=
̄h
72 i
〈 4 ψ 100 + 3ψ 211 −iψ 210 +√
10 ψ 21 − 1 |− 3√
2 ψ 210 −i√
2 ψ 211 +i√
2 ψ 21 − 1 +√
10
√
2 ψ 210 〉=
√
2 ̄h
72 i〈 4 ψ 100 + 3ψ 211 −iψ 210 +√
10 ψ 21 − 1 |− 3 ψ 210 −iψ 211 +iψ 21 − 1 +√
10 ψ 210 〉=
√
2 ̄h
72 i(− 3 i− 3 i+√
10 i+√
10 i) =(−6 + 2
√
10)i√
2 ̄h
72 i=
(2
√
5 − 3
√
2) ̄h
3616.4.2 The Expectation of^1 rin the Ground State
R 10 = 2
(
Z
a 0) (^32)
e−Zr/a^0
〈ψ 100 |
1
r|ψ 100 〉 =∫
Y 00 ∗Y 00 dΩ∫∞
0r^21
rR∗ 10 R 10 dr=
∫∞
0rR 10 ∗R 10 dr = 4(
Z
a 0) 3 ∫∞
0re−^2 Zr/a^0 dr= 4(
Z
a 0) (^3) (
a 0
2 Z
) 2
1!
=
Z
a 0We can compute the expectation value of the potential energy.
〈ψ 100 |−Ze^2
r|ψ 100 〉=−Z^2 e^2
a 0=Z^2 e^2αmc
̄h=−Z^2 α^2 mc^2 = 2E 10016.4.3 The Expectation Value ofrin the Ground State
〈ψ 100 |r|ψ 100 〉=∫∞
0r^3 R∗ 10 R 10 dr= 4(
Z
a 0) 3 ∫∞
0r^3 e−^2 Zr/a^0 dr= 3!1
4
a 0
Z=
3
2
a 0
Z16.4.4 The Expectation Value ofvrin the Ground State
Forℓ= 0, there is no angular dependence to the wavefunction so no velocity except in the radial
direction. So it makes sense to compute the radial component of the velocity which is the full
velocity.
We can find the term for p
(^2) r
2 min the radial equation.
〈ψ 100 |(vr)^2 |ψ 100 〉 =
∫∞
0r^2 R∗ 10− ̄h^2
m^2(
d^2
dr^2+
2
rd
dr)
R 10 dr