Scattering from Yukawa Potential in Quantum Mechanics Masatsugu Sei Suzuki and Itsuko S

Scattering from Yukawa potential in Quantum Mechanics Masatsugu Sei Suzuki and Itsuko S. Suzuki Department of Physics, SUNY at Binghamton Binghamton, New York, U.S.A. (Date: August 18, 2021)

We consider the Yukawa or screened Coulomb potential (short-range order). This potential was proposed by Yukawa as a model for the nucleon -nucleon interaction. Here, we discuss the scattering of particle by the Yukawa potential by using the Born approximation. The scattering amplitude in the first order Born approximation ( f (1) ) is real. According to the optical theorem the forward scattering amplitude has an imaginary part proportional to total cross section. In order to overcome such a difficulty, it is natural to try calculating the second order Born term ( f (2) ). Using the Mathematica, we will get the exact expression for f (2) and will confirm that the optical theorem is valid for the Yukawa potential up to the order of Born second-order approximation. We also discuss the partial shift at low energy limit. Note that this article is a part of the lecture notes of Phys.422 (Quantum Mechanics II). You can find all the mathematics on the theory of Born approximation in my lecture note on Physics 422 (Quantum Mechanics II), Chapter 13. http://bingweb.binghamton.edu/~suzuki/QuantumMechanics2.html For the first time, I encountered the problem of scattering from the Yukawa potential in a book of quantum mechanics problems and solutions, edited by Masao Kotani and Hiroomi Umezawa (Shokabo, 1968, in Japanese), when I was a undergraduate student. I remembered that I struggled to solve this problem since the calculation of the second Born approximation includes the Cauchy theorem, Jordan’s lemma, and complicated integrals. Note that the present article does not include any topics which are something new. I just collect interesting topics related to the scattering from the Yukawa potential, from many textbooks of scattering, which were published long time ago.

1. Hideki Yukawa and the meson theory

Hideki Yukawa (湯川秀樹 , Yukawa Hideki , 23 January 1907 – 8 September 1981) was a

Japanese theoretical physicist and the first Japanese Nobel laureate for his prediction of the pi meson, or pion. https://en.wikipedia.org/wiki/Hideki_Yukawa

(( Meson theory )) The Story of Spin (S. Tomonaga) The meson theory of Yukawa was briefly explained by Prof. S. Tomonaga in his book.

Yukawa arrived at the idea that there should exist a yet-to-be discovered charged boson, the heavy quantum, and that this charged particle shuttles between the neutron and proton. He concluded that if this particle has a mass about 100 times as large as that of the electron, then the effective range of the nuclear force is on the order of 10 -13 cm (=1 fm). From the analogy that the force between charged particles is mediated by the electromagnetic field, he thought that the nuclear force is mediated by an unknown field which might be called the nuclear force field. When he adopted the Klein-Gordon equation as the equation of this field, then instead of the Coulomb potential e2 / r for the electromagnetic field, g2 er / r appeared, and from the de Broglie- Einstein relation, in place of the zero-mass boson (photon), which is the quantized electromagnetic

field, he obtained a boson of mass ℏ / c for the nuclear force field. As e2 / r is the potential for the Coulomb force, he thought, and calculated the boson mass by putting 10 -13 cm for the force range 1/  , obtaining 100 times the electron mass that I (“Tomonaga”) mentioned above. Thus, this boson is lighter than the proton but heavier that the electron and much heavier than the photon. He therefore, name this particle the heavy quantum as opposed to the light quantum.

(( Note )) Serway et al. Modern Physics The mass of pion is roughly estimated from the Heisenberg’s principle of uncertainty (Serway et al. )

Fig.1 Feynman diagram representing a proton interacting with a neutron via the strong force. The pion mediates the strong force. The blue arrow shows the direction of increasing time.

Now consider the pion exchange between a proton and a neutron that transmits the nuclear force according to Yukawa. We can reason that the energy E needed to create a pion of mass 2 m is given by Einstein’s equation E  m c . Again, the very existence of the pion would violate ℏ conservation of energy if it lasted for a time greater than t ≃ , where E is the energy of 2E the pion and t is the time it takes the pion to travel from one nucleon to the other. Therefore,

ℏ ℏ t   2 . (1) 2E 2 m c

Because the pion cannot travel faster than the speed of light, the maximum distance d it can travel in a time is c t . Using Eq.(1) and d c  t , we find this maximum distance to be

ℏ d  . (2) 2m c

We know that the range of the nuclear force is approximately 1 fm = 10 -15 m. Using this value for d in Eq.(2).we calculate the rest energy of the pion to be

ℏc m c 2   100 MeV  2d

This corresponds to a mass of 100 MeV/c 2 (approximately 250 times the mass of the electron), a value in reasonable agreement with the observed pion mass.

2. Yukawa potential Yukawa potential (also called a screened Coulomb potential) is of the form:

V V( r )  0 e r . r

where V0 = constant and attractive for positive V0 and repulsive for negative V0 . The range of the potential is given by a= 1/  . For a Coulomb potential, we see that   0 and the concept of a range has no meaning since the range becomes infinite. For convenience, here we make a plot of

V/ V 0 as a function of kr , where k is the wave number of a particle and  / k is changed as a parameter;

 V( r ) 1  kr  e k , V  0 (kr ) k

Fig.2 Plot of the Yukawa potential as a function of the distance r. For convenience, we

make a plot of V/ V 0 as a function of kr , where k is the wave number of a particle,

the parameter  / k is changed between 0.5 and 4.5, ( /k )  0.5 . V0 = constant. The range of the potential is given by 1/  . For a Coulomb potential, we see that   0 and the concept of a range has no meaning since the range becomes infinite.

3. Scattering amplitude (the first order Born approximation)

 Fig.3 Elastic scattering. k k '  k . q k'  k . q 2 k sin . 2

We now calculate the first-order Born approximation for the scattering amplitude (Yukawa potential),

1 2 m f(1)   deVriq  r ( r ) 4 ℏ2  1 2 m   (2) 3 k ' Vˆ k 4 ℏ2 12m 14V 1   (2 ) 3 0 4ℏ2 (2)  3 22 q 2mV 1   0 ℏ2  2 q 2  1   4  2 q 2 or

 1 f (1)   , 4  2 q 2 where

V  V( r )  0 e r , q k'  k , q 2 k sin r 2 with

2mV  4  (0 ) ℏ2 2

Evidently, the optical theorem is not valid only for the first order approximation since

4 Im[f (1) ] 0 ,  Im[f (1) (0)]  0 tot k

So, we have to calculate the Born second order in order to check the validity of the optical theorem. (( Note )) Calculation of f (1) (detail)

1 kk'Vˆ  deVe rikr'  ( r ) i kr  (2 ) 3  1  dr ei(k '  k )  r V ( r ) (2 ) 3  1 V  dr eiq  r 0 e   r (2 ) 3   r

q k'  k

 iqr cos  1 V e k'Vˆ k  0 2 r2 dr sin  d  e r (2 ) 3  0  r 0  1 V   (2 )0 er rdr sin  d  e  iqr cos  3 0  (2 )  0

Here we note that

 iqr cos   iqr cos  e  sin d  e    0 iqr  0 eiqr e  iqr  iqr 2sin(qr )  qr

1V  2sin()qr k'Vˆ k  (2) 0 erdrr (2 ) 3  0 qr

1V 4    0 drer sin( qr ) (2 ) 3  q 0

Using the formula of the Laplace transformation, we get

1V 4  q k' Vˆ k  0 (2 ) 3 q  2 q 2

14V 1  0 (2 ) 3   2 q 2

The differential cross section:

(1) 2 2 d2 2mV 1   1 f (1) (0 ) 2  dℏ2( 222  q ) 16  2222 (  q )

2  2 1  (1) d  162 (  222 q )

 with d  2 sin  d  ; and q 2 k sin 2

2  2  1 (1)  2  sin d 2  16  2 2 22 0 (  4k sin ) 2 2 2  sin d  2 222 2 160 [  2k (1  cos  )] 2  2 2  2 16222  (   4)k 2  2 1  4  2 4 k 2

In the limit of k   , we have

1 2 1  (1)   . 16  k2 E

(( Mathematica )) Calculation of the integral

4. Rutherford scattering Here we derive the differential cross section for the Rutherford scattering. There is a Coulomb interaction between the -particle ( Z1e = +2 e charge; e  0 ) and the nucleus with positive charge

(+ Z2e). In the expression of the differential cross section

d(1)  2  2 1  , d162 (  2  q 22 )

V we put 0  Z Z e 2 , we get  1 2

d (1) 2 m 1  (Z Z e 2 )( 2 ) 2 . d1 2 ℏ2( 2  q 22 )

In the limit of   0,

d (1) 1 2 m 1  (Z Z )(2 e 4 ) 2 1 2 ℏ2 2  d 16 k sin 4 2

(Z Z )2 e 4 1  1 2 2  16 E sin 4 2 where

 ℏ2k 2 q 2 k sin , E  . 2 2m

This is just the famous Rutherford scattering of the alpha particle (+ Z1e) by a Coulomb potential of charge +Z2e. Note that this differential cross section using quantum mechanics is in complete agreement with that obtained from a classical analysis of Coulomb scattering.

Fig.4 Rutherford scattering. Hyperbolic trajectory. r1  r2  2a . The target nucleus with Ze is located at the focal point F 1. The alpha particle is at the point P on the hyperbola ae  a2  b2 . See further detail in http://bingweb.binghamton.edu/~suzuki/QuantumMechanicsII/13- 2_Rutherford_scattering.pdf

5. Second-order Born approximation for the Yukawa potential We now study the second-order Born approximation for the Yukawa potential’

1 2 m f(2) (2) 3 k ' VGEiVˆˆ (   ) ˆ k 4 ℏ2 0 k

with

k'(VGEiVˆˆˆˆ ) kkkkk  d "'""( V GEi ˆ   )"" kkk V ˆ , 0 k  0 k

1 k"(Gˆ E i  )"" kk  k " 0 k ˆ Ek  H  i  1  k" k " E E  i  k k " 1  Ek E k "  i  2m 1  ℏ2k 2 k " 2  i

14V 1 k' Vˆ k  0 , (2)3   2 (')k  k 2

1 2 m f(2) (2) 3 dVGEiQkQQ 'ˆˆ (   ) QQk V ˆ 4 ℏ2  0 k 12m 14V 1 1 1   ( )(2)2 3 (0 ) 2 dQ 4ℏ2 (2)  6 2 (')()Qk 22   QkkQ 222  i  2  2 1 1 1  dQ 324 2 (Qk ') 22   ( QkQk ) 222  i 

where  is a positive small constant. We now calculate the integral M (Dalitz integral) defined by

1 1 1 M d Q .  2(Qk ') 22   ( QkQk ) 222  i 

We use the formula (Feynman integral representation)

1 1 1 dz  ,  2 0 [azb (1  z )] ab

In general;

1 (mn  1)!1 tm1 (1  t ) n  1  dt (Feynman, 1949) m n  m n ab( m 1)!( n  1)!0 [ atbt  (1 )]

(( Mathematica ))

Here we also use the notations

a  2 (Q  k ') 2 , b  2 (Q  k ) 2 and

az b(1 z )[2 (Qk ')] 22 z  [  ( Qk )](1 2  z )  22(Q  2 Qkk '')( 22z  Q  2 Qkk 2 )(1)  z

 222k'zz  k (1  ) QQkk 2 2 [ ' zz  (1 )]  2k 2Q 2 2 Qk [' z  k (1  z )] where

q k'  k ,

k'  k  k , k' k  k 2 cos 

Thus, we have

azbz(1)[Qkk ' z  (1)] z222  k [' kk z  (1)] z 2 [Qkk 'z  (1  z )] 2  2 k 2 g 2 (Q  g ) 2   2

For convenience, we use the parameters.

gk'z  k (1  z ) qk z ,

2  2 k 2  g 2 .

gz2 [kk '(1  zz )] [ kk '(1  z )] k22z  k' 2 (1  z ) 2 2 kk ' zz (1  ) kz2[2 2  2 z 1 2(1 zz  )cos]  k2[1  2(1 zz  ) 2(1 zz  )(1  2sin2 )] 2  k2[1  4(1 z  z )sin 2 ] 2

g2 2   2  k 2 .

So that, M can be rewritten as

1 1 1 M dz d Q   4[(Qg )2 22 ] Qk 2  2 i  0 1 1 1  dz d Q   2 22 2 2 0 [(Qg ) ] Qk  i  and

1 M  dzL (g ,τ ) , 0 with

1 L(g , )  d Q ,  [(Q g )2 22 ] [ Q 2 k 2 i  ]

1 J(g , )  d Q ,  [(Q g )2 2 ][ Q 2 k 2 i  ] and

1  M  J (g , ) . 2 

First, we need to calculate J (g , ) . Suppose that the vector g is directed along the Qz axis. When the angle between the vector Q and g is , we have a scalar product as

Fig.5 3D Q space. The vector g is directed along the Qz axis. The angle between Q and g is  .

The scalar product:

Q g  Qg cos 

(QgQQggQ )22 2 22 2Qg cos   g 2

  1 J(,)g  Q2 dQ 2sin  d   2 2222 0 0 [QQg 2 cos  g  ][ Qki   ]   Q2  2 sin d  dQ   2 2222 0 0 [QQg 2 cos  g  ][ Qki   ]  Q2 dQ 2 sin  d   22  2 22 0QkiQ 0 [  2 Qg cos   g  ]

We define the function F( Q ) as

 2 sin d  F( Q )   Q2 g 2  2  2 Qg cos  0 g2  2 2 gQQ  2  ln( ) gQ g2 2 2 gQ  Q 2

(( Mathematica ))

So that, we get

 Q2 dQ J(,)g   F () Q  2 2 0 Q k  i   Q2 dQ g 22  2 gQ  Q 2  [ ln( )]  22 22 2 0 Q k i gQ g   2 gQ Q  QdQ g2  2 2 gQ  Q 2  ln( )  22 22 2 gQki0  g   2 gQQ 

We note that the integrand of this integral is an even function of Q. This integral can be rewritten as

1  Q2 dQ J(,)g   F () Q .  2 2 2  Q k  i 

 2 sin d  Using the form of F( Q ) ; F( Q )  , J (g , ) can be obtained as  2 2 2 0 Q g   2 Qg cos 

1 Q2 dQ 2 sin  d  J (g , )  2QkiQg22  222   2 Qg cos   0   Q2 dQ  sin d    22 2222 0  (QkiQg )[(  cos  )  g sin   ]

Fig.6 Complex plane of 2D-Q space. There are two single poles at Q Q 1 and Q Q 3

inside the contour C (counterclockwise). Two single poles ( Q Q 2 and Q Q 4 ) outside the contour C.

Using the Jordan’s lemma (the absolute value of the integrand reduces to zero in the limit of Q   in the complex Q-plane), we extend the region of the integral over the upper half of complex plane. Using the Cauchy theorem , we get

 Q2 dQ I   22 2222  (QkiQg )[(  cos  )  g sin   ] Q2 dQ   22 2222 C (QkiQg )[(  cos  )  g sin   ] Q2 dQ   C (QQQQ1 )(  2 )( QQ  3 )( QQ  4 )

2[Re(i sQQ 1 )Re( sQQ  3 )]

2 Q( Q Q 1 ) Res ( Q Q 1 )  lim Q Q 1 (QQQQ1 )(  2 )( QQ  3 )( QQ  4 ) Q2  lim Q Q 1 (QQ2 )( QQ  3 )( QQ  4 ) Q 2  1 (Q1 Q 21 )( Q  QQ 31 )(  Q 4 ) k 3 1  2k2 g 2  2  2 g cos  k 3 1  2k2 2  k 2  2 g cos  where

g2 2   2  k 2 ,

2 Q( Q Q 3 ) Res ( Q Q 3 )  lim Q Q 3 (Q11 QQQ )(  2 )( QQQQ  3 )(  4 ) Q2  lim Q Q 3 (Q Q1 )( Q  Q 2 )( QQ  4 ) Q 2  3 (Q3 QQ 13 )(  QQ 23 )(  Q 4 ) (g cos i  ) 2  2ig ( cos  ikg )( cos   ik ) (g cos i  ) 2  2ig [( cos  i  )2  k 2 )] and

k 2 1 Res ( Q Q )  3 2kkg ( cos )2  g 22 sin    2

k 1  2k2 g 2  2  2 kg cos 

Where we define  as

g 2sin 2    2

We now calculate

 J(,)g  2 2 i sin  d [Res( QQ  ) Res( QQ )  1 3 0

J1  J 3 where

 J22 i sin  d  [Res( QQ  )] 1  1 0  (g cos i  ) 2  22i sin  d   2 2 0 2ig [( cos  i  )  k ]  (g cos i  ) 2  2 sin d   2 2 0 [(g cos  i  )  k ]

For simplicity, we choose a variable,

xgcos  ig cos   ig 2 sin 2   2 and

g 2 sin cos  dx[sin  g  i ] d  g 2sin 2  2 ig sin  [ig2 sin 2  2  gd cos  ] g 2sin 2  2

ig sin  (g cos  i  ) d  g 2sin 2  2 ig sin   xd  

 dx sin d   ig x

Then we have

g  i  dx x 2 J   2 1  2 2 g i  igx( x k ) i 2 g i  xdx  g x2 k 2 g  i  i 2 g i  1 1  (  ) dx 2gg  i  xkxk  i2 ( ikgi  )(   kg )  ln 2g ( i kgi )(   kg )

We also calculate the integral J3

 J22 i sin  d  [Res( QQ  )] 3  3 0  k 1  22i sin  d   2k2 g 2  2  2 kg cos  0  sin d   k 2 i  2 2 2 0 k g   2 kg cos  2i( k g ) 2   2  ln 2g ( k g ) 2   2

Thus, we have

J J1  J 3 2i( ikgikg )(   )( ikgikg   )(   ) [ln  ln ] 2(g ikgikg )(  )( ikgikg   )(  ) 2i( i   kg  )  ln g( i  kg  )

M is defined by

1  2 1 M J (g , )  . 2 kg2  2  2 2 ik

The scattering amplitude in the second order Born approximation is

1 1 f(2)  22  Mdz 2  32  0 1 1 dz   2  2 2 222 320  (kg    2 ik ) 1 1 dz   2  2 2 2 320  (  2i  k ) where

g2 2  k 2   2 , and

2  2 k 2  g 2   2 4k 2 sin 2 z (1  z ) , 2  2 q 2 z(1  z )

Using the separation of the real part and imaginary part of the integrand

1 (2  2)i  k  (2  2)(i  k  422  4)  k

 2 2ik    (4 4  22k )(  4  4  22 k )

(a) Imaginary part Im[f (2) ] We have the imaginary part,

11  2 kdz Im[f (2) ]    22  2 4 22 320  4  k 1 1 dz  k2  2 2 4 22 160  4  k 1 1 dz  k2  2 2 4 22 22 160  4  k  4 kqz (1)  z 4k 22 q 4  22 k  kq ln( ) 4 22 22 1 2 2 k q 4  k  kq  2 k  16  2kq4 k 22 q  4  22 k

4k 22 q 4  22 k  kq ln( ) 2 2 4 22 22 (2) 1   k q 4  k  kq Im[f ]  2 , 32  q 4k 22 q  4  22 k

where the integral is calculated by Mathematica

4k 22 q 4  22 k  kq ln( ) 1 dz 4k 22 q 4  22 k  kq I   IM  4 22 22 4 22 22 0 4  k  4 kqz (1  z ) 2kq k q  4  k

Note that in the limit of q  0 , we get

1 1 lim IIM  , q0 24k 2  2

leading to the relation

 2 1 Im[f(2) (0)]  k . 162 4 k 2  2

(( Mathematica )) Calculation of I IM

(b) Real part Re[f (2) ] Next, we calculate the real part,

1 1  2dz Re[f (2) ]   22  322  ( 422 4  k ) 0 1 1 dz  4  2 2  22 4 22 22 32  0 qz(1 z )[  4  k  4 kqz (1  z )]

where the solution of this integral is also obtained using the Mathematica

1 dz I  R  22 4 22 22 0 qz(1 z )[  4  k  4 kqz (1  z )] 1 kq(4)2 2  q  422  kq (4)   2  [Arctan 3 q2  4 k 22( q  4  2 ) 2 kq(4)2 2  q  422  kq (4)   2  Arctan ] 2 3 and

1 1 lim I R  . q0 34k 2  2

Thus, we have

(2)1 42 1 Re[f ]  2   32  q2  4 k 22( q  4  2 ) kq(4)2 2  q  422  kq (4)   2 [Arctan 2 3 kq(4)2 2  q  422  kq (4)   2  Arctan ] 2 3

We also have simpler expression as

2 2 (2) 1  1 Re[f ]  2 32  q 4k 22( q  4  2 ) (see the note for this derivation) q 4 k 22( q  4  2 ) Arctan[ ] q2 2 4k 22( q  4  2 )  4

We note that

1 q  lim Arctan[ ]  q0 q 44k 22 4   2  4 k 2

Using this relation, we have

1 1 Re[f (2) (0)]   2  , 322  2 4 k 2 at q  0 .

(( Mathematica )) Calculation of IR .

(( Note )) Simplification of the expression of Re[f (2) ]

We now simplify the expression of Re[f (2) ] using simple trigonometry.

kq(4)2 2  q  422  kq (4)   2 A  Arctan 2 3 kq(4)2 2  q  422  kq (4)   2  Arctan 2 3    where we use  and  as

kq(4)2 2  q  422  kq (4)   2 tan   , 2 3

kq(4)2 2  q  422  kq (4)   2 tan   . 2 3

We note that

tan tan  tan(  )  . 1 tan tan 

Then, we get

q tan tan  4 k 22 ( q  4  2 ) ,  3

kq(4)2 2  q  422  kq (4)   2 1tantan1[    ] 2 3 kq(4)2 2  q  422  kq (4)   2 [ ] 2 3

1 22 22 24 42 222 16 [(4)(kq   q   qk 4 kq  )] 4 1 1 [(8kqq24  22  16)( 4  q 24  qkkq 42 4 222  )] 4 6 1 1 [4(k22 q  4) 2 q 22  ] 4 4 1 q2 [4 k 22 ( q  4  2 )]   4 4 2

Thus, we have q 4k 22( q  4  2 )  3 tan(  )  1 q2 [4k 22 ( q  4)]  2  4 4  2 , q 4 k 22( q  4  2 )  q2 2 [4k 22 ( q  4)]  2  4 or

q 4 k 22( q  4  2 )    Arctan[ ] . q2 2 [4k 22 ( q  4  2 )]  4

6. Optical theorem The scattering amplitude up to the second order, is

f f(1)  f (2) .

At q = 0, we get the scattering amplitude as

1  1 f (1)   , 42 4  and

2 (2)   1 2 k  f222  i   22  32  4k 16   4 k  2   2ik  ( ) 322  2 4 k 2  2 1  322   2 ik

If the Born approximation is to be reliable, we must certainly have the condition

f (2) ≪1. f (1)

For the forward scattering (q = 0), we gave

 2 1 (2) 2 f 32   2 4k 2     ≪1. f (1)  8  2 4k 2 4

When k ≫  (high energy)

(2) f    ≪1. f (1) 16  k

The condition steadily improved as k increases. When k ≪  (low energy, k = 0),

(2) f   ≪1. f (1) 8

By taking into account of the second order Im[f (2) (0)] , we confirm that the optical theorem is valid such that

4 4  Im[f (0)] Im[ f(1) (0)  f (2) (0)] k k 4  Im[f (2) (0)] k  2 1  4  2 4 k 2   (1) where

 2 1  (1)  . tot 4  2 4 k 2

We note that the differential cross section is

d 2  f d 2 f(1) Re[ f (2) ]  Im[ f (2) ] (f(1)  Re[ f (2)2 ])  (Im[ f (2)2 ])

 1 f (1)   , 4  2 q 2

2 2 (2) 1  1 Re[f ]  2 32  q 4k 22( q  4  2 )

q 4 k 22( q  4  2 ) Arctan[ ] q2 2 4k 22( q  4  2 )  4 and

4k 2( q 2  4  )  kq ln( ) 2 2 4 22 2 (2) 1   k( q  4  )  kq Im[f ]  2 32  q 4k 22( q  4  2 )

So that, we have the final form

f f(1)  f (2)  1   4  2 q 2 12  2 1 q 4 k 22( q  4  2 )  [ Arctan[ ] 32  2 q 4 22 2 q2 2 k( q  4  ) 4k 22( q  4  2 )  4 4k 22( q  4  2 )  kq ln( ) 4k 22( q  4  2 )  kq i ] 4k 22( q  4  2 )

7. Validity of the Born approximation (another approach); direct calculation The validity for the Born approximation is satisfied under the condition

m e ikr dr V( r ) 1 , 2 ℏ2  r

m e ikr V dr 0 e r  1. 2ℏ2  r  r

We note that

eikrV e ikr V dr 0 er 4  drr2 0 e   r rr0 rr  V   4 0  dre (ik ) r  0 V 1  4 0    ik

mV01 2 m V 0 1 24  2  1 . 2ℏ  ik ℏ   2 k 2

(a) Low energy limit ( k   )

2m V 0  1  4  , ℏ2 2

(b) High energy limit ( k   )

2m V k 0  1  4  , ℏ2 k 

The inequality becomes easier to satisfy to satisfy as k increases, implying that the Born approximation becomes much accurate at high incident particle energy.

(( Note )) Scattering length a The scattering length a is defined by

 a  , 

4 2mV as will be derived later, with    0 . The condition for the scattering for the high energy ℏ2  2 limit (as described above) is

k  4  . 

This condition can be rewritten as

ka ≫ 1.

The de Broglie wavelength  is related as the wave number as

2   . k

So that, the condition is also expressed as

 a ≫ . 2

The de Broglie wavelength  is much shorter than the scattering length a.

8. Partial phase shift The Born approximation is a good approximation for the high energy incident particle, but is not so for the low energy incident particle.

 fEq(,) (2 l  1) fEPl ( ) l (cos) , l0

1 1 f() E dzP ()(,) z f E q , l l 2 1

z  cos  ,

q2ℏ 2 q 2 1 cos    , 2k2 4 mE

 1 f (1)   4  2 q 2  1   4 2 2k 2 (1  cos  )  1   4  2 2k 2 (1  z )  1   4mE 4  2 (1  z ) ℏ2 For l = 0 (s-wave)

1 1 f() E dzf(1) (,) E q 0  2 1  1 1  dz 8  2 4mE 1  (1  z ) ℏ2  ℏ2 4mE ln[ 2  (1  z )]| 1 8 4 mE ℏ2 1 ℏ2  2  ln[ ] 8mE 32  mE  2  ℏ2   2  ln[ ] 8mE 4E  2  ℏ2 where

V   0 

ℏ2 2 which has the expected branch point at E   . Thus, while the scattering amplitude f (1) has 8m no branch points, the partial amplitude does. This illustrates (which is, in fact, true of all partial waves) that the left-hand branch cut is introduced into the partial-wave amplitude in the process of making the projection (Taylor).

9. Phase shift analysis using the Heine expansion Using the Heine expansion,

1  (2l  1) QxPl ( ) l (cos ) x  cos  l0 we get the expansion of f (1) as

2mV 1 f (1)   0 ℏ2    2 4k 2 sin 2 2 2mV 1   0 ℏ2  22k 2 (1  cos  )

mV 1   0 ℏ2k 2  2 1  cos  2k 2 mV 1   0 ℏ2k 2 x  cos 

 2 x 1  2k 2

mV 1 f (1)   0 ℏ2k 2 x  cos  mV   0 (2l  1) QxP ( ) (cos ) 2 2  l l ℏ k l0  (2l  1) fkPl ( ) l (cos ) l0

mV fk()  0 Qx () . lℏ2k 2 l

10. Problem and Solution (Sakurai and Napolitano) I found a very interesting problem on the phase shift analysis of the scattering due to the Yukawa potential, in famous textbook on quantum mechanics. Problem 6-5 J.J. Sakurai and J. Napolitano Modern Quantum mechanics, 3 rd edition (Cambridge, 2021)

(( Problem ))

(( Solution ))

Yukawa potential

V V r)(  0 er . r

The first Born approximation

2mV 1 f (1) ( )   0 , ℏ2q 2  2

where

q  k  k',

q2  k 2  k'2 2kk'cos

 2k 2 1(  cos)

where the angle between the vctors k and k’ is . k' k

(b)

The scattering amplitude f )( can be expanded in terms of the phase shift l as

 f() (2'1) l  fkPl' () l ' (cos  ) . l ' 0

Note that

1 Id (cos ) P (cos  ) f (  )  l 1 1  d(cos ) (2' lfkP  1) ( ) (cos  ) P (cos  )   l' l l ' 1 l ' 0

with the use of formula

1 2 d(cos)P (cos )P (cos )   .  l l' ll ', 1 2l 1

Then, we get

 2 I(2' l  1) fkl' ( )  l , l ' l ' 0 2l  1 2 (2l  1) f ( k ) l 2l  1

 2fl ( k )

Thus, we have

1 1 fk( ) d (cos ) P (cos  ) f (  ) l l 2 1 11 2mV  1 d(cos ) P (cos  )  0  l ℏ2  2 2 21   2k (1 cos  )   mV1 1 P( z ) (1)   0 dz l ℏ2k 2 2   2 1 1  z 2k 2 mV  2  0 Q (1  ) ℏ2k 2l 2 k 2

where Ql ( ) is the Legendre function of the second kind and is defined by

1 1 P( z ) Qx( )  dz l , l  2 1 x z with

 2 x 1 . 2k 2

We make a plot of Ql ( x ) as a function of x for x>1, where l is changed as a parameter. (numerical integration is done by using the Mathematica).

 2 Fig.8 Q( x ) as a function of x. x 1  . l = 0, 1, 2, 3,…. l k 2

(b) (i) As is shown in the above figure, we have

Ql ( x ) 0 for x  1.

 2 x 1  2k 2

2 1 (ii) de Broglie wave length    (the range of potential) k 

 or  1 k

mV  2 f( k ) 0 Q (1  ) lℏ2k 2 l 2 k 2 mV  2   0 Q ( ) ℏ2k 2l 2 k 2

2 l 1 mV0 l! 2 k    2 2 2  ℏ k(2 l  1)!!    mV2l1 l !   0 k 2l 1 ℏ2 2l 3 (2l  1)!! or

mV2!l1 l k 2! l  1 l k f() k 0 ()21l  () 21 l  l ℏ2 2 (2l 1)!!  8  (2 l  1)!!  which is proportional to k 2l 1 .

 2 Fig.9 Q( x ) as a function of l with x  1.5 and 2. x 1  . l 2k 2

11. The effective range: Phase shift (Capri) The phase shift for l = 0 (s-wave) is given by

2 2m  sin() kr  sin   k drr2 V () r 0 ℏ2    0 kr  2 2m V e r  sin( kr )    k drr 2 0 ℏ2    0 r kr  2 2mkV  er  sin() kr    0 drr 2 ℏ2     0 r kr    4k 2  ln(1  ) 16 k  2   4k 2  ln(1  ) 4 k  2 where

2mV 2mV   4  0 ,  0  , ℏ2 2 ℏ2 2 4  and

2  er sin() kr  1 4 k 2 drr 2 ln(1  )    2 2 0 r kr  4 k 

Thus, we get

  4k 2 sin   ln(1  ) 0 4 k  2   4k2 8 k 4 ≃ (  ) 4 k 2  4 k2 k 2   (1  )   2

Note that

cos 0 kcot 0  k , sin 0 and

2 cos0 1  sin  0  2k 22 k 2 1  (1  ) 2 2  2

 2k 2 ≃ 1  2  2k 2 ≃ 1 2 2

cos 0 kcot 0  k sin 0  2k 2 1 2 2   k k2 k 2  (1 )   2   2k 22 k 2 ≃ (1  )(1  ) 2 2  2  1 4 1 k 2 (  ) 2   1 1    r k 2 a 2 0

where a is the scattering length and r0 is the effective range.

  1 4 a   , r (  ) . 4  0  

Note that our result of r0 is a little different from that reported by Capri. The parameter  is dimensionless. For the low energy limit, the condition for the Born approximation is ≪ 4  ,

1 leading to a ≪ . For the high energy limit, the condition for the Born approximation is  k 1 k  4  , leading to a ≪ .    We note that the total cross section (the first order Born approximation) is given by

 2 1  (1)  4  2 4 k 2

In the limit of k  0 , we have

1  2 (1)(k 0)   4  a 2 4  2 using the scattering length a. Thus, the scattering length determines the low-energy scattering cross section. We see that

fk0() Sk 0 ()1  e2i0 ( k )  1 2i  cot0 (k ) 1 2sini ( k )  0 cos0 ()k sin  0 () k 2sini ( k ) ≃ 0 1 sin0 (k ) 2ka  i ka

 i where sin[ (k )]  k  ka . Thus, f( k ) has a single pole at k  . This pole 0  0 a

ℏ2 2 1 corresponds to a bound state of energy E   and   . Since the bound state is of the b 2m a form exp(r ) / r , its extension is a .

(( Mathematica )) Calculation

12. Optical theorem (Capri) The scattering amplitude for low energy s-waves may be written as

e2i0 1 f( k )  0 2ik 1 ei0( e i  0  e  i  0 ) 2ik 1  2i sin  e i0 2ik 0

sin   0 ke i0 sin   0 kcos0 ik sin  0 1  kcot 0  ik

We note that

1 1 kcot     r k 2 , 0a 2 0

where a is the scattering length and r0 is the effective range. Thus, we get

1 f( k )  0 1 1  rk2  ik a 2 0

a   1 1iak  r ak 2 2 0

The optical theorem states that the total cross-section  is given by

4   Im[f ( k )] k 0 4 sin   Im[0 ] k ke i0 4 1  sin 2  k k 0

On the other hand, the total cross section is given by

d  d   d f( k ) 2 d   0 2  4 f0 ( k ) sin  2  4 0 ke i0 4  sin 2  k 2 0

So, in this, the optical theorem is proved. We now use the expression of

a f( k )   . 0 1 1iak  rak 2 2 0

Thus, we have

4   Im[f ( k )] k 0 4 a2 k  1 k (1r ak22 )  a 22 k 2 0 4 a2  1 (1rak22 )  ak 22 2 0

On the other hands by direct calculation, we find that

d  d   d f( k ) 2 d   0 2  4 f0 ( k ) 4a2  1 (1r ak22 )  a 22 k 2 0

Thus, in this case, we also verify the optical theorem.

13. Conclusion There are so many excellent textbooks on the scattering due to the Yukawa potential. Before I wrote the present article, I read these textbooks, which were mainly written before 1970’s. I tried to calculate the scattering amplitude for the Yukawa potential up to the second-order Born

approximation. I encountered several integrals which seemed to me to be so complicated. Thanks to Mathematica, I succeeded in getting exact results within seconds. It was amazing to me. After that, I checked the validity of my results by comparing with the results reported in the standard textbooks (such as textbook by J.R. Taylor). The reporting of this article in part is motivated in part by our success in obtaining the exact results by Mathematica. ______REFERENCES H. Yukawa, Proc. Phys. Math. Soc. Jpn . 17 , 48 (1935). H. Yukawa, Tabibito (The Traveler) (World Scientific, 1982). S. Tomonaga, The Story of Spin , translated by T. Oka (University of Chicago, 1997). R.A. Serway, C.J. Moses, and C.A. Moyer, Modern Physics , 3 rd edition (Thomson Brook/Cole, 2005). J.J. Sakurai and Jim Napolitano, Modern Quantum Mechanics , 3 rd edition (Cambridge, 2021). V. Galitski, V. Karnakov, V. Kogan, and V. Galitski, Jr, Exploring Quantum Mechanics (Oxford, 2013). J.R. Taylor, Scattering Theory: The Quantum Theory on Nonrelativistic Collisions (John Wiley & Sons, 1972). R.G. Newton, Scattering Theory of Waves and Particles , 2 nd edition (Springer, 1982). M. Kotani and H. Umezawa edited, Quantum Mechanics Problems and Solutions (in Japanese, Shokabo, 1959). C.J. Joachain, Quantum Collision Theory (North-Holland, 1975). H.J.W. Müller-Kirsten, Introduction to Quantum Mechanics; Schrödinger Equation and Path Integral (World Scientific, 2006). A.Z. Capri, Nonrelativistic Quantum Mechanics , 3rd edition (World Scientific, 2002). A. Goawami, Quantum Mechanics , 2 nd edition (WCB. 1997). R.P. Feynman, Phys. Rev. 76 , 769 (1949). APPENDIX. R.H. Dalitz, Proc. Roy. Soc . A 206 , 500 (1951).

APPENDIX-1 The notations used in this article (summary)

V V( r )  0 e r (Yukawa potential) r

2mV  4  0 . ℏ2 2

2mV 1   f (1) 0  . (the first order Born approximation) ℏ222q4  22  q

2 1 4   (1)   Im[f (2) ] . (Optical theorem) 4  2 4 k 2 k

 q 2 k sin . 2

   . 4

2  Re[f (2) (0)]  . 322  2 4 k 2

 2 2k Im[f (2) (0)]  . 322  2 4 k 2

12  2 1 q 4 k 22( q  4  2 ) Re[f (2) ] Arctan[ ] . 32  2 q 4 22 2 q2 2 k( q  4  ) 4k 22( q  4  2 )  4

4k 22( q  4  2 )  kq ln( ) 2 2 4 22 2 (2) 1   k( q  4  )  kq Im[f ]  2 . 32  q 4k 22( q  4  2 )

APPENDIX-II Mathematical formula

1 Px( ) Fnn (  , 1;1; (1  x )) n 2 1 2

LegendreP[ n, x] for x 1 (Mathematica)

The Legendre function of the second kind is

(  1)  1 3 Qz() F (1,;;)  z 2  3 2 1 (  )(2)z  1 2 22 2 2 using the hypergeometric function, where z  1, arg(z )   , and   -1, -2, -3,…

1 1 Pl ( y ) Ql ( x )  for x  1. 2 1 x y

LegendreQ[n,x] for x 1 (Mathematica)

(( Asymptotics ))

Asymptotically for l   ,

 1 P(cos) Jl [(  )]   Ol ()1 l sin 0 2

2 1  cos[(l  ) ] O ( l 3/2 ) 2l sin  2 4

0    and for argument of magnitude greater than 1,

1 1 Pl ( ) IleOl0 ()()  1 e2 l1 1 (1 e ) 2   O( l 1 ) 2le l (1 e ) 2

where J0 and I0 are Bessel functions.

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