16.451 Lecture 19: The deuteron 13/11/2003

Basic properties:

mass: mc2 = 1876.124 MeV

binding energy:

(measured via -ray energy in n + p d + )

RMS radius: 1.963 0.004 fm (from electron scattering)

quantum numbers: (lectures 13, 14)

magnetic moment: = + 0.8573 N

electric quadrupole moment: Q = + 0.002859 0.00030 bn

( the deuteron is not spherical! ....)

Basic properties:

mass: mc2 = 1876.124 MeV

binding energy:

(measured via -ray energy in n + p d + )

RMS radius: 1.963 0.004 fm (from electron scattering)

quantum numbers: (lectures 13, 14)

magnetic moment: = + 0.8573 N

electric quadrupole moment: Q = + 0.002859 0.00030 bn

( the deuteron is not spherical! ....)

H21d

MeV2245731.2 dnpi

i mmmMmB

0,1, TJ

Important because:

• deuterium is the lightest nucleus and the only bound N-N state

• testing ground for state-of-the art models of the N-N interaction.

1

Because the deuteron has spin 1,there are 3 form factors to describeelastic scattering: the “charge” (Gc),“electric quadrupole” (Gq) and “magnetic(GM) form factors. (JLab data)

Because the deuteron has spin 1,there are 3 form factors to describeelastic scattering: the “charge” (Gc),“electric quadrupole” (Gq) and “magnetic(GM) form factors. (JLab data)

Electron scattering measurements: 2

Interpretation of quantum numbers:

.....4,2,0)1()()(

1

L

LSJ

SSS

L

pn

.....4,2,0)1()()(

1

L

LSJ

SSS

L

pn

• Of the possible quantum numbers, L = 0 has the lowest energy, so we expect the ground state to be L = 0, S = 1 (the deuteron has no excited states!)

• The nonzero electric quadrupole moment suggests an admixture of L = 2(more later!)

introduce Spectroscopic Notation:

with naming convention: L = 0 is an S-state, L = 1 is a P-state, L = 2: D-state, etc...

the deuteron configuration is primarily

JS L12

13S

0,1, TJ 3

Isospin and the N-N system: (recall lecture 13)

1,02

1

2

1 TT

The total wavefunction for two identical Fermions has to be antisymmetric w.r.toparticle exchange:

isospinspinspacetotal

),()(),,( LMspace Yrfr

Central force problem:

with symmetry (-1)L given by the spherical harmonic functions

Spin and Isospin configurations:

S = 0 and T = 0 are antisymmetric ;

S = 1 and T = 1 are symmetric 1

3S state can only be T = 0 !

4

Understanding “L” -- the 2-body problem and orbital angular momentum:

Classical Mechanics:

x1r

2r

CM

1v

2v

• set origin at the center of mass, and let the CM be fixed in space

particles orbit the CM with angular speed

• Kinetic energy for each particle and angular momentum L = I:

• Total angular momentum: L = L1 + L2 with I = I1 + I2

• Total kinetic energy: K = L2/2I

iiiiiii ILrmvmK 2/2222

122

1

1m

2m

5

Equivalent one-body problem:

x

CM

r

v

Particles orbit the center of massbecause of an attractive potential V(r)

with no external force, particle kinematicsare related in the CM system and the 2-bodyproblem is equivalent to a 1-body problem:

)/(1,,

21

12121 mm

mrrrrrr

Total energy:

Moment of inertia can be written: 2222

211 rrmrmI

)(2 2

2

rVr

LE

)(

2 2

2

rVr

LE

Note: L = total angular momentum of the 2-particle system.

6

Quantum mechanics version:

)()( rErH

)()( rErH

2

2

2

222

2 ˆ2;)(

2 rrrrrVH

2

2

2

222

2 ˆ2;)(

2 rrrrrVH

)()()(2

)1(

2 2

2

2

22

rRrRrVrrd

Rd

)()()(2

)1(

2 2

2

2

22

rRrRrVrrd

Rd

H = total energy operator, E = eigenvalue)

2

2

22

22

sin

1cotˆ

),()1(),(ˆ 2 lmlm YY

Angular derivatives define an orbital angular momentum operator:

),()(

),,( lmYr

rRr wave function in the relative coordinate

has this generic form, where R(r) depends on V(r):

Analog of L2/2r2 in the classical problem! lowest energy state has minimum total L!

7

Deuteron Magnetic Moment: d = 0.857 N

In general, the magnetic moment is a quantum-mechanical vector; it must be aligned along the “natural symmetry axis” of the system, given by the total angular momentum:

J

~But we don’t know the direction of , only its “length” and z-component as expectationvalues:

J

)....(;)1(2 JJmJJJJ Jz

in a magnetic field, the energy depends on mJ via NJJ BmgBE

Strategy: we will define the magnetic moment by its maximal projection on the z-axis, defined by the direction of the magnetic field, with mJ = J

NJJmJgz

J

ˆ

NJJmJgz

J

ˆ

Use expectation values of operators to calculate the result.....

8

Calculation of :

NJJmJgz

J

ˆ

NJJmJgz

J

ˆ

Subtle point: we have to make two successiveprojections to evaluate the magnetic momentaccording to our definition, and the spin and orbital contributions enter with different weights.

z

J

mJ

1. Project onto the direction of J:

2. Project onto the z-axis with mJ = J:

)1(cosˆ

JJ

JJ

)1(||cos

JJ

J

J

mJ

Next, we need to figureout the operator for

)1(

1

JJJg NJ

)1(

1

JJJg NJ

9

spin and orbital contributions:

We already know the intrinsic magnetic moments of the proton and neutron, so thesemust correspond to the spin contributions to the magnetic moment operator:

83.391.1

58.579.2

,,

,,

nsNnsNn

psNpsNp

gSg

gSg

For the orbital part, there is a contribution from the proton only, correspondingto a circulating current loop (semiclassical sketch, but the result is correct)

p

I

r 1,ˆ2

ggrI N 1,ˆ2

ggrI N

For the deuteron, we want to use the magnetic moment operator:

Npnnspps LSgSg

,, Npnnspps LSgSg

,,

10

Details:

1. because mn mp

2. L = 0 in the “S-state” (3S1 ) but we will consider also a contribution from the “D-state” (3D1) as an exercise

3. The proton and neutron couple to S = 1, and the deuteron has J = 1

LLp

2

1

Nnnspps JLSgSgJ

2

1,,2

1

2

1 Nnnspps JLSgSgJ

2

1,,2

1

2

1

Trick: use and write the operator as: SSS pn

Nnpnspsnsps LSSggSgg

2

1,,2

1,,2

1 )()()(

But the proton and neutron spins are aligned, and

so the second term has to give zero!

JSJS np

11

continued....

So, effectively we can write for the deuteron:

Nnsps JLSgg

)( ,,4

1

Trick for expectation values:

SLSLSLSLJ

JJJSLJ

2)()(

)1(;

222

2

2222

1

2222

122222

1

SLJSLLLJL

SLJSSLJSSLSJS

NNnsps

NnpNnsps

ggD

ggS

310.0)(3)(

880.0)()(

,,4

11

3

,,2

11

3

12

Comparison to experiment:

N

N

D

S

310.0)(

880.0)(

13

13

Nd 857.0 Nd 857.0

This is intermediate between theS-state and D-state values:

Suppose the wave function of the deuteron is a linear combination of S and D states:

1221

31

3 bawithDbSad

Then we can adjust the coefficients to explain the magnetic moment:

)()()1( 132

132 DbSbd

b2 = 0.04, or a 4% D-state admixture accounts for the magnetic moment !

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