130_notes.dvi

ak+1=

k+ℓ+ 1−n

(k+ 1)(k+ 2ℓ+ 2)

ak

a 1 =

0 + 0 + 1− 2

(0 + 1)(0 + 2(0) + 2)

a 0 =

− 1

2

a 0

R 20 (r) =C

(

1 −

Zr

2 a 0

)

e−Zr/^2 a^0

We again normalize to determine the constant.

16.4 Examples

16.4.1 Expectation Values in Hydrogen States

Anelectron in the Coulomb field of a protonis in the state described by the wave function
1
6 (4ψ^100 + 3ψ^211 −iψ^210 +

√

10 ψ 21 − 1 ). Find the expected value of the Energy,L^2 ,Lz, andLy.

First check the normalization.

| 4 |^2 +| 3 |^2 +|−i|^2 +|

√

10 |^2

36

=

36

36

= 1

The terms are eigenstates ofE,L^2 , andLz, so we can easily compute expectation values of those
operators.

En = −

1

2

α^2 μc^2

1

n^2

〈E〉 = −

1

2

α^2 μc^2

16112 + 9 212 + 1 212 + 10 212

36

=−

1

2

α^2 μc^2

21

36

=−

1

2

α^2 μc^2

7

12

Similarly, we can just square probability amplitudes to compute the expectation value ofL^2. The
eigenvalues areℓ(ℓ+ 1) ̄h^2.

〈L^2 〉= ̄h^2

16(0) + 9(2) + 1(2) + 10(2)

36

=

10

9

̄h^2

The Eigenvalues ofLzarem ̄h.

〈Lz〉= ̄h

16(0) + 9(1) + 1(0) + 10(−1)

36

=

− 1

36

̄h

Computing the expectation value ofLyis harder because the states are not eigenstates ofLy. We
must writeLy= (L+−L−)/ 2 iand compute.

〈Ly〉 =

1

72 i

〈 4 ψ 100 + 3ψ 211 −iψ 210 +

√

10 ψ 21 − 1 |L+−L−| 4 ψ 100 + 3ψ 211 −iψ 210 +

√

10 ψ 21 − 1 〉

=

1

72 i

〈 4 ψ 100 + 3ψ 211 −iψ 210 +

√

10 ψ 21 − 1 |− 3 L−ψ 211 −i(L+−L−)ψ 210 +

√

10 L+ψ 21 − 1 〉