Problem set 4. Solutions
1.
 d 2 l (l + 1)

− k 2  ul (r ) = 0
 2−
2
r
 dr

2 m(V ( x) − E )
h2
ul (r ) = rRl ( r )
a) Radial equation k 2 =
 d 2 l ( l + 1) 
For small r the radial equation can be simplified as follows:  2 −
u (r ) = 0
r 2  l
 dr
whose solution for this physical problem is: ul ( r ) = Cr l since we know that in this case
u (r ) = r 2 we deduce that l=2
For r that tends to infinity the radial equation can be simplified as follows:
d2
2
−kr
 dr 2 − k u l ( r ) = 0 whose solution for this physical problem is: ul (r ) = Ce


Knowing that ul (r ) ∝ e
−
r
2a0
; thus k =
1
h2

→ E = V0 −
2
2 a0
8ma0
b) We can write:

A


 d 2 l ( l + 1) 2m V0 − r α  2m (E )  − r 


 2−
  r 2e 2a0  = 0
−
+
2
2
2

dr
r
h
h

 





Differentiating this equation and rewriting it as a polynomial we can easily find:
h2
h2
l = 1 α =1 A =
E = V0 −
ma0
8ma 2
0
c) the solution is very simple the only thing to do here is to prove that cos θ is an
eigenfunction of L2 with corresponding eigenvalue 2h 2 so the equation to prove is:
L2 cos θ = h 2 cos θ
d) the probability of finding the particle in the unit volume (expressed in spherical
coordinates) is:
2
f (r ,θ ) = ψ r sin θ = r e
2
4
−
r
a0
cos2 θ sin θ
∂f
= 0
→ r = 4 a0
∂r
It easy to see that the only maxima for this function are
∂f
 1 
=0
→θ = arcsin 

∂θ
 3
∞, 2π , 2π
Now r =
ψ rψ
ψψ
=
∫
5
−
r
a0
−
r
a0
r e
cos 2 θ sin θdrdθdϕ
= 5a0
0
∞ , 2 π , 2π
∫
4
r e
cos θ sin θdrdθ dϕ
2
0
2.
a) The Hamiltonian for the problem can be written as:
h 2 ∂2
L2z
H=−
+
V
(
ϕ
)
=
+ V (ϕ )
2mR 2 ∂ϕ 2
2mR 2
In this case being nothing else specified we can assume: V (ϕ ) =0
h2 ∂ 2
ψ = Eψ this is simple second order differential equation whose
2 mR2 ∂ϕ 2
solution we know: ψ = Ae ikϕ , we need to impose some boundary condition on the
problem: ψ (ϕ ) = ψ (ϕ + 2π ) 
→ e 2nk = 1 
→ k = 0,±1, ±2,...
b) −
Normalize the function we get: A =
eigenvalue problem we get: E = hk
1
and substituting to the function its value in the
2π
Answer to c is trivial
d) The two problem are similar in the differential equation that solves them and indeed
in classical mechanics they are identical problems. In quantum mechanics the existence
of boundary condition in this problem generates quantization, the main difference.
Problem 3
2
a) Plot r 2 R for each state.
The lowest (1s) energy level, intuitively, will be found closest to the nucleus. The 2s and
2p have probability density peaks close to each other but further from the nucleus.
b)
1s – twice (r a o = 0, ∞ )
2s – three times (r a o = 0, 2, ∞ )
2p – twice (r a o = 0, ∞ )
c) Remember that s, p, and d correspond to l=0, 1, and 2, respectively. From the lecture
notes:
Y00 =
1
4π
Y10 =
3
cos θ
4π
Y20 =
5
3 cos 2 θ − 1
16π
(
)
Problem 4
a,b) Let’s define the barrier location as 0<x<d. If we declare that the particle is traveling
from left to right, we can simplify our expressions to the right of the barrier.
E>Vo :
 Aeikx + Be −ikx ; x < 0

ψ ( x ) = Ce ik 'x + De − ik 'x ;0 < x < d
 Fe ikx ; x > d

k=
2mE
; k' =
h2
2m( E − Vo )
h2
Notice that the energy of the particle does not change, but it’s probability density does.
Our boundary conditions, as usual, are continuity in both ? and its derivative at x=0 and
x=a.
A+ B = C + D
k ( A − B) = k ' (C − D)
Ce ik 'd + De −ik 'd = Fe ikd
k ' (Ce ik 'd − De −ik 'd ) = kFeikd
The reflection R and transmission T are defined as:
B
R=
A
F
T=
A
2
2
Solving the above set of four equations for B/A and F/A:
B
( k 2 − k '2 )(1 − e 2ik 'd )
=
A ( k + k ' ) 2 − (k − k ' ) 2 e 2 ik 'd
F
4kk ' e i( k '− k ) d
=
A ( k + k ' ) 2 − ( k − k ' ) 2 e 2ik 'd
2
 k 'd 
Vo sin 2 

h 

R=
 k 'd 
Vo 2 sin 2 
 + 4 E ( E − Vo )
 h 
4 E( E − Vo )
T=
2
 k 'd 
Vo sin 2 
 + 4 E ( E − Vo )
 h 
E<Vo :
 Ae ikx + Be −ikx ; x < 0

ψ ( x ) = Ce − ρx + De ρx ;0 < x < d
 Fe ikx ; x > d

k=
2mE
;ρ =
h2
2m(Vo − E )
h2
We don’t really have to solve this again. Instead we can substitute k ' = ih
2 m(Vo − E )
h2
into the above expressions for R and T:
R=
T=
Vo 2 sinh 2 ( ρd )
Vo 2 sinh 2 ( ρd ) + 4E (Vo − E)
4 E (Vo − E )
Vo 2 sinh 2 ( ρd ) + 4E (Vo − E)
c) See attached graph. For E>Vo, the particle can resonate within the barrier, resulting
in a sinusoidal variation in transmission with barrier width. For E<Vo, tunneling occurs;
the probability of finding the particle on the far side of the barrier decreases with larger
barrier width. This dependence is often approximated by a decaying exponential
function.
d) The transmissions are unity for sin( k ' d ) = 0 and sinh( ρd ) = 0 respectively. For the
case of E>Vo, d=np/k’=n?’/2 satisfies this condition. We can think about this from an
interference perspective: electron waves within the barrier interfere constructively for
these wavelengths, making the barrier transparent.
For E<Vo, the condition is only satisfied for d à 0, so a potential barrier higher
than the energy of an incident particle will never be completely transparent to that
particle.