Nov 4, 2013 - Quantum mechanics originates from the study of the spectrum of hydrogen atom ..... Euler was the son of a Swiss baker and was sent to school early. ... teacher: wrote me a set of numbers and I give you on the sum and the ...
An elementary introduction to Quantum mechanic Mehdi Hage-Hassan
To cite this version: Mehdi Hage-Hassan. 00879586>
An elementary introduction to Quantum mechanic.
2013.
= ∫ ψ (r ) F ( x, y , z )ψ (r )dr < Δ F > 2 =< ( F − < F > ) 2 >
(2,19)
4.2 The uncertainty relation We consider the integral:
I (λ ) =
∫
xψ + λ h
2
dψ dx
dx ≥ 0 , λ ∈ R
The development of this integral gives
dψ ⎞ ⎛ dψ 2 I ( λ ) =< x 2 > + λ h ∫ ⎜ xψ + ψ x ⎟ dx + < p x > ≥ 0 dx ⎠ ⎝ dx using [x, px] = i ħ and after integration by parts we find:
I ( λ ) =< x 2 > − λ h + λ 2 < p x2 > ≥ 0 The discriminant of the quadratic equation in either negative or zero:
− 4 < x
2
>
+h
2
≤ 0
If we replace x and px by the standard deviations we find:
< Δx >2< Δp
x
>2≥ h
2
/ 4
We derive the Heisenberg uncertainty relations < Δ x >< Δ p x > = h / 2 ,… < Δ z >< Δ p z > = h / 2 (2,20) -This relationship means that if the wave is close to x this implies that the wave in the phase space is very spread out. -There is the uncertainty relation time -Energy: < Δ E >< Δ t > = h / 2 (2,21)
5. The Harmonic oscillator The Hamiltonian of the one dimension harmonic oscillator is: p2 H = + mω2 x 2 2m The Schrödinger equation is: 15
(2,22)
H ψ n (x) = E nψ n (x) .
(2,23)
x = (h / mω ) q, et p = (mωh)P, α = mω / h , E= ћ ω Є Put Thus we find the following equation:
(
)
) ) Hu(q) = P 2 + q 2 u(q) / 2 with H = hω H
(2,24)
If we make the change of function:
u(q) = f (q) e −q
2
/2
We find the equation: d d2 f ( q ) + ( 2 ε − 1) f ( q ) = 0 f (q ) − 2q 2 dq dq The resolution is done by writing f (q) in the form of a series:
f(q) = a 0 + a 1 q + a 2 q 2 + …
(2,25) (2,26)
If the solutions are bounded at infinity we find that {f (q)} are the Hermite polynomials Hn (q) [2,30] and the wave function is:
u n ( q ) = ( π 2 n! ) n
And
−
1 2
e
−
q2 2
u n ( x ) = (( m ω ) / h )
H n (q )
1/ 4
u n (q )
u n ( − x) = (−1) u n ( x) n
(2,27)
1 - These eigenfunctions form the Hilbert basis of L2(R) and (-1)n is the parity. 2 - The eigenvalues of the energy of the oscillator are:
E n = hω (n + 1 / 2)
(2,28)
6. Oscillator generating function and analytic Hilbert space 6.1 The generating function of the harmonic oscillator is known since long time [1-9] and is written as: 1 − zn q2 z2 − } G ( z , q ) = ∑ ∝n = 0 u n ( q ) = π 4 exp{ 2 qz − (2,29) 2 2 n! 6.2 If the parameter z is complex we find that the functions f n ( z ) = z n / n! form a basis of the analytic Hilbert space known by Fock or Fock-Bargmann space [21]. With (2,30) f n f m = ∫∫ f n ( z ) f m ( z ) d μ ( z ) = δ m , n
dμ(z ) Is the cylindrical measure: 2
2
d μ ( z ) = e − ( x + y ) dxdy / π , z = x + iy 6.3 It is easy to show the following useful formulas [25]:
f ( z ) = ∫ f ( z ' ) e z z ' d μ ( z ' ) And ∫ e α z e β z d μ ( z ) = e αβ 16
(2,31)
(2,32)
III- Dirac formalism of quantum mechanics After reading the work of Heisenberg, Dirac observed several unresolved problems: 1 – Definition of the state of the physical system. 2 - x, px are non-commuting observables then we must find the eigenfunctions of x, and px. 3 - The state of the oscillator. 4 - The connection between quantum and classical mechanics. Dirac introduced a new notation for the Euclidean space and the Hilbert space which leads to a new formalism of quantum mechanics and he find a new function: the delta function which is at the origin of the theory of distributions. We present first the Dirac notation in the simplest way, starting from the Euclidean space, and then we extend it to the Hilbert space, space of Schrödinger wave functions then we introduce the delta function and the eigenfunctions of momentum px. .
1. Euclidean Space and Dirac notation The search for eigenfunctions of the observable (x) led Dirac to introduce a new notation for the vector space and the dual space from the scalar product notation. Using this new notation Dirac find the δ-function and the well known Dirac transformation .
1.1 Dirac’s notation The scalar product of two vectors is the bracket:
r r x .y =
bra ↓ ket
(x , y ) =
x∗(y) =
x y
(3,1)
We cut the bracket in two word bra and ket. We denote the vectors of the space by ket { y } and the vectors x* of the dual space by the bra { x }.
a) The Euclidean space E3 In Dirac notation the basis vectors E3 may be written by ( 1 , 2 , 3
,
,
of the three-dimensional Euclidean space
).
*
b) The dual space E 3 By definition, the dual space is the space of linear applications E3 in R.
E 3* Is the dual space and we have the basis
r r e 1* , e
* 2
,
r e
* 3
r r r In Dirac notation: e1* = 1 , e2* = 2 , e3* = 3
If K is the field of the complex number C, c) The identity operator 1-The projector pi is defined by:
r r x must be replaced by x .
pi = i i
2- Completeness or the identity operator:
With p i2 = p i
I = ∑ 17
.
i
i
i
(3,2) (3,3)
1.2 Applications:
x = ∑ xi i
a-Let
= I x
x
We write
= ∑
i
x
i
i
b- In Dirac notation the linear transformation a is written: a i = I × a i = ∑ i a j xi = i x ,
With
i ,
and a i , j = i a j
(3,4)
1.3 Rotation in Euclidean space E (k = R or C) 1 - The rotation matrix which leaves invariant the scalar product in the E3 (R) is known by the orthogonal matrix of SO (3) with the Euler angles ( Ω ) = ( ψ , θ , ϕ ) . (3,5) The representation matrix of spin 1 or Pauli matrices [9] are:
⎛0 ⎜ σz = ⎜ i ⎜0 ⎝ With
−i 0 0
[S
x
,S
0⎞ ⎛ 0 ⎜ ⎟ 0 ⎟, σ y = ⎜ 0 ⎜−i 0 ⎟⎠ ⎝ y
] = i h S , [S x
0 0 0 y
i⎞ ⎛0 ⎜ ⎟ 0⎟ ,σ x = ⎜0 ⎜0 0 ⎟⎠ ⎝
0 ⎞ ⎟ − i⎟ 0 ⎟⎠
0 0 i
]
, S z = i h S x , [S z , S x ] = i h S
(3,6)
y
2- The rotation matrix Rs which leaves invariant the scalar product in E2 (C) is a unitary matrix of SU (2) and the parameters are the Euler angles ψ , θ et ϕ with spin half and
⎛0 σ x = ⎜⎜ ⎝1
1⎞ ⎛0 ⎟⎟, σ y = ⎜⎜ 0⎠ ⎝i
− i⎞ ⎛1 ⎟⎟ , σ z = ⎜⎜ 0 ⎠ ⎝0
0 ⎞ ⎟ − 1 ⎟⎠
is a
(3,7)
2. Hilbert space and Dirac notation 2.1 Hilbert space and linear operators Let H be a space of function with basis { ϕ i } and
( ϕ i , ϕ j ) = ∫ ϕ i ( x ) ϕ j ( x ) p ( x ) dx = δ i , Δ
j
(3,8)
p ( x ) Is the weight function and the measure of integration is: d μ ( x ) = p ( x ) dx The space of wave functions { ϕ i ( x )} is a space of square integrable functions which is pres-Hilbert i.e. not complete .
2.2 Hilbert space and Dirac notation a-The Hilbert space in Dirac notation is: { ϕ b The dual-space is:
{ϕ
i
and
ϕ 18
}
i
i
ϕ
j
= δ
i, j
}
c-The identity operator is: { I = ∑
ϕ
i
ϕ
i
i
} ( completeness condition)
d-The operator which leaves invariant the scalar product in the Hilbert space is a unitary operator U has the form [24]: U = e − i α u , u = e − iF (3,9) −iα F is a Hermitian operator and e is the group U1 and u belongs to the sub-unitary groups of operators su (n). - The irreducible space of the unitary groups are finite spaces, disjoint and invariant under unitary group su (n).
3. Dirac transformation and δ-function 3.1 Continuous states of position and Dirac notation a- Continuous states and Dirac notation Dirac noted that the measurement of the position and the momentum px = p implies that x and px= p are the eigenvalues of the observable (x) and (p) that have continuous eigenvectors with:
ˆx x = x x , And ˆp x p x = p p ,
(3,10)
The identity operator I in the continuous case can be written as:
I = ∫ x dx x , I = ∫ p dp p ,
(3,11)
b- Dirac distribution or Dirac function The function Dirac δ (x'-x) is the normalization of the continuous space {| x>}
⎧ 0 if x' ≠ x x' x = δ( x'− x) with δ( x'− x) = ⎨ ⎩∝ if x' = x
(3,12)
Dirac introduced this notation by analogy with Kronecker symbol δij. We will determine its expression in the next section. The first important property: x ' xˆ x = x ' x ' x = x x ' x As a result:
( x ' − x ) δ ( x ' − x ) or x δ ( x ) = 0
(3,13) (3,14)
3.2 The transformation of Dirac and its applications We consider the scalar product + ∝
=
g , f
∫ g ( x ) dx
f (x)
− ∝
In Dirac notation the scalar product is: g
f
=
+ ∝
∫ g ( x ) dx
(3,15)
f (x)
− ∝
Using the identity operator (3,11) we write: (g , f ) =
g
I
f
=
+ ∝
∫
g
x
− ∝
19
dx
x
f
dx
(3,16)
After comparison of the two above expressions, Dirac has defined the function f (x) as the application: H → R.
x :
x
⎯⎯ → x
f
g
And
x
= f (x) ∈ R
f
(3,17)
= g (x)
(3,18)
We also deduce the very useful expressions:
= xf ( x ),
x xˆ f
x pˆ x f
** The state of the system is written by ϕ
=
h f '(x) i
(3,19)
in Dirac notation.
3.2.1 Second important property of delta function: We have:
f ( x 0 ) = x 0 I f = ∫ x 0 x dx x f
So
f ( x 0 ) = ∫ δ ( x 0 − x ) f ( x ) dx
(3,20)
3.2.2 The eigenfunctions of the momentum operators We assume that the eigenfunctions of the momentum p is the continuous function p (x). We write: h x pˆ p = p ' ( x ) = p x p = p p ( x ) i Therefore (3,21) p ( x ) = c exp( ipx / h ) = c exp( ikx ) 3.2.3 The {p} representing If the wave function of the system is ϕ ( x ) we write:
x ϕ
= ϕ (x)
(3,22)
The wave function in the {p} representation or the phase space can be written as: +∝
p ϕ = ϕ( p ) = p I ϕ = ∫ p x dx x ϕ −∝
But
−ikx
p x = ce
(3,23)
+∝
As a result we find:
ϕ( p) = c ∫ e −ikx ϕ( x) dx −∝
(3,24)
3.2.4 Analytical expression of Dirac function We write
+∝
x x' = x I x ' = ∫ x p dp p x'
(3,25)
−∝
So we get:
2 +∝
x x' = δ( x'−x) = c ∫ eip( x−x') dp −∝
We shall calculate the normalization constant c in chapter V.
20
(3,26)
IV- Heisenberg representation and the path integral 1. The evolution operator and Heisenberg representation In this chapter we only want to give an idea of the path integral [3,26] and we assume that the linear superposition of states is preserved through time. We write the Schrödinger representation by: ψ (t ) = U (t , t 0 ) ψ (t 0 ) (4,1) For a conservative system the evolution operator U (t, t0) satisfies the Schrödinger equation: d ih U (t , t 0 ) = HU (t , t 0 ) (4,2) dt If we consider the inverse of the transformation, we write: (4,3) ψ (t 0 ) = U * (t , t 0 ) ψ (t ) And any observable A in the Schrödinger representation “ or picture” may be written in the Heisenberg picture of the form: A H = U * ( t , t 0 ) AU ( t , t 0 ) By a simple calculation we derive the Heisenberg equation dA ∂A ih H = [ AH , H H ] + ih H dt ∂t The Heisenberg equations for the coordinates r = ( x1 , x 2 ,..., x n ) and moments (p1, p2, ...) is: dx i 1 ∂H , = [ xi , H ] = dt ih ∂p i dp i 1 ∂H = [p i , H] = , i = 1, 2,.. dt ih ∂x i
(4,4) (4,5)
(4,6)
This system of equations is formally identical to the Hamilton’s canonical system.
2. The path integral and the Feynman propagator Using the unit operator I = ∑ ψ i ψ i we write the Feynman propagator, or the Green's function, in the form:
K ( r ' , t r , t 0 ) = r ' e − i ( t − t 0 ) H r = r ' e − i ( t ) H Ie − i ( − t 0 ) H r Thus we find: K (r ', t r , t0 ) = ∑ ψ i (r ', t ) ψ ( r , t0 )
(4,7) (4,8)
Using the unitary operators (3,11) we find that 3 ∫ d r K ( r , t r ' , t ' ) K ( r ' , t ' r0 , t 0 ) = K ( r , t r 0 , t 0 )
21
(4,9)
This completeness condition is valid for t> t ', t, t' [t0, t1]. The generalization of this relation to n - 1 intermediate point t> tn-1> tn-2> ...> t1> t0 . Then we write:
K ( r , t r0 , t 0 ) = ∫ d 3 r ∫ d 3 r ... ∫ d 3 r K ( r , t r ' , t ' ) K ( r ' , t ' r0 , t 0 )... K ( r , t r ' , t ' )
(4,10)
If the intermediate time is continuous we write the propagator using the symbols of the path integral in general as
⎧ i t ⎫ r (t ) = r K ( r , t r0 , t 0 ) = ∫ r ( t 0n ) = r 0n d [ r ( t )] exp ⎨ ∫ t 0n L c ( r , r& , t ) ⎬ dt ⎩h ⎭
(4,11)
- The Feynman conjecture Feynman's conjecture is to introduce in the integral the classical Lagrangian and to identify the normalization factor by the coefficient of the propagator of the free particle (below) by putting:
d[r(t )] = lim And K ( r f , t
f
n →∞ ⎛ m ⎞ ⎜ ⎟ Δ t → 0 ⎜⎝ 2π ihΔ t ⎟⎠
n−1 2
dx1dx2 ... dxn
⎧ i ⎫ r ,t ri , t i ) = ∫ r i f, t i f d [ r ( t )] exp ⎨ S ( x ( t )) ⎬ dt ⎩h ⎭
(4,12) (4,13)
S ( x(t )) Is the classical Hamilton-Jacobi action.
3. The propagators of a free particle The propagator of free particle is
K ( x ', t ' x , t ) =
i ( t '− t ) H ] x h
x ' exp[
Using the unit operator
I = ∫ p dp
p
x p = exp( ixp / h ) /
And
(4,14)
2πh
(4,15)
We obtain the expression of the propagator:
K ( x', t' x, t) =
i 1 [ p ( x ' − x ) + ( t ' − t ) p 2 / 2 m ]} ∫ dp exp{ h 2πh
The evaluation of the integral is obtained by using the Gauss integral:
K ( x', t' x, t) =
m i m ( x '− x ) 2 exp{ } 2 π ih Δ t 2Δt h
With Δt = (t '−t ) . 22
(4,16)
V-The harmonic oscillator and the Dirac notation We will present at the beginning Dirac method [8-9] which is based on the observation that the expression of the Hamiltonian has the form of product of a complex number and its complex conjugate. We find the expression of the generating function by a new method and then we derive the wave function of the harmonic oscillator. Using the generating function and the Fock-Bargmann space we simply calculate the properties of the δ-function without recourse to the theory of distributions, the normalization of the free wave and the Feynman propagator of the harmonic oscillator.
1. The state of the harmonic oscillator Dirac noted that the Hamiltonian of the oscillator: p2 H = + mω2 x 2 2m is analogous to the product z ⋅ z , z ∈ C and he posed:
(5,1)
2 2 ( Q + iP ), a + = ( Q − iP ) 2 2 h 1/ 2 x =( ) Q p = ( m h ω )1 / 2 P mω
a = With
) H u n (q) = (a + a + 1 / 2) u n (q) = En u n (q)
So
a , a + Are adjoint operators with:
[a , a ] = aa +
+
− a+a = 1.
(5,2) (5.3)
1.1 The basis of the oscillator It is well known that the basis of polynomials P(z) is: 1, z, z2, z3,…, zn,… dz n d with = nz n −1 And 1= 0 dz dz Comparing a , a + = 1 and [( d / dz ), z ] = 1 we simply deduct the basis of the harmonic oscillator: a +n 0 , 1 ,..., n = 0 ,.. and a 0 = 0. (5,4) n!
[
]
a n = n n − 1 , a+ n =
With
n n+1
(5,5)
+
So a and a are the creation and annihilation or the ladders operators of the harmonic oscillator. The wave function is written using the Dirac transformation by:
u n (q ) = q n = q
a +n n n!
23
(5,6)
1.2 From the isomorphism between the space of the oscillator and the Fock space f n ( z ) = z n / n! we deduce the correspondence: and a
1.3 We also write formally:
af ( a + ) 0 =
(5,7) +
∂f (a ) 0 ∂a +
(5,8)
This formula is very useful for fast calculations in the oscillator basis.
2. The generating function of the harmonic oscillator We will determine the generating function at the beginning by a new method [25] and we simply deduce the wave function of the oscillator in the coordinates representation. The generating function of the oscillator can be written with the help of (5,6) in the form: + zn z n (a + ) n G (z, q ) = ∑∝n =0 u n (q) = ∑∝n =0 q 0 = q e za 0 (5,9) n! n! Using the transformation of Dirac and (5,8) we find: + 1 d (5,10) aq ae za 0 = (q + )G ( z , q ) dq 2 +
q ae za 0 = zG( z, q) Comparing the two expressions (a) and (b) we obtain: d G ( z , q ) = ( 2 z − q )G ( z , q ) dq The solution of this equation is: q2 G ( z , q ) = c exp{( 2 qz − ) + ϕ ( z )} 2 To determine ϕ (z) we use the above expressions and: + 1 d ∂ q a + e za 0 = ( q − )G ( z , q ) = G ( z , q ) ∂z dq 2 ' We find: ϕ ( z ) = − z The solution of the above expression gives the generating function q2 z2 G ( z , q ) = c exp{ 2 qz − − } 2 2 For t = 0 we find: G (0 , q ) = q 0 = u 0 ( q ) The normalization of u0 (q) is c = π-1/4: 1 − q2 z2 zn − } u n ( q ) = π 4 exp{ 2 qz − G ( z , q ) = ∑ ∝n = 0 2 2 n! c-The development of the second member and the comparison with the first member gives the harmonic oscillator wave function (2,27) b-
−
1 2
un (q) = ( π 2 n!) e n
−
q2 2
H n (q), un ( x) = ((mω) / h) un (q) 1/ 4
24
(5,11) (5,12)
(5,13)
(5,14)
(5,15) (5,16)
(5,17)
(5,18)
d- From the transformations (2,29-31) we deduce:
∫ G ( z , q )u n ( q ) = f n ( z ), ∫ G ( z , q ) f n ( z )dμ ( z ) =u n ( q ) G ( z ' , q ) G ( z , q ) = e z 'z , And the kernel function is:
(5,19)
e- We note that the representation {p} or phase space of the oscillator is simply calculated using the Gauss integral.
3. Properties of delta function As a direct application of the generating function and the orthogonality of analytic Hilbert space , Fock or Fock-Bargmann, is the derivation of the properties of the delta function (Dirac distribution) and the normalization of the free wave: We write
q q ' = q I q ' = ∑ n u n (q ) u n (q ' )
(5,20)
This expression is invariant under the change (-q, q ') so the delta function is an even more clarifies the convergence in the function space H !!. From the orthogonality of the Fock space (2,30) we find:
q q' = ∫ ∑ u i ( q ) i, j
zi z j u j (q' )dμ( z ) = ∫ G( z, q)G( z, q' ) dμ( z ) i! j!
(5,21)
By performing the integration using the Gaussian integral we obtain:
q q' = δ (q − q' ) 1 1 exp( − (q − q ' ) 2 ) ∫ exp( +ik ( q − q ' )) dk 2π 4 Integrating this expression gives: =
a-
+∝
∫− ∝ δ ( q − q ' ) dq ' = 1
(5,22)
(5,23)
And using the expression q δ ( q ) = 0 we find: b-
q q ' = δ(q − q ') =
1 ∝ i ( q − q ') k dk ∫ e 2 π −∝
(5,24)
c- Then the constant normalization of the free wave is c = 1 / 2π . So we found the properties of the delta without recourse to the theory of distributions function.
4. The Feynman propagator of the oscillator The Feynman propagator of the oscillator was determined by several methods: the first one is the path integral, the second is the method of the Schwinger Green function, the third is the algebraic method and finally the method of direct calculation using the 25
formula of Mehler [4]. Calculations by these methods are complicated and all the books teachings give only the result. We propose to do the calculations by an elementary method: 1- The Feynman propagator of the oscillator is:
K((x, t), (x' , t 0 )) = x e
i − H ( t −t0 ) h
x' = x e
i − H ( t −t0 ) h
I x'
= e -iω(t - t 0 ) / 2 ∑ n u n ( x )e − inω(t - t 0 ) u n ( x ' )
(5,25)
By using the Fock-Bargmann space as above (5,20) we write: 1/ 2
⎛ mω ⎞ - iω (t - t 0 ) / 2 K((x, t), (x' , t 0 )) = ⎜ × ⎟ e ⎝ πh ⎠ -iω (t - t 0 ) / 2 z , q )G (e -iω (t - t 0 ) / 2 z , q ' )dμ( z ) ∫ G (e
(5,26)
And using the expression of the generating function and asking we find: 1/ 2
α
−i q 2 + q' 2 z 2 + z 2 − iα ⎛ mω ⎞ − iα / 2 2 + e ( z q + zq' ) 2 − e ]dμ( z ) ⎟ e ⎜ ∫ exp([− 2 2 ⎝ πh ⎠ 2- After integration we obtain: 1/ 2 1 ⎛ mω ⎞ − iα / 2 × K((x, t), (x' , t 0 )) = ⎜ ⎟ e π ⎝ πh ⎠ i exp[ [(q 2 + q ' 2 ) cos α − 2 xx' ]] × E1 × E 2 2 sin α α −i α 2 With E1 = ∫ exp[ − ( 2e 2 cos )[( x − ( q + q ' )) 2 ]]dx α 2 4 cos 2
E2 = ∫ exp[−(2ie
And
But
2e
−i
α 2
−i
α 2
α sin )[( y + 2
cos( α / 2 ) = (1 + e
− iα
)
2 α 4 sin 2 2ie
−i
α 2
(−q + q' ))2 ]]dy
+ ∝ − az 2
The integral of Gauss is: ∫−∝ e
sin( α / 2) = (1 − e −iα ) ,
dz =
π , with Re (a)> 0 a
3- After integration of E1 and E2 we find: 1 1 × E1 × E 2 = π with 1 ± cos α > 0 2i sin α e + iα Finally we get the Feynman propagator:
mω i exp[ [( q 2 + q ' 2 ) cos α − 2 qq ' ]] 2 π h i sin α 2 sin α 26
(5,28)
e−iα/ 2 = 1/ e+iα .
And
K((x, t), (x' , t 0 )) =
(5,27)
(5,29)
(5,30)
VI- Central potential and hydrogen atom The problem of hydrogen (AH), has played a central role in the development of quantum mechanics [27-29]. Schrödinger has resolved this problem and found the wave function in term of the coordinate representation and the energy spectrum.
1. Central Potential The Hamiltonian is of the form: p2 − h2 H = + V (r ) = Δ + V (r ) (6,1) 2μ 2m With μ is the reduced mass. We calculate the expression of the Laplacian by using the formula of Laplace-Beltrami and the metric on a variety is: (6,2) ds 2 = dx 2 + dy 2 + dz 2 = ∑ g i , j dx i dx j i,j
The expression of the Laplace-Beltrami is given by the formula ∂ ⎛⎜ ∂ ⎞⎟ 1 ( g i, j ) Δ=∑ ⎜ i, j ∂ x j ⎟⎠ ( g i, j ) ∂ xi ⎝
(6,3)
(g i , j ) Is the inverse of the matrix (g i , j ) and (x1,x2,x3)= (ρ, θ, ). A simple calculation gives: ds 2 = r 2 sin 2 θ dϕ 2 + dr 2 + r 2 dθ 2 (6,4) v ⎡ h2 1 ∂2 ⎤ L2 Hψ ( r ) = ⎢ − + + V ( r ) ⎥ ψ ( r ) = Eψ ( r ) (6,5) And 2 2μ ⎣ 2μ r ∂ r ⎦ We seek a solution by the method of separation of variables or by searching the solutions of all the commuting observables. v v . The invariance by rotation of the Hamiltonian implies that [ L2 , H ] = 0 but [ L2 , l z ] = 0 . v so the observables are H , L2 , L z .
v2
1.1 The eigenfunctions of L , L z In spherical polar coordinated we write: ∂ h ∂ h L x = ( − cos ϕ cot θ − sin ϕ ) L y = ( − cos ϕ cot θ i ∂ϕ i ∂θ r ⎡ 1 ∂ 1 ∂ h ∂ sin θ + L2 = −h 2 ⎢ Lz = ∂θ sin 2 θ i ∂ϕ ⎣ sin θ ∂θ
r2 L , L z are the spherical harmonics The eigenfunctions of
∂ ∂ ) + sin ϕ ∂ϕ ∂θ
(6,6)
∂2 ⎤ ⎥ ∂ϕ 2 ⎦
(6,7)
1/ 2
⎡ ( 2 l + 1 ) ( l − m )! ⎤ Y l m ( θ , ϕ ) = θϕ l m = ( − 1 ) m ⎢ Pl m (cos θ ) e im ϕ ⎥ ( l + m )! ⎦ ⎣ 4π m Where Pl (cos θ ) are the associated Legendre functions [A6].
27
(6,8)
In Dirac notation we write: r L 2 lm = h 2 l ( l + 1 ) lm
,
L
z
lm
= h m lm
(6,9)
l is the azimuthal quantum number and m the magnetic quantum number. l and m are integers. We deduce the dimension of the space E l = { lm }, and dim( E l ) = 2 l + 1 . If we introduce the ladder operators L+ and L- or the raising and lowering operators: (6,10) L+ =Lx+i Ly and L-=Lx-i Ly with L ± l ± l = 0 we find: L ± lm = l (l + 1) − m ( m ± 1) lm (6,11)
1.2 The radial equation The eigenvalue of the equation of energy becomes
⎡ h 2 1 ∂ 2 l (l + 1)h 2 ⎤ Hψ ( r ) = ⎢ − + + V (r )⎥ Rl (r ) = ERl (r ) 2 2μ ⎣ 2μ r ∂r ⎦
(6,12)
We note that the approximation of the potential in nuclear physics, interactions between nucleons (protons and neutrons), is the potential of the harmonic oscillator.
2. The hydrogen atom The Hamiltonian for a particle in atomic physics can be approximated by the Hamiltonian of the hydrogen atom.
2.1 The wave function and the energy of the hydrogen atom The potential of hydrogen is: e2 k V (r ) = − = , 4πε 0 r r + Since the charge of the proton is e and the electron charge eFor the bound states E } are eigenfunctions The matrix elements of the tensor operator are defined by: 1 ρjm Tqk ρ' j ' m' = ρj T k ρ' j ' j ' m' , kq ( j ' k ) jm 2 j +1
(7.43)
2
and
. (7,44)
ρj T k ρ' j ' Is a coefficient which depends of the tensor operator. 3-The rules of selection of the operator tensor: The matrix elements of the tensor operator is different from zero if:
j − j ' ≤ k ≤ j + j ' , with q = m − m ' .
Applications: The transitions of hydrogen atom are given in the table 2.
39
(7,45)
VIII- Classification of elementary particles and Quarks model 1. Introduction Dirac predicted the existence of antiparticles starting from his famous equation of the electron [A1] and in nuclear physics Elliot approximate in 58 the nuclear interaction by the quadrupole-quadrupole interaction [20] to interpret the energy levels of Mg and found that the wave function is developed on the basis of the representation of SU (3). But the discovery of new particles and resonance particle decays quickly led to the classifications of particles performed by Gell-Mann and Ne'eman [33] with the group SU (3) in 64. But the study of the scattering of electrons by protons and neutrons show that these particles have a complex structure and the discovery of strange particles has caused confusion and to break the impasse Gell-Mann was assumed that the elementary particles are formed from particles called quarks. And the introduction of the Quarks model by Gell-Mann has helped to solve the defects of this model. At first we want to give a reminder of the work of Yukawa and Heisenberg in nuclear physics as an introduction to the quark model knowing that protons and neutrons are elementary particles and are part of baryons family.
2. Yukawa potential and the particle state in nuclear physics We know that the gravitational and electromagnetic interactions have infinite range (potential 1 / r) and the binding energy of hydrogen atoms is low which means that electromagnetic interactions are also low. a- In nuclear physics, the interaction is short-range (10-12cm) and the binding energy of nuclei is very high, for example, the binding energy of deuteron energy is 2.2 MeV where the name of strong interactions. b- Yukawa proposed a potential of the strong interaction and predicted the existence of the particle (meson) assumed as the quantum field of strong interaction. c- Experience shows that the intensity its range of interaction between proton and neutron are the same and this has led Heisenberg to introduce the nucleon and the isospin.
2.1 Yukawa nuclear potential By analogy with the derivation of the Coulomb potential and that consists the spherical solution of the known equation Δ ϕ ( r , t ) = − 4 πρ ρ is the density of the charges. The solution is : ϕ (r ) = q / r
40
(9,1)
(9,2)
Where is q = ∫ ρ( r ) dv the charge of the source Yukawa start from the Klein-Gordon equation
1 ∂2 − μ 2 )ϕ ( r , t ) = 0, 2 2 c ∂t
(Δ −
(9,3)
The potential must be independent of time and directions so we seek solution of the radial part of 1 d ⎛ 2 dϕ ⎞ 2 (Δ − μ 2 )ϕ(r , t ) = 0 Æ 2 (9,4) ⎜r ⎟ − μ ϕ(r , t ) = 0 r dr ⎝ dr ⎠ Solving the last equation leads to a physically acceptable solution and that is the well known Yukawa nuclear potential: e −μr ϕ(r ) = g (9,5) r Where μ is the mass with (c = ħ = 1) and 1 / μ is range.
2.2 The state of a particle in the shell model of nuclei Heisenberg by analogy with the theory of spin considered that protons and neutrons are two aspects of a single particle, the nucleon and thus he introduced the isospin.
r
The notations in isospin particle I = ( I x , I y , I z ), and ( I , I z ) . Nucleons are in two isospin states: I , I z = 1 / 2 is the proton state. I , I z = −1 / 2 is the neutron state.
So the nucleons are fermions and the basis are the products of: The basis of the oscillator, the spin state and isospin
{n n n
s, ms I , I z
x y z
} with
nx ny nz = nx ny nz
(9,6)
ψ [ n x ,n y ,n z ] ( x, y, z) = ψ n x ( x )ψ n y ( y)ψ n z (z)
And The energy is:
E N = h ω ( N + 3 / 2 ), N = n x + n y + n z The total degeneracy of a state energy EN is given by: d
N
= (N
+ 1 )( N
(9,7) (9,8)
+ 2)
The spherical basis of the harmonic oscillator is ψ nlm ( r , θ, ϕ) = N nl ρ l Lln+1 / 2 (ρ 2 ) exp( −
With
ρ = r mω / h ,
N nl
ρ2 )Ylm (θ, ϕ) 2
⎡ 2Γ ( n + 1) ⎤ =⎢ ⎥ ⎣ Γ ( n + l + 3 / 2) ⎦
1/ 2
The principal quantum number N is connected to the radial quantum number n and azimuthal quantum number l by N = 2 ( n − 1 ) + l . 41
(9,9)
3. The Quarks model A-The elementary Particles: Elementary particles are numerous and can be classified as follows: 1-The hadrons are particles capable of strong interactions and consist of particles called quarks. Hadrons are: The Baryons particles with spin half (p, n, ...) The mesons Particles with integer spin (π ±, π0, ...) 2-The leptons are particles capable of weak and electromagnetic interactions. These particles are: The electron , muon , tau (charge negative ) (no charge) And neutrinos , , 3- Photon massless and spin 1 4 - Gravitons and massless particle of spin 2 5- Antiparticles: At any charged particle is a particle associated with the same mass and opposite charge. B-Quarks: The Quarks are characterized by the quantum numbers are summarized in the table: Quark Charge Q Charm C strange S Top T Bottom B
u 2/3 0 0 0 0
d -1/3 0 0 0 0
c 2/3 +1 0 0 0
s -1/3 0 -1 0 0
t 2/3 0 0 +1 0
b -1/3 0 0 0 -1
Table 3
4. The quantum numbers of Quarks In what follows we will be interested in the quantum numbers of quarks [33]. 1- Strangeness and hypercharge It is well known that the electromagnetic interaction is due to the electric charge Q of the particle and by analogy the field of strong interaction is due to a baryonic charge B. Similarly, the baryon charge should be additive such as the electric charge. We note that the charge: For nucleons: Q=Iz+1/2 for the proton Q=Iz-1/2 for the neutron For mesons: Q=Iz=1 for π+ Q=Iz=0 for π0 Q=Iz=-1 for π-1 Gell-Mann proposed an additive quantum number S, which he called strangeness S is equal zero for mesons and nucleons, but S = -1 for K-, K0, K+, ∑+, ∑0, ∑-. 42
The corresponding antiparticles have opposite strangeness S = +1.
S =Y −B
(9,10)
Y is the hypercharge. So the nucleons and mesons are non-strange particles. We can also deduce the relation of Gell-Mann-Nishijima:
Q = Iz + Y /2
(9,11)
2- parity Attributing intrinsic parity appears as an experimental necessity under the assumption of parity conservation in strong interaction processes. - The parity of fermions and anti fermions is negative. 3- color Baryons consist of three quarks with spin ½ and the wave function must obey the Pauli principle, but this is not the case. To eliminate this contradiction color was introduced as a new quantum number to save the Pauli principle. The color has since become the basis of the modern theory of strong interactions between Quarks. The quanta of these fields are called gluons whose numbers is eight. 4- Quarks, leptons and fundamental interactions
Quarks
u d
c s
t b
Leptons
νe e
νμ μ
ντ τ
Gluons
Photons
bosons Intermediaries
Gravitons
strong
Electromagnetic
weak
Gravitons
Table 4
5- The symmetries Laws All the interactions conserve energy, momentum, angular momentum, charge, the number of quarks, the baryon number and lepton numbers. But the weak interactions do not conserve the number of quarks. 6- The mass formula of Gell-Mann-Okubo The mass of the formula: M = m0 + aY+b[I(I+1)-Y2/4].
43
(9,12)
7- The search for Higgs bosons The theories of electroweak and strong interactions are called standard model. But this theory predicts zero mass to all particles and Higgs proposed a mechanism to give mass to these particles, but this mechanism provides for the existence of a heavy particle the Higgs boson. 8- Spectrum masses of the 8 baryons Jp=1/2+ and the 10 baryons Jp=3/2+ Masse(GeV/c2)
S=-3
1,6S=-2
1,4S=-1
Ξ0, Ξ- ‐
0
S=-2 S=0 S=-1
+
∑ ,∑ ,∑
1S=0
n, p
Jp=1/2+
p
Jp=3/2+
Table 5
5- Tables of Hadrons
p n Λ ∑+ ∑0 ∑Ξ0 Ξ-
Q
S
I
+1 0 0 +1 0 -1 0 -1
0 0 -1 -1 -1 -1 -2 -2
1/2 1/2 0 1 1 1 1/2 1/2
Quarks content uud udd uds uus uds dds ssu ssd
Baryon Jp=1/2+ Lifetime Mass 2 s MeV/c 938,27 ∞ 938,57 888,6 1115,7 2,63 10-10 1189,4 0,810-19 1192,5 7,410-20 1197,4 1,510-10 1314,9 2,9-10 1321,3 1,610-10 Baryon Jp=3/2+ 44
Decay modes
peρ pπ-, nπ0 nπ+, pπ0 Λγ nπΛπ0 Λπ-
Δ++ Δ+ Δ0 Δ∑++ ∑*0 ∑*Ξ*0 Ξ*Ω
Q
S
I
2 +1 0 -1 +1 0 -1 0 -1 -1
0 0 0 0 -1 -1 -1 -2 -2 -3
3/2 3/2 3/2 3/2 1 1 1 1/2 1/2 0
Quarks content uuu uud udd ddd uus uds dds ssu ssd sss
Mass MeV/c2 1231 1235 1234 1234 1383 1384 1387 1532 1535 1672
Lifetime
Decay modes
s 0,6 10-23 0,6 10-23 0,6 10-23 0,6 10-23 2 10-23 2 10-23 2 10-23 7 10-23 7 10-23 0,8 10-23
pπ+ p π0, n π + p π-, n π0 n πΛ π+ Λ π0 Λ πΞ-π+, Ξ0π0 Ξ-π0, Ξ0πΛK-, Ξ0π-, Ξ-π0
Table 6 : Baryons
π+, ππ0 η η‘ K+, KK 0,
Q
S
I
+1,1 0 0 0 +1,1 0
0
1
Mesons JP=0Quarks content Mass MeV/c2 139,57 u ,
0 0 0 +1,1 +1,1
1 0 0 1/2
(u -d )/√2 (u -d -2s )/√6 d s )/√6 (u u, s
1/2
134,97 548,8 957,5 493,65 497,67
d, s
Q
S
I
ρ+, ρρ0 ω φ K*+, K*-
+1,-1 0 0 0 +1,-1
1 1 0 0 1/2
K*0,
0
0 0 0 0 +1,1 +1,1
1/2
Mesons JP=1Quarks content Mass MeV/c2 774 u , 774 (u -d )/√2 781,9 (u +d )/√6 1019,4 s 891,8 u, s 896,1
d, s
Table 7: mesons 45
Lifetime
Decay modes
s 2,6 10-8
μν, eν
8,4 10-17 7 10-19 3 10-21 1,24 10-8
γγ γγ, π+π-, π+π-γ ηπ+π-, π+π-γ μν, π0, π+π-
0,89 10-10 5,17 10-8
Lifetime
π+π-, π0π0 πЄν, πμν, πππ
Decay modes
s 0,4 10-23 0,4 10-23 7 10-23 20 10-23 1 10-23 1 10-23
π0 π+ππ+π-π 0, π0γ K+K-, K0K0 π0, K0π+, K0ππ0, K0π0, K0π0
IX-The N-fermions states and the classification of chemicals elements We start from the Pauli principle, the theory of second quantization and the determination of one particle state to study the many-body system. But the approximation of the single-particle state by the hydrogen atom is very interesting for the study and the classification of chemical elements. We also give an idea of the shell model and the states of a system of fermions in the notations of the second quantization.
- The one-particle states in atomic physics The particle state is the tensor product of the wave function and the state of spin so the wave function of the ‐ th particle is: ψ [ Ei ] ( i ) = ψ
n ili m i
( ri ) sm
i
(8,1)
,
ψ n i l i m i ( ri ) Is the spatial part and the sm s is the spinor part. b-In Dirac notation we write: ψ nlm , sms = nlm sm s ,
(8,2)
d- Usually appears in the Hamiltonian the term . of the spin-orbit interaction and the wave function takes the form.: (8,3) ψ n ( ls ) jm ( r , s ) = ∑ lm l , sm s (ls ) jm ψ nlm ( r ) sm s ml , m s s
1. Pauli principle The Hamiltonian of a system of n-identical particles, such as electrons, is invariant under permutation. In the case of a two-particle system we write: H ψ (1, 2 ) = E ψ (1, 2 )
et H ψ ( 2 ,1) = E ψ ( 2 ,1)
(8,4)
The operator of permutations is defined by P ψ ( 2 ,1) = ψ (1, 2 ) The probability of presence can be written ψ (1, 2 )
2
= ψ ( 2 ,1)
2
= P ψ (1, 2 )
2
= P 2 ψ (1, 2 )
2
So P = ±1 so we find the normalized wave function must be symmetric or antisymmetric: 1 [ψ (1,2) + ψ (1,2)] ψ A = 1 [ψ (1,2) − ψ (1,2)] ψs = 2 2
46
(8,5)
(8,6)
Pauli principle: - The bosons have symmetric wave functions - Fermions have anti-symmetric wave functions - The space of bosons have integer spin: 1, 2,3, ... - The space of fermions have half-integer spin 1/2, 3/2, 5/2, ... - Bosons obey Bose-Einstein statistics - Fermions obey Fermi-Dirac statistics
2. The wave function of the system of N-fermions We assume that the one-particle states are a solution one particle Hamiltonian h, e.g. Hartree-Fock , with: (8,7) hϕ E = Eϕ E And we assume that E1
