1. What is the GROUND STATE?
lowest energy state
What does GROUND STATE mean in
Quantum Field Theory?
Shouldn’t that just be the vacuum state? | 0 
which has an
compared to 0.
Fields are fluctuations about the GROUND STATE.
Virtual particles are created from the VACUUM.
The field configuration of MINIMUM ENERGY is
usually just the obvious   0
(e.g. out of  away from a particle’s location)

2. L can only carry even powers of 
We’ve discussed
Vector fields (spin 1 particles) , g
massless Klein-Gordon (Proca) Lagrangian
Dirac spinor fields (spin ½ particles) q, e, m, t
Dirac Lagrangian
What about Scalar fields (spin 0 particles)?
By definition, to “transform like a scalar” means
'(x')= (x)
under a Lorentz transformation
Note: a Lagrangian's dependence on space-time
exists only through the field and its first derivative
xm doesn't appear explicitly in any Lagrangian
is a 4-vector so only appears in a
Lagrangian contracted with itself!
NOTE:
L can only carry even powers of 
i.e. *, (* )2
to itself be real and preserve U(1) and Lorentz invariance

3. £ £ £ £  cannot be made U(1)-invariant
the simplest Lagrangian
with (xm) and m  dependence is
this is the Klein-Gordon Lagrangain!
from which  / - m [ /  (m ) ] =0
yields
the familiar Klein-Gordon equation
however
 cannot be made U(1)-invariant
if it is real
and the particles to be described are
electrically charged
but as you have seen (Exam 2)
 cannot be made U(1)-invariant
if it is real
and the particles to be described are
electrically charged

4. A U(1) transformation sends:
If  is strictly real, this is not an invariant form
unless q = 0
in which case we have no charge-coupling to photon fields.
Insisting that  be real would mean the
Lagrangian cannot be U(1) invariant.
Insisting on U(1) invariance means  cannot be real.

5. £ £ £ -m2 So instead we try an explicitly COMPLEX FIELD:
Recall: , separately satisfied conjugate field equations
for a Vector Field
to remain invariant.
for a Dirac Field
but for a simple scalar, the usual 4-vector “DOT” product
becomes just the one-dimensional product 
-m2
is this some kind of interaction? m2 m2
This is more like some sort of self-propagation term
self-persistence…
not a Feyman vertex, but a MASS term!

6. £=½½¼
Klein-Gordon Lagrangian for a “complex” scalar field
Scalar (spin=0) particle
(with “self-interaction” term)
£=½½¼
Note: OBVIOUSLY globally invariant under U(1) transformations
  ei “globally” means   constant
Not LOCALLY invariant yet.
That is introduced through a covariant derivative with new fields.
A completely EQUIVALENT form found
by separating out the R() and I() parts:
 =1 + i2



7. £=½½ ¼
Klein-Gordon Lagrangian for a “complex” scalar field
Scalar (spin=0) particle
(with “self-interaction” term)
Then
£=½½ ¼
becomes
£=½11 + ½22
½12 + 22 ¼12 + 22 
Which is ROTATIONALLY invariant under SO(2)!!
1 2
Remember SO(2) U(1) are “isomorphic”
Our Lagrangian yields the field equation:
1 + 12 + 112 + 22  = 0
or equivalently
 2 +  22 +  212 +  22  = 0
Klein-Gordon equation
some sort of
interaction between
the independent states

8. 1 + 12 + 112 + 22  = 0
Obviously as 0 there are no reactions
and we will have only free particle states.
And as (or in regions where)   0
we have the empty state | 0 
representing the lowest possible energy state
and serving as the vacuum.
The exact numerical value of
the energy content/density of | 0 
is totally arbitrary…relative.
We measure a state’s or system’s energy with
respect to it and usually assume it is or set it to 0.
What if the EMPTY STATE did NOT
carry the lowest achievable energy? =
We will call 0| |0 = vev the
“vacuum expectation value” of an operator state.

9. U U    ( , ) Clearly the Lagrangian under consideration
the coefficients of our
general interaction terms
Now, inspired by the classical Lagrangian where L = T - V
we separate the “kinetic” from the “potential” terms U U
where all the dependence is carried by   (U )
And as U  0 we have free, massless particles
What is the minimum U possible?
Keeping just the mass term (a free particle)?
Where  vanishes (empty space)?
Let’s try looking at extrema in U

10. and I guess we’ve sort of been assuming
some calculus of variations U
1
Letting  = *
= 0 U
2
= 0
Extrema occur not only for
but also for
But since  = *
> 0
This must mean -2 > 0
2 < 0
and I guess we’ve sort of been assuming
 was a mass term!

11. defines a circle of radius2
defines a circle of radius
| | / 
in the 2 vs 1 plane 1
Furthermore, from
not x-y “space” U
1
we note
2U
12
which at  = 0 gives just 2
< 0
making  = 0 the location of a local MAXIMUM!

12.  2  1 Lowest energy states exist in this
circular valley/rut of radius v =
This clearly shows the U(1)SO(2)
symmetry of the Lagrangian
But only one final state can be “chosen”
Because of the rotational symmetry all are equivalent
We can chose the one that will simplify our expressions
(and make it easier to identify the meaningful terms)
“subtract off” the
vacuum’s energy
expanding the field about the ground state: 1(x)=+(x)

13. L=½11 + ½22
Scalar (spin=0) particle Lagrangian
L=½11 + ½22
½12 + 22 ¼12 + 22 
with these substitutions:
v =
becomes
L=½ + ½
½2 +2v+v2+ 2 
¼2 +2v+v2 + 2 
L=½ + ½
½2 + 2 v ½v2
¼2 +2 ¼22 + 22v+v2
¼2v+v2

14. Explicitly expressed in
L=½ + ½ ½(2v)2
v2 + 2¼2 +2  + ¼v4
½  ½(2v)2
Explicitly expressed in
real quantities  and v
this is now an ordinary
mass term!
 “appears” as a scalar (spin=0)
particle with a mass
½
 “appears” as a massless scalar
There is NO  mass term!