nuclear forces - lecture 3 -

R. Machleidt Nuclear Forces - Lecture 3 QCD & Nucl. Forces (CNSSS13)

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Nuclear Forces- Lecture 3 -

R. MachleidtUniversity of Idaho

QCD and Nuclear Forces;Effective Field Theory (EFT) for

Low-Energy QCD

12th CNS International Summer School (CNSSS13)28 August to 03 September 2013 -- CNS Wako Campus

• In Lecture 2, we have seen how beautifully the meson theory of nuclear forces works. This may suggest that we are done with the theory of nuclear forces.

• Well, as it turned out, the fundamental theory of the strong interaction is QCD (and not meson theory). Thus, if we want to understand the nuclear force on the most fundamental level, then we have to base it upon QCD.

• So, we have to start all-over again; this time from QCD.


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Prologue


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Lecture 3: QCD and Nuclear Forces; EFT for Low-Energy QCD

• The nuclear force in the light of QCD

• QCD-based models (“quark models”)

• The symmetries of low-energy QCD• An EFT for low-energy QCD: Why?• An EFT for low-energy QCD: How?• Power counting• The Lagrangians• The Diagrams: Hierarchy of nuclear forces


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The nuclear force in the light of QCD

If nucleons do not overlap, ….

then we have two separated colorless objects. In lowest order, they do not interact. This is analogous to the interaction between two neutral atoms. Like the Van der Waals force, the nuclear force is a residual interaction. Such interactions are typically weak as compared to the “pure” version of the force (Coulomb force between electron and proton, strong force between two quarks, respectively).

?


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The analogy

Van der Waals force

Residual force between two electric charge-neutral atoms

Nuclear force

Residual force between two color charge-neutral nucleons

Note that residual forces are typically weak as compared to the “pure” version of the force (order of magnitude weaker).


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Van der Waals force

+-

+-

Residual force between two electric charge-neutral atoms

Nuclear force

Residual force between two color charge-neutral nucleons

How far does the analogy go?

What is the mechanism for each residual force?

Two-photon exchange (=dipole-dipole interaction)

The perfect analogy would be a two-gluon exchange.

g

g

No!

No!

Because it would create a force of infinite range (gluons are massless), but the nuclear force is of finite range.


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But they could interact by quark/anti-quark exchange:

Non-overlapping nucleons, cont’d

This is meson exchange!


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When nucleons DO overlap, we have a six-quark problem with non-perturbative interactions between the quarks

(non-perturbative gluon-exchanges). A formidable problem!

?Attempts to calculate this by lattice QCD are under way; see e.g., T. Hatsuda, Prog. Part. Nucl. Phys. 67, 122 (2012); M.J.

Savage, ibid. 67, 140 (2012).


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QCD-based models (“quark models”)

To make 2- and 3-quark (hadron spectrum) or 5- or 6-quark (hadron-hadron interaction) calculations feasible, simple assumptions on the quark-quark interactions are made. For example, the quark-quark interaction is assumed to be just one-gluon-exchange or even meson-exchange.

This is, of course, not real QCD. It is a model; a “QCD-inspired” model.

The positive thing one can say about these models is that they make a connection between the hadron spectrum and the hadron-hadron interaction.

Conclusion

• Quark models are not QCD. So, they are not the solution.

• Lattice QCD is QCD, but too elaborate for every-day nuclear physics.

• What now? • Let’s take another look at QCD.


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The symmetries of low-energy QCD

• The QCD Lagrangian

• Symmetries of the QCD Lagrangian

• Broken symmetries


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The QCD Lagrangian

with

and the gauge-covariant derivative

where r=red, g=green, and b=blue,

and the gluon field tensor


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Symmetries of the QCD Lagrangian

,0, du mm

Quark masses

Assuming

the QCD Lagrangian for just up and down quarks reads


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Symmetries of the QCD Lagrangian, cont’d

Introduce projection operators

“Right-handed”

“Left-handed”

Complete

Idempotent

Orthogonal

Properties


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Consider Dirac spinor for a particle of large energy or zero mass:

Thus, for mass-less particles, helicity/chirality eigenstates exist

uuup 5)ˆ(

uPuu

uPuu

LL

RR

and the projection operators project them out


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quark field of right-handed chirality

quark field of left-handed chirality:

For zero-mass quarks,

the QCD Lagrangian can be written

symmetry )2()2( RL SUSU = “Chiral Symmetry”(Because the mass term is absent; mass term destroys chiral symmetry.)


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Interim summary

(approximate) Chiral Symmetry


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Chiral symmetry and Noether currents

symmetry)2()2(Global RL SUSU

Noether’s Theorem Six conserved currents

3 left-handed currents

3 right-handed currents


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Alternatively, …

3 vector currents

3 axial-vector currents

6 conserved “charges”

Note: “vector” is Isospin!


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Explicit symmetry breaking due to non-zero quark masses

.)2( VSU

The mass term that we neglected breaks chiral symmetry explicitly:

with

Both terms break chiral symmetry, but the 1st term is invariant under

if , then there is (2) symmetry

(=Isospin symmetry).

u d Vm m SU


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Spontaneous symmetry breaking

From the chiral symmetry of the QCD Lagrangian, one would expect that “parity doublets” exist; because:

Let


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Conclusion: Parity doublets are not observed in the low-energy hadron spectrum.

Chiral symmetry is spontaneously broken.

What would parity doublets look like?

Nucleons of positive parity: p(1/2+,938.3), n(1/2+,939.6), I=1/2;

nucleons of negative parity: N(1/2-,1535), I=1/2.

But, the masses are very different: NOT a parity doublet!

A meson of negative parity:

the “same” with positive parity:

But again, the masses are very different: NOT a parity doublet!

Spontaneous symmetry breaking, cont’d

.1),770,1( I

.1),1260,1(1 Ia


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Classical examples

Definition:

When the ground state does not have the same symmetries as the Langrangian, then one speaks of “spontaneous symmetry breaking”.

Lagrangian and ground state have both rotational symmetry.

No symmetry breaking.

Lagrangian has rot. symmetry;

groundstate has not.

Symmetry is spontaneously broken.


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Here: The 3 axial generators are broken, therefore, 3 pseudoscalar bosons exist: the 3 pions

This explains the small mass of the pion. The pion mass is not exactly zero, because the u and d quark masses are no exactly zero either.

)3,2,1( aQAa

,, 0

Goldstone’s Theorem

When a continuous symmetry is broken, then there exists a (massless) boson with the quantum numbers of the broken generator.


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Goldstone bosons interact weakly.

The breaking of the axial symmetry in the QCD ground state (QCD vacuum) implies that the ground state is not invariant under axial transformations, i.e.

Thus, a physical state must be generated by the axial charge,

More about Goldstone Bosons

| 0 0.AaQ

| | 0 .Aa aQ

0

0 0 | | 0 0.

AQCD a

AQCD a a QCD

H Q

H Q H

Since commutes with , we have

The energy of the state is zero. The state is energetically degenerate with the vacuum.

It is a massless pseudoscalar boson with vanishing interaction energy.


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Summary

• explicitly broken, because the u and d quark masses are not exactly zero;

• spontaneously broken: i.e., axial symmetry is broken, while isospin symmetry is intact.

• There exist 3 Goldstone bosons: the pions

QCD in the u/d sector has approximate chiral QCD in the u/d sector has approximate chiral symmetry; but this symmetry is broken in two symmetry; but this symmetry is broken in two

ways:ways:


27R. Machleidt Nuclear Forces - Lecture 3 QCD & Nucl. Forces (CNSSS13)

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Knowing the symmetries, we can now apply …

Weinberg’s

“Folk Theorem”

Weinberg


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Weinberg’s Folk Theorem is the stepping stone to what has become known as an Effective Field Theory (EFT).

Here, in short, the reasoning:

QCD at low energy is non-perturbative and, therefore, not solvable analytically in terms of the “fundamental” degrees of freedom (dof); quarks and gluons are ineffective dof at low energy.

Below the chiral symmetry breaking scale pions and nucleons are the appropriate dof.

Moreover, pions are Goldstone bosons which typically interact weakly. Thus, there is hope that a perturbative approach might work.

To ensure the connection with QCD, we must observe the same symmetries as low-energy QCD, particularly, spontaneously broken chiral symmetry, such that our chiral EFT can be indentified with low-energy QCD.

1 GeV


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How to do this EFT program? Take the following steps:

• Write down the most general Lagrangian including all terms consistent with the assumed symmetries, particularly, spontaneously broken chiral symmetry. (Note: There will be infinitely many terms.)

• Calculate Feynman diagrams. (Note: There will be infinitely many diagrams.)

• Find a scheme for assessing the importance of the various diagrams (because we cannot calculate infinitely many diagrams).


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The organizational scheme

“Power Counting”

Organize the contributions in terms of ;

where denotes a momentum (derivative) or a pion mass ( );

is the chiral symmetry breaking scale, 1 GeV;

and 0.

Q

Q m

Chiral Perturbation Theory (ChPT)


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The Lagrangian


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The pi-pi Lagrangian

At order two (leading order)• Derivative part

with (any number of pion fields).

Goldstone bosons can only interact when they carry momentum → derivative coupling. Lorentz invariance → even number of derivatives.

• Plus symmetry breaking mass term:


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The pi-N Lagrangian with one derivative (=lowest order or “leading” order)

Non-linear realization of chiral symmetry; Callan, Coleman, Wess, and Zumino, PR 177, 2247 (1969).


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pi-N Lagrangian, cont’d

The term is a problem.

Perform a 1/ expansion to remove it.

"Heavy Baryon Chiral Perturbation Theory"

N

N

M

M


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pi-N Lagrangian with two derivatives(“next-to-leading” order)



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The N-N Langrangian (“contact terms”)

• At order zero (leading order):

• At order two:

• At order four:


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Why contact terms?

• Renormalization: Absorb infinities from loop integrals.

• A physics reason (see next slide).

Two reasons:


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What is the physics of contact terms?

Contact terms take care of the short range structures without resolving them.

Consider the contribution from the exchange of a heavy meson

+ + + …



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Remember the homework we have to do

• Write down the most general Lagrangian including all terms consistent with the assumed symmetries, particularly, spontaneously broken chiral symmetry. (Note: There will be infinitely many terms.)

• Calculate Feynman diagrams. (Note: There will be infinitely many diagrams.)

• Find a scheme for assessing the importance of the various diagrams (because we cannot calculate infinitely many diagrams).


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The Diagrams

Power counting for Feynman diagrams

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2N forces 3N forces 4N forces

Leading Order

Next-to Leading Order

Next-to-Next-to Leading Order

Next-to-Next-to-Next-to Leading Order

The Hierarchy of Nuclear

Forces all vertices

Power 2 2 2 2

with 22

i

ii i

A C L

nd



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This finishes Lecture 3.In Lecture 4, we will discuss the

diagrams in more detail

A pedagogical review article about all this:R. Machleidt and D.R. Entem, Chiral Effective Field Theory and Nuclear Forces,Physics Reports 503, 1 (2011).

nuclear forces - lecture 3 -

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