763312A QUANTUM MECHANICS I - Solution set 10 - Autumn 2016
1. The electron of a hydrogen atom is in the 4f state. To be more precise,
its wave function is ψ431 (r, θ, ϕ) = R43 (r)Y31 (θ, ϕ). Find the eigenvalues
of operators H, L2 and Lz in this state.
Solution
The wave function
ψ431 (r, θ, ϕ) = R43 (r)Y31 (θ, ϕ)
(1.1)
corresponds to the energy eigenvalue E4 = E1 /42 = −0.85 eV (cf. lecture
notes sec. 5.4).
We recall that the Lz operator involves only differentiation with respect
to the azimuth angle. In addition, we recall that a spherical harmonic
function Ylm is an eigenfunction of Lz with the eigenvalue m~ (cf. lecture
notes sec. 5.5). Thus
Lz ψ431
=
Lz R43 Y31
=
R43 Lz Y31
=
R43 ~Y31
=
~ψ431 .
(1.2)
We recall that L2 operates only on angles. In addition, we recall that a
spherical harmonic function Ylm is its eigenfunction with the eigenvalue
l(l + 1)~2 (cf. lecture notes sec. 5.5). Thus
L2 ψ431
= L2 R43 Y31
= R43 L2 Y31
= R43 3(3 + 1)~2 Y31
12~2 ψ431 .
=
1
(1.3)
2. A hydrogenic atom consists of a single electron orbiting a nucleus with Z
protons (Z = 1 would be hydrogen, Z = 2 ionized helium, Z = 3 doubly
ionized lithium, and so on). Determine the bohr energies En (Z), the
binding energy, the Bohr radius and the Rydberg constant for a hydrogenic
atom. Express your answers as appropriate multiples of the hydrogen
values. Where in the electromagnetic spectrum would the Lyman series
fall, for Z = 2 and Z = 3?
Solution
For hydrogenic atoms the Coulomb potential is
V (r) = −
Ze2
= ZVH (r),
4π0 r
(2.1)
where VH (r) is the Coulomb potential for hydrogen. This is because now
the nucleus has a charge Ze and the electron, still, −e.
We can obtain the wanted quantities for hydrogenic atoms by substituting
e2 with Ze2 in the formulae for hydrogen. Thus,
"
2 2 #
m
Ze
1
En (Z) = −
= Z 2 En ,
(2.2)
2~2 4π0
n2
a(Z)
=
R(Z)
=
4π0 ~2
a
= ,
me2 Z
Z
2 2
m
Ze
= Z 2 R.
4π0 c~3 4π0
(2.3)
(2.4)
The binding energy, Eb (Z), is the additive inverse of the ground state
energy, Eb (Z) = −E1 (Z) = −Z 2 E1 .
Lyman series corresponds to transition from an excited state, n ≥ 2 to the
ground state. Te wavelength range can be obtained from (lecture notes p.
39 Eq. (318))
!
1
1
1
(2.5)
= R(Z)
− 2 .
λ
n2f
ni
Upper boundary is determined from the transition from the first excited
state to the ground state,
1
1
1
3
4
= R(Z)
− 2 = R(Z) ⇔ λmax =
.
(2.6)
λmax
12
2
4
3R(Z)
Lower boundary is determined from the transition from the continuum
(n = ∞) to the ground state,
1
1
1
1
= R(Z)
− 2 = R(Z) ⇔ λmin =
.
(2.7)
λmin
12
∞
R(Z)
2
For Z = 2 we have
1
≈ 2.28 · 10−8 m = 22.8 nm
22 R
(2.8)
4
≈ 3.04 · 10−8 m = 30.4 nm,
3 · 22 R
(2.9)
λmin =
and
λmax =
which are in the ultraviolet region (Extreme Ultraviolet, EUV).
For Z = 3 we obtain
λmin
λmax
≈ 1.01 · 10−8 m = 10.1 nm,
≈ 1.35 · 10
−8
m = 13.5 nm,
(2.10)
(2.11)
which are also in the ultraviolet region (Extreme Ultraviolet, EUV).
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3. Consider Earth-Sun system as a gravitational analogue to the hydrogen
atom.
a) What is the potential energy function? Let m be the mass of the
Earth, and M the mass of the Sun.
b) What is the ”Bohr radius”, ag , for this system? Work out the actual
number.
c) Write down the gravitational ”Bohr formula”, and by equating En to
the classical
p energy of a planet in a circular orbit of radius ro , show
that n = ro /ag . From this, estimate the quantum number of the
Earth.
d) Suppose the Earth made a transition to the next lower level (n − 1).
How much energy (in Joules) would be released? What would the
wavelength of the emitted photon (or, more likely, graviton) be? Express your answer in light years. Is the remarkable answer a coincidence?
Solution
a) Potential energy function for gravitational potential is
V (r) = −G
m1 m2
.
r
(3.1)
mM
.
r
(3.2)
For Earth-Sun system this reads
V (r) = −G
So by replacing e2 /(4π0 ) with GmM and the mass of the electron with
the mass of the Earth, in the equations for hydrogen, we get the results
for the gravitational system.
b) For hydrogen
a=
4π0
e2
~2
.
me
(3.3)
Using the recipe stated above we obtain for the gravitational analogue
ag =
~2
.
GM m2
(3.4)
Pluging in accepted values for the quantities gives
ag ≈ 2.34 · 10−138 m.
This is an unfathomably small number.
4
(3.5)
c) Bohr formula for hydrogen is
"
me
= −
2~2
En
e2
4π0
2 #
1
,
n2
(3.6)
thus,
Eng = −
G2 M 2 m3
.
2~2 n2
(3.7)
Classical energy for a planet on a circular orbit with radius ro is
1
mM
mv 2 − G
.
2
ro
Ec =
(3.8)
The gravitational force exerted to a planet in an orbit is GmM/ro2 , this is
equal to the centripetal force mv 2 /ro . From this we get that
1
GmM
mv 2 =
2
2ro
(3.9)
mM
1
GmM
mM
GmM
mv 2 − G
=
−G
=−
.
2
ro
2ro
ro
2ro
(3.10)
and, thus,
Ec =
Now we set Ec = Eng :
−
G2 M 2 m3
2~2 n2
=
⇔ n2
=
⇔n =
GmM
2ro
GM m2
ro
2
r~
ro
.
ag
−
(3.11)
|ag =
~2
GM m2
(3.12)
(3.13)
For Earth-Sun system ro = 1.496 · 1011 m, so
n ≈ 2.53 · 1074 .
(3.14)
d) Energy released in the transition is
G2 M 2 m3
∆E = Ei − Ef = −
2~2
1
1
−
2
n
(n − 1)2
.
(3.15)
We estimate
1
1
= 2
(n − 1)2
n
1
1 − 1/n
5
2
1
≈ 2
n
2
1+
n
,
(3.16)
where we have used the Taylor series of 1/(1 − x)2 in the vicinity of x = 0,
to first order in x. With this the energy released becomes
2
G2 M 2 m3
G2 M 2 m3 1
1
−
1
−
.
(3.17)
∆E ≈ −
=
2
2
2~
n
n
2~2 n3
Plugging in the values gives ∆E ≈ 2.09 · 10−41 J. The corresponding
wavelength is
λ=
hc
≈ 9.52 · 1015 m ≈ 1 ly.
∆E
(3.18)
Is this a coincidence? Let us have a closer look.
From part c) we get
n2 =
GM m2 ro
.
~2
(3.19)
With this we have
hc
~2 n 3
λ=
= 2πc~ 2 2 3 = c 2π
∆E
G M m
r
ro3
GM
!
.
(3.20)
Again from c) we get
r
v=
GM
2πro
=
,
ro
T
where T is the period of the orbit (for the Earth 1 a). Thus,
r
ro3
T = 2π
GM
(3.21)
(3.22)
and
λ
=
cT.
(3.23)
So, the result is not a coincidence, but a property of transitions between
extremely high energy levels.
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4. Calculate the expectation value of potential energy for the hydrogen atom
energy eigenstates. Calculate
also hv 2 i using hV i + hT i = En . Finally,
p
2
calculate the speed v = hv i in SI units.
Solution
The hydrogen atom is discussed in the lecture notes. There it is found out
that the potential is
V (r) = −
e2 1
.
4π0 r
(4.1)
Moreover, the energy eigenfunctions are obtained at p. 38. Employing
them, we obtain
hψnlm |V |ψnlm i
=
−
e2
2
N
4π0 nl
Z π Z 2π
×
∞
Z
r
0
∗
l
2
2r
2
2l+1
r dr
Ln−l−1
na
l
0
{z
|
}
=1
2
=
e
Ym (θ, ϕ) Ym (θ, ϕ) dϕ dθ
0
=
2l−1 −2r/na
2
∞
2r
e
2r
2
2l+1 −2r/na
2l+1
Nnl
r
e
Ln−l−1
−
dr ||t =
(4.2)
4π0
na
na
0
Z
i2
na 2l+2 ∞ 2l+1 −t h 2l+1
e2
2
N
Ln−l−1 (t) dt
||Ex. 9.5
t
e
−
4π0 nl 2
0
{z
}
|
Z
Γ(l+1+n)/(n−l−1)!
=
2En .
(4.3)
Clearly,
hψnlm |H|ψnlm i = hψnlm |T |ψnlm i + hψnlm |V |ψnlm i = En .
(4.4)
On the other hand,
2
hv i =
=
=
p2
m2
2
p2
m 2m
2
hT i .
m
(4.5)
Thus
2
hψnlm |T |ψnlm i
m
2
(En − hψnlm |V |ψnlm i)
m
2En
−
.
m
hψnlm |v 2 |ψnlm i =
=
=
7
(4.6)
Energy eigenvalues of hydrogen were obtained in the lecture notes, (p. 37)
4
En = − 32π2me
. Thus,
2 ~2 n2
0
hψnlm |v 2 |ψnlm i =
e2
4π0 ~
2
1
.
n2
(4.7)
The speed
v
=
=
p
hv 2 i
e2 1
.
4π0 ~ n
(4.8)
If one compares the ground state (n = 1) speed with the speed of light,
one obtains
1
e2
≈
(4.9)
α=
4π0 ~c
137.
This is the so-called fine-structure constant, a natural constant that characterizes the strength of interactions between matter and radiation.
8