GATE 2026 PH – Question 59
The Hamiltonian of two interacting spin-1/2 particles is $H=(A/\hbar^2)\mathbf S_1\cdot\mathbf S_2$, where $\mathbf S_1$ and $\mathbf S_2$ 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 _____
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Correct answer: 10.51 to 10.61
Explanation
**Total spin and the exchange interaction.** For two spin-½ particles, with total spin $\mathbf S=\mathbf S_1+\mathbf S_2$:
$$\mathbf S_1\cdot\mathbf S_2=\tfrac12\left(S^2-S_1^2-S_2^2\right).$$
In units of $\hbar^2$, $S_1^2=S_2^2=\tfrac34$ and $S^2=S(S+1)$:
- **Triplet ($S=1$):** $\mathbf S_1\cdot\mathbf S_2=\tfrac12\left(2-\tfrac34-\tfrac34\right)=\tfrac14$, so $E_T=\dfrac A4$.
- **Singlet ($S=0$):** $\mathbf S_1\cdot\mathbf S_2=\tfrac12\left(0-\tfrac32\right)=-\tfrac34$, so $E_S=-\dfrac{3A}{4}$.
With $A=10.56$ eV $>0$, the singlet is the ground state.
**Excitation energy:**
$$E_T-E_S=\frac A4+\frac{3A}4=A=\mathbf{10.56\ eV}.$$