Transcript “probability current” “probability density” Klein Gordon equation

Lagrangian formulation of the Klein Gordon equation
L   L d 3 x, L
Klein Gordon field
lagrangian density
 ( x)
L =    ( x)    ( x)  m2 ( x)† ( x)
†
}
}
T
V
L
L


0


(  )
(     m2 )  0
Manifestly Lorentz
invariant
Euler Lagrange equation
Klein Gordon equation
The Lagrangian and Feynman rules
Associate with the various terms in the Lagrangian a set of propagators
and vertex factors


The propagators determined by terms quadratic in the fields, using the Euler
Lagrange equations.
The remaining terms in the Lagrangian are associated with interaction vertices.
The Feynman vertex factor is just given by the coefficient of the corresponding
term in iL
e.g.  (   ieA ) ( x)  (   ieeA ) ( x)   ieA ( *   (  * ) )
†
k, 
iL  ie( pi  p f ) A *
}
p
p'
New symmetries
L =    ( x)    ( x)  m2 ( x)† ( x)
†
Is invariant under
 ( x)  ei  ( x)
…an Abelian (U(1)) gauge symmetry
A symmetry implies a conserved current and charge.
e.g.
Translation
Rotation
Momentum conservation
Angular momentum conservation
What conservation law does the U(1) invariance imply?
Noether current
L =    ( x)    ( x)  m2 ( x)† ( x)
†
Is invariant under
 ( x)  ei  ( x)
i
…an Abelian (U(1)) gauge symmetry
i 
L
L
0  L =
 
 (  )  (   † )

(  )
0 (Euler lagrange eqs.)
 L
 L  
 L

 i 
  
  i  
   (   † )


  (  )  
  (  ) 
 







 j  0,

ie  L
L
†
j  

 
†

2   (   )
 (  ) 
Noether current
The Klein Gordon current
L =    ( x)    ( x)  m2 ( x)† ( x)
†
Is invariant under
 ( x)  ei ( x)

 j  0,
…an Abelian (U(1)) gauge symmetry

ie  L
L
†
j  

 
†

2   (   )
 (  ) 
j  ie      
KG
*
*

This is of the form of the electromagnetic current we used for the KG field
c.f.our original interpretation as probability density …
Physical interpretation of Quantum Mechanics
i t  21m  2  0
Schrödinger equation (S.E.)

t
i ( S .E.)  i (S .E.)
*
*
= 
 .j  0 continuity eq.
j   2im ( *   * )
2
“probability density”
“probability current”
2
 t 2  2  m2
Klein Gordon equation
  i( * t   t )
*
  Neip. x ,   2 E N
Negative probability?
j  i( *   * )
j   (  , j)
2
f p  e
ip. x
1
2 p0V
Normalised free particle solutions
The Klein Gordon current
L =    ( x)    ( x)  m2 ( x)† ( x)
†
Is invariant under
 ( x)  ei ( x)

 j  0,
…an Abelian (U(1)) gauge symmetry

ie  L
L
†
j  

 
†

2   (   )
 (  ) 
j  ie      
KG
*
*

This is of the form of the electromagnetic current we used for the KG field
Q   d 3 x j 0 is the associated conserved charge
Suppose we have two fields with different U(1) charges :
1,2 ( x)  ei Q 1,2 ( x)
1,2
L =   1 ( x)   1 ( x)  m21 ( x)† 1 ( x)
†
   2 ( x)   2 ( x)  m22 ( x)† 2 ( x)
†
..no cross terms possible (corresponding to charge conservation)
U(1) local gauge invariance and QED
 ( x)  ei ( x )Q ( x)
L =    ( x)    ( x)  m2 ( x)† ( x)
†
not invariant due to derivatives
     ei ( x )Q  ei ( x )Q   iQei ( x )Q   ( x)
To obtain invariant Lagrangian look for a modified derivative transforming covariantly
D  ei ( x ) D
Need to introduce a new vector field
D     iQA
A  A   
 ( x)  eiQ ( x ) ( x)
D  ei ( x ) D
A  A   
L =  D ( x)  D  ( x)  m2 ( x)†  ( x)
†
Note :
   D     iQA
is invariant under local U(1)
is equivalent to
p   p   eA
universal coupling of electromagnetism follows from local gauge invariance
L =    ( x)    ( x)  m2 ( x)†  ( x)  j  A  O(e2 )
†
The electromagnetic Lagrangian
0

 E1
 E2

 E3
F    A   A
F  F ,
A  A   
LEM   14 F F   j  A
M 2 A A
 E1
0
B3
 E2
 B3
0
 B2
B1
 E3 

B2 
 B1 

0 
Forbidden by gauge invariance
The Euler-Lagrange equations give Maxwell equations !
L
L



0

 
A
( A )
  F `  j

.E   ,
.B  0,
B
0
t
E
B 
j
t
E 
The photon propagator

The propagators determined by terms quadratic in the fields, using the Euler
Lagrange equations.
  F       A   (  A )  j
Choose as
1
   A

Gauge ambiguity
A  A   
(gauge fixing)
  A    A   2
i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve
1
1
    A  (1  ) (  A )  ( g  2  (1  )   ) A  j


In momentum space
1
p p 
  2
1  
i 
g
p

(1

)
p
p


g

(1


)
 




p2 
p2 


(‘t Hooft Feynman gauge ξ=1)