1

1

INTRODUCTION

Introduction

1.1

Nuclear Properties

Some important and known properties of nuclei are given in the table 1 . No.

Property

Value

01 02 03 04 05 06 07

Mass ⇐⇒ Energy Charge Size (radii) Shape ⇐⇒ Quadrupole moment Spin Parity Magnetic moment

∼ 10−27 kg Ze ∼ 10−15 m = fm ∼ 10−31 m2 ? ? e~ ∼ µN ≈ 2m = 5.0508 × 10−27 J/T. p

Table 1: Properties of nuclei.

1.2

Nuclear forces

Nuclear forces differ from all other known types of forces. They cannot be of electrical origin and magnetic origin. Magnetic moments of the nucleons is extremely weak. Expressions for Coulomb force FC and gravitational force FG are simple of the form: FC = K

q1 q2 r2

and

FG = G

m1 m2 r2

The main interest here is to determine an expression for nuclear force. Certainly this expression cannot be of the type given above, as nuclear force is of short range and any expression for it must account the nuclear force properties listed in table 2. No.

Property

Value

01 02 03 04 05 06 07

Short range force. Strongest attractive force. (Approximately) Charge independent. Spin dependent. Saturated force. Non-central force. Exchange force.

b ∼ 10−14 m 1039 × FG within b Fn−n = Fn−p = Fp−p ? ? ? Tensor force. ? Yukawa’s theory

Table 2: Properties of nuclear force.

In order to find an expression for the nuclear force, two simple approaches could be followed. 1. Systematic study of all nuclei that exists in this nature starting from hydrogen to say, Uranium nuclei. or 2. Thorough study of an isolated simple nucleus or simple system where nuclear force exists, on all properties and understanding them through a suitable theory. Simple system which exhibits the effect of nuclear force is two nucleon system. The three possible cases are given in table 3. Out of the three only n − p system exists in bound state. A prior nuclear reaction is required, in order to use neutrons as a projectile in any experiment. Although neutrons can be used as a target, in reality this experiment is highly difficult to carry out. Therefore, an 1

1.3

Schr¨odinger equation for a nucleon

No.

01 02 03

System n−n n−p p−p

1

Interactions

Nuclear Nuclear Nuclear + Coulomb

State

unbound unbound + bound unbound

INTRODUCTION

Preferences Theorist

Experimentalist

I I II

III II I

Table 3: Two nucleon system.

experimentalist will give last preference to use neutrons as a target irrespective of whether it is a two body problem or many body problem. On the other hand, a theorist will chose either n − n system or n−p system first, to avoid the inclusion of Coulomb interactions in the theory. Thorough understanding of these interactions in unbound state requires sound knowledge of vacuum physics and ion optics.

1.3

Schr¨ odinger equation for a nucleon

~2 2 ∇ ψ + [E − V (~r)]ψ = 0 (1.1) 2m Where symbols have the usual meaning. Since, nucleus is approximately, spherical in shape, therefore spherical coordinates must be used. In spherical coordinates the wave function is ψ = ψ(r, θ, φ) = R(r)Y (θ, φ) Where, R(r)- only radial wave function and Y (θ, φ)- spherical harmonics which are angular functions. In case of a central potential ‘ V = V (r)’ then it will be sufficient to solve only radial part of Schr¨odinger equation to obtain the solution, which is   d2 R(r) 2 dR 2m `(` + 1)~2 + + 2 E − V (r) − R(r) = 0 dr2 r dr ~ 2mr2 The correction to the potential V (r) is [`(` + 1)~2 /2mr2 ] which is called the centrifugal barrier term, arises due to separation of angular part from total wave-function. If V (r) is assumed as a Coulomb potential [V (r) ∝ (−1/r)], then centrifugal barrier term [∝ `(` + 1)/r2 ] and the sum of two potentials appear as shown in the figure 1. Coulomb potential varies as 1/r where as barrier term varies as 1/r2 , therefore the latter term reduces to zero quicker than the former term. However, this relation is valid only for one particle system and cannot be used for a system which contains two particles. Simple examples for two particle bound system are hydrogen atom and deuteron.

1.4

Schr¨ odinger equation for two particles

There are plenty of examples of two particle system in nature, either in bound state or in free state like two particle collision or scattering processes. Deuteron is a bound system of two nucleons, a neutron ‘n’ and a proton ‘p’ separated by a distance of r. The theory discussed below, is applicable for all such systems. Let m1 and m2 be the masses of two particles with position coordinates ~r1 and ~r2 as shown in figure 2 (a) in lab frame. The interaction potential between the two particles is V = V (~r1 − ~r2 ). The eigen value Schr¨odinger equation ‘Hψ = Eψ’ for a two particle system in lab frame is written as,   ~2 2 ~2 2 − ∇ − ∇ + V (~r1 − ~r2 ) ψ(~r1 , ~r2 ) = E ψ(~r1 , ~r2 ) (1.2) 2m1 1 2m2 2 Where, ∇21 and ∇22 denote second order differential operators, called Laplacian operators, with respect to the position coordinates of particle 1 and 2. In this case, the wave function ‘ψ’ cannot be expressed as a product of two wave functions. i.e., ψ(~r1 , ~r2 ) 6= φ(~r1 ) φ(~r2 ), because potential is a 2

1.4

Schr¨odinger equation for two particles

1

INTRODUCTION

V (r) 1:V ∝

l(l+1) r2

- Centrifual barrier

2 : V ∝ − 1r - Coulomb Potential 3 : V ∝ ( l(l+1) − 1r ) - Sum of the above two r2

1 3 2 r

Figure 1: Plot of Coulomb (. . . ), centrifugal barrier (99K) potentials and their sum (—) as a function of r. function of both ~r1 and ~r2 . i.e. V = V (~r1 − ~r2 ). This is a restriction that hinders further simplification in this frame of reference. To simplify the case, the frame of reference is changed from lab system to center of mass (CM) system with coordinates ~r and ρ ~, as shown in figure 2 (b). The two transformation relations are, ~r1 = ~r1 (~r, ρ ~) and ~r2 = ~r2 (~r, ρ ~). Where, ~r = ~r12 = ~r1 − ~r2

- relative coordinate

(1.3)

which is the vector distance between two nucleons. Here, for simplicity it is denoted as ~r instead of ~r12 , and it should not be confused with the position coordinate of any other point mass. Magnitude of this vector ‘|~r| = r’ provides the separation distance between the two interacting particles. Then V = V (~r). From the definition of center of mass, the CM position coordinate (~ ρ ) is defined as, Mρ ~ = m1~r1 + m2~r2

(1.4)

Then ψ = ψ(~r1 , ~r2 ) = ψ(~r, ρ ~), with both ~r and ρ ~ are independent of each other. In such a case, wave function can be separated as, ⇒

ψ(~r, ρ ~) = φ(~r) φ(~ ρ)

(1.5)

Now, the protocol is to transform the operators ∇1 (~r1 ) and ∇2 (~r2 ) collectively to ∇r (~r) and ∇ρ (~ ρ). From the relations (1.3) and (1.4), these operators can be transformed as follows, ∂ψ ∂ψ ∂~r ∂ψ ∂~ ρ = + ∂~r1 ∂~r ∂~r1 ∂~ ρ ∂~r1 ∇1 ψ(~r, ρ ~) = ∇r ψ (1) + (m1 /M )∇ρ ψ = [∇r + (m1 /M )∇ρ ] ψ(~r, ρ ~) ∂ψ ∂ψ ∂~r ∂ψ ∂~ ρ ∇2 ψ(~r, ρ ~) = = + ∂~r2 ∂~r ∂~r2 ∂~ ρ ∂~r2 ∇2 ψ(~r, ρ ~) = ∇r ψ (−1) + (m2 /M )∇ρ ψ = [−∇r + (m2 /M )∇ρ ] ψ(~r, ρ ~)

∇1 ψ(~r, ρ ~) = and

3

(1.6)

(1.7)

1.4

Schr¨odinger equation for two particles

m1

1

n

m1

p m2

~r1

~r1 ~r2

(a)

n

INTRODUCTION

CM

~r ρ ~

p m2

~r2

(b)

~r = ~r1 − ~r2

Figure 2: Two particle system in (a) lab system and in (b) CM frame of reference. Here, a neutron ‘n’ and a proton ‘p’ separated by a distance of r is considered. Here, the nabla operators, ∇r and ∇ρ operate on the wave functions ψ = ψ(~r, ρ ~). Now consider (∇21 /m1 + ∇22 /m2 ), then using the relations (1.6) and (1.7), ∇21 ∇2 + 2 m1 m2 ⇒ ⇒

∇21 ∇2 + 2 m1 m2

m 1 2 m 2 2 1  1  ∇r + ∇ρ + −∇r + ∇ρ m M m M  2  1 m 1 m 1 1 2 2 2 2 2 ∇ + ∇ + ∇ + ∇ = m1 r M 2 ρ m2 r M 2 ρ !  2 ∇2ρ ∇r = + µ M =

(1.8)

m2 where, µ = mm11+m is the reduced mass of two particle system (like n and p of deuteron), M = 2 m1 +m2 - total mass of the system. Using (1.3) and (1.8) in (1.2), Schr¨odinger equation for two particle system in CM system becomes, " ! # ∇2ρ −~2 ∇2r + + V (~r) ψ(~r, ρ ~) = E ψ(~r, ρ ~) (1.9) 2 µ M

The eigen value ‘E’ is a scalar additive term and therefore it is expressed as ‘E = Er + Eρ ’. Using this and (1.5) in (1.9) and re-arranging the terms, it leads to,  2   2  −~ 2 −~ 2 ∇ φ(~r) φ(~ ρ) + ∇ φ(~ ρ) φ(~r) + V (~r) φ(~r) φ(~ ρ) = (Er + Eρ ) φ(~r) φ(~ ρ) (1.10) 2µ r 2M ρ Dividing (1.10) by φ(~r) φ(~ ρ) and separating the two coordinate equations, leads to,  2  1 −~ 2 ∇ φ(~r) + V (~r ) = Er φ(~r) 2µ r

and similarly

 =⇒

 −~2 2 ∇ + V (~r ) φ(~r ) = Er φ(~r) 2µ r

−~2 2 ∇ φ(~ ρ) = Eρ φ(~ ρ) 2M ρ

(1.11)

(1.12)

The relation (1.11) depends solely on the relative coordinate ~r and the relation (1.12) depends solely on the center of mass coordinate ρ ~. Thus, in quantum mechanics, as well as in classical mechanics, the two particle problem can be reduced to a single particle (moving) in some central potential. Out of the two relations, only the center of mass relation in relative coordinate (1.11) is important as only it contains the interaction potential, V (~r ), between the two particles.

4

1.5

Deuteron

1.4.1

1

INTRODUCTION

Simplified Schr¨ odinger equation for deuteron

To know more about the deuteron and in general about nuclei, it is necessary to identify the interaction potential V that acts between the nucleons. For simplicity, here only central potential V = V (~r) = V (r) is considered. Since the wave function ψ = ψ(r, θ, φ) = R(r) Y (θ, φ), and only the relative radial coordinate ‘r’ in CM system is important. Then, only for relative radial coordinate, the relation (1.11), after the rearrangement, becomes,   d2 R(r) 2 dR 2µ `(` + 1)~2 R(r) = 0 (1.13) + + 2 E − V (r) − dr2 r dr ~ 2µr2 where, ` is the angular momentum quantum number. In general, the radial function R(r) may not be zero at r = 0. For some specific reasons, to be stated later, it is required to have zero R value at r = 0. To achieve this, a modified radial wave function u(r) is used, which is defined as follows: u(r) = rR(r)

=⇒

R(r) = u(r)/r

This ensures that u(r) → 0 as r → 0, irrespective of the value of R(r). =⇒

du dR =R+r dr dr

and

2 dR d2 R 1 d2 u = + 2 r dr2 r dr dr

Using this, the radial Schr¨ odinger equation in CM system with modified radial wave function, becomes a simple relation as,   d2 u(r) 2µ `(` + 1)~2 + E − V (r) − u(r) = 0 dr2 ~2 2µr2   2µ `(` + 1)~2 in simple form : u ¨ + 2 E − V (r) − u = 0 (1.14) ~ 2µr2

1.5 1.5.1

Deuteron Simple observed facts of Deuteron

• Deuteron is the only bound system with mass number A = 2. • In case of deuteron EB = 2.225 MeV, but for most nuclei have EB ≈ 8 MeV. Since, BE/A ≈ (EB )avg ⇒ deuteron is a loosely bounded system. • Deuteron’s angular momentum is J π = 1+ (in units of ~). • Magnetic momentum is µd = 0.85735 ± 0.00003 µN . Where, µN - is nuclear magneton. • Quadrupole moment is Q = 2.82 × 10−27 cm2 . • Charge distribution half value radius is R ≈ 0.1 fermi.

1.6

Ground state of deuteron

Deuteron in ground state, ` = L = 0 and Er = −EB - binding energy of the deuteron. Let ‘r’ be the distance between the neutron ‘n’ and proton ‘p’. Let ‘b’ and ‘Vo ’ be the width of the potential (approximately range of interaction) and strength (depth) of the interaction potential. Different forms of central potential ‘V (r)’ that can be tested are ( −Vo when 0 < r ≤ b 1. Square Well potential: V (r) = 0 when b < r < ∞ 2. Exponential potential: V (r) = −Vo er/b 5

1.6

Ground state of deuteron

1

3. Gaussian potential: V (r) = −Vo er 4. Yukawa potential: V (r) = −Vo

INTRODUCTION

2 /b2

er/b (r/b)

All these potentials are plotted in the figure 3(a) for reference. Among all these, square well potential being the simplest, it is used in the relation (1.14) to study the ground state of the deuteron. V (r)

1 : V = −Vo e−r/b 2 2 2 : V = −Vo e−r /b 3 : V = −Vo e−r/b /(r/b) b 2 3 1 r

V (r)

b r I (r < b)

−Vo

II (r > b)

−Vo (a)

(b)

Figure 3: Schematic representation of the radial dependence of the nuclear potential energy. (a) Relative representation of four potentials. (b) Two regions I and II described by the square well potential with ‘b’ and ‘Vo ’. A square well potential defined above is represented in the figure 3(b) with two regions I and II. With this potential Schr¨ odinger equation must be solved separately in two regions. Region I: r < b In this region, V = −Vo and let u = u1 , then (1.14) becomes, u ¨1 +

2µ [−EB + Vo ] u1 = 0 ~2

(1.15)

2µ [Vo − EB ] = k12 ~2

(1.16)

Let Then the relation simplifies to

u ¨1 + k12 u1 = 0 which is an equation similar to that of a SHM. Solution of this equation can be of the form, u1 = A1 eik1 r + B1 e−ik1 r

(1.17)

Region II: r > b In this region, V = 0 and let u = u2 , then (1.14) becomes, u ¨2 +

2µ [−EB ] u2 = 0 ~2

2µ [EB ] = k22 ~2 u ¨2 − k22 u2 = 0. Solution of this equation will be of the form, Let

=⇒

u2 = A2 ek2 r + B2 e−k2 r

(1.18) (1.19)

(1.20)

Where, A1 , B1 , A2 and B2 are integration constants which are solved using boundary conditions. A well behaved solution (or function) must follow all these conditions. 6

1.6

Ground state of deuteron

1

INTRODUCTION

Boundary conditions: 1. At r = 0, ⇒ u1 = 0. From the 0 = A1 + B 1  relation (1.17),  (1.17) becomes u1 = A1 eik1 r − e−ik1 r u1 = A sin k1 r

where

⇒

A1 = −B1 . Then relation

A = 2iA1

(1.21)

2. As r → ∞ then u2 → 0. Otherwise, the wave function becomes infinite which is absurd. In other words, the wave function must vanish outside the interaction range. With this, the relation (1.20) becomes 0 = A2 (∞) + B2 (0). This is valid only when A2 = 0. Then the relation (1.20) takes the form, u2 = B e−k2 r where B = B2 (1.22) 3. At r = b ⇒ [u1 (r) = u2 (r)]r=b . Because, the two wave functions must be continuous at the interface boundary. Therefore, from the relations (1.21) and (1.22), at this boundary, A sin k1 b = B e−k2 b

(1.23)

4. The continuity condition at the boundary requires that [u˙1 (r) = u˙2 (r)]r=b . This provides, A k1 cos k1 b = −Bk2 e−k2 b

(1.24)

The ratio (1.24) ÷ (1.23) helps to eliminate the integration constants, k1 cot k1 b = −k2

=⇒

cot k1 b = −k2 /k1

cot(k1 b)

Substituting the values of k1 and k2 from the relations (1.16) and (1.19) in the above equation leads to s s (2µ/~2 )EB EB 0 =− cot k1 b = − 2 (2µ/~ )(Vo − EB ) (Vo − EB ) (1.25) Since EB ≈ 2.225 MeV and nuclear force is very strong, it can be expected that, Vo  EB , so that, Vo − EB  0. Also, since both EB > 0 and Vo > 0. ⇒ cot k1 b < 0. π 3π 0 π 2π 2 2 The above condition is true only when, k1 b ≥ π/2 or θ = k1 b 3π/2 or 5π/2 . . ., as shown in the figure 4 for two nodes. Since u(r) cannot have any and all the nodes inside b. Therefore only π/2 is the valid solution. i.e., k1 b ≥ π/2. The minimum of this condition is that k1 b = π/2 ⇒ Figure 4: A plot of f (k1 b) = cot k1 b versus (k1 b)2 = π 2 /4. Using (1.16), it follows that (2µ/~2 )(Vo − θ, displaying the nodes of f (k1 b) at π/2 and at 3π/2. EB )b2 = π 2 /4. Since, Vo − EB  0 =⇒ Vo − EB ≈ Vo =⇒


π2 2µ 2 V b = o ~2 4

(1.26)

µ ≈ mp /2 ≈ mn /2 - reduced mass of deuteron. Therefore Vo b2 ≈ 130 MeV fm2 . This relation is not sufficient to obtain the explicit values of unknowns Vo and b. Therefore, one more relation is needed to obtain their unique values. Let ‘rd ’ be the radius of deuteron. Then expectation value of ‘rd2 ’ is defined as R∞ ∗ 2 u r u dr hu|rd2 |ui 2 hrd i = = 0R ∞ ∗ d (1.27) hu|ui 0 u u dr 7

1.7

Excited states of deuteron

1

If we assume that in deuteron, n and p always stays on the circumference of the sphere about center of mass, as shown in the adjacent diagram, then radius of deuteron is given as rd = r/2. Then from (1.27), rd is defined in terms of separation distance ‘r’ between the two nucleons of deuteron as, # "R b R∞ R 1 ∞ ∗ 2 1 0 u∗1 r2 u1 dr + b u∗2 r2 u2 dr 2 4 R0 u r u dr (1.28) hrd i = = R∞ ∗ Rb ∗ ∞ ∗ 4 u u2 dr u u1 dr + 0 u u dr 0

1

b

INTRODUCTION

n

rd

p

r

2

Square root of (1.28) provides the deuteron radius ‘rd ’ as a function of Vo and b (exercise). Therefore from (1.26) and (1.28) unique values of Vo and b are obtained by numerical analysis, which are Vo ≈ 35 MeV and b ≈ 2 fermi. At 2 fm, Coulomb potential between two protons is ∼ 0.2 MeV  35 MeV. This implies that nuclear force is very strong and it is applicable till about 2 fm (⇒ short range). The constants A and B in (1.21) and (1.22) can be u1 = A sin k1 r solved using the normalization condition: V (r) Z 0

b

u∗1 u1 dr

Z + b

∞

u∗2 u2 dr = 1

(1.29)

u2 = B e−k2 r b

The solution of hthe above i1/2 relation can be shown 2k2 that (exercise) A = 1+k2 b and using this in (1.23)

rd = 1/k2

r

I II the value of B can be obtained as B = A sin k1 b ek2 b . The two wave functions in the two regions behaves −Vo according to (1.21) and (1.22) as shown in the figure 5. In the II region, the wave function is u2 = B e−k2 r . Let the deuteron radius be such that rd = 1/k2 , then u2 = B/e. i.e., when the amplitude of u2 (r) decreases to 1/e times its maximum amplitude at the boundary, Figure 5: Plot of wave functions in the two regions according to (1.21) and (1.22). it provides the radius of the deuteron as, s ~2 rd = 1/k2 = ≈ 4.38 fm ⇒ rd ≈ 2.2 b 2µEB i.e., deuteron’s radius is only approximately 2.2 times the range of interaction. This implies that deuteron is a loosely bound system.

1.7

Excited states of deuteron

Schr¨odinger equation with modified radial wave function u(r) = r R(r) for deuteron is given by the relation (1.14), as   `(` + 1)~2 2µ u ¨(r) + 2 E − V (r) − u(r) = 0 ~ 2µr2 In case of excited states, ` 6= 0. Therefore, the ‘`’ term will be present in the above relation. Again, for simplicity, here also a square well potential is considered. ( −Vo when 0 < r ≤ b V (r) = 0 when b < r < ∞ Case I: In the first region, r < b, V = −Vo and E = −EB . Let u = u1 , then   `(` + 1)~2 2µ =⇒ u ¨1 (r) + 2 (Vo − EB ) − u1 (r) = 0 ~ 2µr2

8

(1.30)

1.7

Excited states of deuteron

1

INTRODUCTION

Using the substitution (1.16), the above relation simplifies to,   `(` + 1) 2 u ¨1 + k1 − u1 = 0 r2

(1.31)

This is a spherical Bessel relation and hence its solution will be of the form, u1 (r) = A1 j` (k1 r) + B1 y` (k1 r)

(1.32)

where j` (k1 r) and y` (k1 r) are spherical Bessel function (of first kind) and Neumann function (or spherical Bessel function of second kind) respectively. For excited states ` > 0, so only higher Bessel functions are the solutions. Case II: In the second region, r > b, V = 0 and E = −EB . Let u = u2 , then   2µ `(` + 1)~2 =⇒ u ¨2 (r) + 2 (−EB ) − u2 (r) = 0 (1.33) ~ 2µr2 Using the substitution (1.19), the above relation simplifies to,   `(` + 1) 2 u ¨2 − k2 + u2 = 0 r2 1

0.6

j0 j1 j2 j3

0.8 0.6

0.2 0 yl (k1 r)

jl (k1 r)

(b)

0.4

0.4 0.2 0

−0.2 −0.4 −0.6

−0.2

y0 y1 y2 y3

−0.8

−0.4 −0.6

(1.34)

−1

(a) 0

2

4

6

−1.2

8 10 12 14

0

2

4

(k1 r)

6

8 10 12 14

(k1 r)

Figure 6: A plot of spherical Bessel functions of (a) first kind ‘j` (k1 r)’ and that of (b) second kind ‘y` (k1 r)’ for ` = 0, 1, 2 and 3. The solution of the above equation will be u2 (r) = A2 j` (k2 r) + (iB2 )y` (k2 r) = Bh` (ik2 r)

(1.35)

where ‘h` (ik2 r)’ is spherical Hankel function, which is also referred as spherical Bessel function of third kind. 1.7.1

Boundary Conditions

1. At r = 0

=⇒

u1 (r) = 0. From (1.32),

0 = A1 (0) + B1 (∞)

⇒

B1 = 0 ∀ ` = 1, 2 · · ·

Therefore, the valid solution is u1 (r) = A1 j` (k1 r) = A j` (k1 r) 9

(1.36)

2  2. At r = b

=⇒

NUCLEON-NUCLEON SCATTERING

 u˙ 1 (r) u˙ 2 (r) = , then from (1.32) and (1.35), it leads to u1 (r) u2 (r) r=b   k1 A j`0 (k1 r) ik2 B h0` (ik2 r) = A j` (k1 r) B h` (ik2 r) r=b

Where, j`0 (k1 r) = j`−1 (k1 r) −

(` + 1) j` (k1 r) - first order derivative of ‘j` (k1 r)’, and k1 r

(` + 1) h` (k2 r) - first order derivative of ‘h` (k1 r)’. ik2 r Using these relations in the above expression, h0` (ik2 r) = h`−1 (ik2 r) −

k1

⇒

For any `,

n j`−1 (k1 b) −

o

(`+1) k1 b j` (k1 b)

j` (k1 b) k1 j`−1 (k1 b) (` + 1) − j` (k1 b) b j`−1 (k1 b) ⇒ j` (k1 b)

ik2 = = =

n h`−1 (ik2 b) −

o

(`+1) ik2 b h` (k2 b)

h` (ik2 b) ik2 h`−1 (ik2 b) (` + 1) − h` (ik2 b) b   ik2 h`−1 (ik2 b) k1 h` (ik2 b)

(1.37)

h`−1 (ik2 b) < h` (ik2 b), Therefore, it leads to v u 2µ h`−1 (ik2 b) k2 u 2 EB < 1 and = t 2µ ~ <1 h` (ik2 b) k1 [V − E ] o B 2 ~

Therefore, the LHS of relation (1.37) is much less than 1. i.e.,

j`−1 (k1 b) 1 j` (k1 b)

⇒

j`−1 (k1 b)  j` (k1 b)

In other words for ` > 0 ⇒ j`−1 (k1 b) ≈ 0. In case of first excited state, ` = 1 then j0 (k1 b) ≈ 0. This implies the possible solutions are zero’s of spherical Bessel function. i.e., k1 b = ±π, ±2π, ±3π . . . In the region I, only k1 b = ±π is the allowed solution. 2µ ⇒ k12 b2 = π 2 ⇒ [Vo − EB ]b2 = π 2 ~2 Since Vo  EB ⇒ Vo − EB ≈ Vo , therefore, Vo b2 = π 2 ~2 /2µ ≈ 420 MeV. Since b ≈ 2 fm, this leads to Vo ≈ 144 MeV. This implies that, in the first excited state of deuteron, the potential depth must be around 144 MeV. In case of even higher excited states, this implies that, the potential depth increases to even higher values. However, experimentally measured value of Vo is around 35 MeV. This implies that deuteron’s do not have any excited states.

2

Nucleon-nucleon scattering

2.1 2.1.1

Scattering theory: a brief revisit Phase shift analysis

Scattering process is describe using wave mechanics. Let ψin and ψsc be the incident and scattered wave functions respectively, then total wave function ψtot is written as ψtot = ψin + ψsc Let ψin = A sin(~k . ~r + δ) and ψsc = A0 sin(~k 0 . ~r + δ 0 ) be the incident and scattered wave from the target respectively. Here, 10

2.2

Partial wave analysis of scattering process

2

NUCLEON-NUCLEON SCATTERING

• A and A0 are amplitudes (and their squares ∝ intensity) of two waves. In quantum picture intensities represent number of particles/unit area/unit time. • k and k 0 are wave vectors that depends on energy • δ and δ 0 are phases of two waves. Scattered wave function (ψsc ) contains the informations of interaction potential V (r). This is because, if ψsc (k 0 , δ 0 ) there is no interaction between projectile and target (X), then incident wave (ψin ) would have propagated in a straight line path, as shown in dotted lines in the adψin (k, δ) jacent figure. In case of elastic scattering, kinetic energy X is conserved and therefore |~k| = |k~0 | = k, A = A0 but δ 6= δ 0 and the phase shift δ arises entirely due to nuclear interaction and therefore contains the informations of interaction potential. If there are no interactions, then phase shift δ = 0. If a relation between phase shift δ and differential scattering cross section is obtained, then cross section for the scattering process can be obtained from δ. This analyzing process is called phase shift analysis. In a scattering process, incident wave, in general, will be a plane wave propagating along z direction with a modΩ k0 mentum ~k (mono energetic) and is represented by a wave 0 ψsc k function ψin as, θ ~ ψin = eik.~r = eikr cos θ = eikz (2.1) k where, θ is the scattering angle between ~k and ~r. ψ k in When this incidents on a target then it will experience a spherically symmetric potential and get scattered in all possible directions with momentum ~k 0 in a spherical symmetry, therefore it is represented by another wave Figure 7: Scattering process in wave mefunction ψsc with a spherical wave as chanics with incident wave ψin and scat0r ik tered wave ψsc . Picture also display the e ψsc = f (k 0 , θ) (2.2) measurement scheme at scattering angle θ r for a solid angle dΩ = dφ sin θ dθ. 0 where, f (k , θ) is called scattering amplitude. Incident particle breaks spherical symmetry to cylindrical symmetry. Therefore, total wave function ψtot of a scattering process is written as 0

ψtot = ψin + ψsc = e

ikz

eik r + f (k 0 , θ) r

(2.3)

In case of elastic scattering, kinetic energy is conserved and therefore |~k| = |k~0 | = k, then the above relation becomes eik r ψtot = ψin + ψsc = eik z + f (k, θ) r

2.2

Partial wave analysis of scattering process

When the de Broglie wavelength (λ = h/p) of a projectile is comparable with the nuclear size (or interaction radius R), then quantum effects becomes important. The interaction region as seen by an approaching projectile, is divided into a set of zones concentric with the head on collision point X with target. The zones are labeled by the orbital angular momentum quantum number `. Although the concept of impact parameter b is imprecise, in a semi-quantum picture it is still used. Classically angular momentum Lpis related to b as, L = mvb, where as in quantum mechanics, L = `~ or in strict quantum sense, L = `(` + 1) ~. Combining the two, p p mvb = `(` + 1) ~ ⇒ b = `(` + 1) ~/p 11

2.2

Partial wave analysis of scattering process

2

√ √ √

NUCLEON-NUCLEON SCATTERING

12λ ¯

6λ ¯

2λ ¯

λ ¯ X `=0 `=1 `=2 `=3 `=4

Figure 8: Sectional view of partial waves of an incident beam. Since, from de Broglie’s relation, p = ~k, therefore ⇒ b = ⇒

b=

p

p `(` + 1)/k

`(` + 1) λ ¯

p Where λ ¯ = 1/k = λ/2π is the reduced wavelength. ∴ Impact parameters ‘b’ scales as `(` + 1). In quantum mechanics only integral values of ` are allowed. Therefore b in units of λ ¯ for first few ` values are shown below. ` b

0 (λ ¯)

1

1 √ 2

2 √ 6

3 √ 12

4 √ 20

5 √ 30

6 √ 42

The innermost (` = 0) zone (I zone) corresponds to particles with impact parameters less than the reduced wavelength λ ¯ of beam particles. It is the region between b = 0 and b = 1λ ¯ and the corresponding partial wave is called s-wave. The next zone (II zone) includes the region between b=λ ¯ √ and b = 2λ called as p − wave and so on. In this way, a plane wave, representing a parallel beam, is ¯ split into a set of ` partial waves each associated with a particular zone as shown in the figure 8. 2.2.1

Relation between energy and impact parameter

Since, the momentum p is related to wave vector k through the relation, p = ~k, it implies that, p2 /~2 = k 2 ⇒p k 2 = 2mE/~2 ⇒ k 2 ∝ E . Where, E is the kinetic energy of the particle. 2 Since b = `(` + 1)λ ¯, therefore b2 = `(` + 1)λ ¯ ⇒ b2 = `(` + 1)/k 2 . Substituting the value of 2 2 k from the above relation, b becomes, `(` + 1)~2 ⇒ b2 ∝ 1/E (2.4) 2mE This informs that, higher the incident energy, lower will be the impact parameter. In case of two particle systems m should be replaced by reduced mass µ of the system. For two elementary nucleons, µ = (m1 m2 )/(m1 + m2 ). `(` + 1)~2 `(` + 1)(~c)2 b2 = ⇒ b2 = 2µE 2(µc2 )E b2 =

12

2.3

Cross-section in terms of phase shift

2

NUCLEON-NUCLEON SCATTERING

With, mn ≈ mp ⇒ µ ≈ mp /2 ⇒ µc2 = 466 MeV and ~c ≈ 3.1637 × 10−27 J m ⇒ ~c ≈ 200 MeV fermi. With all these informations, one can calculate the variation of b as a function of ` for energies, 0.01 MeV, 0.1 MeV, 1 MeV, 10 MeV and so on (exercise). A nucleon will get scattered only if, impact parameter ‘b’, for a given `, is less than the range ‘ro ’ of the nuclear force. i.e., b ≤ ro or equivalently b2 ≤ ro2 . Using (2.4) it leads to, `(` + 1)~2 ≤ ro2 2µE`

⇒

`(` + 1)~2 ≤ E` 2µro2

or E` ≥

`(` + 1)~2 2µro2

Now, considering ro = 2 fm as the range of nuclear force, the RHS of the above relation then depends only on ` values. One can estimate either the number of partial waves involved for a given energy E` of the projectiles or one can estimate the allowed range of energy for a given partial ` wave, using the above relation. For example when ` = 0, s-wave, the allowed energy range (with equal sign) is, `(` + 1)~2 `(` + 1)(~c)2 = ≈ 10.73 MeV 2µro2 2µc2 ro2 Therefore, for all energies of projectiles up to 10 MeV only spherically symmetric s partial wave will be involved in the scattering process. Due to its spherical symmetry, measuring only differential scattering cross section is sufficient to understand it. E` =

2.2.2

Explicit expressions for wave functions: partial wave analysis

From quantum mechanics, asymptotic solutions, in a region far away from scattering center (i.e., r  1/k) are seeked. The incident ‘ψin ’, scattered ‘ψsc ’ and total wave functions ‘ψtot ’ are divided in to a large number of partial waves. If the incident particles interacts with the target, then waves representing them are expected to exhibit some change in energy and some change in phase after the interaction. In case of elastic scattering, only shift in phase ‘δ’ between the two waves is expected. Therefore, this theory also referred as Phase Shift analysis. In this picture the three wave functions are respectively represented by the following equations. ∞ X

 sin(kr − `π/2) (2` + 1)(i) ψin = Pl (cos θ) kr `=0   ∞ X (2` + 1) eikr ψsc = (e2iδ` − 1)Pl (cos θ) 2ik r `



(2.5)

(2.6)

`=0

ψtot

  ∞ X ` 2iδ` sin(kr − `π/2) = (2` + 1)(i) e Pl (cos θ) kr

(2.7)

`=0

On comparing the relations, (2.2) and (2.5) the scattering amplitude is f (k 0 , θ) =

∞ i X (2` + 1) h 2iδ` e − 1 Pl (cos θ) 2ik

(2.8)

`=0

In the above relations, δ` is the phase shift of the ` th partial wave between the incoming wave and the outgoing waves due to the interaction between projectiles and the target. Pl (cos θ) is the ` th order Legendre polynomial.

2.3

Cross-section in terms of phase shift

dσ From quantum mechanics, differential cross section for an elastic scattering process is shown as, dΩ = 2 |f (k, θ)| . Therefore, total cross section ‘σtot ’ of that process can be obtained by integrating the above equation with respect to elementary solid angle (dΩ) over the entire geometry of the process as, Z Z dσ σtot = dΩ ⇒ σtot = |f (k, θ)|2 dΩ (2.9) dΩ

13

2.4

Phase shift determination

2

NUCLEON-NUCLEON SCATTERING

The elementary solid angle has the following values: dΩ = sin θ dθ dφ, in which θ varies from 0 to π and φ varies from 0 to 2π. Substituting (2.8) in the above relation, it leads to 2 Z 2π Z π ∞ X i (2` + 1) h 2iδ` σtot = dφ sin θdθ − 1 Pl (cos θ) e 2ik 0 0 `=0 ∞ 2 X (2` + 1) h i 2 Z π 2iδ` = 2π sin θdθ |Pl (cos θ)|2 −1 e 2ik 0 `=0   Z ∞ π X (2` + 1)2   = 2π 4 sin2 δ` d(cos θ) |Pl (cos θ)|2 (2.10) 2 4k 0 `=0

Let t R= cos θ, ⇒ dt = d(cos θ) then limits changes to 1 → −1. Therefore, the above integral −1 becomes 1 dt |Pl (t)|2 = (2)/(2` + 1) - from orthogonality condition of Legendre polynomial. Using this in (2.10), the total scattering cross section becomes, ` σtot = σtot =

∞ 4π X (2` + 1) sin2 δ` k2

(2.11)

`=0

• Since k 2 ∝ E, therefore any increase of E value decreases σtot . In other words • σtot is the sum of all the partial cross sections. • In case of low energy scattering (E < 10 MeV), only s − wave , ` = 0 will be involved. Then ⇒

2.4

0 = (4π/k 2 ) sin2 δ . σtot o

Determination of phase shift for low energy n − p scattering

The magnitude of phase shift (δo ) depends on the magnitude of interaction potential between projectile and the target. Normally, protons are used as targets and neutrons are used as projectiles in experiments. The energy of thermal neutrons E ∼ 0.0252 eV1 . But chemical binding energy of molecules (say H2 ) is ∼ 4.5 eV. Therefore, Emin > 5 eV, otherwise projectiles do not see isolated protons from H2 molecules. 2.4.1

Square well potential with hard core

Two nucleons never come into contact even in the presence of strongest force, like nuclear force, acting between them. The distance through which the two nucleons comes together is called the core distance ‘c’. At this distance the two nucleons experience a massive repulsive force of magnitude ∞. Therefore, the resulting potential is referred as hard core potential, which is defined in case of square well potentials as follows:   when 0 ≤ r ≤ c ∞ V (r) = −Vo when c < r ≤ (b + c) (2.12)   0 when (b + c) < r < ∞ Therefore the potential has three regions. Total wave function is ψn`m` (r, θ, φ) = Rn` (r)Y`m` (θ, φ) Since the potential is function of r only, therefore only radial part of the Schr¨odinger wave equation needs to be solved. In terms of modified radial wave-function u` (r), the Schr¨odinger wave equation is,   2µ `(` + 1)~2 u ¨(r) + 2 E − V (r) − u(r) = 0 ~ 2µr2 1

According to Maxwell’s distribution of velocities, at room temperature (∼ 293 K), the most probable velocity of neutrons is ∼ 2200 m/s, which corresponds to an energy of ∼ 0.0252 eV. This is called energy of thermal neutrons.

14

2.4

Phase shift determination

2

NUCLEON-NUCLEON SCATTERING

V (r)

In case of s − wave ` = 0, then the above relation simplifies to 2µ u ¨(r) + 2 [E − V (r)] u(r) = 0 (2.13) ~ Region I: u = u1 , V (r) = ∞, then relation (2.13) simplifies to,

b

∞

c

(b + c) r

(b + c)

I II III 2µ [E − ∞] u (r) = 0 (2.14) 1 ~2 −Vo The only possible solution of the above reb lation is that u1 = 0, otherwise there exists finite probability of finding the particle in region I, which is contradictory to the assumption of po- Figure 9: Square well potential with hard core tential in the region. according to (2.12) on right and geometric arrangeRegion II: u = u2 , V (r) = −Vo , then rela- ment of nucleons on the left. tion (2.13) simplifies to, 2µ u ¨2 (r) + 2 [E + Vo ] u2 (r) = 0 (2.15) ~ [E + Vo ] = k12 , then (2.15) simplifies to u ¨2 (r) + k12 u2 (r) = 0, which is a second order linear Let 2µ ~2 differential equation of SHM. Solution could be assumed of the form, u ¨1 (r) +

u2 (r) = A sin k1 r

(2.16)

Region III: u = u3 , V (r) = 0, then relation (2.13) simplifies to, 2µE u3 (r) = 0 (2.17) ~2 Let 2µ [E] = k22 , then (2.17) simplifies to u ¨3 (r) + k22 u3 (r) = 0, which is again an equation of SHM. ~2 Since it is in the region after the interaction potential, therefore its solution should be shifted by δo from the incident one and it could be assumed of the form, u ¨3 (r) +

u2 (r) = B sin(k2 r + δo )

(2.18)

Boundary conditions 1. At r = c → u1 = u2 = 0 ⇒ u2 |r=c = 0 ⇒ A sin k1 r|r=c = 0 as A 6= 0 ⇒ sin k1 r|r=c = 0 This is true when it is written as [sin k1 (r − c) = 0], therefore the u2 becomes, u2 (r) = A sin k1 (r − c) 2. The continuity condition requires that, at r = (b + c) → ⇒

u˙ 2 u˙ 3 = u2 u3

A k1 cos k1 (r − c)|r=(b+c) B k2 cos(k2 r + δo )|r=(b+c) = A sin k1 (r − c)|r=(b+c) B sin(k2 r + δo )|r=(b+c) ⇒

(2.19)

k1 cot k1 (b) = k2 cot[k2 (b + c) + δo ]

Simplifying for the phase shift δo from the above relation,   k1 δo = cot−1 cot k1 b − k2 (b + c) k2

(2.20) (2.21)

(2.22)

The relation (2.22) indicates that phase shift δo is function of wave number k. In other words, it is a function of incident energy E and depth of potential well Vo . If δo is measured from experiment, 0 = (4π/k 2 ) sin2 δ . then from this one can estimate low energy n − p scattering cross section as σtot o 15

2.4

Phase shift determination

2

From deuteron parameters Vo ≈ 35 MeV, b ≈ 1.5 fm and c ≈ 0.4 fm. Then phase shift δo depends only on the energy E of neutron projectiles. The figure 10 displays the theoretical curve of s − wave scattering 0 in which phase shift δ is computed cross section σtot o from (2.22) and it also displays experimentally mea0 . sured values σtot Clearly, there are discrepancies between theoreti0 in the low energy cal and experimental values of σtot range and the two curve meet around 5 MeV. The experimental value show a plateau value of 20.36 barn2 in the energy range 10 eV to 0.5 MeV. The experimental 0 value below 5 eV, value show a sudden rise in the σtot which is of the order of the chemical binding energy of molecules (say H2 ). The projectiles (neutrons), cannot see separate protons, instead they see molecules that 0 in the energy < 1 eV. results in the rise of σtot

NUCLEON-NUCLEON SCATTERING

σ0 experiment 20.36 b theory 1 eV

5 MeV E

Figure 10: Theoretical (dashed line) and experimental (—) values of scattering cross section σ0 as function of projectile energy E for np - scattering process.

Possible reasons for the discrepancies 0 , here, a square well potential with hard core is assumed which may 1. In the determination of σtot not be a realistic one. Other forms of potentials, exponential, Gaussian or Yukawa should be tried. However, later it was noticed that all these potentials yield similar results. Therefore, the potential used is not a source of error.

2. The values of Vo , b and c are obtained from deuteron studies. These parameters have consistently reproduced two important properties namely binding energy EB and radius rd of deuteron. Therefore, they cannot be source of errors. 3. The measured angular momentum of deuteron is J π = 1+ . Deuterons’ do not have any excited states, therefore it exists only in the ground state, in which ` = L = 0. ~ +S ~ ~ = 1. Therefore, S ~ = 1, this means that spins of two nucleons ⇒ J~ = L ⇒ J~ = S are aligned parallel and the corresponding spin multiplicity is 2S + 1 = 3 - a triplet state. i.e., deuteron exists in a triplet state, so, n and p interacts in the triplet state of deuteron. This is in contrast to the case of n − p scattering, in which, orientations of both n and p are random. Therefore, interaction between them could be both due to singlet state as well as triplet state. Therefore, the computed cross-section using deuteron parameters contain only triplet part without singlet part, which could be the reason for the discrepancies between theoretical and experimental cross sections. This idea was first suggested by E. P. Wigner, according to him, total cross section σ0 is 3 1 σ0 = σt + σs . Where, 34 and 41 are weight factors for triplet and singlet states respectively. 4 4 Here, σ0 is measurable, σt is computable from deuteron parameters and σs is not possible to compute at this stage independently. NOTE: However, interaction can be made either only triplet or only singlet, by using polarized projectile beam and polarized target. 2

1 barn = 100 fm2 = 10−24 cm2

16

2.4

Phase shift determination

2

(A) when a > 0 then ⇒ u3 = a − r

NUCLEON-NUCLEON SCATTERING

(B) when a < 0 then ⇒ u3 = a + r

u(r) u2 = A sin k1 r

u(r)

u2

c

u3

(b + c)

c r

(+a)

u3

(b + c) r

(−a)

!htbp Figure 11: Behaviour of Scattering length (A) when a > 0 then ⇒ u3 = a − r and (B) when a < 0 then ⇒ u3 = a + r. 2.4.2

Scattering length

Total scattering cross section for s- wave is, σ0 = 4π sin2 δ0 . At very low energies, i.e., as E → 0 then k2 k → 0. This makes δ0 → 0, otherwise σ0 becomes infinite, which is unphysical.    2 4π δ0 2 sin δ0 = 4π ∴ lim σ0 = lim 2 k→0 k→0 k k Let us define (δ0 /k) = −a. Since ‘a’ has the dimension of length associated with σ0 , therefore it is called Fermi’s “scattering length”. This is identified as the radius of sphere from which a point neutron gets scattered. ⇒ σ0 = 4π a2 − is called scattering cross-section in zero energy approximation (ZEA). Since σ0 = (3/4)σt + (1/4)σs , therefore if as and at are scattering lengths for singlet and triplet scattering respectively, then under ZEA, σ0 = π(3a2t + a2s ) Significance of scattering length Since the interaction potential affects only the outgoing wave, the solution of total wave-function outside the range of the nuclear force is written as X ψtot (r) = (2` + 1)(i)` eiδ` [sin(kr − `π/2 + δ` )/kr] p` (cos θ) `

for s− wave ` = 0

⇒

p` (cos θ)|`=0 = p0 (cos θ) = 1 ⇒

In the ZEA, limk→0 ⇒

⇒

eiδ0 → 1

ψtot (r) = eiδ0 ⇒

sin(kr + δ0 ) kr

limk→0 ψtot (r) ≈

r ψtot (r) = u(r) = r + (δ0 /k)

⇒

(kr+δ0 ) kr

u(r) ≈ r − a

Last equation represents a equation of a straight line. Scattering length ‘a’ can be either positive or negative depending on how u(r) varies with r. Sign of the scattering length indicates whether the system is bound or unbound. If extrapolation of u(r) for r > (b + c) make a positive intercept on the r− axis resulting in a > 0 that indicates system is bound (state). On the other hand, if extrapolation of u(r) joins the r axis on the negative side then it indicates that system is unbound with a < 0. These two conditions are summarized as, • If a > 0 then wave function converges (see figure 11 (A)) ⇒ system has a bound state . • If a < 0 then wave function diverges (see figure 11 (B)) ⇒ system has no bound state . 17

2.5

Coherence

2

NUCLEON-NUCLEON SCATTERING

Experimental Results Total scattering cross section for s− wave n−p scattering: σ0 ≈ 20.36 b and p from deuteron parameters, σt ≈ 2.53 b. p Since σ0 = 3/4σt + 1/4σs ⇒ σs ≈ 73.85 b. In ZEA, as = σs /4π ⇒ as = ± 24.24 fm and at = σt /4π ⇒ at = ± 4.49 fm. From the above calculations it is impossible to say whether as (at ) is bound or unbound. Further investigation is required.

2.5

Coherence

2.5.1

Interference in optics

Interference of two monochromatic light intensities I1 and I2 takes place only when distance between two light sources (slits) are much less than the wavelength (λ) of the light. In technical terms, coherence length must be high. Consider total intensity I = I1 + I2 On the other hand if instead of intensities. 2.5.2

⇒

I = |A1 ± A2 |2

I = |A1 |2 + |A2 |2 − incoherent addition. − then it is coherent addition, in which amplitudes add up

Interference in nuclear physics

Similarly in nuclear physics if de-Broglie wavelength of neutrons is greater than inter-nucleon distance then coherent scattering will results. i.e., λn > inter- nucleon distance, otherwise, if λn < inter-nucleon distance (∼ 10−8 cm) then it leads to incoherent scattering. When incident neutrons strike the randomly distributed protons, which have no definite phase relationship, then it results in incoherent scattering. On the other hand, when the target protons are bound in a system (like those in H2 molecule) then all protons have definite relationship in their spin orientation and hence they are in coherence with each other. At room temperature, molecular hydrogen exists in two forms. viz., (1) Ortho hydrogen, in which spins of two protons are with parallel alignment and (2) Para hydrogen, in which spins of two protons are with anti-parallel alignment in 3:1 ratio. All the possible interactions of single spin-up (↑) and single spin-down (↓) neutrons with ortho (↑ ↑) and para molecules (↑ ↓) are shown in the following schematic. (a) Spin-up neutron with Para Hydrogen ↑

(b) Spin-down neutron with Para Hydrogen

triplet

↑

↑

singlet

↓

triplet

↓ ↓

singlet

(c) Spin-up neutron with Ortho Hydrogen ↑

(d) Spin-down neutron with Ortho Hydrogen

triplet

↑

↑

singlet

↑

singlet

↓ ↑

triplet

This schematic shows that, para hydrogen will result in both singlet and triplet states from single neutron scattering, where as ortho molecule will result in either singlet n − p or triplet n − p scattering from a single neutron. This implies, theoretically, neutron scattering with ortho hydrogen is simple to 18

2.6

Ortho-Para Scattering

2

NUCLEON-NUCLEON SCATTERING

analyze. On the contrary, it is simple to obtain a system of completely para hydrogen molecules. At very low temperature ∼ 20 K, almost all hydrogen will be in the para state. At such low temperature, λn ∼ 7 × 10−10 m = 10 × separation distance between two protons in H2 molecule, which satisfies the condition for coherent scattering.

2.6 2.6.1

Very low energy neutron scattering with Ortho and Para Hydrogen molecules Theory of Coherent Scattering

This is a very low energy scattering theory, originally developed by Schwinger and Teller and later it was improved by Bethe. Let ~sn and ~sp be the spin angular momenta of neutron and proton respectively. ~ of the system during interaction will be S ~ = ~sp + ~sn . Then, Then total spin S ~ 2 = |~sp |2 + |~sn |2 + 2(~sn .~sp ) |S| h i ~ 2 − |~sp |2 − |~sn |2 . This is because ~sn and ~sp commutes with each other. ⇒ (~sn .~sp ) = 1/2 |S| In operator form, their eigen values are, (ˆ sn .ˆ sp ) = 1/2 [S(S + 1) − sp (sp +
1) − sn (sn + 1)]. Where, sn = sp = 1/2 (in units of σ~), therefore sp (sp + 1) = sn (sn + 1) = 21 21 + 1 = 34 .   1 3 (ˆ sn .ˆ sp ) = S(S + 1) − (2.23) 2 2 ~S ~ = (~sp + ~sn ).(~sp + ~sn ) S.

⇒

For triplet state S = 1 and for singlet state S = 0, then relation (2.23) provides (ˆ sn .ˆ sp ) = 1/4 and (ˆ sn .ˆ sp ) = −3/4 respectively for these two cases. Projection operators Let p ˆt and p ˆs be the projection operators for the triplet and singlet states respectively. They are defined such that they produce only one state in a given process, as follows: ( 1 - for triplet state 3 p ˆt = + (ˆ sn .ˆ sp ) ⇒ p ˆt = 4 0 - for singlet state ( 0 - for triplet state 1 sn .ˆ sp ) ⇒ p ˆs = p ˆs = − (ˆ 4 1 - for singlet state Let at and as be the scattering lengths for triplet and singlet states respectively, then total scattering length a is defined as, a = at p ˆt + as p ˆs ⇒

⇒

a = at [3/4 + (ˆ sn .ˆ sp )] + as [1/4 − (ˆ sn .ˆ sp )] a = 1/4[3at + as ] + (ˆ sn .ˆ sp )[at − as ]

(2.24)

A neutron can get scattered by any one of the two protons of a hydrogen molecule at a time and therefore, let a1 and a2 be the scattering lengths for neutron scattering from first (p1 ) and second protons (p2 ) of hydrogen molecules, then from the relation (2.24) they are written as, a1 = 1/4[3at + as ] + (ˆ sn .ˆ sp1 )[at − as ]

a2 = 1/4[3at + as ] + (ˆ sn .ˆ sp2 )[at − as ]

(2.25)

When λn > inter-nucleon distance between two protons, then coherent scattering will result. Then, the total scattering length a is the sum of above two. i.e., a = a1 + a2 . Using (2.25) ⇒

1 sn . (ˆ sp1 + sˆp2 )] a = (3at + as ) + (at − as ) [ˆ 2 19

(2.26)

2.7

Shape independent theory

2

NUCLEON-NUCLEON SCATTERING

1 apara = (3at + as ) . Therefore, under ZEA, 2 total scattering cross section is, σ = 4πa2 , for para molecule it is 2 1 (2.27) σpara = 4π (3at + as ) ⇒ σpara = π|3at + as |2 2 For para hydrogen molecule, (ˆ sp1 + sˆp2 ) = 0

⇒

For ortho hydrogen molecule, (ˆ sp1 + sˆp2 ) 6= 0hand therefore, let (ˆ sp1 + sˆp2 ) = SˆH . Then the relation i (2.26) becomes, aortho = 1 (3at + as ) + (at − as ) sˆn .SˆH . Therefore, under ZEA, total scattering cross 2

section for ortho molecule is

h i 2 1 σortho = 4π (3at + as ) + (at − as ) sˆn .SˆH 2 h i 2 ⇒ σortho = σpara + 4π (at − as ) sˆn .SˆH

(2.28)

Now consider |ˆ sn .SˆH |2 . Since the the protons in different hydrogen molecules are aligned in random directions and incident beam is unpolarized, therefore, |ˆ sn .SˆH |2 must be averaged over all possible orientations of neutron spin. In terms of components, it is, 2 ˆ SˆHx ˆi + SˆHy ˆj + SˆHz k) ˆ |ˆ sn .SˆH |2av = (ˆ snx ˆi + sˆny ˆj + sˆnz k).( 2 = sˆnx SˆHx + sˆny SˆHy + sˆnz SˆHz = (ˆ snx SˆHx )2 + (ˆ sny SˆHy )2 + (ˆ snz SˆHz )2 + cross terms (2.29) Due to random orientation of proton spins and from the properties of spins, the value of cross terms averages to zero. Therefore, only first three terms of (2.29) will survive. Since sˆnx = sˆny = sˆnz = 1/2, therefore (2.29) reduces to, i 1h i 1 1 h ˆ2 2 2 SHx + SˆHy + SˆHz = Sˆ2 = [SH (SH + 1)] |ˆ sn .SˆH |2av = 4 4 H 4

(2.30)

For ortho hydrogen molecule, SH = 1. Therefore |ˆ sn .SˆH |2av = 41 [1(1 + 1)] = 12 . Using this value in 2 (2.28), it reduces to σortho = σpara + 4π (at − as ) 12 σortho = σpara + π |(at − as )|2

(2.31)

From the relation (2.31), If at = as (i.e, triplet scattering length = singlet scattering length), then σortho = σpara . On the other hand if σortho 6= σpara then at 6= as . Experimentally measured values are σortho = 125 b and σpara = 4 b respectively. This clearly suggests that two scattering mechanisms are different. In other words, nuclear force is spin dependent. From the magnitudes of two scattering lengths and from the sign of as in two scattering cross sections, it implies that singlet n − p scattering length is negative and large in magnitude. It can be shown that, at = 5.37 ± 0.04 fm and as = −23.73 ± 0.07 fm This implies triplet scattering is bound and singlet scattering is unbound.

2.7

Potential shape independent theory or Effective range of nuclear fore

A relation between scattering length a and scattering cross section was obtained for zero energy approximation. A more general expression, which is applicable for the entire s − wave (∼ 10 MeV) was derived first by Schwinger and later it was improved by others like Bethe.

20

2.7

Shape independent theory

2

NUCLEON-NUCLEON SCATTERING

Within s − wave let Ea and Eb be the two representative energies of incident particles. Then Schr¨odinger equation with modified radial wave functions ua and ub at Ea and Eb are written as,   2µ `(` + 1)~2 u ¨a + 2 Ea + V (r) − ua = 0 ~ 2µ r2   2µ `(` + 1)~2 u ¨b + 2 Eb + V (r) − ub = 0 ~ 2µ r2 In case of s − wave, ` = 0 and let 2µ Ea = ka2 , ~2

2µ Eb = kb2 ~2

2µ V (r) = U (r) ~2

and

(2.32)

With these substitutions the above relations are simplified as, u ¨a + ka2 ua − U (r)ua = 0 u ¨b +

kb2 ub

(2.33)

− U (r)ub = 0

(2.34)

Multiplying the relation (2.33) by ub and the relation (2.34) by ua and subtracting the one resultant from the other removes the potential U (r) and they reduces to ub u ¨a − ua u ¨b + (ka2 − kb2 )ua ub = 0

ub u ¨a − ua u ¨b = (kb2 − ka2 )ua ub

⇒

Integrating the above equation by parts with respect to dr between the limits 0 → ∞, it results in  ∞  ∞ Z ∞ Z Z ub u˙ a − u˙ a u˙ b dr − ub u˙ a − u˙ a u˙ b dr = (kb2 − ka2 )ua ub dr 0 0 0 Z ∞ ∞ 2 2 ua ub dr (2.35) [ub u˙ a − ua u˙ b ]0 = (kb − ka ) 0

In order to solve the above relation, a comparative function v(r) is defined such that v(r) = u(r) outside the range of nuclear interaction and it represents the extrapolation of u(r) backwards at inside the range of interaction as shown in the adjacent diagram. Then writing Schr¨ odinger equations with this wave function for two energies Ea and Eb and carrying out the above procedure, resultant equation is similar to relation (2.35), as Z ∞ ∞ 2 2 [vb v˙ a − va v˙ b ]0 = (kb − ka ) va vb dr (2.36) 0

Subtracting the relation (2.36) from (2.35) it leads to [ub u˙ a − ua u˙ b − vb v˙ a + • As r → ∞

⇒

va v˙ b ]∞ 0

=

(kb2

−

ka2 )

Z 0

∞

(ua ub − va vb ) dr

(2.37)

ua = ub = va = vb = 0 − all wave functions vanish.

• at r = 0 ua = ub = 0 and va (r) = vb (r) 6= 0. Outside the interaction range comparative function is v(r) = u(r), it contains the informations of nuclear interaction in phase shift δ as, va (r) = A sin(ka r + δa )

atr = 0 ⇒

If va (r = 0) is normalized then 1 = A sin(δa )

⇒

va (r = 0) = A sin(δa )

A = 1/(sin δa ).

Then normalized comparative function at Ea

21

va (r) =

sin(ka r + δa ) sin δa

(2.38)

2.7

Shape independent theory

2

NUCLEON-NUCLEON SCATTERING

Differentiating (2.38) with respect to r it leads to v˙ a (r) = ka

cos(ka r + δa ) sin δa

⇒

v˙ a (r)|r=0 = ka cot δa

(2.39)

Similarly Then normalized comparative function at Eb becomes vb (r) = A sin(kb r + δb ),

v˙ b (r) = kb

cos(kb r + δb ) sin δb

⇒

v˙ b (r)|r=0 = kb cot δb

(2.40)

Therefore, LHS of (2.37) reduces to LHS = upper limit − lower limit = 0 − [0 + va v˙ b − vb v˙ a ]r=0 LHS = −[1 × kb cot δb − 1 × ka cot δa ]

⇒

LHS = ka cot δa − kb cot δb

Therefore LHS comprises of two phase shifts δa and δb at two energies Ea and Eb . Using this in (2.37), it becomes, Z ∞ 2 2 ka cot δa − kb cot δb = (kb − ka ) (ua ub − va vb ) dr (2.41) 0

This is an exact equation and therefore it can be solved for any value (up-to 10 MeV) of energy of s - wave. Approximations 1. Chose Ea to follow ZEA, then Ea = E0 ⇒ ka ≈ k0 and Eb ≈ E ⇒ kb ≈ k. Then ua ≈ u0 , ub ≈ u, va ≈ v0 and vb ≈ f . With this approximation the relation (2.41) changes to Z ∞ k0 cot δ0 − k cot δ = (k 2 − k02 ) (uu0 − vv0 ) dr (2.42) 0

2. As k0 → 0 then δ0 → 0, this makes cot δ0 → ∞ ⇒ k0 cot δ0 = k0 /δ0 or k0 cot δ0 = −(1/a) - reciprocal of scattering length. With all these relation (2.42) changes to Z ∞ 1 2 − − k cot δ = k (uu0 − vv0 ) dr (2.43) a 0 3. Let Eb = E  V (r) - interaction potential then u ≈ u0 and v ≈ v0 . This changes (2.43) to Z ∞ 1 (u20 − v02 ) dr (2.44) − − k cot δ = k 2 a 0 If r0 is the effective range of nuclear interaction then RHS of above relation is written as, Z r0 Z ∞ 1 − − k cot δ = k 2 (u20 − v02 ) dr + (u20 − v02 ) dr a r0 0 Beyond the range r0 of interaction u0 = v0 , then the above relation becomes Z r0 Z r0 1 1 2 2 2 2 − − k cot δ = k (u0 − v0 ) dr ⇒ + k cot δ = k (v02 − u20 ) dr a a 0 0 Z Integral on the RHS has the dimensions of length, therefore 2 0

r0

(v02 − u20 ) dr ≈ r0 - effective

range of interaction. Then the above relation is simplified as, k 2 r0 1 k cot δ = − 2 a

  1 k 2 r0 a cot δ = −1 ak 2

⇒ 22

(2.45)

3

DEUTERON PROPERTIES: REVISIT

Therefore, phase shift δ depends on known quantities: energy E ∝ k 2 , , range r0 of interaction and scattering length a. In case of s - wave, scattering cross section is: σ0 = (4π/k 2 ) sin2 δ0 . Since

sin2 δ =

1 1 + cot2 δ

⇒

1

sin2 δ = 1+

1 (ak)2

h

k2 r0 a 2

−1

⇒

i2

sin2 δ = a2 k 2 +

a2 k 2 h 2

k r0 a 2

Therefore, energy dependent scattering cross section as a function of effective range r0 is            4π  a2 k 2 4π a2 σ0 = 2 ⇒ σ0 = h 2 i2 h 2 i2  k   a2 k 2 + k r0 a − 1    a2 k 2 + k r0 a − 1   2 2

−1

i2

(2.46)

When the scattering process involves both singlet and triplet interactions, then two components of scattering lenghts, viz., as and at and two components of effective ranges, viz., (r0 )s and (r0 )t results, then scattering cross section becomes a function of these four parameters as, σ = (1/4)σs + (3/4)σt          3  1 4π a2s 4π a2t σ= i2 + i2 h 2 h 2  2 2  4  a2 k 2 + k (r0 )s as − 1   4 at k + k (r20 )t at − 1  s 2

3

(2.47)

Deuteron Properties: Revisit

3.1

Magnetic moment in ground state

Experimentally measured magnetic moment of Deuteron ‘~ µd ’ is 0.85735 µN , Where µN ≈ 3.152×10−14 −1 MeV T - nuclear magneton. Theoretically Deuteron’s magnetic moment is computed as ⇒

µ ~d = µ ~p + µ ~n For deuteron ground state: ` = 0

⇒

µ ~ d = (~ µ` + µ ~ s )p + (~ µ` + µ ~ s )n (~ µ` )p = (~ µ` )n = 0

(~ µs )p = gsp . ~s = (5.5855 × 1/2) µN ,

and

(~ µs )n = gsn . ~s = (−3.8270 × 1/2) µN

Where gsp and gsn are gyro-magnetic ratio’s of proton and neutron respectively. µ ~ d in µN Experimental Theoretical

= =

0.85735 0.87925

Discrepancy

=

0.02190

Table 4: Magnetic moment of Deuteron. Note that the experimentally µ ~ d values are measured very accurately up-to 7th decimal place. Therefore the amount of discrepancy at the second decimal place (≈ 2.19%) with the theoretical estimation is very large (see table 4). This must be accounted by suitably modifying the theory.

3.2

Quadrupole moment

~ = 0, therefore corresponding quadrupole moment Q is zero, In case of ground state of deuteron L implying that deuteron is spherical in shape. However, although it is small positive value, it is measurable and its value is ≈ 2.82 × 10−27 cm2 , and therefore out of spherical shape. This suggest that the ground state of deuteron is different, which must be accounted so that theoretical value matches the experimental value. 23

3.3

3.3

Angular momentum

3

DEUTERON PROPERTIES: REVISIT

Angular momentum ~ L

J~ = 1

~ S

π = (−1)L

0

(0+1)=1

1 (↑↑)

+1

valid

1

(1+0)=1 (1+1)=0 =1 =2

0 (↑↓) 1 (↑↓)

-1 -1

not not not not

2 2

(2+0)=2 (2+1)=1 =2 =3

0 (↑↓) 1 (↑↑)

+1 +1

not valid (∵ J~ = 2) valid not valid (∵ J~ = 2) not valid (∵ J~ = 3)

validity

valid (∵ π = −1) valid (∵ π = −1) valid valid

Table 5: Magnetic moment of Deuteron. The experimentally measured angular momentum of deuteron in ground state is J π = 1+ (σ~). ~ either Since there are two nucleons with half integral spin, therefore only two possible values of S ~ > 1 are not allowed. There are various combinations of L ~ ~0 (↑↓) or ~1 (↑↑) in units of σ~, and values S ~ ~ ~ ~ ~ and S that results in J = 1 and the possible cases are listed in the table 5. Note that J = L + S and ~ − S| ~ · · · |L ~ + S|. ~ it can have values from |L

3.4

Deuteron ground state as an admixture

~ = 2 is also a valid state of deuteron, with at least one case From the table 5 it is clear that L results in J~ = 1. Therefore, deuteron ground state can be assumed as made up of a linear combination of two states, viz, ψL=0 and ψL=2 . Therefore, ground state wave function of deuteron is: ψgs = a ψS + b ψD . ⇒ ψgs is an admixture of 3 S1 and 3 D1 components. ~ =0 Here |a|2 and |b|2 provides the probability of finding the deuteron in L ~ (a) S ~ and L = 2 states respectively and if not simultaneously, deuteron can be found ~ is not constant and therefore in both these states at different times. ⇒ L ~s1 ~s2 angular momentum is not a good quantum number at least for deuteron and ↑ ↑n p the corresponding nucleon-nucleon interaction potential does not commute ~r with L2 . ~ is not constant means dL~ 6= 0 ⇒ dL~ = ~τ - torque acting Classically L dt dt ~s1 (b) on the body. Torque is also defined as ~τ = ~r × F~ ⇒ ~τ = r F sin θ , ↑ ~ n S where θ is the angle between the ~r and F~ ⇒ ~τ = ~τ (r, θ). This provides F~ = F~ (r, θ) and the corresponding potential is V = V (r, θ) - a non-central ~r potential. All these discussions provide the evidence that natrue of ncuelar ~s2 force is non-central with the corresponding non-central potential. p↑ In quantum picture, this non-central potential arises due to interaction ~ = ~s1 + ~s2 of the between radial vector ~r (between n and p) and total spin S two nucleons of deuteron. Out of several possibilities, figure 12 displays two possible orientations of these vectors in deuteron at different times. In figure Figure 12: Angle be~ in (a) ~ = θ between two vectors is 90◦ and in figure 12 (b) it tween ~r and S 12 (a) the angle ∠~rS ◦ is 0◦ . Let F~ (90◦ ) and F~ (0◦ ) be the two interaction forces for these two cases. θ = ◦ 90 and in (b) Then the question that arises in the mind is whether F~ (90◦ ) = F~ (0◦ )? If θ = 0 . nuclear force is non-central, then F~ (90◦ ) should be different from F~ (0◦ ).

24

3.5

Nuclear interaction potential in deuteron

3.5

3

DEUTERON PROPERTIES: REVISIT

Nuclear interaction potential in deuteron

From the previous sections it is leant that, nuclear potential and therefore force are spin dependant and non-central. Therefore, nuclear interaction potential (Vnp ) must include all these components, with these informations Vnp can be written as, Vnp (r, S, θ) = V (r) + V (S) + V (r S)

(3.1)

Where, • V (r) = −V0 f (r) - central potential. • V (S) = Vσ (~σ1 ~σ2 ) - spin dependant potential3 . • V (r S) = VT Sˆ12 - non-cetral potential defined by a tensor operator Sˆ12 . 3.5.1

Conditions on nuclear interaction potential

1. Vnp must be a scalar quantity, because potential energy is a scalar quantity. 2. Vnp must be invariant under rotations (i.e., J must be conserved) and reflections (parity π must be conserved i.e., space reversal). 3. Vnp must be symmetric between the particles 1 and 2. It follows froms the conditions 1 and 2 that the vectors ~r and ~σ should appear an even number of times in Vnp . Condition 3 requires that ~σ1 and ~σ2 should appear in a bilinear combination. The two linearly independent possible quantities which can be constructed satisfying these conditions are, (~σ1 . ~σ2 ) and (~σ1 . rˆ)(~σ2 . rˆ). With this knowledge, the tensor operator (Sˆ12 ) is defined as, Sˆ12 = 3(~σ1 . rˆ)(~σ2 . rˆ) − (~σ1 . ~σ2 ) 3.5.2

or

3 Sˆ12 = 2 (~σ1 . ~r)(~σ2 . ~r) − (~σ1 . ~σ2 ) r

(3.2)

Properties of tensor operator

• The average value of Sˆ12 over all directions vanishes. i.e., hSˆ12 i = 0. • Tensor operator Sˆ12 satisfies the followingcommutation relations: 1. [S2 , Sˆ12 ] = [J2 , Sˆ12 ] = [Jz , Sˆ12 ] = 0 2. [Sz , Sˆ12 ] 6= 0 [Lz , Sˆ12 ] 6= 0 [L2 , Sˆ12 ] 6= 0

3.6

Ground state wave function of deuteron

Ground state wave function of deuteron is ψgs = a ψS + b ψD where |a|2 = 1 for pure S - state |b|2 = 1 for pure D - state As the nuclear interaction potential is spin dependant, the ground state wave function is also spin dependant, therefore it can be written as, S − state : ψS = ψS (~r, ~σ ) = RS (r) YS (θ, φ) χS (~σ1 , ~σ2 )

(3.3)

• RS (r) - radial wave function for S - state, • YS (θ, φ) - angular wave function for S - state, • χS (~σ1 , ~σ2 ) - spin wave function for S - state. Let XS = YS (θ, φ) χS (~σ1 , ~σ2 ) - angular spin function, then ψS (~r, ~σ ) = RS (r) XS . Exactly on similar lines D - state wave function can be written as, ψD (~r, ~σ ) = RS (D) XD . 3

 ~σ is a Pauli’s spin matrix with components σx =

0 1

  1 0 , σy = 0 i

25

  −i 1 and σz = 0 0

 0 −1

3.6

Ground state wave function

3.6.1

3

DEUTERON PROPERTIES: REVISIT

Spin wave functions

Spin wave function χS (~σ1 , ~σ2 ) is written for two states separately. Let spin-up and spin-down states are denoted by using Pauli’s spinors α and β defined as column matrices such that,     1 0 α= : Spin-up state (↑) β= : Spin-down state (↓) 0 1 Measured angular momentum is J = 1, what is experimentally measured is mJ = +1, such that mJ = mL + mS . 1. For S - state: L = 0 and J = 1 ⇒ S = 1 - triplet state ⇒ two particles’ states are spin-up states. i.e. χS :

mS = +1. This result is possible only when (↑)1 (↑)2 or α1 α2 .

2. For D - state: L = 2 and J = 1 with mJ = +1 what are the values of mL and mS such that mJ = mL + mS ? With L = 2 the possible values of mL are 2, 1, 0, -1 and 2. Since deuteron has only two half integral spin particles, therefore, maximum mS = +1 and not anything else. Therefore the possible spin functions χD are mJ

mL

mS

Allowed?

+1 +1 +1 +1 +1

2 1 0 -1 -2

-1 0 1 – –

yes yes yes no no

χD (↓ ↓) (↑ ↓ + ↓ ↑) (↑ ↑) – –

β1 β2 α1 β2 + β1 α2 α1 α2 – –

Therefore for two nucleon system, with mL = -1 and -2 it is impossible to find any value of mS such that mL + mS = +1. Hence mL = -1 and -2 are not allowed states. 3.6.2

Angular wave functions q

1 For S - state: L = 0 ⇒ mL = 0. Therefore angular wave function YLmL is : Y00 = 4π For D - state: L = 2 ⇒ mL = 2, 1, 0,-1,-2. Out of these mL = -1 and -2 are not allowed. Therefore the possible angular wave functions are : r 5 Y20 = p20 (θ) where p20 (θ) = (3 cos2 θ − 1)/2 4π r 5 Y21 = − p21 (θ) where p21 (θ) = −3 cos θ sin θ 24π r 5 Y22 = p22 (θ) where p22 (θ) = 3 sin2 θ 96π

3.6.3

Angular-spin wave functions

Here angular-spin wave functions XL are normalized therefore they associated with some factors as follows: 1. For S - state: XS = YS (θ, φ) χS (~σ1 , ~σ2 )

⇒

XS = Y00 α1 α2

(3.4)

2. For D - state: XD = YD (θ, φ) χD (~σ1 , ~σ2 ) r r r 1 3 3 ⇒ XD = Y20 α1 α2 + Y21 (α1 β2 + β1 α2 ) + Y22 β1 β2 10 20 5 26

(3.5)

3.7

Magnetic moment

3

DEUTERON PROPERTIES: REVISIT

Exercise 1. Proove the following: (a) σx α = β,

σy α = iβ,

(b) σx β = α,

σy β = −iα, σz β = −β √ Sˆ12 XS = 8 XD √ Sˆ12 XD = 8 XS − 2XD -PRIZE.

2. Proove that 3. Proove that

3.7

σz α = α

Magnetic moment of deuteron with S and D state admixture

Theoretically Deuteron’s magnetic moment is computed as ⇒

µ ~d = µ ~p + µ ~n

µ ~ d = (~ µ` + µ ~ s )p + (~ µ` + µ ~ s )n

~ = 0 and 2, therefore it is different from the earlier As deuteron ground state is an admixture of L discussion in subsection 3.1.     µ ~ d = g` . ~` + gs . ~s + g` . ~` + gs . ~s n

p

Where g’s are gyromagnetic ratios for orbital and intrinsic spin motion respectively for proton and neutron. ⇒ (g` )p = 1, (g` )n = 0, and (gs )p = 5.5855 µN , (gs )n = −3.8270 µN . The intrinsic spin angular momentum ~s = (1/2) σ in units of ~. Let p and n are respectively represented as 1 and 2 here, then the above relation becomes,     1 1 ~ µ ~ =µ ~ d = ` + g1 . σ1 + g2 . σ2 2 2 Experimentally magnetic moment is measured along z (or field) direction, then µ ~ =µ ~ z . In center of mass system, (`z )lab = (1/2)(`z )cm . Then above relation in CM system becomes µ ~z =

1 [(`z + g1 . σ1z ) + (g2 . σ2z )] 2

(3.6)

Theoretically, the magnetic moment of deuteron in ground state is estimated as its expectation value in quantum mechanics, which is defined as, h~ µz i = hψgs |~ µz |ψgs i

with

ψgs = a ψS + b ψD

= ha ψS + b ψD |~ µz |a ψS + b ψD i

= ha ψS |~ µz |a ψS i + hb ψD |~ µz |b ψD i

= |a|2 hψS |~ µz |ψS i + |b|2 hψD |~ µz |ψD i

h~ µz i =

I

+

II

(3.7)

Where I = |a|2 hψS |~ µz |ψS i and II = |b|2 hψD |~ µz |ψD i. Also using the relation (3.6) in the above relation, it becomes, 1 I = |a|2 hψS | [(`z + g1 . σ1z ) + (g2 . σ2z )] |ψS i 2 In case of S - state L = ` = 0 ⇒ `z = 0, then the above relation is written as, |a|2 hψS |(g1 . σ1z ) + (g2 . σ2z )|ψS i 2 |a|2 = [hψS |g1 . σ1z |ψS i + hψS |g2 . σ2z |ψS i] 2 |a|2 I= [ IA + IB ] 2 I=

27

(3.8)

3.7

Magnetic moment

3

DEUTERON PROPERTIES: REVISIT

Now consider IA, note that ψS = RS (r) XS and using relation (3.4), it is estimated as IA = hRS (r) Y00 α1 α2 |g1 . σ1z |RS (r) Y00 α1 α2 i In the above expression σ1z is a spin operator and therefore it operates only on spin functions, further note that it operates only on first particle spin function |α1 i, then the above relation becomes, IA = hRS (r)|RS (r)i hY00 |Y00 i hα2 |α2 i hα1 |g1 . σ1z |α1 i = (1) (1) (1) g1 hα1 |α1 i

⇒

IA = g1

| ∵ σ1z |α1 i = 1. |α1 i

Above result is obtained because the wave functions are normalized. Now consider IB, IB = hRS (r) Y00 α1 α2 |g2 . σ2z |RS (r) Y00 α1 α2 i Exactly on similar lines it can be shown that IB = g2 , in which spin operator σ2z operates only on |α2 i. Using these results in (3.8) it leads to I=

|a|2 [g1 + g2 ] 2

If the deuteron ground state is a pure S - state, then |a|2 = 1 and |b|2 = 0, this will make the relation (3.7) only for S - state as, g1 + g2 5.5855 + (−3.8270) = ⇒ h~ µz iS = 0.87925 µN (3.9) 2 2 Note that experimentally measured value h~ µz iexpt = 0.85735 µN , is different from the above value, therefore it suggest that ground state of deuteron is not a pure S - state. Now consider II, and using (3.6), ti becomes h~ µz iS =

|b|2 hψD |(`z + g1 . σ1z ) + (g2 . σ2z )|ψD i 2 |b|2 II = [hψD |`z |ψD i + hψD |g1 σ1z |ψD i + hψD |g2 σ2z |ψD i] 2 |b|2 II = [ IIA + IIB + IIC ] 2 Now consider IIA, note that ψD = RD (r) XD and using relation (3.5), it is estimated as * r r r 1 3 3 IIA = RD (r) Y20 α1 α2 + Y21 (α1 β2 + β1 α2 ) + Y22 β1 β2 |`z | 10 20 5 + r r r 1 3 3 RD (r) Y20 α1 α2 + Y21 (α1 β2 + β1 α2 ) + Y22 β1 β2 10 20 5 II = |b|2 hψD |~ µz |ψD i

⇒

II =

(3.10)

The operator `z operates only on angular wave function |YLmL i such that `z |YLmL i = mL |YLmL i and `2 |YLmL i = L2 |YLmL i = L(L + 1) |YLmL i. Since other wave functions are normalized, they simply reduce to 1, then the above relation becomes,   1 3 3 hY20 |`z |Y20 i + hY21 |`z |Y21 i + hY22 |`z |Y22 i IIA = 10 20 5   1 3 3 27 = (0) + (1) + (2) ⇒ IIA = (3.11) 10 20 5 20 Now consider IIB, * IIB =

r

r r 1 3 3 RD (r) Y20 α1 α2 + Y21 (α1 β2 + β1 α2 ) + Y22 β1 β2 |g1 σ1z | 10 20 5 + r r r 1 3 3 RD (r) Y20 α1 α2 + Y21 (α1 β2 + β1 α2 ) + Y22 β1 β2 10 20 5 28

4

PROTON-PROTON SCATTERING

No.

n − p - scattering

p − p - scattering

1 2

Interaction is purely nuclear Interaction is short range, ∴ only s - wave scattering is included. Non-identical particles

Interaction is nuclear + Coulombic Long range Coulombic interaction includes higher partial waves even at low energies. Identical particles

3

Table 6: Differences between n − p and p − p - scattering process. The operator σ1z operates only on first particle’s spin wave function, and other wave functions reduce to one, then   1 3 3 IIB = hα1 |g1 σ1z | α1 i + {hα1 |g1 σ1z | α1 i + hβ1 |g1 σ1z | β1 i} + hβ1 |g1 σ1z | β1 i 10 20 5   1 3 3 g1 IIB = g1 (1) + {1 + (−1)} + (−1) ⇒ IIB = − (3.12) 10 20 5 2 Exactly on similar lines it can be shown that IIC = −

g2 2

(3.13)

2  g1 g2  Using the results (3.11), (3.12) and (3.13) in (3.10) it becomes II = |b|2 27 20 − 2 − 2 . If the deuteron ground state is a pure D - state, then |a|2 = 0 and |b|2 = 1, this will make the relation (3.7) only for D - state as,   1 27 g1 g2 h~ µz iD = − − ⇒ h~ µz iD = 0.2350 µN (3.14) 2 20 2 2

This is different from the experimental value, therefore ground state of deuteron is not pure D state, instead it expected as a linear combination of S - state and D - state. i.e., using (3.10) and (3.14)

h~ µz iexpt = |a|2 h~ µz iS + |b|2 h~ µz iD

0.85735 = |a|2 0.87925 + |b|2 0.2350

(3.15)

From the probability theory, the normalization condition requires that |a|2 + |b|2 = 1 |b|2 = 1 − |a|2 . Using this in (3.15) it leads to   0.85735 = |a|2 0.87925 + 1 − |a|2 0.2350 = |a|2 [0.87925 − 0.2350] + 0.2350

⇒

⇒

|a|2 = 96.6% and |b|2 = 3.4%

Therefore probability of finding deuteron in S - state is 96.6% and that in D - state is 3.4%, from the magnetic moment of deuteron study. ⇒ Ground state wave function of deuteron is predominantly in S - state with a small admixture of D - state.

4

Low energy proton-proton scattering

Proton-proton scattering differs from the n − p scattering in many ways, as given in the table 6:

4.1

Consequences of identical particles

Here both scatterer (target) and scattered particles are identical. Therefore, experimentally the detector cannot distinguish the two. i.e., the detector cannot distinguish the small angle (θ) scattering from the large angle (π − θ) scattering (see the figure 13) in the CM system. 29

4.1

Consequences of identical particles

4

θ

(a)

PROTON-PROTON SCATTERING

θ

(b)

p

p

p

π−θ

p

π−θ

Figure 13: Identical particles: (a) small angle (θ) scattering and (b) large angle (π − θ) scattering. Classically, probability of finding either of the particles after scattering at an angle ‘θ’ and at an dσ(π−θ) angle ‘π − θ’ are given from their differential scattering cross sections dσ(θ) as their sum. dΩ and dΩ   dσ dσ(θ) dσ(π − θ) i.e., = + (4.1) dΩ CM dΩ dΩ In quantum mechanics, the scattering cross sections are derived from the wave functions. As the two particles are moving in opposite directions in CM system, let they are represented as ψ(~r ) and ψ(−~r ) respectively. The total wave function is the sum of these two wave functions. i.e.,

ψ(~r ) ± ψ(−~r )

(4.2)

In relation (4.2) the positive and negative signs are respectively for a symmetric wave function and for an anti-symmetric wave function. From scattering theory, ψtot (~r ) = ψin (~r ) + ψsc (~r )

⇒

ψtot (~r ) = eikz + f (θ)

eikr r

and

ψtot (−~r ) = e−ikz + f (π − θ)

eikr r

Using these two wave functions in (4.2), it becomes     eikr eikr ikz −ikz ψ(~r ) ± ψ(−~r ) = e + f (θ) ± e + f (π − θ) r r   ikr e = eikz ± e−ikz + [f (θ) ± f (π − θ)] r   ikr e = eikz ± e−ikz + f± (θ) r In the above relation f± (θ) = f (θ) ± f (π − θ). Then from the quantum mechanics, differential scattering cross section is defined as,   dσ = |f± (θ)|2 = | f (θ) ± f (π − θ) |2 dΩ QM ⇒

= |f (θ)|2 + |f (π − θ)|2 ± 2< [f (θ)f ∗ (π − θ)]

(4.3)

Therefore, quantum mechanical cross section (relation (4.3)) contains one more extra part called ‘interference term’ as compared to that obtained by classically, (relation (4.1)). This result is due to identical nature of particles. For proper comparison of cross sections between classical and quantum theories, θ = π/2 value is considered here. With this the two cross sections becomes,   dσ dσ =2 (4.4) dΩ CM dΩ (π/2)   dσ dσ =2 ± 2< [f (π/2)f ∗ (π/2)] dΩ QM dΩ (π/2)   dσ dσ dσ =2 ±2 (4.5) dΩ dΩ dΩ QM

(π/2)

30

(π/2)

4.1

Consequences of identical particles

4

PROTON-PROTON SCATTERING

In a constructive scattering (interference), the amplitudes add up whereas in a destructive scattering (interference) they vanish. Therefore, relation (4.5) is double valued. (   dσ 4 dΩ - constructive scattering dσ (π/2) = (4.6) dΩ QM 0 - destructive scattering Experimental results show good agreement with the quantum mechanical cross-section. Protons being fermions, they follow Pauli’s exclusion principle. According to Pauli’s principle, wave function representing a fermion must be antisymmetric. Similarly, for two fermions, total wave function (ψtot ) must be antisymmetric4 . Total wave function (ψtot ) is the product of spatial wave function (φspace ) and spin wave function (χspin ). ψtot = φspace χspin

(4.7)

Scattering of two identical fermions (nucleons here) can take place only in a state where the total wave function is antisymmetric with respect to a permutation of 2 particles. The parity for such a state is given as π = (−1)L+S . Specific cases dσ dΩ QM




dσ dΩ (π/2)



⇒ constructive scattering. This implies a positive space contribu~ = 0, 2, 4 · · · . Then tion from the relation (4.5) ⇒ symmetric space wave function ⇒ L this necessitates spin wave function to be anti-symmetric (↑↓), from Pauli’s principle. This implies singlet scattering. i.e., singlet scattering can take place only in the spatial S, D, · · · states. In other words, when the spatial part of the system of two identical fermions is symmetrical (even ~ then total intrinsic spin must be antisymmetrical (S ~ = 0) and vice versa. L)
dσ 2. When dΩ = 0 ⇒ destructive scattering. This implies a negative space contribution from QM ~ = 1, 3, 5 · · · . Then from the relation (4.5) ⇒ anti-symmetric space wave function ⇒ L (4.7) spin wave function must be symmetric (↑↑). This implies triplet scattering. i.e., triplet scattering can take place only in the spatial P, F, · · · states. 1. When

=4

These two cases are consequences of identical spin half particles. For a lab energy of ≤ 10 MeV, only s - partial wave interaction is significant. Therefore in s - wave (p − p) scattering only singlet scattering contributes to the measured total cross section at θ = π/2 and Pauli’s principle forbids triplet scattering for s - partial wave. Interaction potential In case of p − p scattering, both nuclear and Coulombic interactions are present. Therefore total interaction potential is the sum of two potentials. i.e.,

Vtot = VN + VC

Case I: Consider pure Coulombic interaction (switch off VN ) Then Vtot = VC and the corresponding Schr¨odinger equation becomes, ∇2 ψtot +

2µ [E − VC ]ψtot = 0 ~2

Solving exactly this equation in spherical coordinates becomes exceedingly difficult. However with the use of parabolic coordinates, this Schr¨odinger equation had been solved. The resulting 4

The antisymmetrical form of the wavefunction for fermions is generally taken as a ‘brute fact’, i.e., as a defining characteristic of fermions or as a feature of nature that cannot be otherwise explained.

31

4.1

Consequences of identical particles

System (p − p)nucl (n − p)singlet

4

PROTON-PROTON SCATTERING

Vos

bs

−13.3 MeV −14.3 MeV

2.58 fm 2.50 fm

Table 7: Comparison between (p − p)nucl and (n − p)singlet systems. differential cross section was first derived by MOTT, which is a modified form of Rutherford’s expression for scattering and it is called as Mott’s scattering cross section:     dσ dσ = dΩ Mott dΩ C      Z 4 e4 4 θ 4 θ cosec + sec + = 16E 2 2 2 (4.8)           (−1)2J 2 θ 2 θ 2 θ cosec sec cos η ln tan 2J + 1 2 2 2   dσ = |fC (θ)|2 + |fC (π − θ)|2 ± 2< [fC (θ)fC∗ (π − θ)] (4.9) dΩ Mott Where |fC (θ)|2 is the first term of (4.8) which is the famous Rutherford’s expression for Coulombic scattering, |fC (π −θ)|2 is the second term of (4.8) which is the recoil term arises due to identical particle nature, < [fC (θ)fC∗ (π − θ)] is the interference term due to identical particles. Case II: Consider pure nuclear interaction (switch off VC ) An expression similar to (4.9) is expected even for this case, and therefore   dσ ∗ = |fN (θ)|2 + |fN (π − θ)|2 ± 2< [fN (θ)fN (π − θ)] dΩ N

(4.10)

Case III: Considering the two potentials gives the combined effect and the resulting differential scattering cross section is written as,         dσ dσ dσ dσ = + ± (4.11) dΩ dΩ Mott dΩ N dΩ Mott×N The third term is due to interference of different type, not due to identical particle nature instead due to the presence of two interaction potentials in the scattering process. i h  dσ
 σexp The experimental results of angular distributions dΩ versus E versus θ and energy σMott distributions for s - wave are as shown in figure 14. The cross section is normalized by σMott in energy distribution to identify the three regions. In the lower energy region this ratio is 1.0 meaning measured   σexp total cross section σexp is entirely due to Coulombic interaction. With little rise in energy σMott decreases below 1.0 after reaching a minimum then it again starts toraise  with energy and this region σexp is called interference region. When energy is above 0.7 MeV, then σMott raises above 1.0 and now the measured cross section is due to nuclear interaction only. Due to identical particle nature Pauli’s principle allows only singlet (↑↓) scattering in (p − p) system, whereas in (n − p) system both singlet (↑↓) and triplet (↑↑) scatterings are allowed. Therefore comparison between the two systems must be carried out with singlet state scattering only. Further in the comparison only nuclear interaction of (p − p) system should be considered. The experimentally measured potential depth Vos and interaction range bs values for s - state are as shown in the table 7. From the table it is clear that F (p − p)nucl ≈ F (n − p)singlet .

32

4.1

Consequences of identical particles

4

(a)

PROTON-PROTON SCATTERING

(b)


dσ

σexp σMott

I

dΩ

III

E1 > E2

1.0

II E2

I

E1 35o

II

III E (MeV)

0.7

θ

Figure 14: (a) Angular distribution and (b) Energy distribution of cross section clearly depicting Coulomb interaction region by I, interference region by II and nuclear interaction region by III.

To study the nuclear force in (n − n) system either one has to carryout (n − n) scattering or he has to use other methods. Here, for simplicity, mirror nuclei of 3 nucleons are used and they are Triton (31 H2 ) and Helion (32 He1 ). Each of these nuclei can make three bonds with 2 (n − p) bonds in common. (31 H) nuclei has an extra (n − n) bond whereas (32 He) nuclei has extra (p − p) bond. Therefore binding energies of two nuclei are
B.E. 31 H = E[2(n − p)] + E[1(n − n)]
B.E. 32 He = E[2(n − p)] + E[1(p − p)]

p

n

3H 1

n

n

2 (n − p) + 1 (n − n)

p

3 He 2

p

2 (n − p) + 1 (p − p)

Difference between the two binding energy (∆B) is

∆B = B.E. 32 He − B.E. 31 H = {E[2(n − p)] + E[1(p − p)] − E[2(n − p)] − E[1(n − n)]} ⇒

= {E(p − p) − E(n − n)}

∆B = E(p − p)Coulomb + {E(p − p)nucl − E(n − n)}

(4.12)

2

Since ∆B is a measurable quantity and E(p − p)Coulomb = EC = 35 R eA1/3 can be estimated accuo rately. Therefore, E(p−p)nucl and E(n−n) can be examined. It is found that E(p−p)nucl −E(n−n) ≈ 0. This implies (n − n) interaction and (p − p)nucl interactions are same. ⇒ F (p − p)nucl ≈ F (n − n). Combining all these results, then it can be written as, F (p − p)nucl ≈ F (n − n) ≈ F (n − p)singlet

(4.13)

Relation (4.13) is more popularly known as Charge Independence of nuclear force which includes charge symmetry. n other words nuclear force is BLIND to the charge on the nucleons. Then to distinguish the nucleons n and p, a new quantity called isotopic spin or isospin is defined, such that nucleons differ only in their isospin (t) values. The degeneracy of isospin for nucleons is 2 - an isospin doublet, which is equals to (2 t + 1). ⇒

2t + 1 = 2

⇒

t = 1/2

Both the nucleons have isospin value of 1/2 with t3 = 1/2 for p and t3 = −1/2 for n. Where t3 is ~ or along z. When two nucleons the projection of t about the third axis reference line such as along B ~ ~ ~ ~ are coupled, then total isospin is T = t1 + t2 ⇒ T = 0, 1, which are isospin singlet and isospin triplet respectively. Isospin is a quantum number and it is a conserved quantity in strong interaction. Then Pauli’s principle is generalized for such non-identical particles. 33

4.1

Consequences of identical particles

(N N )

S = 0, T = 1 a fm

(pp) (nn) (np)

4

−17.1 ± 0.2 −16.6 ± 0.6 −23.7 ± 0.02

re

T3

S = 1, T = 0

fm

2.79 ± 0.02 2.84 ± 0.03 2.73 ± 0.03

PROTON-PROTON SCATTERING

a fm +1 −1 0

– – 5.42 ± 0.01

re

fm

– – 1.73 ± 0.02

Table 8: Scattering length ‘a’ and effective range ‘re ’ for (pp), (nn) and (np) systems obtained from scattering. Generalized Pauli’s principle Thi principle holds mainly non-identical particles like nucleons, n and p. The total wave function (ψtot ) must be antisymmetric for such fermions. This wave function is a product of thre wave functions, viz., spatial (φspace ), spin (χspin ) and (τisospin ) ψtot = φspace χspin τiso

(4.14)

where 1. (φspace ) is characterized by total angular momentum L (of 2 nucleons) such that ( 0, 2, 4 · · · − symmetric L= 1, 3, 5 · · · − antisymmetric 2. (χspin ) is characterized by total total intrinsic spin S such that ( ↑ ↑ − spin triplet − symmetric S= ↑↓ − spin singlet − antisymmetric 3. (τiso ) is characterized by total isospin T such that ( ↑ ↑ − isospin triplet − symmetric T = ↑↓ − isospin singlet − antisymmetric The triplet state is symmetric with respect to the exchange of the two nucleons whereas the singlet state is anti-symmetric. Since the total wavefunction of the system has to be anti-symmetric with respect to the exchange of two fermions, L + S + T should be odd, such that parity is π = (−1)L+S+T . Moreover the nucleon-nucleon interaction can bound a system of 2 nucleons only if they are in the L = 0 and S = 1 state. Therefore the generalised Pauli principle imposes T = 0 which is the iso-singlet state: only a (np) (deuteron) system is bound, and not the (nn) nor the (pp) one. This is corroborated with the scattering data for three types of nucleon pairs are as shown in the table 8. This table list (N N ) scattering length ‘a’ and effective range ‘re ’ for two nucleon process. Second and third columns are for spin singlet and isospin triplet and fifth and sixth columns are for S = 1 and T = 0 states respectively. Scattering lengths for (pp) and (nn) systems are nearly same and they differ slightly from (np) scattering system. This is because, in case of (pp) and (nn) scattering only neutral pions (π o ) are exchanged whereas in (np) scattering both (π o ) and (π ± ) are exchanged, which results in small difference in a value. In spin singlet states, the signs of all three scattering lengths are negative which implies no bound state in spin singlet or isospin triplet. In case of (np) scattering in isospin singlet state, the scattering length is positive and it is a bound state also its value is different form that for T = 1 state of (np) system, which is a clear indication of the isospin dependence of nuclear force.

34