Block 2 : Angular Momentum Spin and Scattering

M.Sc. Final (Physics)

Block – 2

Paper - I

UNIT 3 ANGULAR MOMENTUM

UNIT 4 SPIN

UNIT 5 SCATTERING THEORY Angular momentum, Spin and Scattering

UNIT 1 ANGULAR MOMENTUM

Structure

1.0 Introduction

1.1 Objectives

1.2 Eigen values and Eigen vectors of Angular Momentum

1.2.1 Angular momentum 1.2.2 Eigen values and Eigen vectors* of Lz

1.3 Characteristic Algebraic Relations 1.3.1 Commutation relations

2 1.4 Spectrum of J , Jz

2 1.5 Eigen Vectors of J and Jz

1.5.1 Total angular momentum 1.5.2 Ladder operators

1.6 Orbital Angular Momentum and Spherical Harmonics.

1.6.1 Operators in polar form

1.6.2 Eigen values and Eigen functions of Lz

1.6.3 To find Lx operator 1.6.4 Eigen values and Eigen functions of L2

1.7 Angular momentum and Rotations.

1.7.1 Commutation relations for Angular Momentum Operators

2 1.7.3 Eigen function and Eigen values for L

1.8 Rotation Operator

1.9 Rotational Invariance and Conservation of Angular Momentum

2 Angular momentum

1.10 Rotational Degeneracy

1.11 Let Us Sum up

1.12 Check Your Progress: The Key

1.0 INTRODUCTION

Angular momentum is one of the basic features of quantum mechanics. Its conservation is a universal phenomenon independent of the nature of the reference frame that allows calculation of energy values of the system under different conditions. In relativistic domain also it holds good. Rate of change of angular momentum gives torque on the system of particles. In the following discussions we will have exhaustive treatment for angular momentum operator, the operator in spherical polar coordinates and finding Eigen values and Eigen functions. Further we will notice that angular momentum is rotational analogue for the linear momentum.

1.1 OBJECTIVES

After interacting with the material presented here students will be able to i) Relate two kinds of angular momenta with a dynamic system ii) Write angular momentum operator in polar form iii) Use commutation relations iv) Understand features of ladder operators v) Calculate Eigen values for Lz and J vi) Relate angular momentum to the rotation of frame of reference

1.2 EIGEN VALUES AND EIGEN VECTORS OF ANGULAR MOMENTUM

1.2.1 Angular momentum

3 Angular momentum, Spin and Scattering

Angular momentum L for a particle of momentum p is defined through L = r x p (To understand it classically it is to be considered as moment of momentum). In its fully expanded form L = iLx + j Ly + kLz

= (i x + j y + k z) x (i px + j py + k pz)

= i (y pz – z py) + j(z px – x pz) + k(x py – y px) Here we have expressed vectors r, p and L in their components along the Cartesian axes. We have also used the orthogonal properties of vectors viz., i x i = j x j = k x k = 0 ; i x j = k, j x k = i, k x i = j and j x i= - k etc.

Let us concentrate on the z-component of angular momentum (equating coefficients of k) we have Lz = (x py – y px) We can express this relation in operator form by writing a cap (^ symbol) on each of the letter under consideration. Thus Lx represents an Lx operator. Operator form of L is L .

The other symbols can be Lz → etc., x → etc. and px → = - iħ /x etc. Lz x px We find

Lz = - iħ ( x /y - y /x) (1)

Similarly we can find the expressions for Lx and Ly . Hint: Use symmetry to write

抖 Lx= - ih( y - z ) (2) 抖z y 抖 and Ly= - ih( z - x ) (3) 抖x z It is found that Eigen functions for many practical systems when expressed in Cartesian coordinates are very cumbersome and complex in comparison to those expressed in cylindrical polar coordinates. Hence we will express a few occasionally used Eigen functions polar coordinates and then find corresponding Eigen values for them.

1.2.2 Eigen values and Eigen vectors* of Lz (*Eigen functions are also called Eigen vectors)

4 Angular momentum

The relation

Lz= - ih f indicates that the corresponding Eigen function could be  = (). Writing Eigen value equation as

Lz () = E() A trial solution could be () = Ae-im

-imf - im f  -ih (Ae ) = E Ae f Or - iħA(-im) e-im = Ae-im suggesting that E = mħ (4) We find in the following section that  = Ae-im

1.3 CHARACTERISTIC ALGEBRAIC RELATIONS

1.3.1 Commutation relations When two operators A and B are written as A B and B A we have an important relation

A B - B A = [ A , B ] (5)

The operator [ A , B ] is called commutation operator. Example 1 儋 To find what the operator [ ] represents? xpx- p x x We must remember that an operator without an operand hardly conveys any meaning.

儋 Hence we write [ ](x) to find what the operator is like. Writing these xpx- p x x operators in their usual form gives y (x ) y (x ) [x (iħ ) – (iħ )x] (x) = - [x (iħ ) – (iħ )x] x x x x

y (x ) y (x ) = - [x (iħ ) - iħ(x) - xi(x) ] x x = - iħ(x)

Or [x (iħ ) – (iħ )x] = - iħ x x

Corrolary [(iħ ) x - x(iħ )] = iħ (6) x x Check your progress-1 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit

i) Show that [ p,,x] = iħ 抖 Hint p= ih + j h + k h and terms like iħ x = 0 抖x y z y ------

2 ii) Find [x ,px] ------

Example 2 To show that the operator [x, d/dx] = -1 Let us operate this operator on , we have d d d d x  - x. = x  -  - x  = - dx dx dx dx

6 Angular momentum

d Remember x = 1 dx Thus [x, d/dx] = - qued. Can you show that [d/dx, x] = 1? Some commutation rules To solve commutation problems we need to remember following rules so that every time we need not worry to obtain relations from first principles. Those curious enough can easily verify those relations from definition (first principles). 1. [(A+B),C] = [A,C] + [B,C] 2. [AB,C] = [A,C]B + A[B,C] 3. [AB,CD] = A{[B,CD]} + {[A,CD]}B 4. [A,B] = - [B,A] (7)

2 1.4 EIGEN VALUES (OR SPECTRUM) OF JZ AND J

2 Let us assume that the simultaneous Eigen functions for J and Jz are (,m) such that J2 (,m) = ħ2(,m) and Jz (,m) = ħm(,m) to find values of  and m, we use ladder operators with some well known relations given below:

[J+,Jz] = - iħJ+

[J-,Jz] = ħJ-

[J+,J-] = - 2ħJz

† † and J+ = J- and J+ = J- (8) ( † is spelt as dagger and so the above quantities are spelt as J plus dagger equals J minus etc. Dagger represents complex conjugate of the quantity over which it is super scripted.)

We will also use

2 2 J+J- = J – Jz + ħJz

2 2 and J-J+ = J – Jz - ħJz (9)

(Would you like to verify these relations by actually substituting values of J+ and J-?)

1.5 EIGEN VECTORS OF J2 AND Jz

1.5.1 Total angular momentum Total angular momentum is defined as sum of the orbital component and spin component of angular momenta J = L + S J follows the same commutation rules as do L. if that is so S will also behave similarly. Thus

[Jx,Jy] = iħJz ; [Jy,Jz] = iħJx ; [Jz,Jx] = iħJy Example

Find [Jx, Jx]. Since [Jx, Jx] = [Jx, Jy] = Jx Jx,- JxJx it is identically zero.

We can show that Show that [Jy, Jy] = 0 and so is [Jz,Jz]

Check your progress-2 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) Prove that J x J = iħJ

Hint – Write J = iJx + jJy + kJz, then find J x J and find if the sum of all the three commutations produce the desired result. ------

2 ii) Prove that [Jx, J ] = 0 ------

8 Angular momentum

1.5.2 Ladder operators

The operators J+  (Jx + iJy) (10) and J-  (Jx - iJy) (11) are called ladder operators. This will become clear in the following sections when we operate those on appropriate Eigen functions. Let us find properties of tehse operators.

1. What is [Jz,J+]?

[Jz,J+] = JzJ+,-J+Jz

= Jz(Jx + iJy) - (Jx + iJy) Jz

= JzJx + i Jz Jy – JxJz - iJyJz

= JzJx – JxJz + i (Jz Jy – JyJz)

= [Jz,Jx] + i[Jz,Jy]

= iħJy + i(-iħJx) = ħ(Jx + iJy)

= iħJ+ (13)

We can also Show that [Jz,J-] = -ħJ- (14)

2 and that J commutes with J+ and J-

What is the value of [J+,J-]?

[J+,J-] = J+,J- - J-J+

= (Jx +iJy) (Jx - iJy) - (Jx -iJy) (Jx +iJy)

2 2 2 = Jx + iJyJx – iJxJy +Jy +iJyJx – JxJy – Jy

= i[Jy,Jx] + i[Jy,Jx] = 2i [Jy,Jx] (15)

Some important cases Case I Consider

2 2 2 2 (,J ) = (,Jx ) + (,Jy ) + (, Jz )

= (Jx, Jx) + (Jy, Jy) + (Jz, Jz)

 (Jz, Jz) Let us write (,m) then

2 ((,m),J (,m))  (Jz(,m), Jz(,m))

2 2 Or [(,m), ħ J (,m)]  [ħm Jz(,m) , ħm Jz (,m)] i.e. ħ2  ħ2m2 (16) Case II

[Jz,J+] = (JzJ+ - J+ Jz) = ħ J+ 

Or (JzJ+ ) = (J+ Jz) + ħ J+  Again writing (,m) we have

(JzJ+ ) (,m) = (J+ Jz) (,m) + ħ J+ (,m)

= J+mħ (,m) + ħ J+ (,m)

= ħ(m + 1) J+ (,m) (17)

This equation suggests that the Eigen function J+ (,m) is an Eigen vector of Jz, belonging to an Eigen value of (m + 1) ħ . It can be shown that J+ J+ (,m) is also an Eigen vector of Jz with an Eigen value of (m + 2) ħ. Important note- Eigen function and Eigen vector can be used interchangeably. Both convey the same meaning. Similarly Eigen value is nothing but energy value. Case III

2 2 2 [J ,J+] = J J+ - J+ J = 0

2 2 Or (J ,J+ )(,m) = (J+ J )(,m)

2 = J+ ħ (,m)

2 = ħ  J+ (,m) (18)

2 This equation suggests that J+ (, m) is an Eigen vector of J belonging to an Eigen value of ħ2. An Example

To find the maximum value of Jz

We note that J+ J+ J+ (, m) will have an Eigen value of (m+3) ħ for Jz and so on. Let  be the maximum Eigen value possible for a given value of ħ of J2, then

JzJ+(,m) = ( + 1) ħ J+(,m) cannot be expected. Hence

JzJ+(,m) = 0

Or J-J+(,m) = 0

2 2 But J-J+(,m) = J - Jz - ħ Jz

10 Angular momentum

2 2 Hence (J - Jz - ħ Jz) (,m) = 0

2 2 Or J (, m) - Jz (,m) - ħ Jz) (,m) = 0 Or (, m) - ħ22(,m) - ħ2 (,m) = 0 Or ( - ħ22 - ħ2) (,m) = 0 Or ( - ħ22 - ħ2) = 0 Whence  = - ½ + ½ (1 + 4) ½ (19) Case IV

[Jz,J-] = JzJ- - J- Jz = - ħ J-

So JzJ- = J- Jz - ħ J- and JzJ-(,m) = J- Jz (,m) - ħ J-(,m)

= J- ħm (,m) - ħ J- (,m)

Or JzJ-(,m) = (m-1) ħ J- (,m) (20)

This equation indicates that J- (,m) is an eigen vector of Jz belonging to an Eigen value of (m-1). Reiterative Example

JzJ-(,{m-1}) = (m-2) ħ J- (,m)

2 It can further be shown that the minimum value of Jz, belonging to J with the Eigen value of ħ2 is  = - ½ - ½ (1 + 4) ½ (21) An interesting task can be to Show that  = -  hence 2 must be an integer n.

Case V What is ?  +  = 2 = n = 1 + (1 + 4) ½ Or (n-1)2 = 1 + 4 Or  = (1/4)[(n-1)2 - 1] = [(n/2 – 1/2)2 – (1/2)2] n n Or  = (- 1) = J(J+1) where J = 0, 1/2, 3/2… 2 2

1.6 ORBITAL ANGULAR MOMENTUM AND SPHERICAL HARMONICS

1.6.1 Operators in polar form To express various operators in polar form we will use following set of relations: Set I x = r sin  cos  y = r sin  sin  z = r cos  (22) Set II r2 = x2 + y2 + z2 (23) Set III tan2  = [x2 +y2]/z2 (24) Set IV tan  = y/x (25) Now our task is to find r r r , , x y z q q q , , x y z f f f and , , x y z Once this is done their substitution in appropriate formulae would yield angular momentum operator in polar form.

Step A From r2 = x2 + y2 + z2 r 2r = 2x x r Or = x/r x r Or = r sin  cos  / r x

12 Angular momentum

r Or = sin  cos  (26) x It can be shown that r = sin  sin  (27) y r = cos  (28) z Step B From tan2  = [x2 +y2]/z2 we have 2 tan  sec2 [/x] = 2x/z2 q x Or = x z2 tan q sec 2 q q r sinq cos f Or = x z2 cos 2q tan q sec 2 q q cosq cos f Or = (29) x r Similarly q cosq sin f = (30) y r q cosq and = - (31) z r

Step C From tan  = y/x we find that f sec2 = - y/x2 x r sinq sin f = - r2 cos 2 f sin 2 q f r sinq sin f Or = - x r2 sec 2f cos 2 f sin 2 q

f cosecq sin f Or = - (32) x r Proceeding in a similar manner we find that fcos f cosec q = (33) y r f and = 0 (34) z

How to find ? x We know that x is a function of r, , and . In mathematical notation we express this fact as x = x(r, , ) Recognizing this fact we can write 抖r 抖q 抖 f = + + x 抖r x 抖q x 抖 f x r q f Substituting values of , and in this equation we get x x x

cosq cos f cosecq sec f tan f = [sin  cos ] + [ ] + [ - ] x r q r f r

cosq cos f cosecq sin f Or = sin  cos  + - (35) x r r q r f Students are expected to verify that 抖 cosq sin f cos f =sinq sin f + + (36) 抖y r rq r sin q f 抖 sinq and =cosq - (37) 抖z r r q

1.6.2 Eigen values and Eigen functions of Lz From 抖 Lz= - ih[ x - y ] 抖y x we have

14 Angular momentum

抖y y Lzy = - ih[ x - y ] 抖y x 抖yr 抖 y q 抖 y f 抖 y r 抖 y q 抖 y f = -ih[ x ( + + ) - y ( + + )] 抖r y 抖q y 抖 f y 抖 r x 抖 q x 抖 f x 抖y ycos q sin f y cos f cosec q = -ih[ r sinq cos f ( sin q sin f + + ) 抖rq r f r 抖ysin q cos f y cos q cos f y sin f cos ec q -r sinq sin f ( + - )] 抖r1 q r f r y Or Lzy = - ih f It suggests that

Lz= - ih (38) f 1.6.3 To find Lx operator In a similar fashion we can find 抖 Lx= - ih[sinf + cos q cos f ] (39) 抖q f 抖 and Ly= - ih[ - cosf + cot q sin f ] (40) 抖q f

Once we have determined all components of L we can find the operator L (use L = i Lx + j Ly + k Lz ) and also another important operator L2 using the fact that

L2 = Lx 2 + Ly2 + Lz2 抖 2 2 1 2 Or 2 = -ħ [- (r )] L r2 抖r r 2 2 1抖 1 Or 2 = -ħ [ (sinq )+ ] (41) L sinq抖 q q sin2 q q 2

Important Note-In all the above text and text that follows we are using only h line or h cross, which is equal to h (Planck’s constant) divided by 2. It is quite likely that print does not show it distinctly. In some cases the size of h line is not same due to unavoidable reasons but it conveys the same meaning.

Some conventions

In the above discussion we have written  = (). It is more convenient to follow some rules now onwards. Observe change in presentation that follows with use of  for  and  for  . Rule 1: A capital letter will denote function whereas a letter in lower case will represent variable. Thus  or more conveniently  will represent a function of variable . This fact can be expressed as    () or    () Rule 2: For brevity/convenience  () will be written as .

Can now you suggest what would R(r) represent?

1.6.4 Eigen values and Eigen functions of L2 Let us write

2 2 L2 Y(,) = ħ Y(,)  ħ Y (42) Using it for

2 2 1抖 1 2 = ħ [ (sinJ )+ ] L sinJ抖 J J sin 2 J J 2 We obtain 1抖 Y 12 Y (sinJ )+ + l Y = 0 (43) sinJ抖 J J sin 2 J j 2 We can solve this equation by separation of variables writing Y (,) = ()()    Substituting this in (23) we get 1抖 qf 1 2 qf (sinJ )+ + lqf = 0 sinJ抖 J J sin 2 J j 2

sin2J On multiplying it with we get qf

sin2J 1抖 q 1 2 f [ (sinJ )+ lq ] + = 0 qsin J抖 J J f j 2

sin2J 1抖 q 1 2 f Or [ (sinJ )+ lq ] = - = m2 (44) qsin J抖 J J f j 2 One simplest equation among these is

16 Angular momentum

1 2f - = m2 (45) f j 2 This on simplification reads 2f + m2  = 0 (46) j 2 Whose solution is  = Aeim (47) For the function to be single valued the necessary condition is ( +2) = () Which is equivalent to suggesting that ei2m = 1 This implies that m = 0;  1;  2;  3 etc.

What is the value of A? Let us apply normalization condition to the wave function . The normalization condition

2p being f* f d j = 0 0 2p A2 dj = 0 0 1 Or A = (48) 2p

Thus the Eigen function for Lz is 1  = eim (49) 2p Finding  In order to find  we solve the second equation of the relation (24) sin2J 1 抖 q [ (sinJ )+ lq ] = m2 qsin J抖 J J Let us write cos  = u and  () = F (u)  F

dF m2 We get [(1- u2 ) ] + [l - ]F = 0 (50) �u du 1 u 2 General case is clumsy to solve hence we take special case when m = 0. dF [(1- u2 ) ] +lF = 0 (51) u du To solve this equation we use power series for F(u) and write

a ur+ s F (u) = 0 r dF and = (r+s)a ur+ s-1 du 0 r 抖 dF so that [(1- u2 ) ] = [(1 - u 2 ) (r+s)a u r+ s-1 ] 抖u du u 0 r

抖ゥ r+ s-1 r + s+1 =[邋 (r+s)ar u ] - [ (r+s)a r u ] 抖u0 u 0

ゥ =[ (r+s)(r+s-1)a ur+ s-2 ] - [ (r+s)(r+s+1)a u r + s ] 邋0r 0 r So substituting these values in relation (31) we get

[(r+s)(r+s-1)a ur+ s-2 ]- {(r + s )( r + s + 1)a u r + s } +l a u r + s ] = 0 0 r r r (52) Since this equation must be valid for all values of u, coefficients of all powers of u must be independently zero. Collecting the coefficients of ur+s in the above equation we get

(r+s+2)(r+s+1) ar+2 - (r+s)(r+s+1) ar +ar = 0 (r+ s )( r + s + 1 - l ) Or a = a (53) r+2 (r+ s + 2)( r + s + 1) r

This suggests that the power series assumed for F would require its coefficients ar related by above equation. Check your progress-3 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit

i) Find the values of a3; a4 and a4

2 Hint- coefficient a2 is the coefficient of u , for which s = 0; a3 is coefficient of u3 etc. ------

18 Angular momentum

------

But we can have a satisfactory solution only if power series is convergent, i.e., it

r terminates after say aru , where r is some finite number.

r It can happen if there occurs no term after ar u . i.e., (r+s)(r+s+1) -  = 0 Again taking simplest case when s = 0, we have r(r+1) =  the solution of the equation with above condition is called Legendre polynomial Pl(u), with the condition that Pl(1) = 1. Thus

 = B Pl (cos ) The constant B is again obtained from normalization condition for the wave function. Its value is obtained as (should you do the labour of calculating it? Try) 2l + 1 B = [ ]1/2 (54) 2 So that 2l + 1 q= [ ]1/2 P (cos J ) (55) 2 l and the Eigen function for L2 is 1 2l + 1 Y(,) = eim [ ]1/2 P (cosJ ) (56) 2p 2 l In case m were not zero we find that

L2 Yl,m (,) = l(l+1) L2 Yl,m (,)

Where l(l+1) represents Eigen values of L2 and the corresponding Eigen function is Yl,m (,)

Note that for a given value of l, the eigen value is (2l+1) fold degenerate. Since we can associate m = -l; = - l+1; -l+2 etc. with it. Note When a function yields same Eigen value of more than one Eigen functions, the Eigen functions are said to be degenerate

1.7 ANGULAR MOMENTUM AND ROTATIONS

An important fact: Angular momentum is related to rotation of a system It is known that the transformation relations for a rotated frame about z-axis by an  are given by x = x cos  + y sin  y = y cos  - x sin  z = z (57) when   0, we can write cos  = 1 and sin  = 0. Under this limiting conditions x = x + y  y = y - x  and z = z (58)

1.7.1 Commutation relations for angular momentum operators

(1) To find [Lx , x]

Since Lx = (rxp)x = ypz – zpy here subscript suggests that the component along that direction is considered. Thus py is component of p along y direction.

Writing Lx in operator form 儋抖 L = y(- ih ) - z ( - i h ) x 抖z y

But [Lx,x] = Lxx - xLx hence 儋抖儋抖 [Lx,x] = { y(- ih ) - z ( - i h ) } - { y(- ih ) - z ( - i h ) } 抖z y x x 抖z y

儋抖 [Lx,x] = - { y(- ih ) - z ( - i h ) } (59) x 抖z y

20 Angular momentum

As the first bracket is zero because it is either differentiation of x with respect to z or with respect to y.

What is [Lx,x] (x)? It is zero since (x) is function of x and not y or z.

Suggest what should be [Lx,x] (y)?

(2) Rule for [Lx,y] 儋抖儋抖 We have [Lx,y] = { y(- ih ) - z ( - i h ) } - { y(- ih ) - z ( - i h ) } 抖z y y y 抖z y y Second term is meaningless. First term gives 0 + iħz . Hence [L ,y] = iħz. y x It can be shown that

1. [Lx,z] = -iħy

2. [Lx, px] = 0

3. [Lx, py] = iħpz

(3) Let us find [Lx,Ly ]

[Lx,Ly ] = LxLy - Ly Lx

= (ypz-zpy)(zpx-xpz) – (zpx-xpz)(ypz-zpy)

= ypzzpx - zpy zpx - ypzxpz +zpxzpy+zpxzpy - xpzzpy

= y(-iħ) px - 0 -0 -0 -0 -0 -0 - x(-iħ) py

= (-iħ) ( ypx - x py ) = iħ (xpy - ypx)

= iħ Lz (60) Remember we have used the fact that perfect differentials along an axis do not operate on other coordinates along the same axis viz. pzy = 0; pzx = 0 and pzz = - iħ, etc. Proceeding in a similar fashion we can show that

1. [Ly, Lz] = -iħ Lx

2. [Lx, Lx] = 0

2 (4) To find commutation relation for [L ,Lx]

2 2 2 2 [L ,Lx] = [{[L x+L y+L Lz]} ,Lx]

2 2 2 = [ L x ,Lx] + [L y ,Lx] + [L z ,Lx]

2 Let us first compute [L x ,Lx]

2 [L x ,Lx] = Lx [Lx ,Lx] - [Lx, Lx]Lx

= 0 + 0

2 2 Similarly we can show that [L y ,Lx] and [L z ,Lx] are also identically zero.

2 Hence [L ,Lx] = 0

2 2 2 2 Corollary [Lx , L ] = 0 and hence [L ,L ] commute.

The Hermitian operator

† An operator is said to be Hermitian if it is self-conjugate. That is if A = A . Thus for Hermitian operator (A†,) = (,A). Example

2 The operators J , Jx, Jy, Jz are Hermitian. Try to verify.

Schwartz inequality It says that a scalar product of a vector by itself is always greater than or equal to zero. Let ( + ) denote a vector then ( + )* . ( + )  0 (61) Do you know? In bra-ket notation it is written as <( + ) ( + ) > 0

Here  is a number. Note - We may also note that the product is done with the vectors after taking its transpose (i.e., after changing row to column and vice versa). This is necessary as the vector is represented as a column or a row matrix. Hence vector multiplication must follow rules of matrix multiplication.

Eigen function for Lz would be that which satisfies

Lz = c

Or - i ħ = c j So that  = A eic/ħ Since  has to be single valued, if we replace  with + 2, the wave function would not change.

22 Angular momentum i.e., A eic/ħ = A eic(+2)/ħ Or eic2/ħ = 1 or 2c/ħ = 2m Or c = m ħ So that  = A eic/ħ is its Eigen function with an Eigen value of m ħ. Example

To find the Eigen value of (Lx iLy) From

LyLz = i ħ Lx + LzLy and LzLy = - i ħ Lx + LzLy we have Lz ( Lx iLy) = Lz Lx  iLzLy

= Lz Lx  i [- i ħ Lx + LzLy]

=  ħ (Lx iLy) + Lz(Lx  iLy)

= (Lx  iLy)(Lz ħ)

So Lz ( Lx iLy) (m) = (Lx  iLy)(Lz ħ) (m)

But (Lz ħ) (m) = (m  1) ħ (m)

Hence Lz ( Lx iLy) (m) = (m  1) ħ (Lx  iLy) (m) (62)

Thus if (m) is a wave function of Lz with an Eigen value of m ħ, then (Lx  iLy) (m) are also Eigen functions of Lz with Eigen values of (m  1) ħ.

Check your progress-4 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit

i) What are the Eigen values of Lz for Eigen vector (Lx + iLy) (m)

and (Lx - iLy) (m)? ------

1.7.3 Eigen function and Eigen values for L2

Let  represent the largest Eigen value for Lz when operated on (Lx + iLy) and  its maximum negative value for Lz when operated upon (Lx - iLy). Then

Lz(Lx + iLy) () = 0 and Lz (Lx - iLy) () = 0 Consider

2 2 2 2 L () = (Lx + Ly + Lz ) ()

2 = [Lz +(Lx - iLy) (Lx + iLy) – i(LxLy – Ly Lx)]()

2 = [Lz +(Lx - iLy) (Lx + iLy) – i.i Lz]()

2 2 = [ ħ - (Lx - iLy).0 + ħ] () Or L2() = ( + 1) ħ2 () (63) Similarly we can obtain L2() = ( -1) ħ2 () (64) Equations (53) and (54) must be valid at all the time hence ( + 1) must be equal to ( - 1) Two possible situations would be  =  and ( + 1) =  We would disregard second solution as we have assumed that  cannot have larger value than . Hence we take  = , to be the acceptable result. Let us call this value of  as l. then the Eigen value of L2 are l(l+1) ħ2 . What is the magnitude for the Eigen value of L? We can claim that it must be square root of the Eigen value for L2. Recall that quantum number related to L is written as  l(l+1)

1.8 ROTATION OPERATOR

If Rz() represents an operator corresponding to an infinitesimal rotation  about z-axis then

Rz()(x,y,z) = (x,y,z)

24 Angular momentum

= (x+y,y-x,z) Now by Taylor expansion it can be written as

= (x,y,z) +y (x,y,z) - x (x,y,z) x y

Or R ()(x,y,z) = (1 + [y - x ] )(x,y,z) z x y q = [1 + Lz] (x,y,z) ih It suggests that q Rz() = [1 + Lz] (65) ih And similarly we can show that q Rx() = [1 + Lx] ih q and Ry() = [1 + Ly] (66) ih

In other words Lx, Ly and Lz are the generator of rotations about z-axis.

1.8 ROTATIONAL INVARIANCE AND CONSERVATION OF ANGULAR MOMENTUM

Invariance dictates no change under some operation, and conservation dictates commutation with energy operator. Rotational invariance is the property of a system such that after undergoing rotation, the new system still obeys Schrodinger equation. Thus for any rotation R, the rotation operator and energy operator commutes. i.e. [R, (E-H)] = 0 Since rotation does not depend explicitly on the time, it commutes with energy operator. Thus [R, H] = 0 here H represents energy operator or the Hamiltonian of the system. Let the system be rotated (say in x-y plane) by an infinitesimal angle d then the rotation operator is

R = 1 + Jz d

d Then from [R, (E-H)] = 0 we find that [(1 + J d ), ] = 0 z dt d Or (Jz) = 0 (67) dt It suggests that angular momentum in such rotations is conserved.

1.10. ROTATIONAL DEGENERACY

Let us have a further infinitesimal rotation of , then we will have

Rz ( +) = Rz(). Rz() q = [1 + Lx] Rz() ih

dRz (q )lim. R z ( q+ D q ) - R z ( q ) i = = - Lz 禗q q 瓺 0 q h To give

-iLz/ħ Rz () = e (68) Example To find a rotation in spherical polar coordinate system For this we will put r = r;  = ; and  =  -  so that

Rz()(r,,) = (r,,-) q y Or [1 - i Lz] (r,,) = (r,,) -  ih f Comparing the two sides we obtain

Lz = - ih f Degenerate functions Let us find out what happens if rotation about z-axis totals 2? This new state is indistinguishable from the state with no rotation. Such state then belongs to the same energy, representing degenerate functions.

1.11 LET US SUM UP

26 Angular momentum

 Angular momentum is one of the basic properties of nature. Among six universal conservation laws, it has lot of significance in physical world. Essentially we have two kinds of angular momenta namely orbital or rotational angular momentum denoted by L and other is spin angular momentum denoted by S. The total of the two LS is denoted by J. for a physical system it has to be positive. J also determines total energy of the system.  Quantum mechanics being the most reliable description of nature we apply it in solving the physical problems. Solving physical problems entails finding an appropriate Eigen function for the system and then solving it for finding Eigen values related to those functions.  Solving physical problems easy when we use angular momentum in operator form. The mathematic of operators is facilitated by commutation rules. Increase in energy (excitation) and decrease in energy (de-excitation) can be computed by so called ladder operators.  Rotation of frame of reference, which is equivalent to change in angular momentum.

1.12 CHECK YOUR PROGRESS: THE KEY

1. i) [p,x] = [(ipx + jpy + kpz), x] = [ipx,x] + [jpy,x] + [kpz,x] = iħ + 0+ 0

2 ii) (x , px) = x(x, px) + (x, px) x. Substitute values of commutation brackets

namely [ipx,x] = iħ and (x, px) = - iħ for result. 2. i) J x J = i(JyJz-JzJy) + j(JzJx-JxJz) + k(JxJy-JyJx) = i[Jy,Jz] + j[Jz,Jx] + k[Jx,Jy] = i(iħJx) + j(iħJy) + k(iħJz) = iħ(iJx + jJy + kJz) = iħJ. Remember i denotes square root of minus one whereas i represents unit vector along x-axis. ii) [Jx, J2] = [Jx,(Jx2 + Jy2 +Jz2] = [Jx,Jx2] + [Jx,Jy2] + [Jx, Jz2] = 0

(r+ s )( r + s + 1 - l ) 3. From a = a for a put s=0 and r=1to get (2-)/6 r+2 (r+ s + 2)( r + s + 1) r 3

similarly a4 = 2(3-)/12 etc. 5. The values are (m+1) ħ and (m-1)ħ by using equation 62.

UNIT 2 SPIN

Structure 2.0 Introduction 2.1 Objectives 2.2 Electron Spin 2.3 Spin ½ and Pauli matrices 2.3.1 Spin and spin operator 2.3.2 The spin matrices 2.3.3 The Pauli spin matrices 2.4 Observable and Wave functions of Spin ½ particles 2.4.1 Spin matrices and Eigen functions 2.4.2 Electron spin functions 2.4.3 Electron spin functions 2.4.4 Further Pauli spin matrices 2.5 Spin of Fields 2.6 Vector Fields and Particles of spin 1 2.6.1 Vector fields and particles 2.6.2 Spin matrices as operators 2.7 Spin Independent Interactions of Atoms 2.8 Spin Independent Nucleon-Nucleon Interactions 2.8.1 Isospin

28 Angular momentum

2.9 Let Us Sum Up 2.10 Check Your Progress: The Key

2.0 INTRODUCTION

To trace the history of an intensive study of any topic in Physics we find that it was the spectrum of hydrogen atom. Hydrogen atom consists of a proton and an electron located near to it in some permitted orbit. The first excited state lies about 10.4 eV above the ground state. But it was found that even the ground state was not singlet (?). In order to explain this, electrons and protons were assumed to possess spin. This spin was assumed to be responsible for hyperfine splitting of energy levels. Spin states were to be either spin up or spin down to accommodate experimental data. With the development of quantum mechanics various aspects of spin were studied. This unit will enable students to appreciate spin functions, spin matrices and discussion of a few spin half systems of particles.

2.1 OBJECTIVES This learning material is intended to acquaint the students with i) Concept of spin and spin operators ii) Spin Eigen functions and Eigen values iii) Pauli spin matrices iv) Spin operators v) Some examples related to spin of particles

2.2 ELECTRON SPIN 2.2.1 Need to have a spin If electron were a spin less particle and only described an orbital motion about the nucleus, its magnetic dipole moment would have been e m = - L 2m0 where m0 is the mass of the electron. This magnetic moment then should be the source of permanent magnetism of the ferromagnetic substance. But the observed magnetic

e e moment of a magnetic specimen gives a factor and not . This has been m0 2m0

29 Angular momentum, Spin and Scattering attributed partially to gyro magnetic effect. Further for S (l= 0) state atom should have no magnetism. This is again against the observed fact. Fine structure of the spectrum is also inconsistent with the orbital angular momentum concept. Sodium D lines are well known example. Fine structure is possible only if there were additional energy levels. Zeeman and Paschen-Back effects too cannot be explained without attributing spin to electron. For S state there should be no splitting of atomic beam in Stern-Gerlach experiment, but this also goes against the experimental observations. To explain all above anomalies Uhlenback and Goudsmit postulated that electron possesses an intrinsic angular momentum. He named it electron spin, which for brevity we simply call spin. Ultimately it was noted that protons and even neutrons possess spin.

2.3 SPIN ½ AND PAULI MATRICES 2.3.1 Spin and spin operator The spin associated with an electron is quantized. For explaining doublet fine structure of alkali atoms it was sufficient to ascribe a spin of ½ ħ to the electron. Associated spin quantum number, s has only one value for the electron namely ½. We describe the quantum state of electron by ħ. Assuming  to be related to quantum number s, the multiplicity of energy levels would be (2s + 1). According to Stern-Gerlach the atomic beam with S = 0, splits into two beams, i.e., the multiplicity is 2. It suggests that s = ½.

Note- S is total spin magnetic quantum number and is given by S = s1  s2 s3 ..Where s1,s2,s2, are spins component electrons of the system or the state. Incidentally all s have the same value of ½. Remember whenever we talk of angular momentum we add ħ with the quantum number. It is also usual to write angular momentum in units of ħ. In that case again ħ is omitted. Thus we conclude that each electron in a given orbit provides a spin contribution of ½ ħ to the angular momentum. Spin operator S represents spin angular momentum operator. Its components are 3x3 spin matrices. 骣0 0 0 琪 Sx= ih琪0 0 - 1 琪桫0 1 0 骣0 0 1 琪 Sy= ih琪0 0 0 琪桫-1 0 0

30 Angular momentum

骣0- 1 0 琪 and Sz= ih琪1 0 0 (1) 琪桫0 0 0 S commutes with L but not with Hamiltonian H. We must note here that (L + S) together commute with H. and hence (L+S) must be a constant of motion not L or S independently. Let us find S2 from above relation We know that S2 = Sx2 + Sy2 + Sz2 Substituting matrix values in the above we get S2= 2ħ2. It may be noted that in general L2 is not a constant of motion. But S2 is constant of motion and commutes with all the dynamical variables. While computing the energy values S2 can be replaced by s(s+1) ħ2. Here s is called an intrinsic spin angular momentum whose value is ½. Important Note- in this and following discussions we have used ħ, h line (or h cross) everywhere whose value is Planck’s constant divided by 2. It is quite likely due to prints in various fonts it may not become explicitly clear. 2.3.2 The spin matrices

(sx), ( sy) and (sz) are called spin matrices and in the matrix form are represented as 骣0 1 0 h 琪 sx = 琪 1 0 1 2 琪桫0 1 0 骣0-i 0 h 琪 sy =琪i 0 - i 2 琪桫0-i 0 骣1 0 0 h 琪 sz = 琪 0 0 0 (2) 2 琪桫0 0 1 骣1 0 0 2 2 琪 So that s = 2h 琪 0 1 0 (3) 琪桫0 0 1 The necessary condition is that S is not expressed in terms of r and p. Examples  Mesons are spin zero; electrons, protons, neutrons, neutrinos and  mesons are spin half; and photons and gravitons are spin one particles. Try to find what spin a graviton possesses.

If we place a spin half particle in a uniform magnetic field B for a time , its probability

i -mzB - amplitude becomes e h t times what it would be in no field. Here z is either + or – of some number. If the particle is not in pure spin up or spin down state we can describe its condition in terms of its amplitude to be both in the pure up or pure down state. But in a magnetic field those states will have phases changing at different rates. Hence on what time and how long the particle is in the field matters most. Aggregates of particles that are sufficiently tightly bound can be regarded as a composite particle. It can then be characterized by definite magnitudes of their total internal momentum as long as their internal motions and the relative spin orientations of their component particles are not significantly affected by the interactions between the aggregates. The spin of the aggregate is found thus by observing the triangle rule of vector addition: the magnitude of the sum of two angular momentum vectors can have any value ranging from the sum of their magnitudes to the difference of their magnitude, by integer steps. It also indicates that the sum of the z-component of the angular momentum equals that of their resultant. We can show this by applying it to linear combination of products of Eigen states of two commuting angular momentum operators that are Eigen states of total angular momentum. The same kind of addition formula holds for products of rotation or tensor operators and for the states produced when a tensor operator acts on an angular momentum Eigen state. Thus the spin of aggregates (of n particles) each of which has spin half plus any number of particles with spin zero, ignoring internal angular momentum of these particles. The value of s can be from 0 to ½ n. Total orbital angular momentum however can be zero or some integer in general. It can be a good exercise to see what kind of value results if spin and orbital angular momenta are added together. Aggregate of particles that have zero or integral spin are described by symmetric wave functions. Such an aggregate obeys Bose-Einstein statistics. Particles that have half or half of odd integral spins are described by anti-symmetric wave functions. Such particles obey Fermi-Dirac statistics. 2.3.3 The Pauli spin matrices The two state system of electron spin is very useful and hence is conveniently represented as sigma matrices. Anybody working in quantum mechanics needs to memorize these popularly known as Pauli spin matrices. 骣0 1 骣0 -i 骣1 0 x = 琪 ; y = 琪 ; z = 琪 (4) 桫1 0 桫i 0 桫0 1

32 Angular momentum

Product of these matrices taken two at a time also finds use in the analysis of interactions. The three matrices are further analogous to the three components of a vector called sigma () vector

2.4 OBSERVABLE AND WAVE FUNCTIONS OF SPIN ½ PARTICLES 2.4.1 Spin matrices and Eigen functions The spin coordinates take on 2s + 1 values for a particle of spin s. A convenient set of ortho-normal one particle spin functions is provided by the normalized Eigen functions of the J2 and Jz matrices. These Eigen functions are (2s + 1) row and one-column matrices that have all elements zero except one. Example What are the spin functions for s = 3/2? The four spin Eigen functions are 1 0 0 0 0 1 0 0 u(3 / 2) = ; u(1/ 2) = ; u(- 1/ 2) = ; u(- 3 / 2) = (5) 0 0 1 0 0 0 0 1 The corresponding Eigen values are 3ħ/2, ħ/2, -ħ/1 and -3ħ/2. Here we have used double lines to indicate that these denote matrices and not determinants. From simple matrix-matrix multiplication rule one can verify the orthonormality by finding that u1*u1 = 1; u1*u2 = 0 etc. It can be shown that thee are a total of (s+1)(2s+1) symmetric and 2(2s+1) anti- symmetric states for two-particle system with same spin. Check your progress-1 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit

iii) What will be u2* u2 and u2*u4? Assume u1  u(3/2), u2  u(1/2),…. ------iv) How many anti-symmetric spin states are possible with above four u functions? ------

2.4.2 Electron spin functions For brevity the electron spin matrices can be written as S = ½ ħ where

骣0 1 骣0 -i 骣1 0 x = 琪 ; y = 琪 ; z = 琪桫1 0 桫i 0 桫0 1 These matrices are called Pauli spin matrices. For these matrices the normalized Eigen functions are written as 1 0 u(1/2) = ; u(-1/2) = (6) 0 1 these have Eigen value of ½ ħ and -½ ħ respectively. As a corollary it can be shown that the Pauli matrices are both Eigen functions of S2 and have same Eigen value of ¾ ħ2. 2.4.3 Electron spin functions The spin matrices can be written as S = ½ ħ, where  are given be equation (5). The normalized Eigen functions of S 1 0 u(1/2) = ; u(-1/2) = 0 1 are abbreviated as (+) and (-) respectively. So hereafter when we write (+), it would imply that we are concerned with u(1/2) state and when we write (-) it is for u(-1/2) state with Eigen values of ½ ħ and -½ ħ respectively. If we take two electrons system we can have four independent wave functions, viz., (+ +), (+ -), (- +) and (- -). These Eigen functions can be regrouped in a combination that are Eigen functions of

2 (S1 + S2) and S1z and S2z. Check your progress-2 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) xy = - yx = i. ------ii) x(+) = (-); y(+) = i(-) ; z(+) =(+); x(-) = (+) ; y(-) = -i(+) ; z(-) = -(-). ------iii)How would you verify the orthogonality of wave functions for two electrons? ------

34 Angular momentum

iv) How would you write Pauli spin matrix as a 4x4 matrix? ------

2.4.4 Further Pauli spin matrices The two state system of the electron spin is so important that it is very useful to have a neat mode of presenting them. We know that the Hamiltonian for the electron in a magnetic field is

Hij = - ( ijBx + ij By + ij Bz) (7) If we take the magnetic field along z-direction, then

11 = 1; 12 = 0; 21 = 0; and 22 = -1 This gives

骣H11 H 12 Hij = 琪 (8) 桫H21 H 22 For electron in a magnetic field Bz we have

骣 -mBz - m( Bx - iBy ) Hij = 琪 (9) 桫-m(Bx + iBy ) m Bz Terms for x-direction are found to be

11 = 0; 12 = 1; 21 = 1; and 22 = 0 In shorthand notations ’s are represented as in equation (5). Anyone pursuing quantum mechanics need to memorize those matrices.

2.5 SPIN OF FIELDS The spins of electrons and protons are considered to be connected with their moments of motion. Spin is considered to be just a quantum mechanical quantity. Yet we can assume that spin has some physical significance. If in a substance the spins of particles have a preferred direction, then this is interpreted as spin polarization of the substance. Every substance creates a spin field in the space surrounding it when polarized by spins. (This field is also called ‘axion field’). Let us examine this fact: the electric charge manifests as the electric field in the space surrounding it; the magnetic moment manifests itself as magnetic field. From the analogy we can expect that the manifestation of spin result in a hypothetical spin field. The spin of elementary particles can be considered as a source for the spin field. It can safely be assumed that a spin field is produced by spin polarization, i.e., selective orientation of spins in space. The simplest way to achieve a selective spin orientation is through the mechanical rotation of the objects. On such rotations spins are oriented along

35 Angular momentum, Spin and Scattering the axis of rotation. You can seek the details of such a spin field generators from different websites.

2.6 VECTOR FIELDS AND PARTICLES OF SPIN 1 A field is a physical quantity, which takes on different values at different points in space. In other words field is a mathematical function of position and time. For vector fields we can visualize by drawing various vectors at various points in space, each of which gives the field strength and direction of the field at that point.

Vector field is characterized by some amount of a physical quantity called flux, coming in or going out from the space.

2.6.1 Vector fields and particles In the study of elementary particles we also discuss a small family of particles that do not come under leptons, mesons or baryons. These are field particles that are responsible for carrying the forces in order to enable particle interactions. We now believe that inter particle interaction is essentially through the field they establish (the concept of action at a distance was thus replaced by the notion of a field). According to it one particle sets up a field around it and the other particle interacts with the field when it comes under its influence. The quantum field theory further modifies this notion by announcing that the fields are carried by a quantum related to that field. In this point of view first particle emits a quantum the second particle responds to the field then by absorbing the quantum. A force accomplished through the exchange of particles is called an exchange force. Check your progress-3 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) A system of elementary particles was found to have all spin magnetic moments along z-axis. What kind of property must be associated with the space about this system? ------ii) If action at a distance were the central idea for field concept, list at least two fields known to you.

36 Angular momentum

------

The particles responsible for some known fields are summarized below. Interaction force Field particle symbol charge spin Gravitation graviton - 0 2 + - Weak weak bosons W ,W ,Z0  1, 0 1 Electromagnetic photons  0 1 Strong gluon g 0 1

Can you tell which force/s among those listed in the table do not produce a vector field?

2.6.2 Spin matrices as operators We have suggested that a column matrix and a row matrix well represent a vector. Hence we will discuss about matrices associated with the spins to get further insight into this phenomenon.

Let us find  + > s z By multiplying it with < +  , we get

< +   + > = 11 = 1 s z Now let us see the effect of multiplying it by < -  from left

< -   + > = 21 = 0 s z From above two equations we get a clue that if we have a relations like < +  ? > = 1 and < - ? > = 0 There can be only one state vector  + > that satisfies both the relations. It suggests that  + > =  + > (10) s z Example

To find  + > s y s z We have  + > = (i ->) s y s z s x = i ( ->) s x = i  + > Properties of s operators ( +>) = - -> s x ( ->) = +> s x

 + > = i  - > s y  - > - i  - > s y  + > =  + > s z  - > = -  - > (11) s z One can easily show that Prove that = i s x s y s z

2.7 SPIN INDEPENDENT INTERACTIONS Collision of identical particles each one having (2s+1) Eigen functions produce (2s+1)2 independent spin functions for each of the pairs. Any of (2s+1)2 linearly independent combinations of these products can be used to describe them. However they are divided in to three classes. First class consists of one particle spin function in which both particles are in the same spin state. Example: u1.u1 or u2.u2 etc. Second class consists of sum of the products. Obviously these two classes correspond to symmetric functions in which interchange of spins between two particles has no effect on the system. Example: u1.u2 or u3.u4. The third class has differences of products. It produces an anti-symmetric state. Example: u1.u4, u2.u3.

2.8 SPIN INDEPENDENT NUCLEON-NUCLEON INTERACTIONS 2.8.1 Isospin Neutrons and protons in a nucleus possess some mass and are bound in the nucleus with a strong force. They differ in that – one is charged and the other is a neutral particle. Heisenberg suggested that like an electron, a nucleon also possesses spin half magnetic moment. Spin half nucleons also then occupy two states: the isotopic spin up (protons) and isotopic spin down (neutrons) states. Note that just like electron states cannot be identified in the absence of magnetic field (spin up or spin down) so is the case with nucleons. We cannot assess in which spin state a nucleon lies in the absence of a magnetic field. Nucleon is an isospin particle with nuclear spin I = ½ ħ., the Eigen values for protons and neutrons are + ½ and – ½ respectively. The Pauli matrices for nucleons are represented as I = ½ . Those are given below-

38 Angular momentum

骣0 1 骣0 -i 骣1 0 1 = 琪 ; 2 = 琪 ; 3 = 琪 (12) 桫1 0 桫i 0 桫0- 1 These also follow commutation relations as shown below-

[1 , 2 ] = 2i3 etc. (13) p 1 0 We can represent a nucleon as (nucleon) > = = p + n (14) n 0 1

Example 1 0  ½ ½ > = refers to proton and  ½ -½ > = refers to a neutron. 0 1

Can you suggest that if the Eigen function for isospin are  Iml> than Iz  Iml> = ml  Iml> 骣1 0 Hint : Iz = 3 = 琪 ,operate Eigen function with it. Alternatively in bra-ket 桫0- 1 notation the Eigen value ml is contained in the ket.

2.9 LET US SUM UP

 For long the fine structure of spectral lines was debated. With the availability of more and more sophisticated instruments fine structure was established beyond doubt. In the initial period there was no successful and plausible explanation. Uhlenback and Goudsmit tried with a solution giving a revolutionary idea without proof that electrons possessed spin. They claimed that like any celestial body electron also spin about its own axis. Though it is still under debate whether electron spins, yet has been established beyond doubt that electron possesses a spin magnetic moment that is responsible for the splitting of energy levels.

 Once the concept of spin was established all treatment of new theory was obvious. In the effort to find functions and energy values electron spin functions and related matrices were defined. The most important result was the explanation of action at a distance.

 The essentials of action at a distance lied in the spin field that enabled to have field particle interaction. Iso-spin concept enabled to find functions and energy levels for collection of Fermi particles.

2.10 CHECK YOUR PROGRESS :THE KEY

1. i) Substitution of u2* and u2 gives u2* u2 = 1 and similarly substritution of u2*

and u4 gives u2*u4 = 0

ii) For four functions: symmetric functions are u1u1, u2u2, u3u3, u4u4, u1u2, u3 u4.

Other combinations produce anti-symmetric namely u1u3, u1u2-u2u3, etc. 骣0 1 2. i) s s = i s is verified by multiplying s matrix 琪 with s matrix x y z x 桫1 0 y 骣0 -i 骣1 0 琪 . The resultant matrix is i.琪 . Which iss matrix multiplied by i. 桫i 0 桫0 1 z

Similarly changing the order verifies the other part. 骣0 1 1 0 骣0 1 0 1 ii) x. (+) = 琪 = similarly x. (-) = 琪 = , 桫1 0 0 1 桫1 0 1 0 other identities are also verified by applying matrix multiplication rules for corresponding matrices and identifying the resulting matrix. iii) Orthonormality of function is verified by showing that when one function in matrix form multiplied by complex conjugate of other matrix, the result is a null matrix. 骣s 0 iv) Pauli matrix as 4x4 matrix is written as 琪 where  represent simple 桫0 s

骣0 0 2x2 matrix and 0 represents 2x2 null matrix琪 . 桫0 0 3. i) When magnetic moments are aligned it says all spins of the particles are aligned.

ii) Gravitational and Coulombian fields are examples of action at a distance.

40 Angular momentum

UNIT 3 SCATTERING THEORY

Structure 3.0 Introduction 3.1 Objectives 3.2 Definition of Scattering Cross Section 3.2.1 Scattering 3.2.2 Scattering Cross Section 3.3 Stationary Wave Scattering 3.3.1 Collision of identical particles 3.4 Representation of Scattering phenomenon by a bundle of Wave packets 3.5 Scattering of a Wave packet by a Potential 3.6 Calculation of Cross section 3.7 Laboratory system and Centre of Mass system 3.7.1 Connection between center of mass and laboratory system 3.7.2 Relation between scattering angle in the two coordinate systems 3.8 Scattering by a Controlled potential wave: Analysis and Phase Shift method 3.8.1 Scattering by a square-well potential 3.8.2 Phase shift in scattering by a spherically symmetric potential 3.9 Impact Parameter 3.10 Relation between Phase shift and Logarithmic derivative 3.10.1 Effect of spin 3.11 Behaviour of Phase Shift at Low energies 3.11.1 Phase shifts at low energies 3.12 Scattering by a Hard Sphere

3.13 Let Us Sum Up 3.14 Check Your Progress: The Key

3.0 INTRODUCTION

When a stream of particles (called projectiles) strike randomly oriented objects then an incident individual particle is sent in any of the all-possible directions, by the randomly oriented objects (called target). Collectively speaking different particles of incident beam are sent in different directions. However the sum of particles sent by objects of the target is same as the number of incident particles. But in the process of sending projectiles back in different direction (usually called scattering), the phases of particles also undergo a change. The phase change matters most when we discuss situations in microscopic domains. To understand the scattering we first undertake two-particle collision and then develop a consistent theory for the beam of the particles. Ultimately we will take up different systems for scattering with fully quantum mechanical aspect. It is only after the advent of quantum mechanics scattering of micro-particles (like alpha and other sub atomic particle), the probe into the atoms and even nucleus was that led to the better understanding of the matter. Since the properties of the scatterer are reflected in the scattering process scattering experiments has become an important method now for the study of atoms, molecules, nuclei and fundamental particles.

3.1 OBJECTIVES After studying this unit students will be able to i) Understand two body dynamical problems ii) Understand scattering process iii) Know the factors on which scattered beam intensity depends and hence will be able to estimate the scattering cross section iv) Recall scattering under different conditions v) Find relationship between wave associated with projectile and size of the scattering atoms.

42 Angular momentum

3.2 DEFINITION OF SCATTERING CROSS-SECTION 3.2.1 Scattering Scattering is a phenomenon that takes place when a collimated beam of particles with well defined energies and properties (called projectiles), collide a target of atoms and ultimately shatter in all possible directions. When we talk of miniscule matter (refers to electrons, atoms, nuclei and so on) moving with relativistic speed we dare to associate with it a wave. In scattering, wave functions associated with the particles do not vanish at large distances from the scatterer. In short we consider only non-relativistic potential scattering of particles.

Scattering experiments are the standard means of investigation in to structures and inter- particle interactions in atomic, nuclear and particle physics. The energy of the beam and the nature of the particles in it determine the extent of scattering in different directions. The solutions might also correspond to the scattering by a force field, where energy is specified in advance and the behaviour of wave functions is found in terms of energy. The measurements indicate that different number of particle appears in different directions.

3.2.2 Scattering cross section The scattering measurements involve intensity of the beam as a function of angle (or may be some other property of the projectile – target system). Most common measurements are done on cross section. To avoid complexities, we assume that the wavelength associated with the incident beam of particles is small (as is the case with matter waves) as compared to inter atomic distances of the target. Further we will analyze scattering by a single atom or nucleus and then add the separate effects of various scattering centers.

Consider a beam of particles moving along z-direction with I0 particles per unit area per unit time. When particles undergoes scattering then the number of particles received by the detector per unit time is given by

dN = () n I0 d (1) where n gives the number of scatterers in the target.

() is numerically equal to the cross sectional area of the projectiles scattered per scatterer of the target into a unit solid angle in the direction of . Essentially this ratio has the dimensions of an area, and so it is called a cross section. (However, it is pertinent to notice that this is not basically an area, but only a probability). For this treason () is called the differential scattering cross section. If we integrate differential scattering cross section over the solid angle we get total scattering cross section. When the particles undergo scattering through an angle  at an azimuthal angle of , due to a potential, then the angular distributions of particles is related through

(Numberof particles scattered into a solid angle dW per unit time) (,) d = (2) I0

Where (,) is scattering cross section and d is the differential solid angle along (,). See Fig.1

For spherically symmetric potentials (,) is independent of . The cross section of scattering for a spherically symmetric potential is given by

Fig. 1

p2 p  = 蝌[s ( q )sin q ]d q d j 0 0

 = 2p [ s ( q )sin q ]d q (3) 0

In the above analysis we assume that the scattering centers (scatterers) act independently of each other. Further this expression holds well for a wave as for a beam of particles.

44 Angular momentum

Physical interpretation Assume that we have a target, which has N scatterers per unit volume. A beam of particles of cross section A and intensity I0 strikes normally on it. Then the intensity I (x) of the beam after penetrating a distance x into the target is related to the loss of intensity dI(x) on penetration of dx into the target. The relation is given by

s - A dI(x) = NA dx (4) A

[Note that (Na dx) is the number of scatterers in volume (A dx). Here we have ignored multiple scattering].

On integration we find that

dI( x ) s = NA dx 蝌I( x ) A

Nx Or I(x) = I0 e (5)

Here N represents macroscopic cross section and is interpreted as inverse of mean free path.

3.3 STATIONARY WAVE SCATTERING The Schrodinger equation for scattering can be written as {(ħ2/2)2 + V(r)} = E(r) Where E is related to p (= ħk), the momentum of homogeneous particle wave proceeding in positive z-direction, through E = (p2/2). The wave is given by (r) = (z) = eikz 3.3.1 Collision of identical particles Consider a collision of two particles a and b. After collision particle a goes along path1 and particle b goes along path 2 as shown in Fig.2.

If amplitude of happening of this event is f () then the probability P1 of observing it is proportional to [f ( )]2. If however the particle a goes along 2 and particle b goes along

2 1, the probability of occurring this process is P2 = [f (-)] , see Fig.3. 45 Angular momentum, Spin and Scattering

a  b 2

Fig.2 Ignoring the spins of the individual particles and assuming particles a and b to be identical, we will have amplitude that any one particle goes along 1 and the other goes along 2 given by f () + f ( - ) 1

a  2 b

Fig.3 In collision as also in scattering, the forces of mutual interaction acting on two particles need also to be considered. It is usually convenient to consider these problems in center of mass coordinates, so that the relative position vector r = r2 – r1 continuously changes. The position of center of mass is

m1 r 1+ m 2 r 2 rcm = (6) m1+ m 2

If both the particles are of the same mass i.e., m1 = m2 we have r+ r r = 1 2 (7) cm 2 The scattering function is given by 1 u( r )= eikz + f (q , j ) e ikr (8) r r Where r, ,  are the coordinates of the relative position vector r. The asymptotic forms of wave functions can be written as

eikr eikz+ e- ikz -[ f (q , j )] + f ( p - q , j + p ) (9a) r and

eikr eikz- e- ikz -[ f (q , j )] - f ( p - q , j + p ) (9b) r

46 Angular momentum

The differential scattering cross section is given by

s( q , j )= [f ( q , j )]� f ( p + q , j p )]2

=[f (q , j )]2 + [ f ( p - q , j + p )] 2 � 2Re f ( q , j ) f *( p + q , j p )

When the particles are indistinguishable the interference term is zero and so

s( q , j )= [f ( q , j )]2� [ f ( p + q , j p )] 2 (10)

Which combines the probabilities of observing incident and scattered particle. Particles, which interfere with positive sign, are bosons while those interfere with negative sign are fermions.

Check your progress-1 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) Which is the interference term in equation above equation (10)? ------ii) Write equation (10) for electron. ------

3.4 REPRESENTATION OF SCATTERING PHENOMENON BY A BUNDLE OF WAVE PACKETS

The scattering problem is a very difficult problem if the coordinates of two particles are time-dependent. Such that they are initially in the form of separated wave packets, then they move into the common region. There they interact, and finally separate into two wave packets moving in a direction that has different probability depending on several factors. This problem can be handled safely if we consider it as a time-independent, stationary-state problem that is much easier to solve. In the time-independent, stationary- state problem we consider the state of one particle as corresponding to a plane wave at large distances. The other particle (target) is assumed to be stationary. The plane wave interacts with a fixed potential of the other particle. Outgoing waves are identified with the scattered particle. The process when conceived to be a steady one can then very well 47 Angular momentum, Spin and Scattering be represented by the usual interpretations of the wave function. It is usual practice to describe it in the center of mass system, in which a particle of mass interacts with the field of a potential of another particle at the origin. Try to answer ‘What is mean by isotropic scattering?’

3.5 SCATTERING OF A WAVE PACKET BY A POTENTIAL

Strictly speaking scattering problem needs temporal* description (*temporal means as a function of time). But when there is a steady current of particles approaching a target from a very large distance we can use time independent Schrodinger equation. 2(r) + (2/ ħ2 ) [E – V(r)](r) = 0 where  is the mass of the particle in the beam. For more than one particle  is the reduced mass and E is the energy in the center of mass system. Let us consider a plane incident wave and an out going scattered wave. Then the solution of the Schrodinger equation has the form lim. eikr y(r )= eikr + f ( q )[ ] (11) r r where k represents vector along z-axis and the potential is assumed to be spherically symmetric. The first term represents wave incident on the scatterer (scattering centre). Further, we have assumed that the amplitude of the scattered wave is independent of the azimuthal angle . If the incident beam is along z-direction, the equation (12) becomes lim. eikr y(r )= eikz + f ( q )[ ] r r The flux density is given by the second term. If  is proportionality term depending on angle  then the scattered flux and incident wave flux are given respectively by

2 2 eikz eikr  f (q )[ ] and  f (q )[ ] (12) r r respectively.

48 Angular momentum

Let the detector be kept at a distance r and the area dA intercepts those waves, then the flux

ikr 2 e 2 dA  f (q )[ ] dA =  f(q )[ eikr ] r r 2

ikr 2 e 2 Or  f (q )[ ] dA =  f(q )[ eikr ] d (13) r

Equation (13) represents the number of waves passing through the detector of area dA.

dA Here we have written = d, the solid angle subtended by the detector at the scatterer. r 2 Thus the differential scattering cross section is

2 Jf( q )[ eikr ] dW s( q )dW = Jf( q ) 2 dW

2 = [eikr ] dW (14)

3.6 CALCULATION OF CROSS SECTION  = R2 is the general formula for the cross section of a spherical body. But it is only a geometrical cross section. We notice that the cross section defined for scattering does not match with it. The reasons are several. A few of them are- the scattering occurs from the interaction of particle with the potential function of the scatterer which may be different for particles of different natures, energy of the projectile causes the projectile to penetrate the field of scatterer through different extent, the scattering centers may be inhomogeneously distributed as well as they may not in one plane. To have then the rough estimation of cross section we depend on impact parameter, p.

The area πp2 is then the area within which a collision will result in scattering through an angle θ. By differentiating it we can find dp corresponding to scattering between the angle θ and θ + dθ. The corresponding solid angle is d = 2π sinθ dθ. The area corresponding to dp is then

dA = 2πpdp = (π/2)s2sinθ dθ. (15)

The ratio |dA/d)| is the probability for scattering into unit solid angle.

Let us see an example: Nuclear cross sections are expressed in barn (b). One barn is equal to 10-28 m2. If Al27 is bombarded with slow neutrons, the most common reaction is elastic scattering, or an (n,n) reaction. The cross section is about four times the geometrical cross-section of the Al nucleus, about 2-3 b. But if we calculate geometrical cross section for Al27it comes to about 0.7 b only. This result, which does not have any explanation in classical physics, can well be expected on the basis of quantum mechanics.

The target is generally a thin foil in case of solids or a column of material in the case of liquid or gases. The projectiles are atomic or subatomic particles. Hence the scattering cross section does not exceed a few barns.

3.7 LABORATORY SYSTEM AND CENTER OF MASS SYSTEM

What is center of mass system (CM)?

It is the coordinate system in which the center of mass of the interacting particles is at rest and remains at rest during and after interaction. It is an inertial system in which the center of mass does not move.

The collision problem is solved most conveniently by transforming it to the center of mass system. In this system, the result will be that the velocities have changed direction by some angle.

Another important property is that in the CM system, the final momentum must be zero, so the final velocities after collision are in the ratio of the masses. What is laboratory coordinate system? In the laboratory system the target is assumed to be at rest and the interacting particle is in motion.

Note – it is easier to make measurements in laboratory system. Hence we generally make measurements in laboratory system and do necessary calculations to express the result in center of mass system. 3.7.1 Connection between center of mass and laboratory system

50 Angular momentum

Consider masses m1 and m2 are located at r1 and r2 with respect to a laboratory system. We assume that the target is at rest in the laboratory frame. The location of center of mass for two masses m1 and m2 located at r1 and r2 with respect to a laboratory system is obtained from

m1 r 1+ m 2 r 2 rcm = (16) (m1+ m 2 ) Example-1. To find the mass of the composite particle when two particles are involved.

Let the particles undergoing motion have masses m1 and m2 respectively. Then the Schrodinger equations for those particles are

2 2 [ħ /2m] 1(r) = - [E1 – V1(r)]1(r) and

2 2 [ħ /2m] 2(r) = - [E2 – V2(r)]2(r) and that for the composite system it is

2 2 [ħ /2]  (r) = - [E – V (r)] (r) (16) where (r) =1(r) + 2(r), V(r) = V1(r) + V2(r) and E = E1 + E2 Comparison of the quantities on LHS suggests that 1 1 1 = + (17) m m1 m 2

Example-2. Reduced mass of electron in hydrogen atom is calculated using m1 = 9.11 x

-31 -25 –31 10 kg and m2 = 1.67 x 10 kg to give  = 9.105 x 10 kg. It suggests that if two masses are widely different in magnitude, the reduced mass is almost equal to the mass of the lighter particle. That is the reason the center of mass of the solar system lies within the sun itself.

3.7.2 Relation between scattering angle in the two coordinate systems

To find the relation for the scattering angle, we move the laboratory system in the direction of the incident particle with such a speed that the center of mass is relatively at rest. Thus the target which was stationary in the laboratory system now moves towards projectile with velocity v2 such that

v2 = v – v1 and also v2=( m 1 + m 2 ) v 1 where v is velocity of projectile in the laboratory system and v1 is its velocity in the center of mass system see Fig.4).

m1 c m1 v1 cm v2 m2 m2 v2

Fig.4 Center of mass (CM) system

m1 Hence v2 = v - [m1+ m 2 ]

m2 v Or v2 = [m1+ m 2 ] If the scattering is elastic, the velocity remains the same after collision. Thus from the figure 5, for the laboratory system

v2 cos c + v1 = v cos L and v2 sinc = v sinL

v2 sinqc Or tan qL = v2cosqc + v 1 sinq Or = c (18) cosqc + g

Where  = v1/v2 = m1/m2

m1 L

m1 v m2

Fig.5 Laboratory coordinate system

52 Angular momentum

One can find similarly relation for the cross section for the two systems as

2 3/2 [1+g + 2 g cos qc ] sL ( q L )= s c ( q c ) (19) [1+ g cos qc ]

Check your progress-2 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) Find the location of center of mass of composite of particles m1, m2, m3 and m4 located at r1, r2, r3 and r4 from the origin of laboratory system. ------

ii) Scattering cross section depends on a)………….b)……………and inversely on c)………………..

3.8 SCATTERING BY A CONTROLLED POTENTIAL WAVE: ANALYSIS AND PHASE SHIFT METHOD

3.8.1 Scattering by a square-well potential

Square well potential is defined as

V(r) = V0 when 0 < r < a = 0 when r > a where a is the width of the well. E+ V If b is the impact parameter, n = [0 ]1/2 and E = ½ mv 2 then the scattering angle is E 0 given by

 = - 2 (i-r) = 2 [sin-1 (b/na) – sin-1(b/a)] for 0

lim. eikr y(r )= eikz + f ( q )[ ] r r

The flux density is given by the second term. If  is proportionality term depending on angle  then the scattered flux and incident wave flux are given respectively by

2 2 eikz eikr  f (q )[ ] and  f (q )[ ] r r

Let the detector be kept at a distance r and the area dA intercepts those waves, then the flux

ikr 2 e 2 dA  f (q )[ ] dA =  f(q )[ eikr ] r r 2

ikr 2 e 2 Or  f (q )[ ] dA =  f(q )[ eikr ] d (21) r

Equation (21) represents the number of waves passing through the detector of area dA.

dA Here we have written = d, the solid angle subtended by the detector at the scatterer. r 2 Thus the differential scattering cross section is

2 Jf( q )[ eikr ] dW s( q )dW = Jf( q ) 2 dW

2 = [eikr ] dW (22)

3.8.2 Phase shift in scattering by a spherically symmetric potential Let the Schrodinger equation for spherically symmetric potential V(r) be

2(r) + (2m/ħ2)[E – V(r)] (r) = 0

For spherical symmetry this equation has to be independent of 

R( r ) hence y= l P (cos q ) r l

R( r ) where l satisfies r

d2 R [k2 - U ( r ) - l ( l + 1)] l +R( r ) = 0 (23) dr2 r 2 l with k = (2mE/ ħ2) and U( r )= 2 mV ( r ) / ħ2

54 Angular momentum for r =>  we have

d2 R l +[k2 ] R ( r ) = 0 dr 2 l

e ikr So y=邋Plcos q R l ( r ) = P l cos q l l r

ikr- ikr Cl e D l e Or y=Pl cos q [ + ] l r r

If the incident wave is along z-axis we can write

eikr y=eikz + f ( q ) r

For large distances from the scatterer

ikr ikr- ikr ikz e Cl e D l e e+ f(q ) = Pl (cos q )[ + ] rl r r

ikr ikz e lp Or e+ f(q ) = Pl (cos q )[ b l sin( l - + d l )] r l 2

th Where l is the phase shift of the l partial wave. It is a measure by which lp sin(l- + d ) is displaced in  relative to free particle function. 2 l Thus the effect of spherical potential function is to shift the phase of the scattered wave.

l id l The value of the quantity bl is found to be (i ) (2 l+ 1) e this value helps in finding the total scattering cross section.

4p 2 s=(2 ) [2l + 1]sin dl (24) k l

Note- Simple spherical potential fields are Columbian and gravitational. A microscopic projectile when scattered by these must undergo a phase change. But it does not happen with macro projectiles.

3.9 IMPACT PARAMETER

Impact parameter, b is the distance between the initial path of the projectile when it started its journey towards the target and a parallel line that passes through the center of the target particle. See Fig.6.  is the scattering angle. All those particles that lie at a distance between b and (b+db) are in an annular ring of cross section 2bdb would be scattered such that I 2 b db = - () I 2  sin  d b db Or s( q ) = - (25) sinqd q Negative sign indicates that  decreases with increase in b.

m v0 i  b r1

r1 i

Classically, we can identify impact parameter p, as the distance of closest approach if the particles would not interact, and had moved in straight line. This distance is not changed by transformation to the CM system. In quantum mechanics, relative angular momentum of the particles takes up the role of the impact parameter.

3.10 RELATION BETWEEN PHASE SHIFT AND LOGARITHMIC DERIVATIVES

This derivation is rather tricky. Let us start at r = 0 and proceed outward through the region in which V(r) ≠ 0. The equation this situation can be written as

(1/r2)(d/dr)(r2dR/dr) + {2m[E - V(r)]/ħ2 - l(l + 1)/r2}R = 0.

56 Angular momentum

Eventually, we will reach a point where V(r) can be considered zero and the potential does no longer affect the solution. Let's call this distance r = a. Now we have, on the one hand, our solution coming out from r = 0, and on the other, the general form of the radial function when V = 0. The radial function can thus be written as

Rl(k.r) = Al[cos δl jl(k.r) - sin δl nl(k.r)].

Note that this is not the asymptotic solution at r = a, but includes the phase shifts in the coefficients.

If u(r) is a wave function of the form of Rl, then the value of (1/u)(du/dr) will be continuous at the boundary (a point where wave functions meet), since by definition of wave functions both u and (du/dr) are continuous. The continuity of the logarithmic derivative (1/u)(du/dr), should then give us the phase shifts. To understand it, let us assume that γl is the value of the ratio (1/u)(du/dr) at r = a for the wave function for values of r < a. Then, the continuity of the logarithmic derivative entails

J n k[cos δ l (k.a) - sin δ l (k.a)]/[cos δ J (k.a) - sin δ n (k.a)] = γ (26) l r l r l l l l l

Dividing numerator and denominator by cos δl and solving for tan δl, we can determine the phase shifts in terms of the γl.

We get

J n tan δ = [k l (k.a) - γ J (k.a)]/[k l (k.a) - γ n (k.a)] (27) l r l l r l l

From this equation we can find the cross section.

3.10.1 Effect of spin When spin of the interacting particle is taken into account, each particle will have (2s+1) spin Eigen functions.

Case I

When both the particles are in the same spin state with sz = m ħ, different states are given by

U1(m)U2(m) for s  m  -s.

Case II Symmetric functions for the sum of products are

U1(m)U2(m) + U1(m)U2(m) for m  m Case III There are s (2s+1) states with anti-symmetric functions consisting of differences of products

U1(m)U2(m) - U1(m)U2(m) for m  m

The total number of functions are (2s+1) + (s+1)(2s+1) + s(2s+1) = (2s+1)2. Of these (s+1)(2s+1) are symmetric and s(2s+1) are anti-symmetric.

Example 1. Positive sign appears in differential cross section in (s+1)(2s+1)/(2s+1)2 = (s+1/(2s+1) spin states. Thus for s = 1, positive sign appears in ratio 2/3; and for s=2, it is 3/5 etc.

2. If s = 1/2 three states are symmetric and one is anti-symmetric.

Check your progress-3 Note: a. Write your answers in the space provided below. b. Compare your answers given at the end of the unit i) What fraction of total functions will bear magnetic spin in differential cross section for two spin states?. ------ii) The incident particle may not suffer a scattering on traversing a

58 Angular momentum

distance x is related to eNx. How would you claim it? ------

3.11 BEHAVIOUR OF PHASE SHIFT AT LOW ENERGIES

3.11.1 Phase shifts at low energies

From the relation for  we find that the phase shift completely determines the scattering.

The cross section is maximum when

l =  /2,  3/2, 5/2 etc., and becomes zero at l =n.

When phase shifts are small the two functions fl and Fl of the following differential equations

d2 f l( l + 1) l +[ ]f = 0 dl2 l 2 l

d2 F V l( l + 1) l +[1 - ( ) - ]F = 0 dl2 E l 2 l

tend to be equal, i.e., fl = Fl.

So that

a V 2 d= -f( l ) d l = 0 (28) l0 E l where a is the range of potential function. In such cases scattering becomes isotropic.

3.12 SCATTERING BY A HARD SPHERE

Hard sphere is defined by a potential

V (r) =  if r < a and V (r) = 0 if r > a

59 Angular momentum, Spin and Scattering where 2a is the diameter of the sphere. The Schrodinger equation for this sphere is

2(r) + (2m/ħ2)[E – V(r)] (r) = 0

We take the case r > a as the sphere is impermeable. Then this equation reduces to

2(r) + (2m/ħ2)[E] (r) = 0

Or 2(r) + k2 (r) = 0 where k2 = (2m/ħ2)[E]

This equation written in spherical polar coordinates takes the form

1d dR l ( l + 1) [r2 ]+ [ k 2 - ] R = 0 (29) r2 dr dr r 2

To solve it, we substitute  = kr.

So that r = /k and dr = d/k

So the equation takes the form

d dR l( l + 1) [l2 ]+ [ l 2 - ]R = 0 dl d l r 2

Let us assume the probable solution to be R () = v () then

dv dv 1 l2+ l +[ l 2 -l ( l + ) 2 ] v = 0 (30) dl2 d l 2

This is standard Bessel equation of order (l+ ½ ). Its satisfactory solution is J(l+ ½ )()

J (l ) p  R() = (l+ ½ ) J (l ) (31) l 2l (l+ ½ )

where J(l+ ½ ) (l ) is called spherical Bessel function.

Remember that if J(l ) (l ) is a solution then J(-l ) (l ) has also to be a solution.

Asymptotic behaviour of Rl

When , we can write

60 Angular momentum

2 lp J(l+ ½ ) (l ) =sin(l - ) pl 2

1 lp Or J (l )= sin( l - ) l l 2 and corresponding

1 lp J (l )= cos( l - ) -l l 2

So that

A lp B l p R =sin(l - ) - cos( l - ) l l2 l 2

C Or R=[ J (l )cos d - J ( l )sin d ] (32) ll l l- l l

Where tan l = -(B/A)

Boundary condition-

At r = a the wave function vanishes i.e.,

Jl(l )cos d l- J - l ( l )sin d l = 0

Jl ( ka ) Or tandl = J-l ( ka )

When l = 0,

Jl /2 ( ka ) tand0 = = - tan( ka ) J-l /2 ( ka )

Or d0 = ka

4p and s = sin2 ka (33) k 2

For scattering involving extremely low energy ka <<1, so that

s= 4 p a2 or s a2 (34)

This suggests that low energy scattering is mostly spherically symmetric.

When ka >> 1, we have

2 4p [2l+ 1] Jl ( ka ) s = 2 2 2 kl Jl( ka )+ J- l ( ka ) or s2 p a2 (35)

The ratio of scattering cross sections for projectiles of high energy and low energy is

2 2 high/low  2a /a = 2 with the above information you can try to show that ka >> 1 represents high-energy situation and suggest why is r < a condition is forbidden in the case of solid sphere?

Born approximation Whenever scattering effect is confined near the scattering centers, the amplitude of scattered waves is very weak. For obtaining the scattering amplitudes in such cases Born method is used. Born assumed that the potential energy of interaction between projectile and the scatterers is just a perturbation of first order. Born approximation thus can be applied to those cases when colliding particle energy is very large in comparison to energy of interaction. This is thus a supplement method to partial waves. Ramsauer Scattering It is an excellent example of particle scattering theory offered by the scattering of low-energy electrons by atoms and molecules. He assumed atoms as hard spheres.

3.13 LET US SUM UP

 Scattering is essentially a phenomenon resulting from an interaction of at least one moving object with another stationary (or moving) object. After interaction it is not only that energy and momentum of interacting particles undergo a change but the phases also undergo a change.

62 Angular momentum

 For macro objects a different phase is not noticeable and hence remained hidden till the advent of quantum mechanics which discuss all aspects of dynamical systems in detail. The studies can be undertaken essentially in two different ways 1) by considering motion in laboratory frame of reference and 2) investigating the phenomenon in center of mass system. Since most of the observations are taken in laboratory frame of reference an interrelation is studied between two frames.

 Spin of the interacting particles modifies the formulation. Schrodinger proposed a theory for scattering on the basis of wave packets.

 Ultimately we studied simple phase relationships for spherically symmetric potentials at low energies and just cite few other cases.

3.14 CHECK YOUR PROGRESS

1. i) The interference term is additional to individual amplitude components which is �2Ref (q , j ) f *( p+ q , j p ) , in the expression.

ii) since electrons are Fermions we retain negative interference term s( q , j )= [f ( q , j )]2 - [ f ( p - q , j + p )] 2

m1 r 1+ m 2 r 2 + m 3 r 3 + m 4 2. i) rcm = . Simple principle being moment about center of (m1+ m 2 + m 3 + m 4 ) mass must be zero.

2 ii) Scattering depends on a) sin l, b) number of partial wave l and inversely on c) energy of the projectile or k2. 2) Isotropic scattering means potential function is spherically symmetric 3. i) Substitute values of s in expression given in Case III we find that (s+1)(2s+1) are symmetric functions which bear spin magnetism = 3x5 =15 functions

ii) Nx I(x) = I0 e represents that I(x) out of I0 would suffer scattering for one particle it is proportional to eNx

Reference books for further reading 1. Quantum mechanics: theory and applications, A Ghatak and S Loknathan, McMillan

2. Quantum mechanics, S L Gupta, V Kumar, H V Sharma and R C Sharma, Jai Prakash Nath and Sons. 3. Quantum mechanics, L I Schiff, Mc Graw-Hill 4. Quantum mechanics, B K Agrawal and H Prasad, Prentice Hall of India

5. Introductory quantum mechanics and spectroscopy, K M Jain, South Asian Publishers 6. The Feynman Lecturers in Physics, Vol.III, R P Feynman, R B Leighten and M Sands, Narosa Publishing House 7. Physics, D Halliday and R Resnick + Crane/Walker, John Wiley 8. Quantum Mechanics, V K Thankappan, Wiley Eastern Limited 9. Wikipedia, the free encyclopedia.