Q.M3 - Tirgul 7

Q.M3 - Tirgul 7
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
10.12.2014
Contents
1 Review
1
2 Sakurai Q 3.26
1
3 Sakurai Q 3.28
3
4 Sakurai Q 3.27
5
1
Review
Review base on Sakurai.... and on the review of Ben Baragiola
http://info.phys.unm.edu/~ideutsch/classes/phys531f11/sphericaltensors.
pdf
2
Sakurai Q 3.26
¬
Construct a spherical tensor of rank 1 out of two different vectors U =
(Ux , Uy , Uz ) and V = (Vx , Vy , Vz ).
(1)
b)Explicitly write T±1,0 in terms of Uxyz and Vxyz .
c) Construct a spherical tensor of rank 2 out of two different vectors U and V.
(2)
Write down explicitly T±2,±1,0 in terms of Uxyz and Vxyz .
Solution
~ and V
~ are vectors operators, then:
a) Because U
[Ui , Jj ] = i~Uk
(1)
[Vi , Jj ] = i~Vk
(2)
∗ e-mail:
¬ In
[email protected]
the new addition 3.30
1
∗
We have seen that for spherical vectors:
[Jz , Uq(1) ] = ~qUq(1)
p
(1)
[J± , Uq(1) ] = ~ (1 ∓ q)(1 ± q + 1)Uq±1
(3)
(4)
So
(1)
[Jz , Uz ] = [Jz , U0 ] = 0
(5)
so
(1)
Uz = U0
(6)
next:
√ (1)
(1)
[J+ , U0 ] = ~ 2U1 = [J+ , Uz ] = [Jx + iJy , Uz ] = −i~Uy + ii~Ux
(7)
so
(1)
U1
−1
= √ (Ux + iUy )
2
Following the same schema
√ (1)
(1)
[J− , U0 ] = ~ 2U−1 = [J− , Uz ] = [Jx − iJy , Uz ] = −i~Uy − ii~Ux
(8)
(9)
so
1
(1)
U−1 = √ (Ux − iUy )
2
(10)
In the same way we can get that:
(1)
V0
(1)
V1
(1)
V−1
= Vz
−1
= √ (Vx + iVy )
2
1
= √ (Vx − iVy )
2
(11)
(12)
(13)
b)
(1)
(1)
(1)
We can contract form Uq and Vq the T0,±1 , by using the formula
Tq(k) =
X
hk1 , k2 ; q1 , q2 |k1 , k2 ; k, qiXq(k1 1 ) Zq(k2 2 )
(14)
q1 ,q2
In our case k1 = k2 = k = 1 so
X
Tq(1) =
h1, 1; q1 , q2 |1, 1; 1, qiUq(1)
Vq(1)
1
2
q1 ,q2
2
(15)
where we need that q = q1 + q2 , so:
(1)
(1)
(1)
(1)
(1)
= h1, 1; 0, 1|1, 1; 1, 1iU0 V1 + h1, 1; 1, 0|1, 1; 1, 1iU1 V0
1
1
1
(1) (1)
(1) (1)
= √ (U1 V0 − U0 V1 ) = Uz (Vx + iVy ) − Vz (Ux + iUy )
2
2
2
T1
(1)
(1)
(1)
= h1, 1; 0, 0|1, 1; 1, 0iU0 V0
T0
(1)
(1)
(16)
(1)
+ h1, 1; −1, 1|1, 1; 1, 0iU−1 V1
(1)
(1)
(1)
+ h1, 1; 1, −1|1, 1; 1, 0iU1 V−1 = h1, 1; −1, 1|1, 1; 1, 0iU−1 V1
1
1
(1) (1)
(1) (1)
(1) (1)
+ h1, 1; 1, −1|1, 1; 1, 0iU1 V−1 = √ (U1 V−1 − U1 V−1 ) = √ (Ux Vy − Uy Vx )
2
2
(17)
(1)
(1)
T−1 = h1, 1; 0, −1|1, 1; 1, −1iU0
1
(1) (1)
(1) (1)
= √ (U−1 V0 − U0 V−1 ) =
2
(1)
(1)
(1)
V−1 + h1, 1; −1, 0|1, 1; 1, −1iU−1 V0
1
1
Uz (Vx − iVy ) − Vz (Ux − iUy )
2
2
(18)
c)
In the same way we can find higher order:
(2)
T2
(1)
(1)
= h1, 1; 1, 1|1, 1; 2, 2iU1 V1
(1)
(1)
= U1 V1
=
1
(Ux + iUy )(Vx + iVy ) (19)
2
and
(2)
(1)
(1)
(1)
(1)
= h1, 1; 1, 0|1, 1; 2, 1iU1 V0 + h1, 1; 0, 1|1, 1; 2, 1iU0 V1
1
−1
(1) (1)
(1) (1)
= √ (U1 V0 + U0 V1 ) =
(Uz Vx + Vz Ux + i(Uz Vy + Vz Uy ))
2
2
T1
(20)
.....
3
Sakurai Q 3.28
Write xy,xz, and (x2 − y 2 ) as components of a spherical (irreducible) tensor of
rank 2.
b) The expectation value
Q = ehα, j, m = j|3z 2 − r2 |α, j, m = ji
(21)
is known as the quadrupole moment. Evaluate
ehα, j, m0 |x2 − y 2 |α, j, m = ji
(22)
where m0 = j, j − 1, j − 2, · · · , in terms of Q and appropriate Clebsch-Gordan
coefficients.
3
Solution
a)We know that
r
r
15
15
2
±2iφ
=
(sin (θ))e
(sin2 (θ))(cos(φ) ± sin(φ))2
=
32π
32π
r
r
15 (x ± iy)2
15 x2 ± 2ixy − y 2
=
=
32π
r2
32π
r2
Y2±2
(23)
and
r
Y2±1
=∓
r
r
15
15 z(x ± iy)
15 zx ± izy
±iφ
=∓
(24)
(sin(θ) cos(θ))e
=∓
8π
8π
r2
8π
r2
We can see that:
r
xy = i
r
xz = i
2π 2 −2
r (Y2 − Y22 )
15
(25)
2π 2 −1
r (Y2 − Y21 )
15
(26)
and
r
x2 − y 2 =
8π 2 −2
r (Y2 + Y22 )
15
(27)
b)
r
2
2
Q = ehα, j, m = j|3z − r |α, j, m = ji = ehα, j, j|
16π 0 2
Y r |α, j, ji
5 2
using Wigner-Eckart theorem
r
hα, j||Y2 r2 ||α, ji
16π
√
Q=e
hj2; j0|j2; jji
5
2j + 1
(28)
(29)
We have seen that:
r
0
2
2
ehα, j, m |x − y |α, j, m = ji = e
8π
hα, j, m0 |r2 (Y2−2 + Y22 )|α, j, m = ji
15
(30)
and with Wigner-Eckart theorem
ehα, j, m0 |x2 − y 2 |α, j, m = ji
r
8π
hα, j||Y2 r2 ||α, ji
hα, j||Y2 r2 ||α, ji
√
√
=e
(hj, 2; j, 2|j2; jm0 i
+ hj, 2; j, −2|j2; jm0 i
)
15
2j + 1
2j + 1
(31)
So it follows that:
1 hj, 2; j, −2|j2; jm0 i
ehα, j, m0 |x2 − y 2 |α, j, m = ji = √
Q
2 hj2; j0|j2; jji
4
(32)
4
Sakurai Q 3.27
­
Consider a spinless particle bound to a fixed center by a central force potential.
Relate, as much as possible, the matrix elements
1
hn0 , l0 , m0 | ∓ √ (x ± iy)|n, l, mi
2
and
hn0 , l0 , m0 |z|n, l, mi
using only the Wigner-Eckart theorem. Make sure to state under what conditions the matrix elements are non vanishing.
Solution
As we seen before
r
1
3 ±1
∓ √ (x ± iy) =
Y
(33)
4π 1
2
and
r
z=
3 0
Y
4π 1
(34)
So
r
1
3 ±1
0 0
0
hn , l
∓ √ (x ± iy)|n, l, m1 i = hn , l , m1 |
Y1 |n, l, m1 i
4π
2
q
3
hn0 , l0 || 4π
Y1 ||n, li
√
= hl, 1; m1 , ±1|l, 1; l0 , m01 i
2l + 1
0
0
, m01 |
and
r
0
hn , l
0
, m02 |z|n, l, m2 i
0
= hn , l
0
, m02 |
3 0
Y |n, l, m2 i
4π 1
q
3
hn0 , l0 || 4π
Y1 ||n, li
0
0
√
= hl, 1; m2 , 0|l, 1; l , m2 i
2l + 1
So
hn0 , l0 , m01 | ∓
√1 (x ± iy)|n, l, m1 i
2
hn0 , l0 , m02 |z|n, l, m2 i
=
hl, 1; m1 , ±1|l, 1; l0 , m01 i
hl, 1; m2 , 0|l, 1; l0 , m02 i
(35)
We didnt done all. We need to speak on selection roles.
m0 = m + q
0
l = |l ± 1|
­ In
the new addition 3.31
5
(36)
(37)
in our case:
m01 = m1 ± 1
(38)
m02 = m2
(39)
6