V ( f ) = ½   2 + ¼  4

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# V ( f ) = ½   2 + ¼  4 - PowerPoint PPT Presentation

 √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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√λ

√λ

V(f) = ½2 + ¼4

-

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+ ¼u4

plus new self-

interaction terms

The observable field describes

particles of ordinary mass |m|/2.

energy scale

we can neglect

U

U

1

some calculus of variations

Letting = *

= 0

U

2

= 0

Extrema occur not only for

but also for

But since = *

This must mean-2 > 0

2 < 0

> 0

and I guess we’ve sort of been assuming

 was a mass term!

2

defines a circle of radius

|| / 

in the 2vs1 plane

Furthermore, from

1

U

1

not x-y “space”

we note

2U

12

which at = 0gives just2

< 0

making = 0the location of a local MAXIMUM!

Lowest energy states exist in this

circular valley/rut of radius v =

 2

 1

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)

Scalar (spin=0) particle Lagrangian

L=½11 + ½22

½12+22 ¼12+22 

with these substitutions:

v =

L=½ + ½

½2+2v+v2+2 

¼2+2v+v2+2 

becomes

L=½ + ½

½2+ 2 v ½v2

¼2+2 ¼22+22v+v2

¼2v+v2

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!

Of course we want even this Lagrangian to be invariant to

LOCAL GAUGE TRANSFORMATIONS

D=+igG

Let’s not worry about the

higher order symmetries…yet…

free field for the gauge

particle introduced

Recall: F=G-G

 *  12 + i22

again we define:  1 + i2

so, also as before:

with

Note:

v =0

Exactly the same potentialU as before!

L=

[+ v22] + [  ] + [ FF+ GG] -gvG

1

2

g2v2

2

1

2

-1

4

+{

g2

2

gG[-] + [2+2v+2]GG

2

1

2

+ [2v3+v4+2v2

- (4+43v+62v2+4v3+4v2

+ 222+2v22+v4 + 4 ) ]}

L= [+ v22] + [  ] + [ FF+ GG] -gvG

1

2

g2v2

2

1

2

-1

4

+{

g2

2

gG[-] + [2+2v+2]GG

4

2

1

2

- [4v(3+2)

+ [ v4]

+ [4+222+ 4]

and many interactions

between  and 

which includes a

numerical constant

v4

4

The constants , v give the

coupling strengths of each

which we can interpret as:

a whole bunch

of 3-4 legged

vertex couplings

massless

scalar

free Gauge

field with

mass=gv

scalar field 

with

L =

-gvG +

+

+

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

G

• Is this an interaction?
• A confused mass term?
• G not independent?

( 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

Note:

or

/=(cos + isin )(1 + i2)

=(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!

ring of possible ground states

equivalent to

rotating the

system by

angle -

2

1

(x)

(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

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 theSUL(2)×U(1)YLagrangianin such a way as to
• endow W,Zs with mass but leave s massless.

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

Higgs=

+

0

for Q=0Yw = 1

Q=1Yw = 1

couple to EW UY(1) fields: B

+

0

£

£

Higgs= withQ=I3+Yw /2and I3 = ±½

Yw = 1

Consider just the scalar Higgs-relevant terms

with

Higgs

not a single complex function now, but a vector(an isodoublet)

Once again with each field complex we write

+ = 1 + i2 0= 3 + i4

†  12 + 22 + 32 + 42

Higgs

L

Higgs

just like before:

U=½m2† + ¹/4 († )2

-2m2

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

we’d find EXTRANEOUS unphysical fields with the kind of bizarre interactions

once again suggestion non-contributing “ghost particles” in our expressions.

+

0

Higgs=

So let’s pick ONE field to remain NON-ZERO.

1or2

3or4

because of the SO(4) symmetry…all are equivalent/identical

might as well make  real!

v+H(x)

0

0

v+H(x)

Can either choose

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

0

v+H(x)

or

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

v.e.v.!

= v

0

1

This would be disastrous for the choice + = v + H(x)

since0|+ = 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:

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)

+

0

0

v+H(x)

Higgs=

=

which, because of the explicit SO(4) symmetry, the proper

gauge selection can rotate us within the1,2,3,4space,

reducing this to a single observable real field which we

we expand about the vacuum expectation value v.