properties from the wave function
(hermitian matrices have real eigenvalues).

c 
2
j ,m
jm transformed  Tˆ jm
original
Hence, to get the expectation value:
trf
jm Mˆ jm trf org jm Tˆ 1Mˆ Tˆ jm
Mˆ  TˆMˆ Tˆ 1
org
Adding spin (1)
• Projection is known (m quantum number)
• Length of the two spins is known (j1 and j2)
• Several possibilities to construct projection
by adding the two spins
Example: m = jmax - 2
j1 - 2
j2
< j1, m1 | j2, m2 > = < j1, j1 - 2 | j2, j2 >
jmax - 2
m
Adding spin (1)
• Projection is known (m quantum number)
• Length of the two spins is known (j1 and j2)
• Several possibilities to construct projection
by adding the two spins
Example: m = jmax - 2
j1 - 1
j2-1
< j1, m1 | j2, m2 > = < j1, j1 - 1 | j2, j2 - 1 >
jmax - 2
m
Adding spin (1)
• Projection is known (m quantum number)
• Length of the two spins is known (j1 and j2)
• Several possibilities to construct projection
by adding the two spins
Example: m = jmax - 2
j1
j2- 2
< j1, m1 | j2, m2 > = <j1, j1 | j2, j2 - 2 >
jmax - 2
m
Adding spin (2)
• Projection is known (m quantum number)
• Length of the maximum total spin known
jmax = j1 + j2
• Several possibilities to construct projection
from different sizes of total spin j
j1
j2
| j1 –j2 | ≤ j ≤ j1 + j2
j
Note:
j1, j2 and j can be interchanged. However, changing the
composite state with one of the constituent states is not
trivial and requires re-weighting of the constituent states
j = j1 + j2 -2
jmax - 2
m
< j, m > = < j1 + j2, j1 + j2 - 2 >
< j, m > = < j1 + j2 - 1, j1 + j2 - 2 >
< j, m > = < j1 + j2 - 2, j1 + j2 - 2 >
}
m≥j
Adding spin (3)
•
m=jmax
m1=j1
& m2=j2
j=j1+j2
& m=jmax
m=jmax-1
m1=j1 & m2=j2-1
m1=j1-1 & m2=j2
j=j1+j2 & m=jmax-1
j=j1+j2-1 & m=jmax-1
m=jmax-2
m1=j1 & m2=j2-2
m1=j1-1 & m2=j2-1
m1=j1-2 & m2=j2
j=j1+j2 & m=jmax-2
j=j1+j2-1 & m=jmax-2
j=j1+j2-2 & m=jmax-2
Note:
- j1  m1  j1 , thus m1 has (2 . j1+1) possible values and m2 (2 . j2 +1).
Each combination shows up exactly ones in the second column of the table
so the total number of states is (2 . j1+1) (2 . j2 +1).
•
The third column has the same amount of states as the second column.
The quantum number j is a vector addition, thus it will never be lower than |j1 – j2|, which is called jmin.
•
For m < jmin the amount of states is jmax – jmin + 1 = (j1+j2) – |j1 – j2| + 1 for each m.
This situation occurs for - |j1 – j2|  m  |j1 – j2| , thus (2 |j1 – j2| + 1) times.
Number of states:
•
For m  -jmin and m  jmin the amount of states is jmax  m +1. This results in the following sum:
Number of states:
•
(2 |j1 – j2| + 1) . [(j1+j2)-|j1 – j2| + 1]
2
m  j1  j2
 j
1
m  j1  j2 1
 j2   m  1
Together these contributions add up to (2 . j1 + 1) (2 . j2 + 1)
Adding spin – Clebsch Gordan
Example: m = jmax - 2
j = j1 + j2 -2
j1
jmax - 2
j2- 2
jmax - 2
m
JM 
C
J , M , m1 , m2
m1 , m2
m
with : J  M
and : M  m1  m2
m1 , m2
or : JM 
C
J , M , m1
m1 , M  m1
m1 , m2
e.g.
J 1, J  2  C1 j1 , j2  2  C2 j1 1, j2 1  C3 j1  2, j2
From symmetry relations and
ortho-normality, the C coefficients
can be calculated. The first few:
m1,m2
JM
½x½
1, 1
1, 0
0, 0
1, -1
+½, +½
1
0
0
0
+½, -½
0
½
½
0
-½, +½
0
½
½
0
-½, -½
0
0
0
1
Combining spin and boost
Lorentz transformations:
 cosh( ) 0

1
 0
Lz  p   
0
0

 sinh(  ) 0

p
tanh(  ) 
sinh(  ) 
E
0 sinh(  ) 

0
0 
1
0 

0 cosh( ) 
p
E
cosh( ) 
w
w
w 
E 2  p2

 
p  E, p  p  p  R , ,0 zˆ

For:
Jackson (section 11.7) calculates the corresponding operator:



 ip
ˆ
ˆ
Extracting
rotations:
L p   e
L p   Rˆ  , ,0Lˆ z  p Rˆ 1  , ,0


ˆ
Allowing definition of the canonical state: L p  jm  p, jm
Intermezzo: rotation properties
– The total spin commutes
with rotation

 2 i nJ 


 
 i n J  2
 i n  J 
J e
 jm  e
J  jm  j  j  1 e
 jm
– However, the projection is affected with a phase.
Consider the rotation around
the quantization axis




R jm   jm' e
m'
e.g.
jm' e

 i n  J 
 
i nz  J 

jm  jm'
jm  jm' e
iJ z
jm   m'm e
im
– Euler rotations, convention: z y’ z’’
Advantage: quantization axis used twice for rotation
iJ z '' iJ y ' iJ z
ˆ
Re
e
e
– Rotations are unitary operators. The rotation around y’
includes a transformation of the previous rotation.
  UU 1 e
 i J y
 e iJ z e
 i J y
– This results in: R
ˆ  eiJ z eiJ y eiJ z
eiJ z  e
 iJ y '
i.e. Rotations can all be carried out in same
coordinate system when order is inverted.
Intermezzo: the rotation matrix
Summary of the result from the previous page
R  jm   jm' e iJ z e
 iJ y
e iJ z jm  jm'
m'


  Dmj 'm    jm'    Dmj 'm   jm' 
m'
 m'

– Euler z y z convention makes left and right term easy
jm' e iJ z e
e im ' jm' e
 i J y
 i J y
e  iJ z jm
jm e im  e im '  d mj 'm    e im
– The expression for djm´m is complicated, but is used mainly for
the deduction of symmetry relations:
d mj 'm    
m'  m
j
d mm
'  
Note 1: Inverse rotation is accomplished by
performing the rotations through negative
angles in opposite order:
( Djm’m(, , ) )-1 = Djm’m(-,-,-)
d mj 'm    d jm,  m'  
Note 2: Since rotations are hermitic,
the conjugate matrix is:
Djm’m(-,-,-)=Dj*mm’()
Note 3: Combination of all this gives:
Dj*m’m()=(-)m’-mDj-m’,-m()
Rotations, boosts and spin
Describing the rotation of a canonical state:


ˆ
ˆ
ˆ
R1 p, jm  R1L p  jm  Rˆ1Rˆ0 Lˆ z  p Rˆ01 jm
 Rˆ1 Rˆ 0 Lˆ z  p Rˆ 01 Rˆ11 Rˆ1 jm  Rˆ1,0 Lˆ z  p Rˆ1,01 Rˆ1 jm


j
j
ˆ
ˆ
 LR1 p  Dm 'm,1 jm'   Dm 'm,1 LR1 p  jm'
 D
m'
j
m 'm ,1
m'


j
R1 p, jm'  Dm 'm,1 R1 p, jm'

m'
Remember:
R1   jm  Rˆ1   jm   Dmj 'm,1   jm'
m'
i.e. rotation of a canonical state rotates the boost
and affects the spin state in the same way as it
would in the `particle at rest’ state.
Helicity

 ˆ

ˆ
The helicity state is defined with: p, j  L p R0 j  p, j
ˆ  p Rˆ j  Rˆ Lˆ Rˆ 1Rˆ j  Rˆ Lˆ j
Note: L
0
0 z 0
0
0 z
Compared to the same operations on the spin state:
 ˆ

j
ˆ
L p R jm   D
p, jm'
0
m 'm , 0
m

1 ˆ
j
ˆ
ˆ
ˆ
ˆ
ˆ
R0 Lz jm  R0 Lz R0 R0 jm   Dm 'm, 0 p, jm'
m


j
Hence: p, j   Dm ,0 p, jm
m
In other words: helicity () is the
spin component (m) along the
direction of the momentum.
Note that the helicity state does
not change with rotations:


Rˆ1 p, j  Rˆ1Lˆ  p Rˆ 0 j  Rˆ1 Rˆ 0 Lˆ z j


 Rˆ10 Lˆ z Rˆ101 Rˆ10 j  Lˆ R1 p Rˆ10 j  Rp, j
Note that the helicity state does ˆ  
ˆ Lˆ  p Rˆ 1 Rˆ Lˆ  p Rˆ 1 Rˆ j


L
p
p
,
j


R
1
0
0 z
1
0
0 z
0
0
0
not change with boosts (as long
 ˆ

as the direction is not reversed):  Rˆ Lˆ  p Rˆ 1 Rˆ j  Lˆ  p

R
j


R
p
0 z
10
0
0
10
0
10 , j
Discrete symmetries
Parity

x

x
ˆ

  dx 

p     p
   dt 

J  x  p 
J
Commutation
relations:
ˆRˆ  Rˆ ˆ


ˆ
ˆ
ˆL p   L p ˆ
Charge conjugation
Cˆ
particles  antiparticles
i
Dirac:

Conjugating+transposing gives:
i
T
C

T
C 1   

Resulting in:
Left-handed

  M  T  0
C is the matrix doing the transformation:
Mirror analogy
Left-handed
 particle
  M   0
C C i
1

T
   M C 1C T  0

C 1  i     CMC 1 C T  0
i


   M C T  0
e.g. C b  b
 c
antiparticle
Time reversal
e.g.
b
Vcb*
W+
Vcs
s
c
ˆ
c
c
c
b
Vcb*
Vcs
W+
s
Note that time reversal changes t in –t and input states
in output states (in other words: < bra | to | ket > ).
Another way to show this:
i.e. the transformed state does not
obey the description of motion of the
Hamiltonian, it needs an extra ‘–’ sign.

ˆ
H   i 
t


ˆ
ˆH   ˆi   i ˆ 
t
t
The solution is to make
time reversal anti-unitary:
Note: this can also
be shown with the
commutation relation:
ˆ
 
 
 p, x   i  p, x   i
ˆHˆ   ˆi 
    i
 *
 i   t 
t
 '
t


 '

 ˆ 
t 
*
Time reversal continued
Next, the the time reversal
operator is split in a unitary part
and a complex conjugation.
The m states consist of real
numbers, i.e. projections.
Hence a time reversed expectation
~
value can be described with:
 ~
Calculating the expectation
value of operator Â:  A
ˆ
And combining with the
time reversal operator:
 K   K  m  m
m

m  K m    m  m
*
m
m
  

m m'  m'  1 m   
m,m '
    A i.e.   Aˆ * 
1 ~
~
ˆ
ˆ
ˆ
 A    Aˆ   A
Since a second time reversal should
restore the original equation:
 Aˆ      ~ ~
 ~ ˆAˆ *   ~ ˆAˆ *ˆ 1ˆ 
~
 ~ ˆAˆ ˆ 1 
Aˆ  Aˆ *

~  ˆ 

Hermitic
~
~
2 ~
ˆ
 ˆ Aˆ ~   Aˆ   A
~
~
~
~
 
  
2
 ˆ2 Aˆ ˆ2    Aˆ  ˆAˆ ˆ1   Aˆ
Time reversal and spin
From the commutation relation:
J , J   i
i
j
ˆ
ijk


x
Consistent with:

J k   J i , J j  i ijk  J k 

x
ˆ

  dx 

p     p
   dt 

J  x  p 
J
Rˆ ˆ jm  Rˆ jm
ˆRˆ
j
 ˆDmm
'
jm
The equation:
j
mm'
T
j
j
Tmm
D
' m 'm"   
j * j
mm' m 'm"
D
T
 
t t
ˆ
jm'  D
j
m ' m"
j *
mm '
jm'  D
j * j
mm' m 'm"
D
T
Djm,  m' 
j m'
jm
holds if:
 m',  m Dmj 'm"   j m Djm,m"
mm'
ˆ, Rˆ  0
j
j ˆ
j
j
 Rˆ Tmm
jm
'

T
R
jm
'

T
D
'
mm '
mm ' m 'm" jm"
j *
mm '
D
j m
*
And:
*
t  t
j
mm'
T
j * j
mm ' m 'm"
D
T
  
jm"
j m
 m',  m
 d mmj '  
 m",  m'   j m2 m' Djm,m"   j m Djm,m"
Time reversal and spin continued
Time reversal of a canonical state:
ˆ p , jm
 
j m
*



j m
j


 p, jm t t  Tmm

p
,
jm
'




p
, jm'
'
m ',  m


j m 
 p, j,m or : ˆ p ,  , jm   
p   ,    , j,m
Time reversal of a helicity state:



j
j * 
p, j   Dm p, jm  p, jm  Dm p, j
m

j * 
ˆ p, jm  ˆDm p, j  Dmj ˆ p , j and
*
j m 
j m
j
 p   ,    , j,m   Dm,     ,     p    ,    , j,
*
j
m,  
with : D
Hence:
   ,      m Dmj     ,      m ei  j Dmj , 


j m
 m   i  j
Dmj ˆ p, j   
e Dmj ,  p   ,    , j,
Should give:

ˆ
 p, j
j
m
D
j
 i 
m
D e

p   ,    , j
Parity and spin
The parity operation on a canonical state:
ˆ



p, jm    p, jm   p   ,    , jm
The parity operation on a helicity state:



j
j * 
p, j   Dm p, jm  p, jm  Dm p, j
m

j * 
j *
ˆ p, jm  ˆDm p, j  Dm ˆ p , j  m Djm,ˆ p , j


j *
 p   ,    , jm  Dm    ,     p   ,    , j
and
   ,      m Djm,     ,       m ei  j Dmj , 
m Djm,ˆ p , j  m ei  j Dmj , p    ,    , j
j *
m
with : D
Hence:
Should give:
Note:
ˆ

 i  j 
p, j  e
p   ,    , j,


 p, jm  p   ,    , jm
Since helicity states include rotational properties:
but:


 p, j  p   ,    , j

 ˆ
ˆ
p, j  L p R0 j  Rˆ0 Lˆ z j
Composite states
SM 
 mm
1
2
SM m1m2 
m1 , m2
0,0 
s-quark
1
This ground state can decay
to two vector mesons:
Bs-meson
b-quark
c-quark
c-quark
s-quark
Isospin = 0
Spin = 0
Parity = -
m1m2
Note: M = m1 + m2
1 , 1  1  1 , 1
2 2
2
2
2 2
Bs-meson
W+
S , m1 , m2
m1 , m2
Bs ground state:
b-quark
C
s-quark
s-quark
J/-meson
(easy to detect)
Isospin = 0
Spin = 1
Parity = (C-parity = -)
-meson
Isospin = 0
Spin = 1
Parity = (C-parity = -)
Two body decay properties
SM 
 mm
1
2
SM m1m2 
m1 , m2
Spin states of Bs, J/ and :
C
S , m1 , m2
m1m2
m1 , m2
0,0 
1 1,1  1 0,0  1  1,1
3
3
3
Not the complete story… Consider momentum in two particle decay (Bs rest frame):


m1 , m2  a p  | m1  p  | m2
Normalization

=
Hence:
, m1 , m2
 , m m
C
  D     , SM '
 , SM 
Remember:
Rˆ   jm   D
j
m 'm
 jm'
Rˆ    , SM
m'

Still no complete story… Consider angular momentum:
lm, SM   dY   , SM
l
m
With:
S , m1 , m2
1
2
m1 , m2
j
m 'm
m'
2l  1 l *
Y  
Dm0 ,0
4
l
m
Angular Total spin
momentum
Missing: Formalism that describes angular momentum and Yml states.
Rotation with angular momentum
Rˆ   lm, SM   dYml  Rˆ   , SM   dYml  DMS 'M   ' , SM '
  dDMS 'M Yml  ' , SM ' 
2l  1
S
l *


d

D

D
M 'M
m 0 ,0 ' , SM '

4
Split up the Yml state, to match with new rotation:
l *
m0
D
D
,0  D
l
mm'
l *
mm'
? D ' ,0  D
l *
m '0
l *
mm'
R  D ' ,0
1
l *
m '0
4 l

Ym '
2l  1
Rˆ   lm, SM   dDMS 'M  Dml 'm  Yml ' ' , SM '  DMS 'M  Dml 'm   lm ' , SM '
i.e. the transformation is a product of the rotation of two rest states:
Rˆ  lm, SM  Rˆ  lm , SM  DMS 'M Dml 'm  lm' , SM '

C
m1 , m2
S , m1 , m2
, m1m2
Angular momentum and spin
We can also express them as states with a sum of angular momentum and spin:
 lm, SM
JM J , lS 
S
JM J lm, SM S 
m,M S

C
m,M S
J ,m, M S
lm SM S
C
J ,m,M S
lm, SM S
Note: MJ = m + MS
m,M S

   C J ,m, M S lm
m,M S 

m1m2 M S
m1 , m2

   C J ,m , M S  CS ,m1 ,m2  dYml   , m1m2
m,M S 
m1 , m2




Since:
Rˆ  lm, SM S  DMS S 'M S Dml 'm  lm' , SM S '
And:
DMS S 'M S Dml 'm  


, m1m2 

Note: MS = m1 + m2
(see 1 page back)
M J ' M S ' m' M J M S m DMJ J ',M J 
M J ', M J
The state can be rotated with:
Rˆ  JM J , lS  DMS J 'M J  JM J ' , lS
Note that (as expected) the sum of angular momentum and spin (J) is not
affected by the rotation, neither are the angular momentum (l) and the total
spin (S). This result is the equivalent of the non-relativistic L-S coupling.
Two body decay and helicity (1)
As with spin, we need to consider momentum in the helicity state:




 ˆ
 ˆ
ˆ
ˆ
12  a p  | 1  p  | 2   L p R  1 L p R  1





 Rˆ  Lˆ z  p  1 Rˆ  Lˆ z  p  1  Rˆ   Lˆ z  p  1 Lˆ z  p  1
 Rˆ   00 | 12
Angles are zero. Particles are boosted back to
back along the positive and negative Z-axis
The link with previous page is provided via the
relation between canonical and helicity states:


, m1m2  a  p  | m1  p  | m2 


s1 *
s2 *
 a  Dm11  Dm22   p , 1  p , 2
 a
  SM
S ,M S ,

 C
s
S ,M S ,
m1m2 S 12 D
S *
MS
S *
MS
Cm12 , S ,  D

 Rˆ   00 | 12
 = 1 - 2 Why?
MS = m1 + m2
12 , S , M S


  , 12
w
a  4
p
Single state
normalization.
Rest mass: w
Momentum: p
1
 , 12
a
Two body decay and helicity (2)
Expressing states with the sum of angular momentum and spin in helicity states:
JM J , lS 

C J ,m, M S
m,M S


C J ,m,M S
m,M S



m1 , m2
C J ,m,M S
m,M S

l
C
d

Y
 S ,m1 ,m2  m  , m1m2
m1 , m2
CS ,m1 ,m2  dYml  
CS ,m1 ,m2
m1 , m2
 C12 ,S ,M S Cm12 ,S ,  d
S , M S ,
*
lm
C
J ,m,M S
m,M S


m,M S
C
S , m1 , m2
m1 , m2
C J ,m,M S

m1 , m2
*
S , M S ,
DMJ J   DMS S      C

S
C
C

D
 12 ,S ,M S m12 ,S , M S   , 12
*
*
2l  1 l *
2l  1 l *
Dm 0     C
Dm 0  DMS S    DMJ J   
2J 1
2J 1
lm
Intermezzo, check…
 C
12 , S , M S
Cm12 ,S ,
S ,M S ,
CS ,m1 ,m2
*
2l  1 l *
 Dm 0    DMS S    , 12
4
 C12 ,S ,M S Cm12 ,S ,
S , M S ,
2l  1
J *
d f (l , m) DM J    , 12

4

2l  1
2J 1 J *
f ' (l , m)  d
DM J    , 12
2J 1
4

Normalization (later, easier the other way around)
NJ
 JM J , 12
Two body decay and helicity (3)
Check the transformation properties for rotations:
S *
ˆ
R  JM J , 12  N J  d DM J   Rˆ  Rˆ   00 | 12

 N J  d D

S *
MJ
 N J  d  D
 ,M J '
D
S
MJMJ '
 Rˆ     00 | 12
*
S
MJMJ '
  D
*
S
M J '
 N J  d D

*
S
M J '
'Rˆ ' 00 | 12
'Rˆ ' 00 | 12
 DMS J M J '   JM J ' , 12
Transforms as it should…
Remember:
ˆ 

R1 p, j  Rp, j
i.e. Mj transforms, but 1 and 2 do not.
Canonical versus helicity states
JM J , 12   d N J DMJ J   , 12   d
*

 C
, m1m2 

Cm12 ,S ,  DMS S   , 12
*
12 , S , M S
2J 1 J *
DM J   , 12 (2 pages back)
4
(3 pages back)
S ,M S ,
s1 *
m11
D
D
s2 *
m2 2
 , 12
 , 12  Dms111 Dms222  , m1m2
JM J , 12   d N J DMJ J    , 12   d
*
Dms111  Dms222   


 M
S
Combine
2J 1 J *
DM J   Dms111  Dms222   , m1m2
4
m1m2  12 DMS S   
S ,M S ,
DMS S   DMJ J     
*
lm
*
2l  1
M S lmM J  l 0 Dml 0  
2J 1
4 l
D   
Ym  
2l  1
l *
m0
JM J , 12    
 S , M S lm
JM J , 12  
Sl
 JM J , lS
(4 pages back)
Normalization
2l  1 2 J  1 4
M S lmM J  l 0 M S m1m2  12  dYml   , m1m2
2 J  1 4 2l  1
2l  1
 l 0  12
2J 1

M S ,m
M S lmM J M S m1m2  dYml  , m1m2
Decay amplitudes
Remember:
The transition amplitude to a helicity
state is calculated with the matrix element:


A  p , 1 |  p , 2
 a  12
m JM
J
m JM


, 12  a  p, 1 |  p, 2


p, 1 |  p, 2  a  , 12
J
w
 4
12
p
m JM
Momentum of decay products: p
Resonance rest mass: w
J
And:
Complete set
 4
w
12 JM J 12 JM J 12
p
m JM
J
, 12 JM J , 12  , 12  d N J D
With:

  d N J D
J *
MJ

Hence:
J *
MJ
A  NJ D
 , 12
J *
MJ
, 12 N J D
   4 w JM J 12
p
w
a  4
p
 , 12
J *
MJ

m JM
J
Check…
J *
MJ
 NJ D
 FJ
1 2
Helicity amplitude
Helicity amplitude
Remember:
JM J , 12  
Sl
2l  1
 l 0  12 JM J , lS
2J 1
2J 1
And: N J 
4
Switch to canonical states:
J
F12
w
 4
JM J , 12
p
m JM
J
Complete set
 4
w
JM J , 12 JM J , lS JM J , lS
p
m JM
w
2l  1
 4
 l 0  12 JM J , lS

p Sl 2 J  1
N J  F12
J
m JM
J
w 2J 1
2l  1
 4
 l 0  12 JM J , lS

p
4 Sl 2 J  1
  2l  1  l 0  12
Sl
J
w
4
JM J , lS
p
m JM
J
m JM
Partial wave
amplitude: als
Canonical states
J
Bs0 → J/ , tree level
Vcb*
b
Vcs
Bs0
W+
s
b
0
Bs
s
W
W
s
Bs0
b
Oscillation
W-
Vcs
Vcb*
Decay
c
c
J/
s
s

s
s
J/
c
c

The basics
a X0 b X0
d a 
i  a
i     M   
dt  b  
2  b 
 Ma

 0
 a

0
0 

Mb 
 a  b
0

b 
No mixing, just 2 states
1 0
 
0 1
Particle and anti-particle:
Ma  Mb
Schrödinger
Eigen states of the Hamiltonian:  



  iM t 
2


 a   a0  e 
 




  iM t
b 
2
 
Time evolution of eigen states  b0  e 

 0
M
en



0



0
0

M
The basics
a Bs0  b Bs0
d a 
i  a
i     M   
dt  b  
2  b 
Mixing:
Hermitic:
  12 




 21

 M M12 


M
M
 21

*
M 12  M 21
12  *21
 M
 *
 M12
M12 

M 
 
 *
 12
12 

 
Note: to obtain properties with real values, the matrix needs to be Hermitic.
The basics
a Bs0  b Bs0
 M
 *
  M 12
M 12  i  
  *
M  2  12
12    a 
  
  b 

i
i

 * i * 
M



M


12 

12  M 12 
 12
2
2
2






0



0

 a
 b 
i
i
i *   


M     M 12  12  M 12*  12

2
2
2



The matrix equation is 0 if:
a  p b  q
i *
M  12
q
2

i
p
M 12   12
2
*
12
Eigen-states:
 p  p 
   
 q   q 
The basics
Eigen-states:
 p  p 
   
 q   q 
p Bs0  q Bs0
Note: The Hamiltonian describes the quantum mechanical system. Only for
the eigen-states of the Hamiltonian Mass and Decay time have meaning
B1  p Bs0  q Bs0
B2  p Bs0  q Bs0
B1,2 (t )  B1,2  e
p  q 1
2
i


 i  M1,2  1,2 t
2


2
Note 1: No particle and antiparticle. Two different masses
and decay times
Note 2: What is the meaning of time (t) at this point. This
is a composite system, one particle contains two states.
The time is calculated in the rest frame of the particle…
On the next slide it becomes obvious.
The basics
Bs0
t
0
s t
B


p B

 q Bs0  e
p B


q B
0
s
B1  B2
2p
0
s
B1  B2
2q
PB  B
0
s


0
s t
P B B
0
s
0
s
f t   e
2
e
0
s
i 

 i  M1  1 t
2


e

e
i 

 i  M 2   2 t
2


 p Bs0  q Bs0  e

 p B q B
0
s
0
s
2q
 B B t 
t

i 

 i  M 2   2 t
2 

2p
0
s

1 t

i 

 i  M1  1 t
2 

0
s
B B t 
0
s
2 t
e
0
s

1 2
t
2
2
2
 f t 
 f   t  Bs0  f   t 
 f t 
2
q

f t 
p
f 
e
q 0
Bs
p
p 0
Bs  f   t  Bs0
q
i 

 i  M1  1 t
2 

2
cos   M 2  M1   t 
Note: So time (t) is just the decay time of the measured particle.
e
2
i 

 i  M 2   2 t
2 
