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E 1 0 0 0 0 . . . H =0 E 2 0 0 0 . . . 0 0 E 3 0 0 . . . 0 0 0 E 4 0 . . . :. 1 0 0 : ·. 0 0 1 : ·. 0 1 0 : ·. with the “basis set” :. ,. ,. ,. - PowerPoint PPT Presentation
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E1 0 0 0 0 . . .H = 0 E2 0 0 0 . . . 0 0 E3 0 0 . . .
0 0 0 E4 0 . . .:
.100
:
·
010
:
·
001
:
·, , ,
...with the “basis set”:
This is not general at all (different electrons, different atoms require different matrices)Awkward because it provides no finite-dimensional representation
That’s why its desirable to abstract the formalism
azrea
z /23
100
1
aZrea
Zr
a
Z 2/23
200 21
2
1
iaZr erea
Z
sin
8
1 2/25
121
cos2
1 2/25
210aZrre
a
Z
iaZr erea
Z sin
8
1 2/25
211
aZrea
rZ
a
Zr
a
Z 3/2
2223
300 21827381
1
Hydrogen Wave Functions 10000 ::
01000 ::
00100 ::
00010 ::
00001 :: 0
0000
Angular Momentum|nlmsms…>
l = 0, 1, 2, 3, ...Lz|lm> = mh|lm> for m = l, l+1, … l1, lL2|lm> = l(l+1)h2|lm>
Sz|lm> = msh|sms> for ms = s, s+1, … s1, sS2|lm> = s(s+1)h2|sms>
Of course |nℓm> is dimensional again
But the sub-space of angular momentum(described by just a subset of the quantum numbesrs)
doesn’t suffer this complication.
can measure all the spatial (x,y,z) components
(and thus L itself) of vmrL
not even possible in principal !
rixyx
irL
,,
ix
yy
xiLzSo, for
example
azimuthalangle inpolar
coordinates
Angular Momentumnlml…
Lz lm(,)R(r) = mħ lm(,)R(r)
for m = l, l+1, … l1, l
L2lm(,)R(r)= l(l+1)ħ2lm(,)R(r)
l = 0, 1, 2, 3, ...
Measuring Lx alters Ly (the operators change the quantum states).The best you can hope to do is measure:
States ARE simultaneously eigenfunctions of BOTH of THESE operators!We can UNAMBIGUOULSY label states with BOTH quantum numbers
ℓ = 2mℓ = 2, 1, 0, 1, 2
L2 = 2(3) = 6|L| = 6 = 2.4495
ℓ = 1mℓ = 1, 0, 1
L2 = 1(2) = 2|L| = 2 = 1.4142
2
1
0
1
0
Note the always odd number of possible orientations:
A “degeneracy” in otherwise identical states!
Spectra of the alkali metals
(here Sodium)all show
lots of doublets
1924: Pauli suggested electrons posses some new, previously un-recognized & non-classical 2-valued property
Perhaps our working definition of angular momentum was too literal…too classical
perhaps the operator relations
yzxxz
xyzzy
zxyyx
LiLLLL
LiLLLL
LiLLLL
may be the more fundamental definition
Such “Commutation Rules”are recognized by mathematicians as
the “defining algebra” of a non-abelian
(non-commuting) group[ Group Theory; Matrix Theory ]
Reserving L to represent orbital angular momentum, introducing the more generic operator J to represent any or all angular momentum
yzxxz
xyzzy
zxyyx
JiJJJJ
JiJJJJ
JiJJJJ
study this as an algebraic group
Uhlenbeck & Goudsmit find actually J=0, ½, 1, 3/2, 2, … are all allowed!
ms = ± 12
spin “up”spin “down”
s = ħ = 0.866 ħ 3 2
sz = ħ 12
| n l m > | > = nlm12
12
10( )
“spinor”
the most general state is a linear expansion in this 2-dimensional basis set
1 0 0 1( ) = + ( ) ( )
with 2 + 2 = 1
spin : 12p, n, e, , , e , , , u, d, c, s, t, b
leptons quarks
the fundamental constituents of all matter!
SPINORBITAL ANGULAR
MOMENTUMfundamental property
of an individual componentrelative motionbetween objects
Earth: orbital angular momentum: rmv plus “spin” angular momentum: I in fact ALSO “spin” angular momentum: Isunsun
but particle spin especially that of truly fundamental particlesof no determinable size (electrons, quarks)
or even mass (neutrinos, photons)
must be an “intrinsic” property of the particle itself
Total Angular Momentumnlmlsmsj… l = 0, 1, 2, 3, ...
Lz|lm> = mħ|lm> for m = l, l+1, … l1, lL2|lm> = l(l+1)ħ2|lm>
Sz|lm> = msħ|sms> for ms = s, s+1, … s1, sS2|lm> = s(s+1)ħ2|sms>
In any coupling between L and S it is the TOTAL J = L + s that is conserved.
ExampleJ/ particle: 2 (spin-1/2) quarks bound in a ground (orbital angular momentum=0) stateExamplespin-1/2 electron in an l=2 orbital. Total J ?
Either3/2 or 5/2possible
BOSONS FERMIONS
spin 1 spin ½ e,p, n,
Nuclei (combinations of p,n) can have
J = 1/2, 1, 3/2, 2, 5/2, …
BOSONS FERMIONS
spin 0 spin ½
spin 1 spin 3/2
spin 2 spin 5/2 : :
“psuedo-scalar” mesons
quarks and leptonse,, u, d, c, s, t, b,
Force mediators“vector”bosons: ,W,Z“vector” mesonsJ
Baryon “octet”p, n,
Baryon “decupltet”
Combining any pair of individual states |j1m1> and |j2m2> forms the final “product state”
|j1m1>|j2m2>
What final state angular momenta are possible?What is the probability of any single one of them?
Involves “measuring” or calculating OVERLAPS (ADMIXTURE contributions)
|j1m1>|j2m2> = j j1 j2;
m m1 m2 | j m >
j=| j1j2 |
j1j2
Clebsch-Gordon coefficients
or forming the DECOMPOSITION into a new basis set of eigenvectors.
Matrix Representationfor a selected j
J2|jm> = j(j+1)h2| j m >Jz|jm> = m h| j m > for m = j, j+1, … j1, jJ±|jm> = j(j +1)m(m±1) h | j, m1 >
The raising/lowering operators through which we identifythe 2j+1 degenerate energy states sharing the same j.
J+ = Jx + iJy
J = Jx iJy
2Jx = J+ + J Jx = (J+ + J )/2
Jy = i(J J+)/2
adding
2iJy = J+ J
subtracting
The most common representation of angular
momentum diagonalizes the Jz operator:
<jn| Jz |jm> = mmn
1 0 00 0 00 0 -1
Jz =(j=1)
2 0 0 0 00 1 0 0 00 0 0 0 00 0 0 -1 00 0 0 0 -2
Jz =(j=2)
J | 1 1 > =
J±|jm> = j(j +1)m(m±1) h | j, m1 >
J | 1 0 > =
J | 1 -1 > =
J | 1 0 > =
J | 1 -1 > =
J | 1 1 > =
| 1 0 > 2
| 1 -1 > 2
0
| 1 0 > 2
| 1 1 > 2
0
J =
J =
< 1 0 |0 0 0 0 00 02
2
0 00 0 0 0 0
22
< 1 -1 |
< 1 0 |
< 1 1 |
0210
21021
0210
020
202
020
2
1 xJ
020
202
020
020
202
020
2
1
i
ii
i
i
i
i
iJ y
100
010
001
zJ
For J=1 states a matrix representation of the angular momentum operators
Which you can show conform to the COMMUTATOR relationship
you demonstrated in quantum mechanics for the differential operators
of angular momentum
[Jx, Jy] = iJz
Jx Jy Jy Jx =
100
000
001
0
000
0
0
000
0
22
22
22
22i
ii
ii
ii
ii
= iJz
100
010
0011 z
JJ
2/10
02/12/1 zJJ
2/3000
02/100
002/10
0002/3
2/3 zJJ
20000
01000
00000
00010
00002
2 zJJ
x
y
zz′R(1,2,3) =
11
11
cossin0
sincos0
001
1
y′
1
=x′
x
y
zz′R(1,2,3) =
11
11
cossin0
sincos0
001
1
y′
1
=x′
2
2
2 x′′
z′′
=y′′
22
22
cos0sin
010
sin0cos
x
y
zz′R(1,2,3) =
11
11
cossin0
sincos0
001
1
y′
1
=x′
2
2
2 x′′
z′′
=y′′
22
22
cos0sin
010
sin0cos
3
y′′′
z′′′ =
x′′′
3
3
100
0cossin
0sincos
33
33
R(1,2,3) =
11
11
cossin0
sincos0
001
22
22
cos0sin
010
sin0cos
100
0cossin
0sincos
33
33
These operators DO NOT COMMUTE!
about x-axis
about y′-axis
about z′′-axis1st
2nd3rd
Recall: the “generators” of rotations are angular momentum operators and they don’t commute!
but as nn
Infinitesimal rotations DO commute!!
10
010
01
100
01
01
2
2
3
3
100
01
01
10
010
01
3
3
2
2
10
01
1
2
3
23
10
01
1
2
3
23
),,( 321 R 1 3 0-3 1 0
0 0 1
1 0 -2 0 1 0
2 0 1
1 0 0 0 1 1
0 -1 1R(1, 2, 3 ) =
1 3 -2 -3 1 0
2 0 1
1 0 0 0 1 1
0 -1 1
=
1 0 -2 0 1 1
2 -1 1
= 1 3 0-3 1 0
0 0 1
or
1 3 -2 -3 1 1
2 -1 1
=
R(1, 2, 3 ) = 1 3 -2 -3 1 1
2 -1 1
R(1, 2, 3 ) =
100
010
001
1
010
100
000
2
001
000
100
+ 3
000
001
010
1R(1, 2, 3 ) =If we imagine building up to full rotations by applying this repeatedly N times
ℓim [R(1, 2, 3 )]N
N
NN )/1(
ℓim N
R(1, 2, 3 ) NN )/1(
ℓim
N
e
Which we can re-write in the form / ie
by slightly re-writing the “vector” components:
00
00
000
1
i
i
00
000
00
2
i
i
000
00
00
1 i
i
Check THIS out:
3222
1221
000
001
010
000
000
010
000
001
000
i
We have found a “new” representation of Lx , Ly , Lz!!
Our alternate approach to motivating /
ieRwhere the generator was anhonest-to-goodness matrix!
00
00
000
1
i
i
00
000
00
2
i
i
000
00
00
3 i
i
gave
representing 3-dimensional rotations
with the basis:
0
2/
2/1
,
1
0
0
,
0
2/
2/1
ii
A B C
A
B
C
satisifes all thesame arithmetic
as Lx, Ly, Lz
321, iwhich we argue
000
00
00
3 i
i
Notice can try diagonalizing the zth matrix:
0)1(
00
0
02
i
i
1,0 Eigenvalues of 1, 0, 1
We should be able to diagionalize 3 by a SIMILARITY TRANSFORMATION!
U 3U =†
100
000
001
010
0
0
000
00
00
0
100
0
22
21
21
221
221
ii
i
i
i
i
U 3U =†
100
000
001
000
0
0
0
100
0
22
21
21
221
221
ii
i
i
0210
21021
0210
020
202
020
2
1 xJ
020
202
020
020
202
020
2
1
i
ii
i
i
i
i
iJ y
100
010
001
zJ
For J=1 states a matrix representation of the angular momentum operators
