Quantum Mechanics - Homework Assignment 10 Alejandro G´ omez omez Espinosa
∗
December 12, 2012
Shankar Shankar Ex. 14.3.2 (1) Show that the eigenvectors of σ n are given by Eq. (14.3.28). Let us write the vector n in spherical spherical coordinates: coordinates: n = sin θ cos φx + sin θ sin φy + cos θz . The Pauli Pauli matrices matrices are:
σx =
0 1 , 1 0
σy =
· − 0 i
i , 0
σz =
1 0
0 1
−
Then, σ n = σx x + σy y + σz z
· − − − − − − − − − · − − − =
0 1 0 sin θ cos φ + 1 0 i
i 1 sin θ sin φ + 0 0
0 cos θ 1
=
0 sin θ cos φ 0 + sin θ cos φ 0 i sin θ sin φ
=
cos θ sin θ cos φ i sin θ sin φ sin θ cos φ + i sin θ sin φ cos θ
=
cos θ sin θ(cos φ i sin φ) sin θ(cos φ + i sin φ) cos θ
=
cos θ sin θe iφ sin θe iφ cos θ
i sin θ sin φ cos θ + 0 0
−
Now, let us find the eigenvalues:
det(σ det(σ n
λI ) =
cos θ λ sin θeiφ
sin θe cos θ 2
−(cos θ − λ)(cos θ − λ) − sin ⇒ (cos θ − λ ) + sin θ = 0 2
2
iφ
−
λ
=0
θ =0
2
λ2 = 1
Then, the eigenvector using λ = 1:
cos θ sin θe iφ sin θeiφ cos θ
cos θx + sin θe ∗
−
−
iφ
−
y=x
[email protected]
1
x y
=1
sin θeiφ x
x y
− cos θy = y
−
0 cos θ
(1)
− | | · − − − − − − − − − − | − | · − 2sin 2θ cos sin θe iφ y x= = 1 cos θ 2sin2
θ
−
e
iφ y
−
2
θ
=
θ
cos
e θ
sin
2
iφ y
−
2
2
Taking y = 1 we found the eigenvector for this eigenvalue. Finally, normalizing it: ( θ )e iφ θ sin( ) 1 −
cos
θ iφ/2 σ n, n, 1 = sin e 2
cos = sin
2
2
θ
e iφ/2 = n+ eiφ/2 −
2
θ
2
That is the same as (14.3.28a). Following the same procedure for λ = cos θ sin θe iφ sin θe iφ cos θ
x = y
−
cos θx + sin θe
iφ
−
sin θe iφ y = cos θ + 1 −
x=
y=
x
2sin
θ
cos
θ
2cos2
θ
2
e
2
x y
1
sin θeiφ x
iφ y
1:
cos θy =
y
θ
e
sin
−
=
2
cos
2
iφ y
−
θ
2
Taking y = 1 we found the eigenvector for this eigenvalue. Finally, normalizing it: −
θ σ n, n, 1 = cos 2
( θ )e θ cos( ) 1
sin
iφ/2
e
−
iφ
2
=
2
θ
sin cos
2
θ
2
e iφ/2 = n eiφ/2 −
(2) Verify Eq. (14.3.29). (14.3.29).
n
Sn
=
=
2
=
n + S n+
| |
= = = =
2
2
2
2
n+ cos cos
2
2
2
σn
n
σx i + σy j + σz z n
n
k i i j n i + i j k
k i i j n+ i + i j k θ
e
θ
e
iφ/2
−
2
cos
iφ/2
−
2
+ cos2 =
n
± | | ± ± | | ± ± | | ± ± −− ± − − −− −− − − − θ 2
sin
θ 2
θ
sin
2
θ
sin
2
θ 2
sin2
θ
iφ
−
i
k
(sin θ cos φi + sin θ sin φ j + cos θk)
2
cos sin
θ
2
θ
2
e iφ/2 eiφ/2 −
+ (i (i i j)sin j)sin θ2 eiφ/2 θ e iφ/2 k sin θ2 eiφ/2 2 θ θ sin j+ i cos eiφ e iφ j+ 2 2
k cos 2 e j)cos (i + i j)cos
eiφ/2
eiφ + e
θ 2
k i i j i + i j k
eiφ/2
iφ/2
−
−
−
Shankar Shankar Ex. 14.3.4 Derive Eq (14.3.39) in two different ways. (1) Write σi σ j in terms of [σi , σ j ]+ and [σi , σ j ]. Let us recall some definitions: [σi , σ j ]+ = 2I δ ij ij
[σi , σ j ] = 2i
ijk σk
k
Using this relations, let us write σi σ j as the semi-sum of both relations: σi σ j =
1 ([σ ([σi , σ j ]+ + [σ [σi , σ j ]) = I δ ij ij + i 2
ijk σk
k
Hence,
× ·
(A σ ) (B σ ) =
·
Ai B j σi σ j
·
ij
=
Ai B j
Iδ ij ij + i
ij
=
ijk σk
k
Ai B j Iδ ij ij + i
Ai B j
ij
ij
= I
k
Ai B j δ ij ij + i
ijk Ai B j σk
ij
k
= I ( I (A B) + i
ijk σk
ij
(A
B)k σk
k
= I ( I (A B) + i (A
·
(2) Use Eqs. (14.3.42) and (14.3.43). This equations are: M = Let define M : M :
1 mβ = T r (M σβ ) 2
mα σα
)(σ B) = M = (σ A)(σ
·
·
Then, 1 mk = T r 2
if σk = I
× B) · σ
σi Ai σ j B j
ij
σi Ai σ j B j σk
ij
mk = =
1 2 1 2
= I
Ai B j T r(σi σ j )
ij
Ai B j 2I δ ij ij
ij
δ ij ij Ai B j
ij
= I (A B)
·
3
if σk = I
1 2
mk =
ij
1 2
=
= i
Ai B j iijl T r (σl σk )
ij
1 2
=
Ai B j T r (iijl σl σk )
ij
1 2
=
Ai B j T r ((σ ((σi σ j )σk )
Ai B j iijl (2δ (2δ lk lk )
ij
ijk Ai B j
ij
= i(A
× B)
Finally, M =
mk σk =
k
(I (A B) + i(A
·
k
I (A · B) + i (A × B) · σ × B)) σk = I (
2) Let R(α,β,γ ) be the unitary operator associated with the rotation specified by Euler angles α, β and γ , defined as a rotation of γ about the z axis, followed by a rotation of β about the y axis, followed by a rotation of α again about the z axis. a) Find explicitly the 2 2 matrix R(α,β,γ ) = e We can use relation (14.3.44) from Shankar:
×
e
iαJ z /
−
= e
iασz /2
θ 2
α 2
=
e
iβJ y /
−
cos
α I 2
1 0 0 1
α
e
iα 2
−
0
α 2
1 0
α 2
cos
α 2
0 1
0 + i sin
α 2
0 e
iα 2
iβσ y /2
−
= cos
α 2
iσz sin
i sin
i sin
2
θ θ σ 2
i sin
0
= e
=
I
= cos
−
= cos =
= 1/2.
− · − − − − − − − −
U [R(θ)] = cos Then, calculate by parts:
iαJ z / e−iβJ y / e−iγJ z / for j
−
β 2
= cos
β I 2
1 0 0 1
cos
β
sin
β
2
2
4
sin
cos
i sin
β 2
β
2
β 2
iσy sin β 2
0 i
i 0
e
iγJ z /
= e
−
iγσ z /2
=
= cos
−
i γ 2
e
0 γ ei
−
0
2
− γ I 2
γ 2
iσz sin
Then, iαJ z /
R(α,β,γ ) = e
e
−
=
=
iβJ y /
iα 2
e
0
−
0
e
iα 2
e
e
−
i γ 2
e
−
iα 2
iα 2
e e
i γ 2
−
− − cos
β
sin
β
2
−
e e
2
−
0
2
e
i γ 2
e
2
β
iα i γ 2 2
β
β
sin
iα i γ 2 2
e
2
sin
cos
2
β
cos
−
iγJ z /
−
0 γ ei 2
β
sin
2
β
cos
2
11 22
b) Find the spinor obtained by acting on jm jm =
| | with this operator, i.e., R(α,β,γ )
The spinor is: R(α,β,γ )
· 1 0
=
=
c) Show that J =
iα 2
e
−
α
e
ei e iα 2
−
α
i2
sin
cos
−
γ
i2
−
−
cos
i γ 2
e
ei e 2
γ
−
2
e
i γ 2
−
sin
β 2
β
iα i γ 2 2
−
e
e
α
γ
2
2
β
sin
ei ei cos
2
β
2
β
2
1 0
·
2
β
2
(sin β cos β cos α, sin β sin β sin α, cos β ) (as expected) in this state. Let us calculate J by parts:
2
11 11 J x 22 22
1 1 11 σx 22 2 22
=
=
= = = =
2
α
γ
2
2
ei ei cos
e
α
γ
i2 i2
−
e
α
γ
2
2
sin
ei ei cos α
γ
β
T
0 1 1 0
2
β
2
β
2
β
e
T
α
α
α
e
ei e 2
i γ 2
ei e 2
iα 2
−
−
sin
γ
i γ 2
−
γ
i2
−
β
cos
sin
2
β 2
β
2
β
2
e
β β β β cos + e iα sin cos 2 2 2 2 β β sin cos eiα + e iα 2 2
2
2
2
i2 i2
−
e
sin
eiα sin
2
e
i2
−
e
−
−
sin β cos β cos α
5
i2
−
cos
2
·
1 . 0
− − − −
11 11 J y 22 22
=
= = = =
1 1 11 σy 22 2 22
=
2
α
γ
2
2
ei ei cos iα i γ 2 2
e
e
−
α
γ
2
2
sin
ei ei cos iα i γ 2 2
β
T
0 i
2
β 2
β
T
−
iα 2
iei e
i γ 2
iα 2
i γ 2
−
2
β
i γ 2
e
cos
−
e e
α
2
iα 2
e
i 0
i γ 2
−
sin
β
cos
β
sin
β
2
β 2
2
2
e
β β β β cos + ie iα sin cos 2 2 2 2 β β sin cos i e iα eiα 2 2
2
2
2
e
−
sin
ie
−
2
e
−
ieiα sin
2
−
−
sin β sin β sin α
− − −
11 11 J z 22 22
1 1 11 σz 22 2 22
=
=
= = =
2
2
2
2
α
γ
2
2
ei ei cos
e
α
γ
i2 i2
−
e
α
γ
2
2
sin
ei ei cos
e
α
γ
i2 i2
−
cos2
e
sin
β 2
β
6
1 0
2
β
2
β
T
2
e
e
iα 2
−
e
α 2
i γ 2
−
2
β 2
γ
i2
−
i γ 2
e
−
ei e
ei e
2
iα 2
−
0 1
α
β
sin2
cos β
T
i2
−
cos
β
sin
β
2
2
γ
cos
sin
β
2
β
2
Shankar 14.4.3 We would like to study here the evolution of a state that starts out as to the B field given in Eq (14.4.27). This state obeys
1 0
and is subject
d i ψ(t) = H ψ dt
|
(2)
|
where H = γ S B, and B is time dependent. dependent. Since Since classica classicall reasonin reasoningg suggests suggests that in a frame frame rotating at frequency ( ω k) the Hamiltonian should be time independent and governed by Br (Eq. (14.4.29)), consider the ket in the rotating frame, ψr (t) , related to ψ (t) by a rotation angle ωt: ωt :
− ·
−
|
|ψr (t) = e
iωtS / |ψ(t)
|
z
−
(3)
Combine Eqs (2) and (3) to derive Schr¨ odinger’s equation for ψr (t) in the S z basis and verify that the classic classical al expectati expectation on is borne out. Solve for ψr (t) = U r (t) ψr (0) by computing U r (t), the propagator in the rotating frame. Rotate back to the lab and show that
|
|
−−−−−→ cos
|ψ(t)
|
S z
ωr t 2
basis
+ i ω ωr ω sin
ωr t
0−
iγB ωr
sin
ωr t 2
e
e+iωt/2
2
(4)
iωt/2
−
Compare this to the state n, + and see what is happening to the spin for the case ω case ω0 = ω. Calculate µz (t) and verify that it agrees with Eq. (14.4.31).
|
Let us combine (2 (2) and (3 (3):
| d i e iωtS / |ψ(t) dt iωtS / |ψ + i e iωtS / dtd |ψ d ωS z |ψ + i |ψ dt
−
ωS z e
d i ψr (t) dt
z
−
According According to (14.4.27), (14.4.27), B = B cos ωti ωt i d ωS z ψ + i dt d ωS z ψ + i dt d i dt d i dt d i dt d i e iωtS z / dt d i e iωtS z / dt
|
|ψ
=
|
|ψ |ψ |ψ |ψ |ψ |ψ
=
−
−
= = = = =
− γ
z
−
| (−γ S · B) e iωtS / |ψ (t) −γe iωtS / S · B|ψ −γ S · B|ψ
=
−
z
−
= =
z
ωt j + B k. Therefore: − B sin ωt j 0
2
z
= H ψr (t)
·
(σx i + σy j + σz k)
(B cos ωti ωt i
ωt j + B k)|ψ − B sin ωt j 0
−γ 2 (B cos ωtσx − B sin ωtσy + B σz ) |ψ −γB 2 (σx cos ωt − σy sin ωt) ωt ) |ψ − γB σz |ψ − ωS z |ψ 2 ω (σx cos ωt − σy sin ωt) ωt ) |ψ + ω S z |ψ − ωS z |ψ 2 ω e iωtS / σx eiωtS / |ψ + (ω (ω − ω )S z |ψ 2 ω σx e iωtS / |ψ + (ω (ω − ω )S z e iωtS / |ψ 2 ω σx |ψr + (ω (ω − ω )S z |ψr 2 0
0
0
z
−
−
z
z
0
0
0
7
−
z
d i ψr dt d i ψr dt d i ψr dt
| | |
where the Hamiltonian Hamiltonian H r = ωS x + (ω (ω0
= ωS x ψr + (ω (ω0
|
= (ωS x + (ω (ω0 = (ωS x + (ω (ω0
− ω)S z |ψr − ω)S z )|ψr − ω)S z )|ψr
− ω)S z does not depend upon time.
In the rotating frame, the time evolution operator is given by U r (t) = e
|ψr (t) = e
iH r t
Therefore:
iH r t
|ψr (0) = e
−
iH r t/ .
−
|ψ(0)
−
and using (3 (3):
|ψ(t) = eiωtS / e z
iH r t
|ψ(0)
−
Now, let arrange the variables in the Hamiltonian just to look compatibles with the expected result:
− − − − · · · − − · | | | − − − − − − − − − − − − H r = ωS x + (ω (ω0
where B = B x + B0
ω )S z =
ω γ
ψ(t) = eiωtS z / e
where n = ψ (t) = = = where
ωr ωr .
ω γ
γBS γB S x + γ B0
iwr tσ/2
γ B
1 ωr σ 2
. Then,
ωr t I 2
ψ(0) = eiωtσz /2 cos
S = ωr S =
r
2
ω γ
z and ωr = γ B 2 + B0
−
=
i sin
ωr t n σ 2
ψ (0)
With this result, let us represent it in the S z basis:
eiωt/2 0 iωt/ 0 e 2 eiωt/2
0
0
iωt/ 2
e
ωr t
cos
0
cos
0
2
ωr t 2
i sin
cos
ωr t
i sin
ωr t
ωr t 2
i sin
2
nz
2
ωr t
i sin
cos
nx
2
ωr t 2
nx
ωr t 2
i sin
0 1 1 + nz 1 0 0
nx
ωr t 2
nz
0 1
1 0
1 0
eiωt/2 cos ω2r t i sin ω2r t nz e iωt/2 i sin ω2r t nx −
nx =
ωrx = ωr
iγB , ωr
nz =
ωrz = ωr
iγB 0 + ω ω0 ω = ωr ωr
Hence,
ψ(t) =
ωr t
eiωt/2 cos e
2
+ i ω ωr ω sin
iωt/2 γB ωr
−
0−
ωr t
sin
ωr t 2
(5)
2
that is equal to (4 (4). In the case when ω0 = ω and γB = ωr : eiω t/2 cos e iω t/2 sin 0
ψ(t) =
−
0
ωr t 2
ωr t
that is the same as the state n+ if θ = ωr t and φ = ω0 t.
|
8
2
(6)
Finally, let us calculate µz (t) :
µz (t)
= = = =
γ γ σz (t) = ψ(t) σz ψ(t) 2 2 1 0 µz (0) ψ(t) ψ (t) 0 1
| | | − | µz (0) cos ω2r t + (ω ω− ω) sin ω2r t − γ ωB sin r r µz (0) cos ωr t + (ω − ω) − γ B sin ωr t
− − − − − − − − − − − − − − − − − − − − 2
0
2
2
2
2
=
µz (0)
=
µz (0)
=
µz (0)
=
µz (0)
2
2
0
2
2
2
2
ωr t 2
2
2
2
2 ωr t (ω0 cos2 + 2 (ω0 ω )2 ω 2 ωr2 (ω0 ω )2 ω 2 ωr2 (ω0 ω )2 ω 2 ωr2
ωr 2 ω )2 ω 2 (ω0 ω )2 ω 2 ωr t cos2 2 2 ωr ωr 2 (ω0 ω )2 ω 2 + ωr2 ωr t 2 cos ωr2 2 (ω0 ω )2 ω 2 + (ω (ω0 ω )2 + ω 2 ωr t cos2 2 ωr 2 2(ω 2(ω0 ω )2 ωr t 2 cos ωr2 2
4) A beam of j = 12 particles passes through one Stern-Gerlach apparatus aligned along the (111) direction, and then a second Stern-Gerlach apparatus aligned along the (111) dire direction. ction. (That (That is, the direc direction tion of ”measur ”measurement ement” ” of the spin has been been rotate rotated d by the tetrahe tetrahedr dral al angle.) angle.) Each apparatus apparatus only transmits ”spin-up” electrons (as defined in its own frame). What fraction of the beam leaving the first apparatus is passed by the second apparatus? From equation (14.3.28), we know that general states of the particles with j =
√
1 S + = 2
|
cos sin
θ
2
θ
2
iφ 2
e φ ei −
,
|S − −
2
1 2
are:
√ −
1 = 2
θ
sin cos
2
θ
2
e i φ ei −
φ 2
(7)
2
Since Since each each appara apparatus tus only only transm transmits its ”spin”spin-up” up” electr electrons ons,, let us calcul calculate ate the state state S + in each direction. If the Stern-Gerlach apparatus is aligned along the (111) direction: φ = π4 and θ 0.95; therefore: φ π 1 cos θ2 e i 1 0.58 58ee i π S 1+ = (8) φ 1+ = 81eei 2 sin θ2 ei 2 0.81
√ √ − ≈ − √ √ −
|
−
2
|S
2+ 2+
π 4
1 = 2
and θ cos sin
8
8
2
Then, in the direction (111): φ =
≈
π + 0. 0.58. Hence,
θ
2
θ
2
φ
e i φ ei −
2
2
9
1 = 2
π
0.58 58eei π 0.81 81ee i 8
−
−
8
(9)
To find the probability, lets use (8 (8) and (9 (9):
|S |S | 2+ 2+
1+ 1+
2
− − 1 2
=
0.58 58ee
π
i8
−
− − −
i8
0.81 81ee
2
π
0.58 58ee i π 0.81 81eei
π
8
8
2 1 iπ iπ ( 0.34 34ee 0.66 66ee ) 2 π π π π 1 0.34 34eei + 0. 0.66 66ee i 0.34 34ee i + 0. 0.66 66eei 4 π π 1 0.12 + 0. 0.22 22ee i + 0. 0.22 22eei + 0. 0.44 4 π π 1 0.56 + 0. 0.22 e i + ei 4 1 π 0.56 + 0. 0.22 cos cos 2 4 0.35
= = = = = =
−
8
8
−
8
−
−
8
4
−
8
8
4
4
4
Therefore, only 0. 0.35% of the beam leaving the first apparatus is passed by the second apparatus. 5) Find all Clebsh-Gordan coefficients for 32
⊗
1 2
=2
⊕ 1.
First let us calculate the number of bases: 3 1 =2 1 4 2=5+3=8 2 2 i.e., there are 8 states for our system. Let us enumerate them:
⊗
⊕ ⇒ ×
− − | − − − − − − | | | | | − | − | | | − | − | − | − | | | | − − √ − 33 11 , 22 22
j j1 m1 , j2 m2 =
,
33 1 1 , 22 2 2
3 1 11 3 1 11 , , , 2 2 22 2 2 2 2 jm jm = 22 , 21 , 20 , 2
,
31 11 , 22 22
,
31 1 1 , 22 2 2
,
3 3 11 3 3 1 1 , , , 2 2 22 2 2 2 2 1 , 2 2 , 11 , 10 , 1 1
,
Let us calculate the coefficients. It is easy to start from the highest state: 33 11 , 22 22
= 22
(10)
Then, using the lower operator:
J jm jm = ( j + m)( j )( j
m + 1) jm jm
−
1
Let us calculate the other values:
J 22 = (2 + 2)(2 −
|21
1 1 (J 1 + J 2 ) J 22 = 2 2 1 33 11 = = J 1 , + J 2 2 22 22 =
−
−
−
−
= =
1 2
= =
1 2
3
−
3 3 + 2 2
31 11 , 22 22
3 2
+
2 + 1) 21 = 2 21
33 11 , 22 22 33 11 , 22 22
3 +1 2
31 11 , 22 22
33 1 1 , 22 2 2
10
+
1 1 + 2 2
1 2
1 +1 2
−
33 1 1 , 22 2 2
(11)
Since 21 is orthogonal with 11 :
|
| √ |11 = 12 3
− − √ | | √ √ | √ | √ √ − √ √ √ √ − − √ √ − √ − √ − − √ | | √ − | − −
33 1 1 , 22 2 2
31 11 , 22 22
(12)
Repeating the same procedure for the next states:
J 21 = 6 20 −
20
= = =
=
− − −
1 1 1 31 11 33 1 1 J 21 = (J 1 + J 2 ) 3 , + , 2 22 22 22 2 2 6 6 1 31 11 33 1 1 31 11 3J 1 , + J 1 , + 3J 2 , + J 2 22 22 22 2 2 22 22 2 6 1 3 1 11 31 1 1 31 1 1 2 3 , + 3 , + 3 , 2 2 22 22 2 2 22 2 2 2 6 3 1 11 31 1 1 31 1 1 +2 3 , + 3 , + 3 , 2 2 22 22 2 2 22 2 2 2 3 1 11 31 1 1 , + , 22 2 2 2 2 2 22 −
−
−
−
−
−
Since 20 is orthogonal with 10 :
10 =
31 1 1 , 22 2 2
2
3 1 11 , 2 2 22
−
33 1 1 , 22 2 2
−
(13)
(14)
√
J 20 = 2 2 −
|2 − 1
= = = =
Since 2
|
| − 1 √ 3 1 1 −1 2 ,
− √ | √ − − − − − − − − − − − − − − | − − − − − | − −
−
1 1 3 1 11 J 20 = (J 1 + J 2 ) + , 22 2 2 2 2 22 2 2 1 31 1 1 3 1 11 31 1 1 3 1 11 J 1 , + , + J 2 , + , 22 2 2 2 2 22 22 2 2 2 2 22 3 1 1 1 3 1 1 1 3 1 1 1 3 1 11 2 , + , +2 , + , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 3 1 1 1 3 1 1 1 2 2 , + , 2 2 2 2 2 2 2 2 −
−
−
−
−
| − 1 is orthogonal with 1
1
(15)
1:
1 =4
3 1 1 1 , 2 2 2 2
2
3 1 1 1 , 2 2 2 2
(16)
√ | − 1 = 2|2 − 2
J 2 −
|2 − 2
= =
√ 12 J |2 − 1 −
√
= 0
− − − −
2 3 1 1 1 (J 1 + J 2 ) 2 , 2 2 2 2 2 −
−
+
3 1 1 1 , 2 2 2 2
(17)
The coefficients can be found easily from relations (10 (10)) to (17 (17)) and the rest are zero, according to (15.2.10). 11
Shankar Shankar Ex. 15.2.3 Argue that 12
1 2
⊗ ⊗
1 2
=
3 2
1 2
1 2
⊕ ⊕
. (This is just a counting argument.)
It is known that every space has (2n (2 n + 1) elements, then: 1 2
⊗ 12 ⊗ 12 2×2×2
3 1 1 2 2 2 = 4+2+2
=
⊕ ⊕
8 = 8
Or in other words, 1 2
1 2
⊗ ⊗
1 2
⊗ ⊗ ⊕⊗ ⊗ ⊕ ⊗ 1 2
=
= (1 =
1 2
3 2
1 2
0)
1 2
1
=
1 2
1 2
0
⊕ 12 ⊕ 12
7) Show explicitly that the state j j1 m1 , j2 m2 having maximally up angular momenta (that is, the state j j1 j1 , j2 j2 ) is an eigenfunction of J 2 with label j = j1 + j2 . Also Also show that the same same holds for the 2 2 2 state j j1 j1 , j2 j2 . Hint: Derive J = J 1 + J 2 + 2J 2J 1z J 2z + J 1+ 1+ J 2 + J 1 J 2+ 2+ and then use it.
|
|
| −
−
−
−
Using the hint, let us calculate each term separately: J 12 j j1 j1 , j2 j2 = 2 j1 ( j1 + 1) j j1 j1 , j2 j2 = 2 ( j12 + j1 ) j j1 j1 , j2 j2
| | | J | j j j , j j = j ( j + 1)| j j j , j j = ( j + j )| j j j , j j 2J z J z | j j j , j j = 2J z ( j | j j j , j j ) = 2 j j | j j j , j j √ j j , j j ) = √ ...√ ...| j j j + 1,1, j j − 1 J | j j j , j j = J ( ...| j 2 2
1 1
1
J 1+ 1+
2−
2
1 1
2
2 2
1 1
2
2
2 2
2 2
1 1
1
1+ 1+
2
2 2
2
1 1
1 1
2 2
2 2
2
2
2 2
1 1
1 2
2
2 2
1 1
2 2
1 1
2 2
But since j1 is the maximum angular momentum, the state j1 + 1 does not exist. Hence, J 1+ j1 j1 , j2 j2 = J 1 J 2+ j1 j1 , j2 j2 = 0 1+ J 2 j 2+ j −
|
|
−
Using this results: J 2 j j1 j1 , j2 j2
|
| | |
J 12 + J 22 + 2J 2J 1z J 2z j j1 j1 , j2 j2
=
2
=
=
2
( j1 + j2 ) + j1 + j2 j j1 j1 , j2 j2
2
(( j1 + j2 )( j )( j1 + j2 + 1)) j j1 j1 , j2 j2
2
( j( j ( j + 1)) j j1 j1 , j2 j2
=
=
j12 + j1 + j22 + j2 + 2 j 2 j1 j2 j j1 j1 , j2 j2
2
|
|
⇐⇒
j1 + j2 = j
That indicates that the state j j1 j1 , j2 j2 is an eigenfunction of J 2 .
|
Now, let us repeat the same procedure for the case of the minimum angular momentum: J 12 j j1
2
| − j , j − j = j ( j J | j j − j , j − j = j ( j 2 2
1
1
2
2
1
2
2
2
1
1
+ 1) j j1
1
2
2
2
2
1
1
2
2
2
2 1
+ j1 ) j j1
1
2
2
2
2 2
2
1
2
2
| − j , j − j = ( j + 1)| j j − j , j − j = ( j 12
| − j , j − j + j )| j j − j , j − j 1
2J 1z J 2z j j1
2
2 J z ( j | j j − j , j − j ) = 2 j j | j j − j , j − j | − j , j − j = 2J √ j − j , j − j − 1) J J | j j − j , j − j = J ( ...| j But since − j is the minimum angular momentum, the state − j − 1 does not exist. Hence, J J | j j − j , j − j = J J | j j − j , j − j = 0 1
1+ 1+
2
2
2−
1
1
1
2
2
1
2
1
2
1+ 1+
2
1
1
2
1 2
2
1
1
2
2
2
2
1+ 1+
2−
1
1
2
2
1− 2+ 2+
1
1
2
2
Using this results: J 2 j j1
| − j , j − j 2
=
| −
J 12 + J 22 + 2J 2J 1z J 2z j j1
− j = (( j + j )( j )( j + j + 1)) | j j − j , j − j = ( j( j ( j + 1)) | j j − j , j − j ⇐⇒ j + j That indicates that the state | j j − j , j − j is an eigenfunction of J . 1
2
1
1
2
1
2
2
1
1
2
2
1
2
j1 , j2 1
2
1
2
2
1
2
2
13
2
2
=j
