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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