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√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 ). - PowerPoint PPT Presentation
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V() ½ 2 + ¼ 4
A translation (x) = 0 + u(x) → u(x) ≡ (x) – 0
V() V(u +) ½(u +)2 + ¼ (u +)4
selects one of the minima by moving into a new basisredefining the functional form of in the new basis
(in order to study deviations in energy from the minimum 0)
√λ
√λ
V +u2 + √ u3 + ¼u4
energy scalewe can neglect
The observable field describes particles of ordinary mass |/2.
plus new self-interaction terms
some calculus of variations UU1
UU22
2
2
1412
2
2
1
2
21 )()(
)( 2
2
2
111
2 Letting = *
1
2
11
2 )( UU2
2
2
22
2 )(
= 0
= 0
Extrema occur not only for )0,0(021
but also for
22
2
2
1
But since = * > 0 This must mean 2 > 0 2 < 0
and I guess we’ve sort of been assuming was a mass term!
02
22
21
defines a circle of radius | | /
in the 2 vs 1 plane
1
2
not x-y “space”
Furthermore, from
2UU1
2
2
1
2
2
2
1
2 2)(
UU1
)( 2
2
2
111
2
we note
2
2
2
1
2 3
which at = 0 gives just 2 < 0
making = 0 the location of a local MAXIMUM!
1
2
Lowest energy states exist in thiscircular valley/rut of radius v = /2
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)
vxx )()(1
)()(2
xx shift to the
selected ground state
expanding the field about the ground state: 1(x)=+(x)
vxx )()(1
)()(2
xx v = /2
Scalar (spin=0) particle Lagrangian
with these substitutions:
becomes
L=½11 + ½22 ½1
2 + 2
2 ¼12 + 2
2
L=½ + ½ ½2
+2v+v2+ 2 ¼2
+2v+v2 + 2
L=½ + ½ ½2
+ 2 v ½v2
¼2 +2 ¼22
+ 22v+v2 ¼2v+v2
L=½ + ½ ½(2v)2
v2 + 2¼2
+2 + ¼v4
Explicitly expressed in real quantities and v
this is now an ordinary mass term! “appears” as a scalar (spin=0)
particle with a mass 22 vm
½ ½(2v)2
22
½ “appears” as a massless scalar
There is NO mass term!
Of course we want even this Lagrangian to be invariant to
LOCAL GAUGE TRANSFORMATIONS D=+igGLet’s not worry about the higher order symmetries…yet…
FFigGigG4
1*
4*
2
1*
2
1 22 L
free field for the gauge particle introduced
Recall: F=GG
FFGGg
Gig
Gig
4
1*
4*
2
1*
2
*2
*2
*2
1
222
L
again we define: 1 + i2 * 12 + i2
2
FFGGg
Gig
4
1
422
**22
1
2
1
22
2
2
1
2
2
2
1
22
2
2
1
2
2211
L
FFGGg
Gig
4
1
422
**22
1
2
1
22
2
2
1
2
2
2
1
22
2
2
1
2
2211
L
212121212
iiiiGig
22211211
22211211
2
ii
iiG
ig
1221
22
Gg
FFGGg
gG
4
1
422
2
1
2
1
22
2
2
1
2
2
2
1
22
2
2
1
2
12212211
L
Exactly the same potential U U as before!
so, also as before:v
1
2
with22 v
Note:v =0
L = [ + v22 ] + [ ] + [ FF+ GG] gvG
12
12
-14
g2v2
2
+{ gG[] + [2+2v+2]GG g2
2
+ [2v3+v4+2v22
12vv2v3v
v22v4 ) ]}
L = [ + v22 ] + [ ] + [ FF+ GG] gvG12
12
-14
g2v2
2
+{ gG[] + [2+2v+2]GG g2
2
+ [ v4]2
4v( 22 ]
12
which includes a numerical constant
v4
4
and many interactionsbetween and
The constants , v give thecoupling strengths of each
which we can interpret as:
L = scalar field with
22 vm
masslessscalar
free Gaugefield withmass=gv
gvG + a whole bunch of 3-4 legged
vertex couplings
But no MASSLESS scalar particle has ever been observed
is a ~massless spin-½ particleis a massless spin-1 particle
plus gvG seems to describe
G
Is this an interaction?A confused mass term?G not independent? ( some QM oscillation between mixed states?)
Higgs suggested: have not correctly identified thePHYSICALLY OBSERVABLE fundamental particles!
spinless , have plenty of mass!
NoteRemember L is U(1) invariant rotationally invariant in , (1, 2) space –
i.e. it can be equivalently expressedunder any gauge transformation in the complex plane
)(xie /
or
/=(cos + i sin )(1 + i2)
=(1 cos 2 sin ) + i(1 sin + 2 cos)With no loss of generality we are free to pick the gaugefor examplepicking:
121tan
/2 0 and
/ becomes real!
cossin21
1
2
ring of possible ground states
equivalent torotating thesystem byangle
1
2tan
2
2
2
1
2sin
2
2
2
1
1cos
2
2
2
1
21212
2
2
1
2
2
2
1
i
(x) (x) = 0
£
With real, the field vanishes and our Lagrangian reduces to
443222
2222
44
24
2
1
2
1
vvv
vv
GGgGGg
GGg
FF
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 have vanished
Can we employ this same technique to explain massive Z and W vector bosons?
Now apply these techniques: introducing scalar Higgs fields with a self-interaction term and then expanding fields about the ground state of the broken symmetry
to the SUL(2)×U(1)Y Lagrangian in such a way as to
endow W,Zs with mass but leave s massless.
+
0
These two separate cases will follow naturally by assuming the Higgs field is a weak iso-doublet (with a charged and uncharged state)
with Q = I3+Yw /2 and I3 = ±½
Higgs= for Q=0 Yw = 1
Q=1 Yw = 1
couple to EW UY(1) fields: B
Higgs= with Q=I3+Yw /2 and I3 = ±½
Yw = 1
+
0
Consider just the scalar Higgs-relevant terms
£Higgs
22 ) (4
1
2
1) (
2
1 02 with† † †
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 + 2
2 + 32 + 4
2
£Higgs
244114411
2
442211
)(4
1)(
2
1
)(2
1
† † †
† † † †
U =½2† + ¹/4 † )2
LHiggs
244114411
2
442211
)(4
1)(
2
1
)(2
1
† † †
† † † †
just like before:
12 + 2
2 + 32 + 4
2 = 22
Notice how 12, 2
2 … 42 appear 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 termswe’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.
1 or 2 3 or 4
because of the SO(4) symmetry…all are equivalent/identical
might as well make real!
+
0Higgs=
Can either choose v+H(x)
00
v+H(x)or
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.
v+H(x)0
0v+H(x)
or
000
00 000
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.
0)(000 xHv )(000 xHv
0)(000 xHv )(000 xHv
)(000 xHv 0
1
= v a non-zerov.e.v.!
This would be disastrous for the choice + = v + H(x)since 0|+ = v implies 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: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
•The scalar field ends up with a mass term; a 2nd (extraneous) apparently massless field (ghost particle) can be gauged away.
•Any GAUGE FIELD coupling to this scalar (introduced by local invariance) acquires a mass as well!
We then applied these techniques by introducing the scalar Higgs fields
through a weak iso-doublet (with a charged and uncharged state)
+
0Higgs=
0v+H(x)
=
which, because of the explicit SO(4) symmetry, the proper
gauge selection can rotate us within the1, 2, 3, 4 space,
reducing this to a single observable real field which we we expand about the vacuum expectation value v.
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