Hamiltonian Analysis

The Hamiltonian describing the interaction between two spin-1/2 particles is given:

$H = \frac{A}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$

Where $A = 10.56 \text{ eV}$ is the coupling constant, and $\vec{S}_1$ and $\vec{S}_2$ are the spin angular momentum operators for particles 1 and 2.

Spin States and Energy Eigenvalues

The total spin $\vec{S} = \vec{S}_1 + \vec{S}_2$. Squaring this gives $S^2 = S_1^2 + S_2^2 + 2\vec{S}_1 \cdot \vec{S}_2$. For spin-1/2 particles, $S_1^2 = S_2^2 = \frac{1}{2}(\frac{1}{2}+1)\hbar^2 = \frac{3}{4}\hbar^2$. Rearranging for the dot product term yields:

$2\vec{S}_1 \cdot \vec{S}_2 = S^2 - S_1^2 - S_2^2 = S^2 - \frac{3}{2}\hbar^2$

The Hamiltonian becomes:

$H = \frac{A}{\hbar^2} \frac{1}{2} \left( S^2 - \frac{3}{2}\hbar^2 \right)$

Singlet Ground State ($S=0$)

The singlet state has total spin $S=0$, so $S^2 = 0$. The energy is:

$E_{ground} = E_{singlet} = \frac{A}{\hbar^2} \frac{1}{2} \left( 0 - \frac{3}{2}\hbar^2 \right) = -\frac{3A}{4}$

Triplet Excited State ($S=1$)

The triplet state has total spin $S=1$, so $S^2 = 1(1+1)\hbar^2 = 2\hbar^2$. The energy is:

$E_{excited} = E_{triplet} = \frac{A}{\hbar^2} \frac{1}{2} \left( 2\hbar^2 - \frac{3}{2}\hbar^2 \right) = \frac{A}{2} \left( \frac{1}{2} \right) = \frac{A}{4}$

Excitation Energy Calculation

The energy required for excitation from the ground state (singlet) to the excited state (triplet) is the difference between their energies:

$\Delta E = E_{excited} - E_{ground}$

$\Delta E = \frac{A}{4} - \left(-\frac{3A}{4}\right) = \frac{A}{4} + \frac{3A}{4} = A$

Final Energy Value

Substituting the given value $A = 10.56 \text{ eV}$:

$\Delta E = 10.56 \text{ eV}$

The required excitation energy is 10.56 eV.