√λ. √λ. V ( f ) = ½ 2 + ¼ 4. -. A translation f (x) = f 0 + u (x) → u (x) ≡ f (x) – f 0. selects one of the minima by moving into a new basis. redefining the functional form of f in the new basis ( in order to study deviations in energy from the minimum f 0 ).
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V(f) = ½2 + ¼4
A translation f(x) = f0 + u(x)→ u(x) ≡ f(x) – f0
selects one of the minima by moving into a new basis
redefining the functional form of f in the new basis
(in order to study deviations in energy from the minimum f0)
V(f) = V(u+f0) = ½(u+f0)2 + ¼ (u+f0)4
= V0+u2+ √ u3+ ¼u4
plus new self-
The observable field describes
particles of ordinary mass |m|/2.
we can neglect
some calculus of variations
Letting = *
Extrema occur not only for
but also for
But since = *
This must mean-2 > 0
2 < 0
and I guess we’ve sort of been assuming
was a mass term!
defines a circle of radius
in the 2vs1 plane
not x-y “space”
which at = 0gives just2
making = 0the location of a local MAXIMUM!
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)
shift to the
selected ground state
expanding the field about the ground state:1(x)=+(x)
L=½11 + ½22
with these substitutions:
L=½ + ½
L=½ + ½
½2+ 2 v ½v2
L=½ + ½ ½(2v)2
v2+2¼2+2 + ¼v4
Explicitly expressed in
real quantities and v
this is now an ordinary
“appears” as a scalar (spin=0)
particle with a mass
“appears” as a massless scalar
There is NO mass term!
Of course we want even this Lagrangian to be invariant to
LOCAL GAUGE TRANSFORMATIONS
Let’s not worry about the
higher order symmetries…yet…
free field for the gauge
* 12 + i22
again we define: 1 + i2
Exactly the same potentialU as before!
[+ v22] + [ ] + [ FF+ GG] -gvG
gG[-] + [2+2v+2]GG
+ 222+2v22+v4 + 4 ) ]}
L= [+ v22] + [ ] + [ FF+ GG] -gvG
gG[-] + [2+2v+2]GG
+ [ v4]
+ [4+222+ 4]
and many interactions
which includes a
The constants , v give the
coupling strengths of each
a whole bunch
of 3-4 legged
But no MASSLESS
scalar particle has
ever been observed
is a ~massless spin-½ particle
is a massless spin-1particle
spinless,have plenty of mass!
plus - gvG seems to describe
( some QM oscillation
between mixed states?)
Higgs suggested:have not correctly identified the
PHYSICALLY OBSERVABLE fundamental particles!
RememberLisU(1)invariant - rotationally invariant in , (1, 2) space –
i.e. it can be equivalently expressed
under any gauge transformation in the complex plane
/=(cos + isin )(1 + i2)
=(1cos-2sin ) + i(1sin+ 2cos)
With no loss of generality we are free to pick the gaugea ,
for example, picking:
/2 0 and/ becomes real!
(x) = 0
With real, the field vanishes and our Lagrangian reduces to
introducing a MASSIVE Higgs scalar field, ,
and “getting” a massive vector gauge field G
Notice, with the field gone,
all those extra
, , and interaction terms
Can we employ this same technique to explain massive Z and W vector bosons?
These two separate cases will follow naturally by assuming the Higgs field
is aweak iso-doublet(with a charged and uncharged state)
withQ = I3+Yw /2and I3 = ±½
for Q=0Yw = 1
Q=1Yw = 1
couple to EW UY(1) fields: B
Higgs= withQ=I3+Yw /2and I3 = ±½
Yw = 1
Consider just the scalar Higgs-relevant terms
not a single complex function now, but a vector(an isodoublet)
Once again with each field complex we write
+ = 1 + i2 0= 3 + i4
† 12 + 22 + 32 + 42
just like before:
U=½m2† + ¹/4 († )2
12 + 22 + 32 + 42=
Notice how 12, 22 … 42appear interchangeably in the Lagrangian
invariance to SO(4) rotations
Just like with SO(3) where successive rotations can be performed to align a vector
with any chosen axis,we can rotate within this 1-2-3-4 space to
a Lagrangian expressed in terms of a SINGLE PHYSICAL FIELD
Were we to continue without rotating the Lagrangian to its simplest terms
we’d find EXTRANEOUS unphysical fields with the kind of bizarre interactions
once again suggestion non-contributing “ghost particles” in our expressions.
So let’s pick ONE field to remain NON-ZERO.
because of the SO(4) symmetry…all are equivalent/identical
might as well make real!
Can either choose
But we lose our freedom to choose randomly. We have no choice.
Each represents a different theory with different physics!
Let’s look at the vacuum expectation values of each proposed state.
Aren’t these just orthogonal?
Shouldn’t these just be ZERO?
Yes, of course…for unbroken symmetric ground states.
If non-zero would imply the “empty” vacuum state “OVERLPS with”
or contains (quantum mechanically decomposes into) some of + or 0.
But that’s what happens in spontaneous symmetry breaking:
the vacuum is redefined “picking up” energy from the field
which defines the minimum energy of the system.
a non-zero proposed state.
This would be disastrous for the choice + = v + H(x)
since0|+ = vimplies the vacuum is not chargeless!
But 0| 0 = v is an acceptable choice.
If the Higgs mechanism is at work in our world,
this must be nature’s choice.
Let’s recap: proposed state.
We’ve worked through 2 MATHEMATICAL MECHANISMS
for manipulating Lagrangains
Introducing SELF-INTERACTION terms (generalized “mass” terms)
showed that a specific GROUND STATE of a system need
NOT display the full available symmetry of the Lagrangian
Effectively changing variables by expanding the field about the
GROUND STATE (from which we get the physically meaningful
ENERGY values, anyway) showed
We then applied these techniques by introducing the proposed state.
scalar Higgs fields
through a weak iso-doublet (with a charged and uncharged state)
which, because of the explicit SO(4) symmetry, the proper
gauge selection can rotate us within the1,2,3,4space,
reducing this to a single observable real field which we
we expand about the vacuum expectation value v.