16.451 Lecture 20: More on the deuteron 18/11/2003 1

Analysis so far: (N.B., see Krane, Chapter 4)

Quantum numbers: (J, T) = (1+, 0) favor a 3S1 configuration (with S = 1, L = 0) as the lowest energy n-p bound state

Magnetic moment: = 0.857 N is 2.6% smaller than for a pure 3S1 state and is consistent with a linear combination of L = 0 and L = 2 components:

04.01 2221

31

3 bandbawithDbSad

If this is correct, we can draw two conclusions about the N-N force:

1. The S = 1 configuration has lowest energy (i.e., observed deuteronquantum numbers) , so the N-N potential must be more attractive for total spin S = 1 than for S = 0.

2. The lowest energy solution for a spherically symmetric potential is purely L = 0, and the L = 2 wave functions are orthogonal to L = 0 wave functions, so a physical deuteron state with mixed symmetry can only arise if the N-N potential is not exactly spherically symmetric!

deuteron:

Quadrupole moment: independent evidence for the D-state

Formalism for electrostatic moments: (see also D.J. Griffiths, `Introduction to Electodynamics’ Ch. 3, and 16.369)

• Recall that an electric charge distribution of arbitrary shape can be described by an infinite series of multipole moments (spherical harmonic expansion), with terms of higher L reflecting more complicated departures from spherical symmetry

• The energy of such a distribution interacting with an external electric field can be written as:

where the charge distribution is located at the origin and has a symmetry axis along z;

q is the electric charge or `monopole moment’

pz is the electric dipole moment

Qzz is the electric quadrupole moment

and in general, higher order multipole moments couple to higher derivatives of the external potential at the origin.

VE

...4

1)0(

02

2

0int

zzz Qz

Vp

z

VqVE ...

4

1)0(

02

2

0int

zzz Qz

Vp

z

VqVE

2

Formalism, continued.... (axially symmetric case)

The multipole moments are defined in general by: rdErE 3ˆ)(

with corresponding multipole operators proportional tospherical harmonic functions of order : )(ˆ

0

YrcE

The first few multipole operators and associated moments are:

)()1cos3(2

)(cos1

)(410

ˆ operator,SymbolMoment,order

20222

10

000

516

34

YrrQquadrupole

Yrzpdipole

Yrqmonopole

E

zz

z

)()1cos3(2

)(cos1

)(410

ˆ operator,SymbolMoment,order

20222

10

000

516

34

YrrQquadrupole

Yrzpdipole

Yrqmonopole

E

zz

z

3

Connection to nuclei and the deuteron problem:

For a nuclear system, the multipole moments are expectation values of the multipole operators, e.g. the electric charge:

ZerdrErerdrq i

Z

ii

30

1

*3 )(ˆ)()(

We already know the electric charge of the deuteron, but what about higher moments?

Electric dipole moment, general case:

0cos||ˆ 3231

* rdrrdEpz 0cos||ˆ 3231

* rdrrdEpz

total wave functionof the system

even underspace reflection

odd,l = 1

integralmust vanish!

The electric charge density is proportional to the probability density, i.e. the wave function squared, summed up for all the protons in the system !!!

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continued...

Basic symmetry property: no quantum system can have an electric dipole moment if its wave function has definite parity, i.e. if ||2 is even.

(Incidentally, the possibility of a nonzero electric dipole moment of the neutron is of great current interest: so far the measured upper limit is |pn| < 10-25 e – cm ....)

In fact, all odd multipole moments must vanish for the same reason

First nontrivial case: Electric Quadrupole Moment Qzz (Krane, sec. 3.5)

rdrrQzz322 )1cos3()(

rdrrQzz322 )1cos3()(

example: uniform density object with ellipsoidal shape

z

a

b

)( 2252 abZeQzz Q > 0 if b > a: “prolate”

Q < 0 if b < a: “oblate”

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How to relate this to a nucleus, e.g. the deuteron?

Assume that the charge distribution is an ellipsoid of revolution with symmetry axisalong the total angular momentum vector J:

As for the magnetic dipole moment,we specify the intrinsic electricquadrupole moment as the expectationvalue when J is maximally aligned with z:

JmJ

EQ

2intˆ

JmJ

EQ

2intˆ

Jz

)1(||cos

JJ

J

J

mJ“Quantum geometry”:

If an electric field gradient is applied along the z-axis as shown, the observableenergy will shift by an amount corresponding to the intrinsic quadrupole momenttransformed to a coordinate system rotated through angle to align with the z-axis

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Result:

Let Qlab be the electric quadrupole moment we measure for the ellipsoidal charge distribution with mJ = J:

intint2

lab1

2/1

2

11cos3 QQQ

J

J

standard, classicalprescription for rotated coordinates

“Quantum geometry”for the rotation function

How do we apply this to anything?

1. A spherically symmetric state (L = 0) has Qint = 0 (e.g. deuteron S-state)

2. Even a distorted state with J = ½ will not have an observable quadrupole moment

3. J = 1, L = 2 is the smallest value of total angular momentum for which we can observea nonzero quadrupole moment

these are the quantum numbers for the deuteron D-state!

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Deuteron intrinsic quadrupole moment:

13

13 DbSad 1

31

3 DbSad

recall our model for the deuteron wave function:

DEDbbDESbaSESaa

DbSaEDbSaEQJmJ

|ˆ||ˆ|2|ˆ|

ˆˆ

2*

2*

2*

13

13

213*

13*

2int

we have:

0

Quadrupole moment contains both pure D-stateand the S-D cross term contributions.

Evaluation requires a model of the deuteron wave function in the S and D states, e.g:

rdrYrrDES DS3

202*

2 ),,()(5

16),,(|ˆ|

8

Result:

bn00003.000286.0

|| 222*int

20

1

10

2

DDSD

rbrbaQ

bn00003.000286.0

|| 222*int

20

1

10

2

DDSD

rbrbaQ

With a good model for the nucleon-nucleon potential, the space dependence of the deuteron wave function and hence the quadrupole moment can be predicted .

note cancellation here

)()( 132

132 DbSad )()( 1

321

32 DbSad

Compare to the information from the magnetic dipole moment:

State of the art: N-N potential models are able to predict both the magneticdipole and electric quadrupole moments of the deuteron with the same values ofa and b! (a2 = 0.96, b2 = 0.04) we must be doing something right

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Basic features of the N-N potential, via the deuteron, etc.:

1. Independent of the value of L, the state with intrinsic spins coupled to S = 1 has lower energy

this implies a term proportional to:

0,4/3

1,4/122

21

22

121

S

SSSSSS

2. The deuteron quadrupole moment implies a non-central component, i.e. the potential is not spherically symmetric. Since the symmetry axis for Q is along J, Q > 0 means that the matter distribution is stretched out along the J – axis:

1S

2S

Q > 0, observed

This implies a “tensor” force, proportional to:

212

2112 3 SS

r

rrS

SS

compare: magnetic dipole-dipoleinteraction, see Griffiths problem 6.20

10

N-N interaction continued....

3. There is also a spin-orbit term, as deduced from N-N scattering experiments with a polarized beam:

SLV OS

~

(This plays a very important role also in determining the correct order of energy levels in nuclear spectra – more later!)

4. Finally, all contributions to the N-N interaction are based on a microscopic meson exchange mechanism, as explored in assignment 3,

N1

N2

M

Where M is a , , ... meson, etc.

and each term has a spatial dependenceof the form:

)()( functionspinr

rmegrV

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State of the art N-N interaction model: (not to scare anybody...)

(89 page exposition of one of only ~3 state-of-the-art models of the N-Ninteraction worldwide – constantly refined and updated since first release.)

12

Brace yourself – it takes 2 pages just to write all the terms down!!!

(they use “” for spin)

(tensor operator)

(spin-orbit interaction)

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more of this...

“Yukawa functions”and derivatives...

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Parameters and Predictions:

deuteron properties:

low energy scattering parameters, etc:

Impressively good agreement for ~ 10,000 experimentaldata points in assorted n-p and p-p scattering experiments plus deuteron observables: 2/d.f <~ 1.07 !

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