Today’s Outline - October 29, 2014mypages.iit.edu/~segre/phys405/14F/lecture_17.pdf · Today’s Outline - October 29, 2014 ... Spinors Reading Assignment: Chapter 4.4 Homework

Today’s Outline - October 29, 2014

• Problem 4.17

• Eigenfunctions of angular momentum

• Problem 4.21

• Intrinsic angular momentum

• Spinors

Reading Assignment: Chapter 4.4

Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, November 05, 2014


C. Segre (IIT) PHYS 405 - Fall 2014 October 29, 2014 1 / 17


• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors






• Problem 4.17


• Problem 4.21


• Spinors





Problem 4.17

Consider the earth-sun system as a gravitational analog to the hydrogenatom.

(a) What is the potential energy function?

(b) What is the “Bohr radius,” ag , for this system?

(c) Write down the gravitational “Bohr formula,” and by equating En tothe classical energy of a planet in a circular orbit of radius ro , showthat n =

√ro/ag . From this estimate the quantum number n of the

earth.

(d) Suppose the earth made a transition to the next lower level (n − 1).How much energy would be released? what would be the wavelengthof the emitted photon (graviton) be?


Problem 4.17 – solution

(a) The potential for a gravitationalsystem is

thus, by analogy with the hydrogenatom

(b) The real Bohr radius is

a =

(4πε0e2

)~2

m

V (r) = −GMm

re2

4πε0→ GMm

the gravitational Bohr radius is thus

ag =~2

GMm2= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

r

e2

4πε0→ GMm


ag =~2

GMm2= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

r

e2

4πε0→ GMm


ag =~2

GMm2= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

re2

4πε0→ GMm


ag =~2

GMm2= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

re2

4πε0→ GMm


ag =~2

GMm2= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

re2

4πε0→ GMm


ag =~2

GMm2

= 2.34× 10−138m






a =

(4πε0e2

)~2

m

V (r) = −GMm

re2

4πε0→ GMm


ag =~2

GMm2= 2.34× 10−138m



(c) The energy for the hydrogenatom is

En = −

[m

2~2

(e2

4πε0

)2]1

n2

the analogous energy for the gravi-tational quantum system is

En = −[ m

2~2(GMm)2

] 1

n2

Equating this gravitational quantum energy with the classical energy of acircular orbit

Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro

1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro

= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2

−→ n2 =GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro

=roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074




En = −

[m

2~2

(e2

4πε0

)2]1

n2


En = −[ m

2~2(GMm)2

] 1

n2


Ec =1

2mv2 − G

Mm

ro

= GMm

2ro− G

Mm

ro= −GMm

2ro

GMm

r2o=

mv2

ro1

2mv2 =

GMm

ro

− GMm

2ro= −

[ m

2~2(GMm)2

] 1

n2−→ n2 =

GNm2

~2ro =

roag

= 2.53× 1074



(d) The energy released in a transition to the next lower state is

∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly

This makes sense because λ = cT and for the earth-sun system, T = 1 y




∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]

1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2

≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)

∆E ≈ −[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]

=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J

λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E

= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m

≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly





∆E = −[G 2M2m3

2~2

] [1

(n − 1)2− 1

n2

]1

(n − 1)2=

1

n2(1− 1/n)2≈ 1

n2

(1 +

2

n

)∆E ≈ −

[G 2M2m3

2~2

] [1

n2

(1 +

2

n

)− 1

n2

]=

[G 2M2m3

2~2

]2

n3

∆E ≈ 2.09× 10−41J λ =hc

∆E= 9.52× 1015m ≈ 1ly



Angular momentum eigenfunctions

We know that the eigenfunctions ofangular momentum must have twovalid quantum numbers, l and m

start by rewriting the componentsof ~L in spherical coordinates

~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0

, r × θ = φ, r × φ = −θ

~L =~i

[

φ∂

∂θ− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ

, r × φ = −θ

~L =~i

[φ∂

∂θ

− θ 1

sin θ

∂

∂φ

]





~L =~i

(~r × ~∇)

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ

~r = r r

~L =~i

[r(r × r)

∂

∂r+ (r × θ)

∂

∂θ+ (r × φ)

1

sin θ

∂

∂φ

]

r × r = 0, r × θ = φ, r × φ = −θ

~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]


Angular momentum in spherical coordinates

~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]

φ = −x sinφ+ y cosφ

θ = x(cos θ cosφ) + y(cos θ sinφ)− z sin θ

~L =~i

[(−x sinφ+ y cosφ)

∂

∂θ

−(x cos θ cosφ+ y cos θ sinφ− z sin θ)1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂

∂θ− cot θ cosφ

∂

∂φ

)Ly =

~i

(+ cosφ

∂

∂θ− cot θ sinφ

∂

∂φ

)Lz =

~i

∂

∂φ



~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]φ = −x sinφ+ y cosφ


~L =~i


∂

∂θ


sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ



~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ



~L =~i


∂

∂θ


sin θ

∂

∂φ

]

Lx =~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ



~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ



~L =~i


∂

∂θ


sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)

Ly =~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ



~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ



~L =~i


∂

∂θ


sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)

Lz =~i

∂

∂φ



~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ



~L =~i


∂

∂θ


sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φC. Segre (IIT) PHYS 405 - Fall 2014 October 29, 2014 7 / 17

Raising and lowering operators in spherical coordinates

The raising and lowering operators become

L± = Lx ± iLy

=~i

[(− sinφ± i cosφ)

∂

∂θ− (cosφ± i sinφ) cot θ

∂

∂φ

]= ~

[(± cosφ+ i sinφ)

∂

∂θ+ i(cosφ± i sinφ) cot θ

∂

∂φ

]= ±~(cosφ± i sinφ)

[∂

∂θ± i cot θ

∂

∂φ

]

However, cosφ± i sinφ = e±iφL± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]This is worked out in Problem 4.21




L± = Lx ± iLy

=~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ

]However, cosφ± i sinφ = e±iφ

L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]

= ~[

(± cosφ+ i sinφ)∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ

]

= ±~(cosφ± i sinφ)

[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ

]

However, cosφ± i sinφ = e±iφL± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]

L+L− = −~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ





L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]

This is worked out in Problem 4.21




L± = Lx ± iLy =~i


∂


∂

∂φ

]= ~


∂


∂

∂φ


[∂

∂θ± i cot θ

∂

∂φ


L± = ±~e±iφ

[∂

∂θ± i cot θ

∂

∂φ

]L+L− = −~2e iφ

(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ



Problem 4.21

(a) Derive

L+L− = −~2(∂2

∂θ2+ cot θ

∂

∂θ+ cot2 θ

∂2

∂φ2+ i

∂

∂φ

)(b) Derive

L2 = −~2[

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]


Problem 4.21(a) – solution

L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]

=− ~2[

∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

[

∂2

∂θ2+ i(csc2θ − cot2θ)

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)Now it is possible to construct the spherical coordinates form of L2



L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[

∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2

+ i csc2θ∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ

− i cot θ∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)

+ i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ

+ cot2θ∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2

∂θ2

+ i(csc2θ − cot2θ)∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ

+ cot θ∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ

+ cot2θ∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)

Now it is possible to construct the spherical coordinates form of L2



L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2



Problem 4.21(b) – solution

From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)

− ~2∂2

∂φ2− ~

(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2

]but this is just the separated angular equation for the hydrogen atom andthe solutions are the spherical harmonics

L2Yml = ~2l(l + 1)Ym

l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)

− ~2∂2

∂φ2− ~

(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2

− ~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ i

∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2− ~

(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)

= −~2[

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2

]

but this is just the separated angular equation for the hydrogen atom andthe solutions are the spherical harmonics


l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

��

i∂

∂φ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l LzYml = ~mYm

l


Intrinsic angular momentum

In quantum mechanics, we deal with small particles, which can, like anelectron be described as “point” particles. We have seen the orbitalangular momentum, characterized by l and m. These particles also carryanother intrinsic form of angular momentum which we call “spin” angularmomentum even though it is not associated with any physical variables ofrotation

Spin behaves in all ways as orbital angular momentum

the spin operator obeyscommutation relations

there exist eigenvectors ofthe spin operator

and there are ladder oper-ators

there are no restrictionsforcing integer spin

[Sx ,Sy ] = i~SzS2|s,m〉 = ~2s(s + 1)|s,m〉Sz |s,m〉 = ~m|s,m〉

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s









[Sx ,Sy ] = i~Sz

S2|s,m〉 = ~2s(s + 1)|s,m〉Sz |s,m〉 = ~m|s,m〉

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s









[Sx ,Sy ] = i~Sz

S2|s,m〉 = ~2s(s + 1)|s,m〉Sz |s,m〉 = ~m|s,m〉

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s









[Sx ,Sy ] = i~SzS2|s,m〉 = ~2s(s + 1)|s,m〉

Sz |s,m〉 = ~m|s,m〉

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0,1

2, 1,

3

2, . . .

m = −s,−s + 1, . . . , s − 1, sC. Segre (IIT) PHYS 405 - Fall 2014 October 29, 2014 12 / 17

Spin 12

simplest system to work out, only 2states

a better representation is that ofa two element vector called a“spinor”

with a general state represented by

operators are 2× 2 matrices

consider the S2 operator

S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up

∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

)

, χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


Spin 12






S2 =

(c de f

)

∣∣∣∣12 +1

2

⟩, spin up∣∣∣∣12 − 1

2

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± =3

4~2χ±


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)

(ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)

c =3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)

(df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)

d = 0, f =3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)

(ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)

c =~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)

(df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)

d = 0, f = −~2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sx =~2

(1 00 −1

)


Spin ladder matrices

We can now proceed to find the matrix representation of the spin raisingand lowering operators. We take the general definition and apply it to thecase where S = 1

2

S±|s,m〉 = ~√s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+

= ~√

3

4+

1

4

χ+

= ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0

thus the two operators become

S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+

= ~√

3

4+

1

4

χ+

= ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+

= ~√

3

4+

1

4

χ+

= ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+

= ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ−

= ~√

3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ−

= ~χ−(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)

(c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)

d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

)

, S− = ~(

0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~

√1

2

(1

2+ 1

)−(−1

2

)(−1

2+ 1

)χ+ = ~

√3

4+

1

4χ+ = ~χ+

S−χ+ = ~

√1

2

(1

2+ 1

)−(

1

2

)(1

2− 1

)χ− = ~

√3

4+

1

4χ− = ~χ−(

c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)


Pauli matrices

Recall that the raising and lower-ing operators were related to the x-and y-components of angular mo-mentum. This is the same for spin.

the matrix representations are thus

Since each component of ~S has aprefactor of ~/2, it is more commonto write the components of the spinusing the Pauli spin matrices

S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)Sy =

1

2i(S+ − S−)

=~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−)

=~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)

σy ≡(

0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)

σz ≡(

1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


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Today’s Outline - October 29, 2014mypages.iit.edu/~segre/phys405/14F/lecture_17.pdf · Today’s Outline - October 29, 2014 ... Spinors Reading Assignment: Chapter 4.4 Homework