Physics 828
Sketch of Solution to Set 2
Shankar 12.3.7
(1) The two-dimensional harmonic oscillator is obviously invariant under rotations
about the z-axis: the magnitude of the position and momenta are unaltered by rotations around the z-axis. Therefore, the Hamiltonian commutes with the generator of
rotations about the z-axis, Lz .
(2) So we write ψ(ρ, φ) = eimφ REm (ρ) where m is an integer, positive or negative.
In two dimensions since we know the Laplacian we have
Ã
h̄2
1 0
00
−
REm
+ REm
2µ
ρ
!
à 2
+
1
h̄ m2
+ µω 2 ρ2
2
2µρ
2
!
REm (ρ) = EREm .
For small ρ assuming that REm (ρ) ∝ ρk (derivatives decrease the power by unity
increasing its importance for small ρ etc.) we can neglect the potential energy term
and the constant term on the right-hand side. Thus we have
00
REm
+
0
REm
m2
∝ 2 REm ⇒ [k(k − 1) + k] = m2 ⇒ k 2 = m2 .
ρ
ρ
For the wave function to be normalizable, for
ρ→0
limit k ≥ 0. Thus we have REm (ρ) −→ ρ|m| .
R∞
0
2
dρ ρ REm
(ρ) to be finite at the lower
(3) For large ρ, terms with inverse powers of ρ including the centrifugal term and
also the constant term on the right-hand side can be neglected. So we have
µ2 ω 2 ρ2
REm ;
h̄2
this is identical to the one-dimensional oscillator equation. See the careful analysis on
page 191. Up to powers of ρ the solution is1
00
REm
=
ρ→∞
µω 2
REm (ρ) −→ e− 2h̄ ρ .
So we write as instructed
1
µω − µω ρ2
µω ³
µω 2 ´ − µω ρ2
00
ρ e 2h̄ and REm
=
−1 +
ρ e 2h̄ .
h̄
h̄
h̄
00
Thus we can neglect the constant term in R and recover the solution. One can use this to check
that terms we have neglected are indeed small compared to the terms we have retained. When you
neglect terms it is a good idea to substitute the solution you have obtained and check that the terms
you have neglected are indeed smaller.
0
REm
= −
1
µω 2
REm (ρ) = Um (ρ) ρ|m| e− 2h̄ ρ .
E
(4) Use the dimensionless variable ² = h̄ω
and y 2 =
equation by h̄ω we have dropping the annoying subscripts
"
1
−
2
Ã
d2
1 d
m2
+
−
dy 2
y dy
y2
!
µω 2
ρ.
h̄
Dividing the radial
#
1
+ y 2 R(y) = ² R(y) .
2
(5) We do the substitution and do elementary calculus and obtain the result given:
"Ã
00
U +
2|m| + 1
y
!
#
− 2y U 0 + (2² − 2|m| − 2) U = 0 .
P
r
r
(6) We substitute U (y) = ∞
r=0 Cr y and collect the coefficient of y . The second
derivative reduces the power by two and so we use the Cr+2 term etc.
(r + 2)(r + 1)Cr+2 + (2|m| + 1)(r + 2)Cr+2 − 2rCr + (2² − 2|m| − 2)Cr = 0
yielding a two-term recursion relation.
(7) We write this as
Cr+2
(2(² − |m| − r − 1)
= −
.
Cr
(r + 2)(2|m| + r + 2)
First if C0 is given C2 and the other even terms can be computed. As r → ∞ we have
Cr+2
2
→ .
Cr
r
2
2
This implies that U (y) grows as ey which overwhelms the e−y /2 in R pushing it out
of the Hilbert space. So the series must terminate. Thus the boundary condition at
infinity leads as usual to energy quantization.
What about the odd terms? A series only with odd terms (set C0 = 0 so that all even
terms vanish) is inconsistent since then U (y) ∼ y for small y and thus R(ρ) ∼ ρ|m|+1
inconsistent with our earlier result in (2). This appears to be suggested as an argument. What if one starts with C0 and C1 non-zero? Substituting into the equation for
U we find that the (1/y)(dU/dy) leads to the term C1 /y and there is no other source
of y −1 terms. Thus C1 = 0 and therefore, all odd terms vanish.
2
Therefore, we have r = 2k and the termination of the series condition yields
² = 1 + r + |m| = |m| + 2k + 1 ≡ (n + 1)
with k = 0, 1, , 2, · · ·.
(8) Since |m| = n − 2k for a given n (i.e., for a given energy) the maximum value
of m is n which occurs for k = 0, The azimuthal quantum number m decreases by steps
of 2 until m reaches the value −n. It is easy to se that there are n + 1 allowed values
of m yielding a degeneracy of n + 1. In Cartesian coordinates the energy is nx + ny + 1
in units of h̄ω. So the degeneracy corresponds to the number of ways in which we can
choose two non-negative integers to add up to n. We can choose nx to be any integer
from 0 to n and ny = n − nx . This yields the same degeneracy.
Shankar 12.6.1
(1) Since there is no angular dependence ` = 0.
(b) We have R(r) ∝ e−r/a0 and for large r (retaining only the dominant terms)
−
h̄2 00
R = ER(r) .
2m
2
h̄
Substituting the given form we obtain E = − 2ma
2 .
0
(c) Clearly the R00 term and the energy term cancel for all r. If the equation is
valid for all r we must have
h̄2 2 0
−
R + V (r) R(r) = 0 .
2m r
Substituting R(r) = e−r/a0 we obtain
h̄2
h̄2
+ V (r) = 0 ⇒ V (r) = −
.
ma0 r
ma0 r
Shankar 13.1.1 and 13.1.3 You should be able to fill in the steps. Here are some
steps dropping some subscripts for notational simplicty.
v =
∞
X
Ck ρk+`+1 .
k=0
3
Substituting into
Ã
00
0
v − 2v +
e2 λ
`(` + 1)
−
ρ
ρ2
!
v = 0
(1)
we extract the coefficient of ρk+l carefully. Since two derivatives reduce the power of
ρ by two we should start from the term with ρk+`+2 with coefficient Ck+1 for the first
term and similarly for the last term. For the second and third terms which reduce the
power of ρ by unity we can start with the ρk+`+1 term with coefficient Ck . Thus we
have
(k + ` + 2)(k + ` + 1)Ck+1 − 2(k + ` + 1)Ck + e2 λCk − `(` + 1)Ck+1 = 0
which yields
Since λ2 =
−q 2 λ + 2(k + ` + 1)
Ck+1
=
Ck
(k + ` + 2)(k + ` + 1) − `(` + 1)
2m
h̄2 W
(2)
(3)
from 13.1.9. For the numerator to vanish we have
e4 λ2 = −
2me2
= 4(k + ` + 1)2 .
2
h̄ E
This yields 13.1.14.
Mathematical aside: Here is a different representation using classical mathematP
k
ical physics. Note that the function L(ρ) ≡ ∞
k=0 Ck ρ obeys the equation
ρL00 (ρ) + [ 2(` + 1) − 2ρ ]L0 (ρ) + (q 2 λ − 2(` + 1))L(ρ) = 0 .
Substituting q 2 λ = 2n (where n is an integer eventually) we have
1 00
ρL + [ (` + 1) − ρ ]L0 + (n − (` + 1))L = 0 .
2
Let z = 2ρ we have
z
d2 L
dL
+ [2(` + 1) − z]
− (` + 1 − n)L = 0 .
2
dz
dz
We know (with a good mathematical methods course) that the general solution to
zw00 + (c − z) w0 − aw = 0
is given by the confluent hypergeometric function w = 1 F1 (a; c; z). Thus we find that
the solution L(ρ) is 1 F1 (`+1−n; 2`+2; 2ρ) . The particular terminating (for integer n)
4
confluent hypergeometric function can be related to the associated Laguerre polynomial.
Shankar 13.3
We are considering the case n = 2 and ` = 1 and thus k = 0 in the notation of the
4
text. So we have W = me
from 13.1.4 and form 13.1.6
8h̄2
s
ρ =
2mW
me2
r
r
=
2
2 r =
2a0
h̄
2h̄
using 13.1.24. From 13.1.10 since k = 0 and ` = 1 we have v = C0 ρ2 and thus
R(r) =
U (r)
− r
= C e 2a0 r
r
q
3
where C is an overall constant. We know that Y10 = 4π
cos θ from 12.5.39 which is
−r/(2a0 )
normalized. So all we need is that Ce
r is notmalized when integrated over the
radial coordinate. We have
C
2
Z ∞
0
− ar
dr r2 r2 e
0
= C 2 24 a50 .
q
1
Thus C = 24a
3 × 1/a0 and including the normalization from the spherical harmonic
0
yields the quoted answer.
Shankar 13.5
Since we are asked to compute hΩi for stationary states |n`mi its time derivative
vanishes. Thus we have h[Ω, H]i = 0 by Ehrenfests’ theorem. So we compute the
~ · P~ as ordered. We calculate (using the summation convention)
commutator for Ω = R
·
Pi Pi
Rj P j ,
2m
¸
·
¸
P~ · P~
Pi Pi
= Rj ,
Pj = 2ih̄
= 2ih̄ T .
2m
2m
We have used
[Rj , Pi Pi ] = Pi [Rj , Pi ] + [Rj , Pi ]Pi = 2ih̄ Pj .
We consider the potential energy term next:
Ã
[Rj Pj , V (R)] = Rj [Pj , V (R)] = Rj
5
!
d
−ih̄
V (R) .
dRj
Note that [Pi , V (R)] can be evaluated in the coordinate representation (in Cartesian
coordinates) by acting on a function f (r):
Ã
#
∂
∂
∂
~ ∇V.
~
[Pi , V (r)]f (r) = −ih̄
V (r)f (r) − V (r)
V (r)f (r) = −ih̄
V (r) = −ih̄R·
∂ri
∂ri
∂ri
Thus we can write formally [Pi , V (R)] = −ih̄∂V (R)/∂Ri . So we need to evaluate
~ · ∇V
~ (R). In the coordinate representation using spherical coordinates for central
R
potentials this is just rV 0 (r). For the Coulomb potential we obtain −V (r) and including
the factor of −ih̄ we obtain ih̄ V (r). Substituting into the basic relation we have
h2T + V i = 0
as asserted. If V (R) ∝ Rn , rV 0 (r) = nV and thus we obtain hT i =
6
n
2
hV i.