22.101 Applied Nuclear Physics (Fall 2006)

Lecture 8 (10/4/06) Neutron-Proton Scattering

References :

M. A. Preston, Physics o f the Nucleus (Addison-Wesley, Reading, 1962).

E. Segre, Nuclei and Particles (W . A. Benjam in, Ne w York, 1965), Chap. X.

We continue the study of the neutron- proton system by taking up the well-known problem of neutron scatteri ng in hydrogen. The scattering cross section has been carefully m easured to be 20.4 barns over a wide energy range. Our intent is to apply the m e thod of phase shifts summ arized in the preced ing lecture to this problem. W e see very quickly that the s-wave approxi m a tion (the condition of intera ction at low energy) is very well justified in the neutron energy range of 1 - 1000 eV. The scat tering-state solution, with E > 0, gives us the phase shift or equiva lently the scattering length. T h is calculation yields a cross section of 2.3 barns which is c onsiderably different from the experim e ntal value. The reason for the discrepancy lies in th e fact that we have not taken into account the spin-dependent nature of the n-p intera ction. The neutron and proton spins can form two distinct spin configurations , the two spins being parallel (t riplet s t ate) or anti-parallel (singlet), each giving rise to a scattering leng th. When this is take n into account, the new estim ate is q u ite close to the experimental v a lue. The conclusion is theref ore that n-p interaction is spin-depende nt and that the anom alously large value of the hydrogen scattering cross section for neutrons is really due to this aspect of the nuclear force.

For the scattering problem our task is to solve the radial wave equation for s-wave for solutions with E > 0. The interior and exterior solutions have the form

u ( r ) B s i n ( K ' r ) , r < r o (8.1)

and

u ( r ) C s i n ( k r o ) , r > r o

(8.2)

1

m E

o

where K ' m ( V E ) / ħ and k / ħ . Applying the interface condition we obtain

K 'c o t ( K ' r o ) k c o t ( k r o o ) (8.3))

o

which is th e rela tion that allows th e p h ase shif t to be dete rm ined in te rm s of the potential param e ters and the incom i ng energy E. W e can sim p lify the task of estimating the phase shif t by rec a lling that the phase shif t is si m p ly rela ted to the sc atte ring leng th by ak

o

(cf. (7.22)). Assum i ng the scatte ring length a is larger than r o , we see the R H S of (8.3) is approxim a tely k co t ( ) . For the LHS, we wil l ignore E relative to V o in K' , and at the sam e ti m e ignore E B relative to V o in K. Then K' ~ K and the LHS can be set equal to

by virtue of (6.4). Notice that this series of approxim a tions has enabled us to m a ke

use of the dispersion relation in the b ound-state problem , (6.4), for the scattering calculation. As a result, (8.3) becom e s

k cot ( o ) (8.4)

which is a relation between the phase shift and the binding energy.

o

Once the ph ase shift is known, the differential scat tering cross section is then given by (7.20),

o

( ) ( 1 / k 2 ) s i n 2 (8.5)

o

A sim p le wa y to m a ke use of (8.4) is to note the trigonom etric relation sin 2 x 1 /( 1 c ot 2 x ) , or

Thus,

si n 2

1

o

1 c o t 2

1

1 2 / k 2

(8.6)

2

( )

1 ħ 2 1 ħ 2

(8.7)

k 2 2 m E E m E

B B

The last step follows because we are mostly interested in estim ating the scattering cro ss section in th e energy ran g e 1 - 100 eV. Putting in the num erical values of the cons tants, ħ 1.055 x 10 -2 7 erg sec, m = 1.67 x 10 -2 4 g, and and E B = 2.23 x 10 6 x 1.6 x 10 -1 2 ergs, we get

B

4 ħ 2 / mE ~ 2.3 barns (8.8)

This value is consider ably lower than the experimental value of the scattering cross section of H 1 , 20.4 barns, as shown in Fig. 8.1.

Fig. 8.1 . Experim e ntal neutron scattering cross section of hydrogen, showing a constant value of 20.4 barns over a wide range of neutr on energy. The rise in the cross section at energies below ~ 0.1 eV can be explained in term s of che m i cal binding effects in the scattering sample.

The explanation of this well-known disc repancy lies in the neglect of spin- dependent effects. It was suggested by E. P. W i gner in 1933 that neutron-proton scattering should depend on whether the neut ron and proton spins are oriented in a parallel configurati on (the triplet state, total sp in angular m o m e ntum equal to ħ ) or in an anti-parallel configuratio n (singlet s t ate, to tal sp in is zero ) . In each cas e th e interaction

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potential is different, and ther efore the phase shifts also w ould be different. Following this id ea, on e can write instead of (8. 7 ),

( )

1 1 s i n 2

k 2 4

os

3 s i n 2

4

o t

(8.9)

We have already m e ntioned that the ground state of the deuteron is a triplet state at E = - E B . If the singlet state produces a virtual state of energy E = E*, then (8.9) would becom e

ħ 2 3 1

B

m E

E *

(8.10)

Taking a value of E* ~ 70 keV, we fi nd from (8.10) a value of 20.4 barns, thus bringing the theo ry in to agre em ent with expe rim e nt.

In summ ary, experim e ntal m easuremen ts have given the following scattering lengths for the two types of n-p interactions, triplet and singlet configurations, and their corresponding potential range and w e ll depth.

Inter action

Scatterin g lengt h a [F]

r o [F]

V o [MeV}

n-p (triplet)

5.4

2

36

n-p (singlet)

-23.7

~ 2.5

18

Notice th at the scatter i ng length f o r the tr iple t sta t e is pos itive , while tha t f o r the s i ngle t state is negative. This illu strates the point of Fig. 7.3.

As a final rem a rk, we note that experi m e nts have shown that the total angular mom e ntum (nuclear spin) of the de uteron ground state is I = 1, where I = L + S , with L being the orbital angular m o m e ntum , and S the in trins i c sp in, S = s n + s p . It is also known that the ground state is m o stly 1s ( ℓ =0), therefore for this state we have S = 1

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(neutron and proton spins are parallel). W e have seen from Lec 6 that the deuteron ground state is barely bound at E B = 2.23 Mev, so all the high er energy states are not bound. The 1s state with S = 0 (neutron and proton spins antiparallel), is a virtual state; it is unbound by ~ 60 Kev. An i m portant im plication is that nuclear inte raction varies with S, or, nuclea r forces a r e spin-depend e nt .

Effects o f P a uli Exclus ion Princip l e

One m i ght ask why are the di-neutron and the di-proton unstable. The answer lies in the indistinguishability of particles and the Exclusion Principle (no tw o fer m ions can occupy the sam e state). Consider the two electrons in a helium atom . Their wave function m a y be written as

( 1 , 2 ) 1 ( r 1 ) ( r 2 )

= A sin k 1 r 1 sin k 2 r 2 (8.11)

r 1 r 2

where 1 ( r ) is the wave functrion of electron 1 at r . B u t since we cannot distinguish between ele c trons 1 and 2, we m u st get the sa m e probability of f i nding these electron s if we exchange their positions (or exchange the particles),

( 1 , 2 ) 2 ( 2 , 1 ) 2 ( 1 , 2 ) ( 2 , 1 )

For ferm ions (electrons, neutrons, p r otons ) we must choose the (-) sign; because of Ferm i-Dirac statistics the wa ve function m u st be anti-sym m e t r ic under exchange. Thus we should modify (8.11) and write

( 1 , 2 ) 1 ( r 1 ) 2 ( r 2 ) - 2 ( r 1 ) 1 ( r 2 )

+

5

If we now include th e spin, then an acceptab l e anti-symm etric wave function is

( 1 , 2 ) 1 ( ) 2 ( )

so that und e r an in terch a nge of partic les, 1 2 , ( 1 , 2 ) ( 2 , 1 ) . This corresponds to S = 1, symmetric state in spin space. But another acceptable anti-symm e t r ic wave function is

( 1 , 2 ) [ 1 ( ) 2 ( ) 1 ( ) 2 ( ) ] which corresponds to S = 0, anti-symmetric state in spin.

have

For the symm etry of the wave function in config urational sp ace we recall that we

( r ) ~ u ℓ ( r ) P m (cos ) e im

r ℓ

which is even (odd) if ℓ is even (odd). Thus, since has to be anti-symm etric, one can have two possibilities,

ℓ even, S = 0 (space sym m e tric, spin anti-symm e t r ic)

ℓ odd, S = 1 (space anti-symm e tric, spin symm etric)

These are called T = 1 s t ates (T is iso b aric sp in ), available to the n-p, n-n, p-p system s.

By contras t , states which are symm etric (T = 0) are

ℓ even, S = 1

ℓ odd, S = 0

These are av ailab l e only to the n - p sy stem f o r which ther e is n o Pauli Exclusion Princip l e.

The ground state of the deuteron is therefor e a T = 0 state. T h e lowest T = 1 state is ℓ = 0, S =0. As m e ntioned above, this is known to be unbound (E ~ 60 Kev). W e

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should therefore expect that th e lowest T = 1 state in n-n and p-p to be also unbound, i.e., there is no s t able di-n eutron or di-proton.

Essential F e atures of N u clear Forces

In closing we summ arize a num ber of im portant features of the nucleon-nucleon inte raction p o tential, s e v e ral of whic h are bas i c to the stud ies in this cla s s [ M eyerhof , Chap. 6].

1. There is a dom i nant short-range part, wh ich is central and which provides the overall shell-m odel potential.

2. There is a p a rt whose range is m u ch sm a ller than the nuclear radius, which tends to m a ke the nucleus spherical and to pair up nucleons.

3. There is a part whose range is of the or der of the nuclear radius, which tends to distort th e n u cleus.

4. There is a sp in-orb it inte r action.

5. There is a sp in-spin interaction.

6. The force is charge independent (Co u lom b intera ction ex clud ed).

7. The force saturates.

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