Today’s Outline - October 23, 2013csrri.iit.edu/~segre/phys405/13F/lecture_16.pdf · C. Segre (IIT) PHYS 405 - Fall 2013 October 23, 2013 1 / 16. Today’s Outline - October 23,

Today’s Outline - October 23, 2013

• Problem 4.12

• Hydrogen atom spectrum

• Angular momentum

• Algebraic determination of eigenvalues

• Angular momentum eigenfunctions

Please fill out the mid-term survey!

Reading Assignment: Chapter 4.4

Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013

C. Segre (IIT) PHYS 405 - Fall 2013 October 23, 2013 1 / 16


• Problem 4.12










• Problem 4.12










• Problem 4.12










• Problem 4.12










• Problem 4.12










• Problem 4.12










• Problem 4.12










• Problem 4.12









Problem 4.12

(a) Using equation 4.88, work out the first four Laguerrepolynomials

(b) Using equations 4.86, 4.87, and 4.88 find v(ρ) forthe case n = 5, l = 2.

(c) Find v(ρ) again for the case n = 5, l = 2, but withthe recursion relation.

v(ρ) = L2l+1n−l−1(2ρ) (4.86)

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x) (4.87)

Lq(x) = ex(

d

dx

)q (e−xxq

)(4.88)


Problem 4.12




v(ρ) = L2l+1n−l−1(2ρ) (4.86)

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x) (4.87)

Lq(x) = ex(

d

dx

)q (e−xxq

)(4.88)


Problem 4.12




v(ρ) = L2l+1n−l−1(2ρ) (4.86)

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x) (4.87)

Lq(x) = ex(

d

dx

)q (e−xxq

)(4.88)


Problem 4.12




v(ρ) = L2l+1n−l−1(2ρ) (4.86)

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x) (4.87)

Lq(x) = ex(

d

dx

)q (e−xxq

)(4.88)


Problem 4.12 (cont.)

(a) We previously generated the first three Laguerre polynomials

, nowdetermine the fourth one.

Lq(x) = ex(

d

dx

)q (e−xxq

)

L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)

= ex(−x3e−x + 3x2e−x + 6x2e−x − 12xe−x − 6xe−x + 6e−x)

L3 = −x3 + 9x2 − 18x + 6





Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1

L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6





Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1

L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6





Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6



(a) We previously generated the first three Laguerre polynomials , nowdetermine the fourth one.

Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6




Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)

= ex(

d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6




Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)= ex

(d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6




Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)= ex

(d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6




Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)= ex

(d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6




Lq(x) = ex(

d

dx

)q (e−xxq

)L0 = 1 L1 = −x + 1 L2 = x2 − 4x + 2

L3 = ex(

d

dx

)3 (x3e−x

)= ex

(d

dx

)2

(−x3e−x + 3x2e−x)

= ex(

d

dx

)(x3e−x − 3x2e−x − 3x2e−x + 6xe−x)


L3 = −x3 + 9x2 − 18x + 6



(b) With n = 5 and l = 2, wehave p = 2l + 1 = 5, q − p =n − l − 1 = 2, and q = 7.Thus we want v(ρ) = L52(2ρ).

This means we need the La-guerre polynomial L7

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)



(b) With n = 5 and l = 2, wehave p = 2l + 1 = 5, q − p =n − l − 1 = 2, and q = 7.Thus we want v(ρ) = L52(2ρ).This means we need the La-guerre polynomial L7

Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)




Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)




Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)




Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)




Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2)

= 5040(21− 14ρ+ 2ρ2)




Lpq−p(x) ≡ (−1)p(

d

dx

)p

Lq(x)

L7(x) = 5040− 35280x + 52920x2

− 29400x3 + 7350x4 − 882x5

+ 49x6 − x7

L52(x) = (−1)5(

d

dx

)5

(−882x5 + 49x6 − x7)

= 2520(42− 14x + x2)

v(ρ) = 2525(42− 14(2ρ) + (2ρ)2) = 5040(21− 14ρ+ 2ρ2)



(c) Recall that we have n = 5and l = 2. v(ρ) is simply a poly-nomial sum

with coefficient recursion rela-tion

starting with c0 which can bedetermined by normalization, weget

putting it all together

v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0

= −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2

= 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2 = 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2 = 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2 = 0







v(ρ) = c0 −2

3c0ρ+

2

21c0ρ

2

=c021

(21− 14ρ+ 2ρ2)

v(ρ) =∞∑j=0

cjρj

cj+1 =2(j + l + 1− n)

(j + 1)(j + 2l + 2)cj

c1 =2(2 + 1− 5)

(1)(4 + 2)c0 = −2

3c0

c2 =2(1 + 2 + 1− 5)

(2)(1 + 4 + 2)c1

= −1

7c1 =

2

21c0

c3 =2(2 + 2 + 1− 5)

(3)(2 + 4 + 2)c2 = 0


Hydrogen spectrum

Energy of a hydrogen state

En =− m

2~2n2

(e2

4πε0

)2

=−13.6 eV

n2

transitions between states aregiven by

Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series

1 Lyman2 Balmer3 Paschen4 Brackett5 Pfund6 Humphreys


Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2

=−13.6 eV

n2


Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef

= hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef = hν

= −13.6 eV

(1

n2i− 1

n2f

)

nf Series



Hydrogen spectrum


En =− m

2~2n2

(e2

4πε0

)2=−13.6 eV

n2


Eγ = Ei − Ef = hν

= −13.6 eV

(1

n2i− 1

n2f

)nf Series



Angular momentum

Two of the three quantum numbersof the hydrogen atom, l and ml arerelated to angular momentum

classically, the angular momentumis defined as

as quantum operators, these be-come explicitly

these three operators do not com-mute with each other

~L = ~r × ~pLx = ypz − zpy

Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]

= Lx Ly−Ly Lx = (y pz−z py )(z px−x pz)−(z px−x pz)(y pz−z py )

= y pz z px

− y pz x pz − z py z px + z py x pz

− z px y pz + z px z py + x pz y pz − x pz z py

= y px [pz , z ]

+ x py [z , pz ] = −i~y px + i~x py = i~Lz


Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum





~L = ~r × ~p

Lx = ypz − zpy

Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ]


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ] = Lx Ly−Ly Lx

= (y pz−z py )(z px−x pz)−(z px−x pz)(y pz−z py )

= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x

[Lx , Ly ] = Lx Ly−Ly Lx = (y pz−z py )(z px−x pz)−(z px−x pz)(y pz−z py )

= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px



= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px − y pz x pz

− z py z px + z py x pz


= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px − y pz x pz − z py z px

+ z py x pz


= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px − y pz x pz − z py z px + z py x pz


= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x



− z px y pz

+ z px z py + x pz y pz − x pz z py

= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x



− z px y pz + z px z py

+ x pz y pz − x pz z py

= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x



− z px y pz + z px z py + x pz y pz

− x pz z py

= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x




= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x




= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px −��y pz x pz − z py z px + z py x pz

− z px y pz + z px z py +��x pz y pz − x pz z py

= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x


= y pz z px −��y pz x pz −��z py z px + z py x pz

− z px y pz +��z px z py +��x pz y pz − x pz z py

= y px [pz , z ]



Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x




= y px [pz , z ] + x py [z , pz ]

= −i~y px + i~x py = i~Lz


Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x




= y px [pz , z ] + x py [z , pz ] = −i~y px + i~x py

= i~Lz


Angular momentum






Ly = zpx − xpz

Lz = xpy − ypx

Lx = −i~y ∂

∂z+ i~z

∂

∂y

Ly = −i~z ∂∂x

+ i~x∂

∂z

Lz = −i~x ∂

∂y+ i~y

∂

∂x




= y px [pz , z ] + x py [z , pz ] = −i~y px + i~x py = i~Lz


Commutators and uncertainty relations

The commutation relations for allthree operators are

and according to the generalizeduncertainty principle

σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2

[Lx , Ly ] = i~Lz

[Ly , Lz ] = i~Lx[Lz , Lx ] = i~Ly

σLxσLy ≥~2|〈Lz〉|

σLyσLz ≥~2|〈Lx〉|

σLzσLx ≥~2|〈Ly 〉|





σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2

[Lx , Ly ] = i~Lz

[Ly , Lz ] = i~Lx[Lz , Lx ] = i~Ly








σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2

[Lx , Ly ] = i~Lz[Ly , Lz ] = i~Lx

[Lz , Lx ] = i~Ly








σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2

[Lx , Ly ] = i~Lz[Ly , Lz ] = i~Lx[Lz , Lx ] = i~Ly








σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2









σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2









σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2









σ2Aσ2B ≥

(1

2i

⟨[A, B

]⟩)2






Total angular momentum

While it is impossible to find simultaneous eigenfunctions of any twocomponents of the angular momentum, it is a different story with the totalangular momentum

using the property

[AB,C ] = A[B,C ] + [A,C ]B

and noting that [Lx , Lx ] = 0

and similarly for [L2, Ly ] and[L2, Ly ]

L2 ≡ L2x + L2y + L2z

[L2, Lx ] = ��[L2x , Lx ] + [L2y , Lx ] + [L2z , Lx ]

= Ly [Ly , Lx ] + [Ly , Lx ]Ly+

Lz [Lz , Lx ] + [Lz , Lx ]Lz

= Ly (−i~Lz) + (−i~Lz)Ly+

Lz(+i~Ly ) + (+i~Ly )Lz

[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z

[L2, Lx ]

= ��[L2x , Lx ] + [L2y , Lx ] + [L2z , Lx ]

= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z

[L2, Lx ] = [L2x , Lx ] + [L2y , Lx ] + [L2z , Lx ]

= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z

[L2, Lx ] = [L2x , Lx ] + [L2y , Lx ] + [L2z , Lx ]

= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0




using the property

[AB,C ] = A[B,C ] + [A,C ]B



L2 ≡ L2x + L2y + L2z


= Ly [Ly , Lx ] + [Ly , Lx ]Ly+




[L2, Lx ] = 0


Eigenfunctions of angular momentum

Since [L2,~L] = 0, we can find si-multaneous eigenfunctions of bothL2 and Lz

these two equations can be solvedfor f using the ladder operators

L± ≡ Lx ± iLy

and [L2, L±] = 0

suppose we make a function L±fwhere f is an eigenfunction of L2

L±f is also an eigenfunction of L2

with the same eigenvalue λ

L2f = λf , Lz f = µf

[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]

= i~Ly ± i(−i~Lx)

= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f )

= L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )





L± ≡ Lx ± iLy

and [L2, L±] = 0





[Lz , L±] = [Lz , Lx ]± i [Lz , Ly ]


= ~(±Lx + iLy )

= ±~(Lx ± Ly )

[Lz , L±] = ±~L±

L2(L±f ) = L±(L2f ) = L±(λf )

L2(L±f ) = λ(L±f )


Ladder operator properties

When we apply the operator Lz tothis new function

as expected, it is an eigenfunctionof Lz as well

Lz(L±f ) = (LzL± − L±Lz)f + L±Lz f

= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )

Thus L+ creates a new eigenfunction of Lz whose eigenvalue has beenincreased by ~, conversely, L− creates a new eigenfunction with eigenvaluereduced by ~.

There must be upper and lower limits to the eigenvalues of Lz since itmust be bounded by the total angular momentum. This leads to the twolimiting relations

L+ft = 0, L−fb = 0

where ft and fb are the top-most and bottom-most eigenfuctions






= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )


There must be upper and lower limits to the eigenvalues of Lz since itmust be bounded by the total angular momentum.

This leads to the twolimiting relations

L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0







= ±~L±f + L±(µf )

Lz(L±f ) = (µ± ~)(L±f )



L+ft = 0, L−fb = 0



Eigenvalues of Lz and L2

Suppose that ~l is the maximumeigenvalue of Lz

we can calculate the eigenvalue λusing the construction L±L∓

rearranging, we have a way of com-puting the eigenvalue of L2

thus we have the eigenvalue of L2 interms of the maximum eigenvalueof Lz .

Lz ft = ~lft , L2ft = λft

L±L∓ = (Lx ± iLy )(Lx ∓ iLy )

= L2x + L2y ∓ i(LxLy − LyLx)

L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz

L2ft = (L−L+ + L2z + ~Lz)ft

= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ft

L2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft

= λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft










L±L∓ = L2 − L2z ± ~Lz

L2 = L±L∓ + L2z ∓ ~Lz


= (0 + ~2l2 + ~2l)ftL2ft = ~2l(l + 1)ft = λft


More about eigenvalues

The same computation can beperformed with the bottom-mosteigenstate of the Lz operator, fb

but this gives us two expressions forλ from the top and bottom states

which can only hold if l = −l

Thus the eigenvalues of Lz are m~and m vary in integer steps from −lto +l , such that

which means that m are either in-tegers or half-integers

Lz fb = ~l fb, L2fb = λfb

L2fb = (L+L− + L2z − ~Lz)fb

= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb

= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb

= λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N

→ l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2










= (0 + ~2l2 − ~2l)fb= ~2l(l − 1)fb = λfb

l(l + 1) = l(l − 1)

l = −l + N → l = N/2


Interpretation of Lz

The eigenvalues for the two opera-tors are, therefore

L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l

a common way of depicting this isto draw a sphere of radius |~L| sothat the projection of ~L onto thez-axis is Lz

Why is |Lz | < |~L|?

Recall that Lx and Ly do not com-mute with Lz , so they are undeter-mined and graphically can be any-where on the Lz = ~m circle

0

m

+1

+2

-1

-2

Lz

Ly

Lx

L

it is better to consider ~L as beingspread over all its values consistentwith Lz .




L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l



Recall that Lx and Ly do not com-mute with Lz

, so they are undeter-mined and graphically can be any-where on the Lz = ~m circle

0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l



Recall that Lx and Ly do not com-mute with Lz , so they are undeter-mined

and graphically can be any-where on the Lz = ~m circle

0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L





L2f ml = ~2l(l + 1)f ml

Lz fml = ~mf ml , m = −l , · · · ,+l




0

m

+1

+2

-1

-2

Lz

Ly

Lx

L



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 2013 October 23, 2013 16 / 16

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Today’s Outline - October 23, 2013csrri.iit.edu/~segre/phys405/13F/lecture_16.pdf · C. Segre (IIT) PHYS 405 - Fall 2013 October 23, 2013 1 / 16. Today’s Outline - October 23,