Question

The Hamiltonian of two interacting spin-1/2 particles is \(H = \dfrac{A}{\hbar^2}\vec{S}_1\cdot\vec{S}_2\), where \(\vec{S}_1\) and \(\vec{S}_2\) are the spin angular momenta of particles 1 and 2, respectively. Here, \(A = 10.56\) eV. The energy in eV required to induce an excitation from the ground state to the excited state (rounded off to two decimal places) is ______

Hint:

Hint:
Use \(\vec{S}_1\cdot\vec{S}_2 = \frac{1}{2}(S^2-S_1^2-S_2^2)\) with total spin \(S=0\) (singlet, ground state) or \(S=1\) (triplet, excited state), then subtract the two energies.

Answer 1

Correct Answer: 10.56

Step 1: Rewrite the Hamiltonian using Pauli matrices.
For a spin-1/2 particle, $\vec S_i = \frac{\hbar}{2}\vec\sigma_i$, where $\vec\sigma_i$ are the Pauli matrices. So
\[ \vec S_1\cdot\vec S_2 = \frac{\hbar^2}{4}\,\vec\sigma_1\cdot\vec\sigma_2, \qquad H = \frac{A}{4}\,\vec\sigma_1\cdot\vec\sigma_2 \]

Step 2: Get the eigenvalues of $\vec\sigma_1\cdot\vec\sigma_2$ from $(\vec\sigma_1+\vec\sigma_2)^2$.
Since each $\vec\sigma_i^2 = 3$ (as $\sigma_x^2=\sigma_y^2=\sigma_z^2=1$),
\[ (\vec\sigma_1+\vec\sigma_2)^2 = \vec\sigma_1^2+\vec\sigma_2^2+2\vec\sigma_1\cdot\vec\sigma_2 = 6 + 2\vec\sigma_1\cdot\vec\sigma_2 \]
The left side is the total spin operator built from Pauli matrices, with eigenvalue $4S(S+1)$ for total spin $S$: this is $0$ for the singlet ($S=0$) and $8$ for the triplet ($S=1$).

Step 3: Solve for $\vec\sigma_1\cdot\vec\sigma_2$ in each case.
Singlet: $0 = 6+2\vec\sigma_1\cdot\vec\sigma_2 \Rightarrow \vec\sigma_1\cdot\vec\sigma_2 = -3$.
Triplet: $8 = 6+2\vec\sigma_1\cdot\vec\sigma_2 \Rightarrow \vec\sigma_1\cdot\vec\sigma_2 = 1$.

Step 4: Turn these into energies.
\[ E_{singlet} = \frac{A}{4}(-3) = -\frac{3A}{4}, \qquad E_{triplet} = \frac{A}{4}(1) = \frac{A}{4} \]
With $A=10.56$ eV $>0$, the singlet sits lower, so it is the ground state and the triplet is the excited state.

Step 5: Find the gap.
\[ \Delta E = \frac{A}{4} - \left(-\frac{3A}{4}\right) = A = 10.56\text{ eV} \]

Final Answer:
Rounded to two decimal places, \[ \boxed{\Delta E = 10.56\text{ eV}} \]