PY452 HW7
Guillem Cucurull
October 2021
1 Problem 8.11
The second derivative of F (x) , at the equilibrium point, can be used to esti-
mate the natural frequency of vibration ω of the two protons in the hydrogen
molecule ion (see Section 2.3). If the ground state energy h̄ω
2 of this oscillator
exceeds the binding energy of the system, it will fly apart. Show that in fact
the oscillator energy is small enough that this will not happen, and estimate
how many bound vibrational levels there are. Note: You’re not going to be able
to obtain the position of the minimum—still less the second derivative at that
point —analytically. Do it numerically, on a computer.
We need to find the second derivative of F (x) at equilibrium, so we need x
when F (x)min . The following calculations are done using M athematica:
F (x) defined in x > 0, where x = R/a
F (x) is minimum at x = 2.493
F (2.493) = −1.129
F ′′ (2.493) = 0.126
Also since F ′′ (2.493) > 0 we know it is a minimum
We know F ′′ (x) is also:
E1 ′′
V ′′ (x) = − F (x)
a2
Now let’s find the second derivative to V (x) for an oscillator:
1
V (x) = mω 2 x2
2
V ′ (x) = mω 2 x
V ′′ (x) = mω 2
1
, Now we plug in the results to get:
E1 ′′
V ′′ (x)min = − F (x)min
a2
E1
mω 2 = − (0.126)
a2
where E1 = −13.6 eV, a is the Bohr radius
r
1 −(0.126)(−13.6)
ω=
a m
where m is the reduced mass of a proton, therefore:
mp mp mp 3.126
m= = = = 1.563eV
mp + mp 2 2
Therefore,
ω = 3.45 · 1014 s−1
This also means that our ground state energy is:
h̄ω h̄ · 3.45 · 1014
Egs = = = 0.114eV
2 2
We need to find if E exceeds the binding energy E:
E = (1 + F (x)min )E1 = (1 + (−1.129))(−13.6) = 1.75
Since E >> Egs , the system will not fly apart or be out of bound.
Now to find how many bound vibrational levels there are, we find the high-
est level using:
1
(n + )h̄ω = E
2
1
(n + )h̄ω = 1.75
2
1.75 1 1.75 1
n= − = − = 7.18
h̄ω 2 2 · 0.114 2
We notice there are 8 states as we also count n = 0 state.
2 Problem 8.28
In Yukawa’s original theory (1934), which remains a useful approximation in
nuclear physics, the “strong” force between protons and neutrons is mediated
by the exchange of π-mesons. The potential energy is
−r
e r0
V (r) = −r0 V0 (1)
r
2
Guillem Cucurull
October 2021
1 Problem 8.11
The second derivative of F (x) , at the equilibrium point, can be used to esti-
mate the natural frequency of vibration ω of the two protons in the hydrogen
molecule ion (see Section 2.3). If the ground state energy h̄ω
2 of this oscillator
exceeds the binding energy of the system, it will fly apart. Show that in fact
the oscillator energy is small enough that this will not happen, and estimate
how many bound vibrational levels there are. Note: You’re not going to be able
to obtain the position of the minimum—still less the second derivative at that
point —analytically. Do it numerically, on a computer.
We need to find the second derivative of F (x) at equilibrium, so we need x
when F (x)min . The following calculations are done using M athematica:
F (x) defined in x > 0, where x = R/a
F (x) is minimum at x = 2.493
F (2.493) = −1.129
F ′′ (2.493) = 0.126
Also since F ′′ (2.493) > 0 we know it is a minimum
We know F ′′ (x) is also:
E1 ′′
V ′′ (x) = − F (x)
a2
Now let’s find the second derivative to V (x) for an oscillator:
1
V (x) = mω 2 x2
2
V ′ (x) = mω 2 x
V ′′ (x) = mω 2
1
, Now we plug in the results to get:
E1 ′′
V ′′ (x)min = − F (x)min
a2
E1
mω 2 = − (0.126)
a2
where E1 = −13.6 eV, a is the Bohr radius
r
1 −(0.126)(−13.6)
ω=
a m
where m is the reduced mass of a proton, therefore:
mp mp mp 3.126
m= = = = 1.563eV
mp + mp 2 2
Therefore,
ω = 3.45 · 1014 s−1
This also means that our ground state energy is:
h̄ω h̄ · 3.45 · 1014
Egs = = = 0.114eV
2 2
We need to find if E exceeds the binding energy E:
E = (1 + F (x)min )E1 = (1 + (−1.129))(−13.6) = 1.75
Since E >> Egs , the system will not fly apart or be out of bound.
Now to find how many bound vibrational levels there are, we find the high-
est level using:
1
(n + )h̄ω = E
2
1
(n + )h̄ω = 1.75
2
1.75 1 1.75 1
n= − = − = 7.18
h̄ω 2 2 · 0.114 2
We notice there are 8 states as we also count n = 0 state.
2 Problem 8.28
In Yukawa’s original theory (1934), which remains a useful approximation in
nuclear physics, the “strong” force between protons and neutrons is mediated
by the exchange of π-mesons. The potential energy is
−r
e r0
V (r) = −r0 V0 (1)
r
2
