Today’s Outline - October 31, 2012

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Today’s Outline - October 31, 2012

• Problem 4.5

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Today’s Outline - October 31, 2012

• Problem 4.5
• Problem 4.15

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Today’s Outline - October 31, 2012

• Problem 4.5
• Problem 4.15
• Problem 4.18

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Today’s Outline - October 31, 2012

• Problem 4.5
• Problem 4.15
• Problem 4.18
• Spin

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Today’s Outline - October 31, 2012

• Problem 4.5
• Problem 4.15
• Problem 4.18
• Spin

Homework Assignment #09:

Chapter 4: 20,23,27,31,43,55
due Wednesday, November 7, 2012

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 1 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are
s
(2l + 1) (l − |m|)! imφ m
Ylm (θ, φ) =  e Pl (cos θ)
4π (l + |m|)!

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are
s
(2l + 1) (l − |m|)! imφ m
Ylm (θ, φ) =  e Pl (cos θ)
4π (l + |m|)!
 |m|  l
m 2 |m|/2 d 1 d
Pl (x) = (1 − x ) Pl (x), Pl (x) = l (x 2 − 1)l
dx 2 l! dx

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are
s
(2l + 1) (l − |m|)! imφ m
Ylm (θ, φ) =  e Pl (cos θ)
4π (l + |m|)!
 |m|  l
m 2 |m|/2 d 1 d
Pl (x) = (1 − x ) Pl (x), Pl (x) = l (x 2 − 1)l
dx 2 l! dx
for m = l we can simplify a bit to

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are
s
(2l + 1) (l − |m|)! imφ m
Ylm (θ, φ) =  e Pl (cos θ)
4π (l + |m|)!
 |m|  l
m 2 |m|/2 d 1 d
Pl (x) = (1 − x ) Pl (x), Pl (x) = l (x 2 − 1)l
dx 2 l! dx
for m = l we can simplify a bit to
s
(2l + 1) 1 ilφ l
Yll (θ, φ) = (−1)l e Pl (cos θ)
4π (2l)!

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5
Use Equation 4.32 to construct Yll (θ, φ)
The general form for the spherical harmonics and the associated Legendre
functions are
s
(2l + 1) (l − |m|)! imφ m
Ylm (θ, φ) =  e Pl (cos θ)
4π (l + |m|)!
 |m|  l
m 2 |m|/2 d 1 d
Pl (x) = (1 − x ) Pl (x), Pl (x) = l (x 2 − 1)l
dx 2 l! dx
for m = l we can simplify a bit to
s
(2l + 1) 1 ilφ l
Yll (θ, φ) = (−1)l e Pl (cos θ)
4π (2l)!
 2l
l 1 2 l/2 d
Pl (x) = l (1 − x ) (x 2 − 1)l
2 l! dx

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 2 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

noting that (x 2 − 1)l = x 2l + ..., the 2l th derivative of this is only

dependent on the first term as the others will always be zero. Consequently

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

noting that (x 2 − 1)l = x 2l + ..., the 2l th derivative of this is only

dependent on the first term as the others will always be zero. Consequently
 2l
1 d
Pll (x) = l (1 − x 2 )l/2 x 2l
2 l! dx

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

noting that (x 2 − 1)l = x 2l + ..., the 2l th derivative of this is only

dependent on the first term as the others will always be zero. Consequently
 2l
1 d (2l)!
Pll (x) = l (1 − x 2 )l/2 x 2l = (1 − x 2 )l/2
2 l! dx 2l l!

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

noting that (x 2 − 1)l = x 2l + ..., the 2l th derivative of this is only

dependent on the first term as the others will always be zero. Consequently
 2l
1 d (2l)!
Pll (x)= l (1 − x )2 l/2
x 2l = l (1 − x 2 )l/2
2 l! dx 2 l!
s
(2l + 1) 1 ilφ (2l)!
Yll (θ, φ) = (−1)l e (1 − cos2 θ)l/2
4π (2l)! 2l l!

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.5 (cont.)

 2l
1 d
Pll (x) = l (1 − x 2 )l/2 (x 2 − 1)l
2 l! dx

noting that (x 2 − 1)l = x 2l + ..., the 2l th derivative of this is only

dependent on the first term as the others will always be zero. Consequently
 2l
1 d (2l)!
Pll (x)= l (1 − x )2 l/2
x 2l = l (1 − x 2 )l/2
2 l! dx 2 l!
s
(2l + 1) 1 ilφ (2l)!
Yll (θ, φ) = (−1)l e (1 − cos2 θ)l/2
4π (2l)! 2l l!
r  l
l 1 (2l + 1)! 1 iφ
Yl (θ, φ)= − e sin θ
l! 4π 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 3 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.
1  
Ψ(~r , t) = √ ψ211 e −iE2 t/~ + ψ21−1 e −iE2 t/~
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.
1  
Ψ(~r , t) = √ ψ211 e −iE2 t/~ + ψ21−1 e −iE2 t/~
2
1
= √ (ψ211 + ψ21−1 ) e −iE2 t/~
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.
1  
Ψ(~r , t) = √ ψ211 e −iE2 t/~ + ψ21−1 e −iE2 t/~
2
1
= √ (ψ211 + ψ21−1 ) e −iE2 t/~
2

ψ21±1 = R21 Y1±1

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.
1  
Ψ(~r , t) = √ ψ211 e −iE2 t/~ + ψ21−1 e −iE2 t/~
2
1
= √ (ψ211 + ψ21−1 ) e −iE2 t/~
2
r !
±1 1 1 −r /2a 3
ψ21±1 = R21 Y1 = √ 2
re ∓ sin θe ±iφ
6a 2a 8π

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15
A hydrogen atom starts out in the following linear combination of the
stationary states n = 2, l = 1, m = 1 and n = 2, l = 1, m = −1:
1
Ψ(~r , 0) = √ (ψ211 + ψ21−1 )
2
Find Ψ(~r , t) and the expectation value of the potential energy hV i.
1  
Ψ(~r , t) = √ ψ211 e −iE2 t/~ + ψ21−1 e −iE2 t/~
2
1
= √ (ψ211 + ψ21−1 ) e −iE2 t/~
2
r !
±1 1 1 −r /2a 3
ψ21±1 = R21 Y1 = √ 2
re ∓ sin θe ±iφ
6a 2a 8π
1 1 −r /2a 
iφ −iφ

ψ211 + ψ21−1 = − √ re sin θ e − e
πa 8a2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 4 / 12

Problem 4.15 (cont.)

1 1 −r /2a 
iφ −iφ

ψ211 + ψ21−1 = − √ re sin θ e − e
πa 8a2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 5 / 12

Problem 4.15 (cont.)

1 1 −r /2a 
iφ −iφ

ψ211 + ψ21−1 = − √ re sin θ e − e
πa 8a2
i
= −√ re −r /2a sin θ sin φ
πa4a2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 5 / 12

Problem 4.15 (cont.)

1 1 −r /2a 
iφ −iφ

ψ211 + ψ21−1 = − √ re sin θ e − e
πa 8a2
i
= −√ re −r /2a sin θ sin φ
πa4a2
i
Ψ(~r , t) = − √ re −r /2a sin θ sin φ e −E2 t/~
2πa

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 5 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r
Z ∞ Z π Z 2π
1 ~2 3 −r /a 3
=− r e dr sin θ dθ sin2 φ dφ
32πa5 ma2 0 0 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r
Z ∞ Z π Z 2π
1 ~2 3 −r /a 3
=− r e dr sin θ dθ sin2 φ dφ
32πa5 ma2 0 0 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r
Z ∞ Z π Z 2π
1 ~2 3 −r /a 3
=− r e dr sin θ dθ sin2 φ dφ
32πa5 ma2 0 0 0
~2 4
=− (3!a4 ) π
32πma6 3

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r
Z ∞ Z π Z 2π
1 ~2 3 −r /a 3
=− r e dr sin θ dθ sin2 φ dφ
32πa5 ma2 0 0 0
~2 4 4 ~2 1
=− (3!a ) π = − = E1 = −6.8eV
32πma6 3 4ma2 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.15 (cont.)

Now calculate the expectation value of the potential energy for this
time-dependent state

e2 1
Z  
hV i = |Ψ|2 − d 3r
4π0 r
e2
Z 
1 2 −r /a 2 2
1
=− r e sin θ sin φ r 2 sin θdr dθ dφ
(2πa)(16a4 ) 4π0 r
Z ∞ Z π Z 2π
1 ~2 3 −r /a 3
=− r e dr sin θ dθ sin2 φ dφ
32πa5 ma2 0 0 0
~2 4 4 ~2 1
=− (3!a ) π = − = E1 = −6.8eV
32πma6 3 4ma2 2
not dependent on time!

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 6 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

conjugate of L± .

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i
= h(Lx ∓ iLy )f |g i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i
= h(Lx ∓ iLy )f |g i
= hL∓ f |g i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i
thus (L± )† = L∓ = h(Lx ∓ iLy )f |g i
= hL∓ f |g i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i
thus (L± )† = L∓ = h(Lx ∓ iLy )f |g i
= hL∓ f |g i
now starting with eq. 4.112,
calculate hL± L∓ i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18

The raising and lowering operators change the value of m by one unit:
m±1
L± fl m = (Am
l )fl

what is Am
l if the eigenfunctions are normalized.

First determine the hermitian

hf |L± g i = hf |Lx g i ± ihf |Ly g i
conjugate of L± .
= hLx f |g i ± ihLy f |g i
thus (L± )† = L∓ = h(Lx ∓ iLy )f |g i
= hL∓ f |g i
now starting with eq. 4.112,
calculate hL± L∓ i
L± L∓ = L2 − L2z ∓ ~Lz

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 7 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]
hfl m |L± L∓ fl m i = hL± fl m |L± fl m i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]
m±1 m m±1
hfl m |L± L∓ fl m i = hL± fl m |L± fl m i = hAm
l fl |Al fl i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]
m±1 m m±1
hfl m |L± L∓ fl m i = hL± fl m |L± fl m i = hAm
l fl |Al fl i
2 m±1 m±1
= |Am
l | hfl |fl i

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]
m±1 m m±1
hfl m |L± L∓ fl m i = hL± fl m |L± fl m i = hAm
l fl |Al fl i
2 m±1 m±1
= |Am
l | hfl |fl i = |Am
l |
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Problem 4.18 (cont.)

hfl m |L± L∓ fl m i = hfl m |(L2 − L2z ∓ ~Lz )fl m i

= hfl m |[~2 l(l + 1) − ~2 m2 ∓ ~2 m]fl m i
= ~2 [l(l + 1) − m(m ± 1)]hfl m |fl m i
= ~2 [l(l + 1) − m(m ± 1)]
m±1 m m±1
hfl m |L± L∓ fl m i = hL± fl m |L± fl m i = hAm
l fl |Al fl i
2 m±1 m±1
= |Am
l | hfl |fl i = |Am
l |
2

p
Am
l = ~ l(l + 1) − m(m ± 1)

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 8 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of
the spin operator

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator Sz |s, mi = ~m|s, mi

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator Sz |s, mi = ~m|s, mi
and there are ladder opera-
tors

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator Sz |s, mi = ~m|s, mi
p
and there are ladder opera- S± |s, mi = ~ s(s + 1) − m(M ± 1)|s, mi
tors

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12

Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator Sz |s, mi = ~m|s, mi
p
and there are ladder opera- S± |s, mi = ~ s(s + 1) − m(M ± 1)|s, mi
tors
there are no restrictions forc-
ing integer spin
C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12
Intrinsic angular momentum
In quantum mechanics, we deal with small particles, which can, like an
electron be described as “point” particles. We have seen the orbital
angular momentum, characterized by l and m. These particles also carry
another intrinsic form of angular momentum which we call “spin” angular
momentum even though it is not associated with any physical variables of
rotation
Spin behaves in all ways as orbital angular momentum
the spin operator obeys com-
mutation relations [Sx , Sy ] = i~Sz
there exist eigenvectors of S 2 |s, mi = ~2 s(s + 1)|s, mi
the spin operator Sz |s, mi = ~m|s, mi
p
and there are ladder opera- S± |s, mi = ~ s(s + 1) − m(M ± 1)|s, mi
tors 1 3
s = 0, , 1, , . . .
there are no restrictions forc- 2 2
ing integer spin m = −s, −s + 1, . . . , s − 1, s
C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 9 / 12
 1
Spin 2

simplest system to work out, only 2

states

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
2 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor”

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1

χ+ =
0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1
with a general state represented by

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1
with a general state represented by
 
a
χ= = aχ+ + bχ−
b

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1
with a general state represented by
operators are 2 × 2 matrices  
a
χ= = aχ+ + bχ−
b

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1
with a general state represented by
operators are 2 × 2 matrices  
a
χ= = aχ+ + bχ−
consider the S 2 operator b
 
c d
S2 =
e f

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

 1
Spin 2

simplest system to work out, only 2 1 1

states | + i, spin up
2 2
1 1
| − i, spin down
a better representation is that of 2 2
a two element vector called a
“spinor” 
1
  
0
χ+ = , χ− =
0 1
with a general state represented by
operators are 2 × 2 matrices  
a
χ= = aχ+ + bχ−
consider the S 2 operator b
 
c d
S2 = 3
e f S 2 χ± = ~2 χ±
4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 10 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

    
c d 1 3 1
= ~2
e f 0 4 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   

c d 3 2 1
1
= ~
e f 0
4 0
  3 2
c ~
= 4
e 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   

c d 3 2 1
1
= ~
e f 0
4 0
  3 2
c ~
= 4
e 0
3
c = ~2 , e = 0
4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   
     
c d 3 2 1
1 c d 0 3 0
= ~ = ~2
e f 0
4 0 e f 1 4 1
  3 2
c ~
= 4
e 0
3
c = ~2 , e = 0
4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   
   
 
c d 3 2 1
1 c d 3 2 0
0
= ~ = ~
e f 0
4 0 e f 41 1
  3 2    
c ~ d 0
= 4 = 3 2
e 0 f 4~
3
c = ~2 , e = 0
4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   
    
c d 3 2 1
1 c d 03 2 0
= ~ = ~
e f 0
4 0 e f 14 1
  3 2    
c ~ d 0
= 4 = 3 2
e 0 f 4~
3 3
c = ~2 , e = 0 d = 0, f = ~2
4 4

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

S 2 matrix

3 3
S 2 χ+ = ~2 χ+ S 2 χ− = ~2 χ−
4 4

   
     
c d 3 2 1
1 c d 0 3 2 0
= ~ = ~
e f 0
4 0 e f 1 4 1
  3 2    
c ~ d 0
= 4 = 3 2
e 0 f 4~
3 3
c = ~2 , e = 0 d = 0, f = ~2
4 4
 
3 1 0
S = ~2
2
4 0 1

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 11 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

    
c d 1 ~ 1
=
e f 0 2 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

    
c d ~ 1
1
=
e f 2 0
0
  ~
c
= 2
e 0

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

    
c d ~ 1 1
=
e f 2 0 0
  ~
c
= 2
e 0
~
c= , e=0
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

         
c d ~ 1 1 c d 0 ~ 0
= =−
e f 2 0 0 e f 1 2 1
  ~
c
= 2
e 0
~
c= , e=0
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

        

c d ~ 1 1 c d 0~ 0
= =−
e f 2 0 0 e f 12 1
  ~    
c d 0
= 2 =
e 0 f − ~2
~
c= , e=0
2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

         
c d ~ 1 1 c d 0 ~ 0
= =−
e f 2 0 0 e f 1 2 1
  ~    
c d 0
= 2 =
e 0 f − ~2
~ ~
c= , e=0 d = 0, f = −
2 2

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12

Sz matrix

~ ~
Sz χ+ = χ+ Sz χ− = − χ−
2 2

         
c d ~ 1 1 c d 0 ~ 0
= =−
e f 2 0 0 e f 1 2 1
  ~    
c d 0
= 2 =
e 0 f − ~2
~ ~
c= , e=0 d = 0, f = −
2 2
 
~ 1 0
Sx =
2 0 −1

C. Segre (IIT) PHYS 405 - Fall 2012 October 31, 2012 12 / 12