HW #8

1. Coupled Harmonic Oscillators

The Hamiltonian is
\(TraditionalForm\`H\  = \ 
        p\_1\^2\/\(2  M\) + p\_2\^2\/\(2  M\) + 
          p\_3\^2\/\(2  m\) + \(1\/2\) \(k(x\_3 - x\_1 - d)\)\^2 + \(1\/2\) \(k(x\_2 - x\_3 - d)\)\^2\)
For the sake of simplicity, we redefine (Subscript[x, 1]) -> (Subscript[x, 1]) - d, (Subscript[x, 2]) -> (Subscript[x, 2]) + d and the Hamiltonian is
\(TraditionalForm\`H\  = \ 
        p\_1\^2\/\(2  M\) + p\_2\^2\/\(2  M\) + 
          p\_3\^2\/\(2  m\) + \(1\/2\) \(k(x\_3 - x\_1)\)\^2 + \(1\/2\) \(k(x\_2 - x\_3)\)\^2\)

(a) Born-Oppenheimer Approximation

Assume (Subscript[x, 1]) and (Subscript[x, 2]) are fixed.  Then the potential energy for (Subscript[x, 3]) is 
(1/2) k(
            ((Subscript[x, 3])^2) - 2  (Subscript[x, 3]) (Subscript[x, 1]) + ((Subscript[x, 1])^2) + ((Subscript[x, 2])^2) - 
              2  (Subscript[x, 2]) (Subscript[x, 3]) + ((Subscript[x, 3])^2)) = (1/2) k(
            2  ((Subscript[x, 3])^2) - 2 ((Subscript[x, 1]) + (Subscript[x, 2])) (Subscript[x, 3]) + ((Subscript[x, 1])^2) + ((Subscript[x, 2])^2))
=(1/2) 
            2 (( k((Subscript[x, 3]) - (((Subscript[x, 1]) + (Subscript[x, 2]))/2)))^2) - 
          k (((((Subscript[x, 1]) + (Subscript[x, 2])))^2)/4) + (1/2) k(((Subscript[x, 1])^2) + ((Subscript[x, 2])^2))
\(TraditionalForm\`\(\(=\)\(\(1\/2\) 
            2 \( k(x\_3 - \(x\_1 + x\_2\)\/2)\)\^2 + \(1\/4\) \(k(x\_1 - x\_2)\)\^2\)\)\)
The energy levels for the (Subscript[x, 3])degree of freedom is therefore (n + (1/2)) ℏ (Sqrt[((2  k)/m)]).

Taking the ground state for the (Subscript[x, 3]) degree of freedom, the Hamiltonian reduces to 
\(TraditionalForm\`H\_eff = 
        p\_1\^2\/\(2  M\) + 
          p\_2\^2\/\(2  M\) + \(1\/2\) ℏ \@\(\( 2  k\)\/m\) + \(1\/4\) \(k(x\_1 - x\_2)\)\^2\).
Separating the center-of-mass motion with P = (Subscript[p, 1]) + (Subscript[p, 2]), X = (((Subscript[x, 1]) + (Subscript[x, 2]))/2), p = (((Subscript[p, 1]) - (Subscript[p, 2]))/2), x = (Subscript[x, 1]) - (Subscript[x, 2]), it becomes 
(Subscript[H, eff]) = 
        ((P^2)/(4  M)) + 
          ((p^2)/M) + (1/4) 
            k (x^2) + (1/2) ℏ (Sqrt[(( 2  k)/m)]).
Leaving the trivial translational energy aside, the energy levels are therefore
\(TraditionalForm\`E = \((n' + 
                1\/2)\) ℏ \@\( k\/M\) + \(1\/2\) ℏ \@\(\( 2  k\)\/m\)\).

(b) Exact Solution

We identify the normal modes of the coupled harmonic oscillators.

One way is to go back to the Lagrangian,
\(L = \(1\/2\) M\ \(x\&·\)\_1\^2 + \(1\/2\) 
                  M\ \(x\&·\)\_2\^2 + \(1\/2\) 
                  m\ \(x\&·\)\_3\^2 - \(1\/2\) \(k(x\_3 - x\_1)\)\^2 - \(1\/2\) \(k(x\_2 - x\_3)\)\^2\).
We rescale the coordinates to make the kinetic terms the same for all three, (Subscript[x, 1]) = (1/(Sqrt[M])) (Subscript[q, 1]), (Subscript[x, 2]) = (1/(Sqrt[M])) (Subscript[q, 2]), (Subscript[x, 3]) = (1/(Sqrt[m])) (Subscript[q, 3]).  Then
\(L = 
              1\/2\ \(q\&·\)\_1\^2 + 
                1\/2\ \(q\&·\)\_2\^2 + 
                1\/2\ \(q\&·\)\_3\^2 - \(1\/2\) \(k(\(1\/\@m\) q\_3 - \(1\/\@M\) q\_1)\)\^2 - \(1\/2\) \(k(\(1\/\@m\) q\_3 - \(1\/\@M\) q\_2)\)\^2\).
We can write the potential term as 
V=(1/2)k((Subscript[q, 1]),  (Subscript[q, 3]), (Subscript[q, 2]))(({{1/M, -1/(Sqrt[m M]), 0}, {-1/(Sqrt[m M]), 2/m, -1/(Sqrt[m M])}, {0, -1/(Sqrt[m M]), 1/M}}))(({{(Subscript[q, 1])}, {(Subscript[q, 3])}, {(Subscript[q, 2])}})).

Eigenvalues[{{1/M, -1/(Sqrt[m M]), 0}, {-1/(Sqrt[m M]), 
          2/m, -1/(Sqrt[m M])}, {0, -1/(Sqrt[m M]), 1/M}}]

Therefore, it is given by three independent harmonic oscillators of frequencies 0, (Sqrt[k/M]), and (Sqrt[k (2  M + m)/m M]).  The zero-frequency mode corresponds to the overall translation, while the other two give the harmonic oscillator levels,
E = (n' + (1/2)) ℏ (Sqrt[(k/M)]) + (n + 
                (1/2)) ℏ (Sqrt[((k(2  M + m))/(m M))]).

(c) Comparison

Taking n = 0 in the exact result, we find
\(TraditionalForm\`E = \((n' + 
                1\/2)\) ℏ\ \@\(k\/M\) + \(1\/2\) ℏ\ \@\(\(k(2  M + m)\)\/\(m\ M\)\)\),
while the Born-Oppenheimer approximation gives
\(TraditionalForm\`E = \((n' + 
                1\/2)\) ℏ \@\( k\/M\) + \(1\/2\) ℏ \@\(\( 2  k\)\/m\)\).
Therefore the Born-Oppenheimer approximation gives the correct result up to a correction suppressed by m/M as expected.

2. A=14

Looking at the nitrogen, the levels with spin/parity that appear together in carbon and nitrogen with similar excitation energies are identified as I = 1, wihle those that appear only in nitrogen as I = 0.

From the ground state and up,

(J^P)	I
\(TraditionalForm\`\(1\^+\)\)	0
\(TraditionalForm\`\(0\^+\)\)	1
\(TraditionalForm\`\(1\^+\)\)	0
\(TraditionalForm\`\(0\^-\)\)	0
\(TraditionalForm\`\(2\^-\)\)	0
\(TraditionalForm\`\(1\^-\)\)	0
\(TraditionalForm\`\(3\^-\)\)	0
\(TraditionalForm\`\(1\^+\)\)	0
\(TraditionalForm\`\(3\^+\)\)	0
\(TraditionalForm\`\(2\^+\)\)	0
\(TraditionalForm\`\(2\^-\)\)	0
\(TraditionalForm\`\(1\^-\)\)	1	8062 - 2313 = 5749, in rough accordance with the excitation energies in carbon and oxygen
\(TraditionalForm\`\(4\^-\)\)	0
\(TraditionalForm\`\(0\^+\)\)	1
\(TraditionalForm\`\(0\^-\)\)	0
\(TraditionalForm\`\(3\^-\)\)	1
\(TraditionalForm\`\(5\^-\)\)	0
\(TraditionalForm\`\(2\^+\)\)	0	appears too close to \(TraditionalForm\`\(3\^-\)\) with I = 1 compared to carbon and oxygen
\(TraditionalForm\`\(3\^+\)\)	0
\(TraditionalForm\`\(2\^+\)\)	1	9172 - 8907 = 265, in rough accordance with the \(2\^+\)-\(3\^-\) splittings in carbon and oxygen

etc