2014/5/8 fundamentals in nuclear physics - freeishiken.free.fr/english/lectures/fnp05.pdf ·...

Fundamentals in Nuclear Physics (Kenichi ISHIKAWA) for internal use only (Univ. of Tokyo)

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Fundamentals in Nuclear PhysicsKenichi Ishikawa ()

http://ishiken.free.fr/english/lecture.html

[email protected]

1

2014/5/8


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Elements of

2

Material will be downloadable from:http://ishiken.free.fr/english/lecture.html

Special Relativity Quantum Mechanics


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Special relativity

3

1905


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Special Principle of Relativity

Physical laws should be the same in every inertial frame of reference.

Special relativity is derived from two principles.

Principle of Invariant Light Speed

There is at least one inertial frame of reference where Maxwells equations hold.


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Maxwells equations

5

1864 D =

H = J

E + Bt

= 0

B = 0

in vacuum

1c2

2Et2

2E = 0

c =1

p00

vacuum velocity of light is uniquely derived from Maxwells equations


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Principle of Invariant Light Speed There is at least one inertial frame of reference where Maxwells equations hold.

light in vacuum propagates with the speed c in one inertial frame of reference, regardless of the state of motion of the light source.


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Principle of Invariant Light Speed There is at least one inertial frame of reference where Maxwells equations hold.

light in vacuum propagates with the speed c in any inertial frame of reference, regardless of the state of motion of the light source.


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Relativity is used in nuclear physics primarily through the expressions for the energy and momentum of a free particle

rest mass velocity m v

E =mc2p

1 v2/c2= mc2 = Mc2energy

momentum p =mvp

1 v2/c2= mv = Mv

=1p

1 2 = v/c


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E =mc2p


non-relativistic limit usually applies for nucleiv c

E mc2 + 12mv2

Mass is a form of energy

nuclear energy


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E =mc2p


non-relativistic limit usually applies for nucleiv c

E mc2 + 12mv2

Mass is a form of energy The faster the particle is, the larger its

observed mass isThe kinetic energy is also observed as a part of the (observed) mass


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E =mc2p


Any form of energy is observed as mass.


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energy momentumE =mc2p

1 v2/c2p =

mvp1 v2/c2

E2 = m2c4 + p2c2v

c=

pc

E

nuclei: non-relativistic limit v c

E mc2 + p2

2mp mv

neutrinos and photons: mc p

v c1 m

2c2

2p2

especially for photons: m = 0

E = pc v = c

E pc1 +

m2c2

2p2


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For two particles A and B

is independent of inertial frames of reference (Lorentz invariant)

Especially

EAEB c2 pA pB

E2 c2p2 = m2c4


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Decay A ! B+ C

AB CmC ,pC mB ,pBmA,pA

Energy conservation EA = EB + EC

Momentum conservation pA = pB + pC

In the rest frame of A

pA = 0 pB = pC

mAc2 =

qp2c2 +m2Bc

4 +q

p2c2 +m2Cc4

find pB ,pC

not easy to solve ...


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EC = EA EB pC = pA pB

E2C c2p2C = (EA EB)2 c2(pA pB)2

m2Cc4 = m2Ac

4 +m2Bc4 2(EAEB c2pA pB)

In the rest frame of A

Lorentz invariant

m2Cc4 = m2Ac

4 +m2Bc4 2mAc2

qm2Bc

4 + p2c2

p2 =

"m2A +m

2B m2C

2mA

2m2B

#c2

can be evaluated with a convenient inertial frame of reference

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Elements of Quantum Mechanics


17


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Wave-particle duality to the Schrdinger equation

18

ei(krt)plane wave p = k, E =

= h2

= 1.055 1034 J s reduced Planck constant

h = 6.626 1034 J s Planck constant

how to extract p and E from ?

In quantum mechanics, a physical quantity is represented by an operator (or matrix) in general.

E = i t

p = i x

,

(x, t) = ei(kxt)

p(x, t) = i x

(x, t)

E(x, t) = i t

(x, t)


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E = i t

p = i x

,E =p2

2m+ V (x)

kinetic energy

potential energy

the time-dependent Schrdinger equation (1D)

3D

or interpreted as probability density to find the particle at x

i t

(x, t) =

2

2m2

x2+ V (x)

(x, t)

i t

(r, t) =

2

2m2 + V (x)

(r, t)

|(x, t)|2 |(r, t)|2

wave function


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Solution with energy E =

(x, t) = (x)eit

the (time-independent) Schrdinger equation (1D)

wave function

(x) continuous functions

and

E : energy eigenvalue

2

2md2

dx2+ V (x) (x) = E (x)

d

dx


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Free particle

21

V (x) = 0

d2

dx2+

2mE2 (x) = 0

(x) = eikx, eikx

k =

2mE2

plane wave

x

eikxeikx

general solution (x) = A eikx + B eikx

(x, t) = A ei(kxt) + B ei(kxt)


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Barrier potential

x = 0 x = a (a > 0)

V0eikx

eikx

eikx

incident

reflection

transmission

E =2k22m

< V0 100 % reflection in classical mechanics 100%

1 = eikx + Aeikx 3 = Deikx

K =

2m(V0 E)/2

continuous,d

dx

2 = BeKx + CeKx

T = |D|2 = 11 + V

20

4E(V0E) sinh2 Ka

transmission coefficient


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E =2k22m

< V0 100 % reflection in classical mechanics 100%

T = |D|2 = 11 + V

20

4E(V0E) sinh2 Ka

transmission coefficient

transmission is nonzero in quantum mechanics

tunneling effect

sinhx =ex ex

2

0.2 0.4 0.6 0.8 1.0

0.05

0.10

0.15

0.20T

E/V0

mV0a2

2 = 8

important in alpha decay

0 1 2 3 4 5 6

0.2

0.4

0.6

0.8

1.0

mV0 a

E/V0 = 0.8


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Parity

2

2md2

dx2+ V (x) (x) = E (x)

if V (x) = V (x) (x) = (x) satisfies

2

2md2

dx2+ V (x) (x) = E (x)

in the absence of degeneracy

(x) = P(x) |P | = 1

(x) = P 2(x) P = 1

P = 1 (x) = (x) even or + parity

P = -1 (x) = (x) odd or - parity

Nuclear states can be assigned a definite parity, even or odd.

Important in the discussion of beta decay


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Lx

= i~y@

@z z @

@y

25

Angular momentum

Or

p

L = r p

In the spherical coordinate Lx = i(sin

+

cos tan

)

Ly = i( cos

+

sintan

)

Lz = i

L2 = 2

1sin

sin

+

1sin2

2

2

Ly = i~z

@

@x

x @@z

Lz = i~x

@

@y

y @@x


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commutation relation

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0

[Lx, Ly] = iLz [Ly, Lz] = iLx [Lz, Lx] = iLy

common eigenfunction of L2 and Lz Ylm(, )spherical harmonics

L2 Ylm(, ) = 2l(l + 1)Ylm(, )

lz Ylm(, ) = mYlm(, )

l = 0, 1, 2, 3,

m = l,l + 1, , l 1, l

Y00 =

14

Y10 =

34

cos

examples

Y1,1 =

38

sin ei

(2l+1) eigenfunctions for given l (2l+1)


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how about spin?

[Lx, Ly] = iLz [Ly, Lz] = iLx [Lz, Lx] = iLy

Lets use the commutation relation

as a definition of angular momentum (operator)L = (Lx, Ly, Lz)

O

r

p

L = r p

forget

For given l, there are (2l+1) eigenfunctions (eigenvectors) (2l + 1) (2l + 1) matrix


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Lz

ll 1

l 2

l + 1l

L+ = Lx + iLy

0

1 2l 0 0 00 0

2 (2l 1) 0 0

0 0 0

3 (2l 2) 0 0 0 0 0

2l 1

0 0 0 0 0

L = Lx iLy

0 0 0 0 02l 1 0 0 0 00

(2l 1) 2 0 0 0

0 0

(2l 2) 3 0 0 0 0 0

1 2l 0

l = 1

1 0 00 0 00 0 1

0

2 00 0

2

0 0 0

0 0 02 0 0

0

2 0


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Lets consider 22 matrices?

Lz 2

1 00 1

L+

2

0 20 0

L

2

0 02 0

Lx 2

0 11 0

Ly

2

0 ii 0

eigenvalues = 2

L2 2 12

12

+ 1

1 00 1

l =

12

spin particle 12

x =

0 11 0

y =

0 ii 0

z =

1 00 1

Pauli matrices

2014/5/8 fundamentals in nuclear physics - freeishiken.free.fr/english/lectures/fnp05.pdf ·...

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