# Homework Assignment 5 Physics 501 Due October19,2009 by ezm24188

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```									                                  Homework Assignment 5
Physics 501
Due October 19, 2009
1. a) Solve the radial time-independent Schrodinger equation for the two-dimensional isotropic
harmonic oscillator.

¯2
h      d2    1 d                m2 ¯ 2
lh     1
−            2
+            +            2
+ mω 2 ρ2 − E REml (ρ) = 0
2m     dρ    ρ dρ               2mρ     2

[Hint: Find the asymptotic form of the solution and let the solution have the form of a
power series times the asymptotic solutions. Find the recursion relation for the coeﬃcients
of the power series and ﬁnd the quantization condition required in order to get a physically
meaningful solution.]

¯2
h        d2    1 d                  m2 ¯ 2
lh
as ρ → 0                 −                +                  +          REml (ρ) = 0
2m       dρ2   ρ dρ                 2mρ2

assume REl (ρ) ∝ ρα        ⇒       α(α − 1) + α − m2 = 0
l                          ⇒       α 2 = m2
l            ⇒   α = ±|ml |

∞                               ∞

α = −|ml | is not allowed since then                |REl (ρ)| ρ dρ ∝2
ρ1−|ml | dρ = ∞
0                              0

¯2
h         d2    1 d                  m2 ω 2 2
as ρ → ∞                  −                 +                 −          ρ REml (ρ) = 0
2m        dρ2   ρ dρ                  ¯2
h

2
The solution is:                   REml (ρ) ∝ e−mωρ                    h
/2¯

E                              mω           1/2
Let               =                            ξ =                          ρ
h
¯ω                              ¯
h

∞
2                                   2
REml (ξ) = ξ |m| e−ξ           /2
ak ξ k = ξ |m| e−ξ       /2
U (ξ)
k=0

d2          2|ml | + 1               d
+                 − 2ξ             + 2 ( − |ml | − 1) U (ξ) = 0
dξ 2           ξ                     dξ

2(ν − + |ml | + 1)
This leads to the recursion relation                  aν+2 =                                              aν
(ν + 2)(ν + 1) + (2|ml | + 1)(ν + 2)
b) Show that the quantization condition is:

E = hω(2k + |ml | + 1) = ¯ ω(n + 1)
¯                    h                           k = 0, 1, 2, · · ·

The recursion relation must terminate. Otherwise we get

aν+2   2
→              as         ν → ∞
aν    ν
2
⇒          U (ξ) → eξ            which is not allowed

⇒            − |ml | − 1 = n              where             n = 0, 2, 4, · · ·

E
=       = n + |ml | + 1 = 2k + ml + 1                      k = 0, 1, 2, · · ·
¯ω
h

E = hω(2k + |ml | + 1) = hω(n + 1)
¯                    ¯                                n = 2k + |ml |

c) Determine the degeneracy of the nth eigenstate.

For a given n,           k = 0, 1, 2, · · · , n/2         and      |ml | = 0, 1, 2, · · · , n
n   ml   k      degeneracy
0    0   0          1
1    1   0          2
-1   0
2    0   1          3
2   0
-2   0
3    1   1          4
-1   1
3   0
-3   0
degeneracy = n + 1
d) Show that the n = 0 eigenstate is the same as the n = 0 eigenstate that we found in
Cartesian coordinates
2
/2¯ −mωy 2 /2¯
u0 (x, y) ∝ e−mωx                   h
e         h

and that the n = 1 eigenstates are linear combinations of the two n = 1 states that we
found in Cartesian coordinates
2
/2¯ −mωy 2 /2¯                                                                  2
/2¯ −mωy 2 /2¯
u1 (x, y) ∝ xe−mωx        h
e         h
and                 u1 (x, y) ∝ ye−mωx                 h
e         h

n/2
|m| −mωρ2 /2¯
h
unm (ρ, φ) ∝ ρ             e                             ak ρk eiml φ
k=0

2                        2
+y 2 )/2¯                      2
/2¯ −mωy 2 /2¯
u00 (ρ, φ) ∝ e−mωρ            h
/2¯
= e−mω(x               h
= e−mωx              h
e         h

2                              2
/2¯ −mωy 2 /2¯
u11 (ρ, φ) ∝ ρ e−mωρ            /2¯ iφ
h
e     = ρ e−mωx               h
e       h
(cos φ + i sin φ)

2
/2¯ −mωy 2 /2¯                               2
/2¯ −mωy 2 /2¯
= x e−mωx             h
e      h
+ iy e−mωx                     h
e         h

2                                  2
/2¯ −mωy 2 /2¯
u1−1 (ρ, φ) ∝ ρ e−mωρ             /2¯ −iφ
h
e      = ρ e−mωx                  h
e      h
(cos φ − i sin φ)

2
/2¯ −mωy 2 /2¯                               2
/2¯ −mωy 2 /2¯
= x e−mωx             h
e      h
− iy e−mωx                     h
e         h

2. Problem 5 on page 99 of Schiﬀ. Obtain an approximate analytic expression for the energy
levels of a spherical potential well (l = 0) when V0 a2 is slightly greater than π 2 ¯ 2 /8m.
h

3. Problem 7 on page 99 of Schiﬀ. Assume that the interaction between the neutron and proton
that make up a deuteron can be represented by a spherical potential well with a = 2.00×10−13
cm. If the lowest (l = 0) energy level of the system is -2.23 MeV, calculate V − 0 in MeV to
three signiﬁcant ﬁgures. How does this answer compare with that which would be obtained
from the approximate formula derived in problem 2 above? [Hint: be sure to use the proton-
neutron reduced mass.]
4. Problem 12 on page 99 of Schiﬀ. Find the energy levels of a three- isotropic harmonic oscilla-
tor by solving the time-independent Schrodinger equation in Cartesian coordinates. What is
the degeneracy of each level? Show that this equation can also be separated in spherical and
cylindrical coordinates.

¯2
h     ∂2    ∂2     ∂2             1
−             +      +            +     mω 2 (x2 + y 2 + z 2 ) − E uE (x, y, z) = 0
2m    ∂x2   ∂y 2   ∂z 2           2

¯ 2 ∂2
h       1
−          + mω 2 x2 − Ex uEx (x) = 0
2m ∂x2  2

¯ 2 ∂2
h        1
−         2
+ mω 2 y 2 − Ey uEy (y) = 0
2m ∂y    2

¯ 2 ∂2
h        1
−           + mω 2 z 2 − Ez uEz (z) = 0
2m ∂z 2  2

1                            1                             1
¯
Ex = hω nx +                           ¯
Ey = hω ny +                       ¯
Ez = hω nz +
2                            2                             2

3                  3
¯
E = Ex + Ey + Ez = hω nx + ny + nz +                        ¯
= hω n +
2                  2

n = nx + ny + nz

n   nx   ny       nz     degeneracy
0    0   0        0          1
1    1   0        0          3
1    0   1        0
1    0   0        1
2    2   0        0           6
2    0   2        0
2    0   0        2
2    1   1        0
2    1   0        1
2    0   1        1
3    3   0        0          10
3    0   3        0
3    0   0        3
3    2   1        0
3    2   0        1
3    1   2        0
3    0   2        1
3    1   0        2
3    0   1        2
3    1   1        1

(n + 1)(n + 2)
The degenracy is:
2
In spherical coordinates:

¯2
h        1 ∂          ∂                      1     ∂                ∂              1     ∂2    1
−                    r2             +                           sin θ        +           2        + mω 2 r2 − E uE (r, θ, φ) = 0
2m       r2 ∂r        ∂r             r2      sin θ ∂θ               ∂θ       r2   sin θ ∂φ2    2

¯2
h             ∂             ∂            1 ∂                ∂          1    ∂2
−                          r2         +                sin θ         +      2
2m            ∂r            ∂r         sin θ ∂θ             ∂θ       sin θ ∂φ2

1
+ mω 2 r4 − r2 E uE (r, θ, φ) = 0
2

uE (r, θ, φ) = REl (r)Ylm (θ, φ)

d             d              2mr2    1
r2             +       2 E − mω 2 r2               REl (r) = l(l + 1)REl (r)
dr            dr              h
¯      2

1 ∂                       ∂               1     ∂2
sin θ               +              Ylm (θ, φ) = −l(l + 1)Ylm (θ, φ)
sin θ ∂θ                    ∂θ            sin2 θ ∂φ2

In cylindrical coordinates:

¯2
h        ∂2    1 ∂    1 ∂2   ∂2                                     1
−                +      + 2    +                                  +       mω 2 (ρ2 + z 2 ) − E uE (ρ, φ, z) = 0
2m       ∂ρ2   ρ ∂ρ  ρ ∂φ2   ∂z 2                                   2

uE (ρ, φ, z) = REml ,z (ρ)Φml (φ)Znz (z)

¯2
h             ∂2     ∂     ∂2      ∂2                                 1
−            ρ2     2
+ρ    +    2
+ ρ2 2                             +     mω 2 (ρ4 + z 2 ρ2 ) − ρ2 E uE (ρ, φ, z) = 0
2m            ∂ρ     ∂ρ   ∂φ       ∂z                                 2

d2
Φml (φ) = − ml Φml (φ)
dφ2

¯2
h         ∂2     ∂      ∂2                             1                                                  ¯ 2 m2
h l
−        ρ2     2
+ρ    + ρ2 2                        +      mω 2 (ρ4 + z 2 ρ2 ) − ρ2 E Rnρ ml (ρ)Znz (z) = −        Rnρ ml (ρ)Znz (z)
2m        ∂ρ     ∂ρ     ∂z                             2                                                   2m

¯2
h         ∂2    1 ∂    ∂2                           1                    ν
−               2
+      +                        +       mω 2 (ρ2 + z 2 ) + 2 − E Rnρ ml (ρ)Znz (z) = 0
2m        ∂ρ    ρ ∂ρ   ∂z 2                         2                   ρ

¯2
h     ∂2    1 ∂                           1              ν
−           2
+      +                   +      mω 2 (ρ2 ) + 2 − E Rnρ ml (ρ) = −nz Rnρ ml (ρ)
2m    ∂ρ    ρ ∂ρ                          2             ρ

¯ 2 ∂2
h        1
−           + mω 2 (ρ2 z 2 ) Znz (z) = nz Znz (z)
2m ∂z 2  2
5. Start with the expression for Lx and Ly that I gave you in lecture

∂         ∂                                               ∂         ∂
Lx = −i¯ − cot θ cos φ
h                      − sin φ                    Ly = −i¯ − cot θ sin φ
h                       + cos φ
∂φ         ∂θ                                             ∂φ         ∂θ

Deﬁne the operators:           L+ = Lx + iLy               and           L− = Lx − iLy

Show that the states        L+ Ylm (θ, φ)          and        L+ Ylm (θ, φ)        are eigenstates of   Lz

∂                       ∂         ∂
h
Lz Lx = (−i¯ )        (−i¯ ) − cot θ cos φ
h                    − sin φ
∂φ                      ∂φ         ∂θ

∂         ∂      ∂                       ∂         ∂
= −¯ 2
h       − cot θ cos φ      − sin φ          +      cot θ sin φ      − cos φ
∂φ         ∂θ    ∂φ                      ∂φ         ∂θ

h
= (Lx Lz + i¯ Ly )

∂                       ∂         ∂
Lz Ly = (−i¯ )
h          (−i¯ ) − cot θ sin φ
h                    + cos φ
∂φ                      ∂φ         ∂θ

∂         ∂      ∂                         ∂         ∂
= −¯ 2
h       − cot θ sin φ      + cos φ          +      − cot θ cos φ      − sin φ
∂φ         ∂θ    ∂φ                        ∂φ         ∂θ

= (Ly Lz − i¯ Lx )
h

¯
Lz (Lx + iLy )Ylm (θ, φ) = [(Lx + iLy )Lz + h(iLy + Lx )] Ylm (θ, φ)

h h                                   h
= (m¯ + ¯ )(Lx + iLy )Ylm (θ, φ) = (m + 1)¯ (Lx + iLy )Ylm (θ, φ)

Lz (Lx − iLy )Ylm (θ, φ) = [(Lx − iLy )Lz + h(iLy − Lx )] Ylm (θ, φ)
¯

= (m¯ − ¯ )(Lx − iLy )Ylm (θ, φ) = (m − 1)¯ (Lx − iLy )Ylm (θ, φ)
h h                                   h

What are the eigenvalues?

The eigenvalues are (m + 1)¯ and (m − 1)¯
h            h

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