Chapter 2 Angular Momentum, Hydrogen Atom, and Helium Atom

28 Pages·2011·0.32 MB·English

Chapter 2
Angular Momentum, Hydrogen
Atom, and Helium Atom
Contents
2.1 Angular momenta and their addition ..................24
2.2 Hydrogenlike atoms ...................................38
2.3 Pauli principle, Hund’s rules, and periodical table .....42
2.4 Ground state of helium atom: 1st approximation ......48
Basic Questions
A. What are the eigenvalues of angular momentum operator?
B. What are the quantum numbers of a state of the single
electron in hydrogen atom?
C. What is total electron spin of ground-state helium atom,
and the spin eigenstate?
23
24CHAPTER2. ANGULARMOMENTUM,HYDROGENATOM,ANDHELIUMATOM
2.1 Angular momentum and addition of two an-
gular momenta
2.1.1 Schro(cid:127)dinger Equation in 3D
Consider the Hamiltonian of a particle of mass m in a central potential V (r)
h(cid:22)2
H^ = 2 +V (r) :
(cid:0)2mr
Since V (r) depends on r only, it is natural to express 2 in terms of spherical
r
coordinates (r;(cid:18);’) as
1 @ @ 1 @ @ 1 @2
2 = r2 + sin(cid:18) + :
r r2@r @r! r2sin(cid:18)@(cid:18) @(cid:18)! r2sin2(cid:18)@’2
Operators in Spherical Coordinates
A. Laplacian operator
By coordinate transformation between Cartesian (x;y;z) and spherical coordi-
nates (r;(cid:18);’)
x = rsin(cid:18)cos’; r = x2 +y2 +z2
q x2 +y2
y = rsin(cid:18)sin’; tan2(cid:18) =
z2
y
z = rcos(cid:18); tan’ =
x
we have
@r @r @r
= sin(cid:18)cos’; = sin(cid:18)sin’; = cos(cid:18)
@x @y @z
@(cid:18) 1 @(cid:18) 1 @(cid:18) 1
= cos(cid:18)cos’; = cos(cid:18)sin’; = sin(cid:18)
@x r @y r @z (cid:0)r
@’ 1sin’ @’ 1cos’ @’
= ; = ; = 0 :
@x (cid:0)r sin(cid:18) @y (cid:0)r sin(cid:18) @z
Using the derivative rule
@ @r @ @(cid:18) @ @’ @
= + +
@x @x@r @x@(cid:18) @x @’
2.1. ANGULARMOMENTUMANDADDITIONOFTWOANGULARMOMENTA25
@ 1 @ 1sin’ @
= sin(cid:18)cos’ + cos(cid:18)cos’
@r r @(cid:18) (cid:0) r sin(cid:18) @’
@ @ 1 @ 1cos’ @
= sin(cid:18)sin’ + cos(cid:18)sin’
@y @r r @(cid:18) (cid:0) r sin(cid:18) @’
@ @ 1 @
= cos(cid:18) sin(cid:18)
@z @r (cid:0) r @(cid:18)
Therefore
@2 @2 @2 @ 1 @ 1sin’ @
+ + = sin(cid:18)cos’ + cos(cid:18)cos’
@x2 @y2 @z2 @r r @(cid:18) (cid:0) r sin(cid:18) @’!
@ 1 @ 1sin’ @
sin(cid:18)cos’ + cos(cid:18)cos’
(cid:2) @r r @(cid:18) (cid:0) r sin(cid:18) @’!
@ 1 @ 1cos’ @
+ sin(cid:18)sin’ + cos(cid:18)sin’
@r r @(cid:18) (cid:0) r sin(cid:18) @’!
@ 1 @ 1cos’ @
sin(cid:18)sin’ + cos(cid:18)sin’
(cid:2) @r r @(cid:18) (cid:0) r sin(cid:18) @’!
@ 1 @ @ 1 @
+ cos(cid:18) sin(cid:18) cos(cid:18) sin(cid:18)
@r (cid:0) r @(cid:18)!(cid:2) @r (cid:0) r @(cid:18)!
and this is equal to, after some considerable algebra
@2 1 @2 1 @2 2 @ cot(cid:18) @
2 = + + + +
r @r2 r2@(cid:18)2 r2sin2(cid:18)@’2 r@r r2 @(cid:18)
1 @ @ 1 @ @ 1 @2
= r2 + sin(cid:18) +
r2@r @r! r2sin(cid:18)@(cid:18) @(cid:18)! r2sin2(cid:18)@’2
1 @ @ 1
= r2 + A^((cid:18);’)
r2@r @r! r2
with de(cid:12)nition
1 @ @ 1 @2
^
A((cid:18);’) sin(cid:18) + :
(cid:17) sin(cid:18)@(cid:18) @(cid:18)! sin2(cid:18)@’2
B. Angular momentum operators
Consider (cid:12)rst the z-component, using the above formulas for @ and @
@x @y
@ @
^
L = ih(cid:22) x y
z
(cid:0) @y (cid:0) @x!
@ 1 @ 1cos’ @
= ( ih(cid:22))rsin(cid:18)cos’ sin(cid:18)sin’ + cos(cid:18)sin’
(cid:0) @r r @(cid:18) (cid:0) r sin(cid:18) @’!
26CHAPTER2. ANGULARMOMENTUM,HYDROGENATOM,ANDHELIUMATOM
@ 1 @ 1sin’ @
( ih(cid:22))rsin(cid:18)sin’ sin(cid:18)cos’ + cos(cid:18)cos’
(cid:0) (cid:0) @r r @(cid:18) (cid:0) r sin(cid:18) @’!
@
= ih(cid:22)
(cid:0) @’
similarly we can derive the x- and y-components as
@ @
L^ = ih(cid:22) sin’ cot(cid:18)cos’
x
(cid:0) (cid:0) @(cid:18) (cid:0) @’!
@ @
L^ = ih(cid:22) +cos’ cot(cid:18)sin’ :
y
(cid:0) @(cid:18) (cid:0) @’!
>From the de(cid:12)nition of the raising and lowering operators
L^ L^ iL^
(cid:6) x y
(cid:17) (cid:6)
it is straightforward to obtain
@ @
L^ = ih(cid:22)e i’ i cot(cid:18) :
(cid:6) (cid:6)
(cid:0) (cid:6) @(cid:18) (cid:0) @’!
C. Angular momentum square
In order to obtain the square of angular momentum operator in the spherical
^ 2
coordinates, consider L
x
(cid:16) (cid:17)
L^x 2 @ @ 2
= sin’ +cot(cid:18)cos’
(cid:16) h(cid:22)(cid:17)2 @(cid:18) @’!
(cid:0)
@ @ @ @ @ @
= sin’ sin’ +sin’ cot(cid:18)cos’ +cot(cid:18)cos’ sin’
@(cid:18) @(cid:18) @(cid:18) @’ @’ @(cid:18)
@ @
+cot(cid:18)cos’ cot(cid:18)cos’
@’ @’
@2 @ @ @
= sin2’ +sin’cos’ cot(cid:18) +cot(cid:18)cos2’
@(cid:18)2 @(cid:18) @’ @(cid:18)
@2 @ @2
+cot(cid:18)cos’sin’ cot2(cid:18)cos’sin’ +cot2(cid:18)cos2’
@’@(cid:18) (cid:0) @’ @’2
L^y 2 @ @ 2
= cos’ cot(cid:18)sin’
(cid:16) h(cid:22)(cid:17)2 @(cid:18) (cid:0) @’!
(cid:0)
@2 @ @ @
= cos2’ sin’cos’ cot(cid:18) +cot(cid:18)sin2’
@(cid:18)2 (cid:0) @(cid:18) @’ @(cid:18)
@2 @ @2
cot(cid:18)cos’sin’ +cot2(cid:18)sin’cos’ +cot2(cid:18)sin2’
(cid:0) @’@(cid:18) @’ @’2
2.1. ANGULARMOMENTUMANDADDITIONOFTWOANGULARMOMENTA27
now
L^2 L^2 +L^2 +L^2 @2 @ @2
= x y z = +cot(cid:18) + cot2(cid:18)+1
h(cid:22)2 h(cid:22)2 @(cid:18)2 @(cid:18) @’2
(cid:0) (cid:0) (cid:16) (cid:17)
1 @ @ 1 @2
= sin(cid:18) + = A^((cid:18);’)
sin(cid:18)@(cid:18) @(cid:18) sin2(cid:18)@’2
1 @ @ 1 L^2
= sin(cid:18) z
sin(cid:18)@(cid:18) @(cid:18) (cid:0) sin2(cid:18) h(cid:22)2
where A^((cid:18);(cid:30)) is as de(cid:12)ned before in the Laplacian 2.
r
The Schro(cid:127)dinger eq. becomes
h(cid:22)2 1 @ @ 1 @ @ 1 @2
r2 + sin(cid:18) + +V (r) (cid:9)(r;(cid:18);’)
((cid:0)2m "r2@r @r! r2sin(cid:18)@(cid:18) @(cid:18)! r2sin2(cid:18)@’2# )
= E(cid:9)(r;(cid:18);’)
or
h(cid:22)2 1 @ @ L^2
r2 +V (r) + (cid:9)(r;(cid:18);’) = E(cid:9)(r;(cid:18);’):
("(cid:0)2mr2@r @r! # 2mr2)
This eq. is separable. Let
(cid:9)(r;(cid:18);’) = R(r)Y ((cid:18);’)
where Y ((cid:18);’) is assumed to be the eigenstate of L^2 with eigenvalue (cid:21)
L^2Y ((cid:18);’) = (cid:21)Y ((cid:18);’)
we have the equation for the radial part of wavefunction
h(cid:22)2 1 @ @ (cid:21)
r2 +V (r) + R(r) = ER(r) :
("(cid:0)2mr2@r @r! # 2mr2)
The solution of this equation depends on the given potential V (r). But the solution
for the angular part of wavefunction Y ((cid:18);’) is universal and can be discussed in
general.
2.1.2 Operators and their algebra
Classically, angular momentum is de(cid:12)ned as
L = r p
(cid:2)
28CHAPTER2. ANGULARMOMENTUM,HYDROGENATOM,ANDHELIUMATOM
or in component form
L = yp zp ; L = zp xp ; L = xp yp
x z y y x z z y x
(cid:0) (cid:0) (cid:0)
in QM, they all become operator
L^ = y^p^ z^p^ ; L^ = z^p^ x^p^ ; L^ = x^p^ y^p^ :
x z y y x z z y x
(cid:0) (cid:0) (cid:0)
Their commutation relations are given by
L^ ; L^ = [y^p^ z^p^ ; z^p^ x^p^ ]
x y z y x z
(cid:0) (cid:0)
h i = [y^p^ ; z^p^ ]+[z^p^ ; x^p^ ]
z x y z
= y^[p^ ; z^]p^ +[z^; p^ ]p^ x^
z x z y
= ih(cid:22)y^p^ +ih(cid:22)p^ x^
x y
(cid:0)
^
= ih(cid:22)L
z
L^ ; L^ = ih(cid:22)L^
y z x
h^ ^ i ^
L ; L = ih(cid:22)L
z x y
h i
we can memorize these relations by
L^ L^ = ih(cid:22)L^ :
(cid:2)
In spherical coordinates, these operators are expressed as
@ @
L^ = ih(cid:22) sin’ cot(cid:18)cos’
x
(cid:0) (cid:0) @(cid:18) (cid:0) @’!
@ @
L^ = ih(cid:22) +cos’ cot(cid:18)sin’
y
(cid:0) @(cid:18) (cid:0) @’!
@
^
L = ih(cid:22)
z
(cid:0) @’
and
1 @ @ 1 @2
L^2 = h(cid:22)2 sin(cid:18) + :
(cid:0) sin(cid:18)@(cid:18) @(cid:18) sin2(cid:18)@’2!
Hence the Laplacian operator 2 and Hamiltonian take the following simple form
r
1 @ @ L^2
2 = r2
r r2@r @r (cid:0) h(cid:22)2r2
h(cid:22)2 1 @ @ 1 L^2
H^ = r2 +V (r)+
(cid:0)2mr2@r @r 2m r2
2.1. ANGULARMOMENTUMANDADDITIONOFTWOANGULARMOMENTA29
It is easy to see L^ commutes with L^2
z
L^ ; L^2 = 0
z
h i
and L^2 commute with operators H^ ,
L^2; H^ = 0 :
h i
(note: we also have L^ ;L^2 = L^ ;L^2 = 0) Therefore we can (cid:12)nd a wavefunction
x y
which is eigenfunctionhs to boith H^h;L^2; aind L^ . Furthermore, since they are separable,
z
the eigenfunction of H^ can be written in general as
(cid:9) (r;(cid:18);’) = R (r)(cid:2) ((cid:18))(cid:8) (’) = R (r)Y ((cid:18);’)
n:l;m nl l m nl l;m
with
@
L^ (cid:8) (’) = (cid:21) (cid:8) (’) i (cid:8) (’) = (cid:21) (cid:8) (’)
z m m m m m m
! (cid:0) @’
h(cid:22)2 @ @ (cid:21)2
L^2Y ((cid:18);’) = (cid:21) Y ((cid:18);’) sin(cid:18) + m (cid:2) ((cid:18)) = (cid:21) (cid:2) ((cid:18))
l;m l;m l;m ! (cid:0)sin(cid:18)@(cid:18) @(cid:18) sin2(cid:18)! l l;m l
H^(cid:9) (r;(cid:18);’) = (cid:21) (cid:9) (r;(cid:18);’)
n:l;m n;l;m n:l;m
h(cid:22)2 1 @ @ 1 (cid:21)
r2 +V (r)+ l;m R (r) = (cid:21) R (r)
! (cid:0)2(cid:22)r2@r @r 2(cid:22) r2 ! nl n;l;m nl
In Dirac notation
(cid:8) (’) m
m
! j i
Y ((cid:18);’) l;m
l;m
! j i
(cid:9) (r;(cid:18);’) n;l;m :
n:l;m
! j i
^ ^ ^
Instead ofdealing with L ;L , and L , one can de(cid:12)ne angular raising and lowering
x y z
operators L^+ and L^ as,
(cid:0)
@ @
L^ L^ iL^ = ih(cid:22)e i’ i cot(cid:18)
(cid:6) x y (cid:6)
(cid:17) (cid:6) (cid:0) (cid:6) @(cid:18) (cid:0) @’!
and we have the equivalent set L^ ;L^ , and L^ or L^ ;L^+ and L^ . The algebra for
x y z (cid:0) z
this 2nd set is more convenient. It is easy to show the commutations between three
operators L ; L are given by
z (cid:6)
L^ ; L^ = h(cid:22)L^ ; L^+; L^ = 2h(cid:22)L^ :
z (cid:6) (cid:6) (cid:0) z
(cid:6)
h i h i
30CHAPTER2. ANGULARMOMENTUM,HYDROGENATOM,ANDHELIUMATOM
Note: Compare this algebra with harmonic oscillator a^; a^ = 1. These algebra will
y
determine the eigenfunctions and eigenvalues of L^2. Ithis alsio easy to derive
L^+L^ = L^2 L^2 +h(cid:22)L^ ;
(cid:0) (cid:0) z z
L^ L^+ = L^2 L^2 h(cid:22)L^ ;
(cid:0) (cid:0) z (cid:0) z
and
1
L^2 = L^2 +h(cid:22)L^ +L^ L^+ = L^2 h(cid:22)L^ +L^+L^ = L^2 + L^+L^ +L^ L^+ :
z z (cid:0) z (cid:0) z (cid:0) z 2 (cid:0) (cid:0)
(cid:16) (cid:17)
Note: compare this with harmonic oscillator H^ = h(cid:22)! a^ a^+ 1 .
y 2
(cid:16) (cid:17)
2.1.3 Eigenvalues and eigenfunctions
(i) Solution for (cid:8) (’) is simple
m
@
ih(cid:22) (cid:8) (’) = (cid:21) (cid:8) (’)
m m m
(cid:0) @’
1
(cid:8) (’) = ei(cid:21)m’=h(cid:22)
m
p2(cid:25)
and using spherical symmetry of the system, we must have
(cid:21)
m
(cid:8) (’+2(cid:25)) = (cid:8) (’) = m = 0; 1; 2;
m m
! h(cid:22) (cid:6) (cid:6) (cid:1)(cid:1)(cid:1)
therefore
L^ m = mh(cid:22) m ; m = 0; 1; 2; :
z
j i j i (cid:6) (cid:6) (cid:1)(cid:1)(cid:1)
(ii) Consider the eigenequations
L^2 l;m = (cid:21) l;m
l;m
j i j i
it is obvious l;m is also eigenfunction of L^
z
j i
L^ l;m = h(cid:22)m l;m
z
j i j i
we next examine the e(cid:11)ect of L^+ and L^ when acting on l;m . We divide the whole
(cid:0)
j i
process by following steps.
(a) Using the commutation relationship L^ ; L^ = h(cid:22)L^ , we have
z (cid:6) (cid:6)
(cid:6)
h i
^ ^ ^ ^ ^ ^ ^ ^
L L l;m = L L L l;m = h(cid:22)mL h(cid:22)L l;m = h(cid:22)(m 1)L l;m
z (cid:6) (cid:6) z (cid:6) (cid:6) (cid:6) (cid:6)
j i (cid:6) j i (cid:6) j i (cid:6) j i
(cid:16) (cid:17) (cid:16) (cid:17)
2.1. ANGULARMOMENTUMANDADDITIONOFTWOANGULARMOMENTA31
therefore, L^+ e(cid:11)ectively increases m by a unity and L^ e(cid:11)ectively decreases m by a
(cid:0)
unity, hence we write
L^ l;m = C (l;m) l;m 1
(cid:6)
j i (cid:6) j (cid:6) i
where C (l;m) are constants. Therefore, the states l;m with m = 0; 1; 2;
can be re(cid:6)lated by raising or lowering operator L^ . j i (cid:6) (cid:6) (cid:1)(cid:1) (cid:1)
(cid:6)
(b) L^2 degeneracy. The eigenvalue of L^2 is independent of m, (cid:21) = (cid:21)
l;m l
L^2 l;m = (cid:21) l;m :
l
j i j i
Proof: using L^2; L^ = 0; we have
(cid:6)
h i
L^2L^ l;m = C L^2 l;m 1 = (cid:21) C l;m 1 = (cid:21) L^ l;m
(cid:6) l;m 1 l;m 1 (cid:6)
j i (cid:6) j (cid:6) i (cid:6) (cid:6)j (cid:6) i (cid:6) j i
= L^ L^2 l;m = (cid:21) L^ l;m
(cid:6) l;m (cid:6)
j i j i
hence
(cid:21) = (cid:21) (cid:21)
l;m 1 l;m l :
(cid:6) !
(c)Foragiven(cid:21) (orl), thevaluesofmmustbebounded. Because L^2 +L^2 >
l x y
0, L^2 = L^2 +L^2 +L^2 L^2 +L^2 +h(cid:22)2m2. Let this maximum value ofD(cid:16)m be M(cid:17)E l,
x y z ! x y j j l (cid:17)
remember l is only an index for labeling. Hence
l;m ; m = 0; 1; 2; ; l :
j i (cid:6) (cid:6) (cid:1)(cid:1)(cid:1) (cid:6)
Altogether, there are (2l+1) values of m.
(d) Consider state l;l . Since its m is at maximum, cannot increase any more,
j i
we must have
L^+ l;l = 0
j i
hence
L^2 l;l = L^2 +h(cid:22)L^ +L^ L^+ l;l = h(cid:22)2 l2 +l l;l
j i z z (cid:0) j i j i
= h(cid:22)(cid:16)2l(l+1) l;l : (cid:17) (cid:16) (cid:17)
j i
Using the degeneracy of L^2, we have
L^2 l;m = h(cid:22)2l(l+1) l;m ; m = 0; 1; 2; ; l :
j i j i (cid:6) (cid:6) (cid:1)(cid:1)(cid:1) (cid:6)
(e) We can now determine the coe(cid:14)cients C (l;m)
(cid:6)
^
L l;m = C (l;m) l;m 1 :
(cid:6)
j i (cid:6) j (cid:6) i
32CHAPTER2. ANGULARMOMENTUM,HYDROGENATOM,ANDHELIUMATOM
Using the fact that L^ y = L^+, assuming l;m and l;m 1 have already been
(cid:0)
j i j (cid:6) i
normalized, we have(cid:16) (cid:17)
L^+ l;m = C (l;m) l;m+1
+
j i j i
C2 (l;m) = l;m L^ L^+ l;m = l;m L^ L^+ l;m
+ h j (cid:0) j i h j (cid:0) j i
= (cid:16)l;m L^2 (cid:17)(cid:16)L^2 h(cid:22)L^ (cid:17) l;m = h(cid:22)2[l(l+1) m(m+1)]
h j (cid:0) z (cid:0) z j i (cid:0)
(cid:16) (cid:17)
or
C (l;m) = h(cid:22) l(l+1) m(m+1)
+
(cid:0)
q
similarly
C (l;m) = h(cid:22) l(l+1) m(m 1) :
(cid:0) (cid:0) (cid:0)
(f) From the de(cid:12)nition of L^+ in spqherical coordinate
@ @
L^+ = ih(cid:22)ei’ i cot(cid:18)
(cid:0) @(cid:18) (cid:0) @’!
and using the fact that
L^+ l;l = 0
j i
we derive the equation for the highest-order spherical harmonic function, l;l =
j i
Y ((cid:18);(cid:30))
ll
@ ( 1)l (2l+1)!!
lcot(cid:18) Y = 0; Y = ceil(cid:30)sinl(cid:18); c = (cid:0)
@(cid:18) (cid:0) ! ll ll p4(cid:25) vu (2l)!!
u
t
where c is the normalization constant, determined by
(cid:25) (cid:25)=2 (cid:25) (cid:25)=2 (cid:25)=2
2(cid:25) sin2l+1(cid:18)d(cid:18) = 2(cid:25) + sin2l+1(cid:18)d(cid:18) = 2(cid:25) sin2l+1(cid:18)d(cid:18)+ cos2l+1(cid:18)d(cid:18)
Z0 Z0 Z(cid:25)=2! Z0 Z0 !
(cid:25)=2 (2l)!!
= 4(cid:25) sin2l+1(cid:18)d(cid:18) = 4(cid:25) :
(2l+1)!!
Z0
By applying L repeatedly, we obtain the general spherical harmonic, Y ((cid:18);’).
(cid:0) lm
(iii) For a general angular momentum L^, all above derivations are still valid,
except that we do not have the representation L^ = ih(cid:22)@=@’, etc. And the corre-
z
(cid:0)
sponding rotational invariance (cid:8)(’+2(cid:25)) = (cid:8)(’) m = 0; 1; 2; are hence
! (cid:6) (cid:6) (cid:1)(cid:1) (cid:1)
not necessarily true. Hence we derive the following general angular momentum
theorem,
L^2 l;m = h(cid:22)2l(l+1) l;m
j i j i
L^ l;m = h(cid:22)m l;m
z
j i j i
m = l; l +1; ;l 1;l
(cid:0) (cid:0) (cid:1)(cid:1)(cid:1) (cid:0)