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 ______
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 ______
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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.
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.
Updated On: Jul 28, 2026
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Correct Answer: 10.56
Solution and Explanation
Step 1: Rewrite \(\vec{S}_1\cdot\vec{S}_2\) using the total spin.
Let \(\vec{S} = \vec{S}_1 + \vec{S}_2\) be the total spin. Squaring, \(S^2 = S_1^2 + S_2^2 + 2\vec{S}_1\cdot\vec{S}_2\), so
\[ \vec{S}_1\cdot\vec{S}_2 = \frac{1}{2}\left(S^2 - S_1^2 - S_2^2\right) \]
Step 2: Find the eigenvalues.
For each spin-1/2, \(S_1^2 = S_2^2 = \hbar^2 s(s+1)\) with \(s=1/2\), giving \(\hbar^2(3/4)\) each. Combining two spin-1/2 particles gives total spin \(S=0\) (the singlet) or \(S=1\) (the triplet), with \(S^2 = \hbar^2 S(S+1)\). So
\[ \vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left[S(S+1) - \frac{3}{4} - \frac{3}{4}\right] \]
Step 3: Evaluate for the singlet and the triplet.
Singlet, \(S=0\): \(S(S+1)=0\), so \(\vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left(0-\frac{3}{2}\right) = -\frac{3\hbar^2}{4}\).
Triplet, \(S=1\): \(S(S+1)=2\), so \(\vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left(2-\frac{3}{2}\right) = \frac{\hbar^2}{4}\).
Step 4: Get the energy levels.
\[ E_{singlet} = \frac{A}{\hbar^2}\left(-\frac{3\hbar^2}{4}\right) = -\frac{3A}{4}, \qquad E_{triplet} = \frac{A}{\hbar^2}\left(\frac{\hbar^2}{4}\right) = \frac{A}{4} \]
Since \(A = 10.56\) eV is positive, \(E_{singlet} < E_{triplet}\), so the singlet is the ground state and the triplet is the excited state.
Step 5: Compute the gap.
\[ \Delta E = E_{triplet} - E_{singlet} = \frac{A}{4} - \left(-\frac{3A}{4}\right) = A \]
So the excitation energy is simply \(A\) itself:
\[ \Delta E = 10.56\text{ eV} \]
Final Answer:
Rounded to two decimal places, \[ \boxed{\Delta E = 10.56\text{ eV}} \]
Let \(\vec{S} = \vec{S}_1 + \vec{S}_2\) be the total spin. Squaring, \(S^2 = S_1^2 + S_2^2 + 2\vec{S}_1\cdot\vec{S}_2\), so
\[ \vec{S}_1\cdot\vec{S}_2 = \frac{1}{2}\left(S^2 - S_1^2 - S_2^2\right) \]
Step 2: Find the eigenvalues.
For each spin-1/2, \(S_1^2 = S_2^2 = \hbar^2 s(s+1)\) with \(s=1/2\), giving \(\hbar^2(3/4)\) each. Combining two spin-1/2 particles gives total spin \(S=0\) (the singlet) or \(S=1\) (the triplet), with \(S^2 = \hbar^2 S(S+1)\). So
\[ \vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left[S(S+1) - \frac{3}{4} - \frac{3}{4}\right] \]
Step 3: Evaluate for the singlet and the triplet.
Singlet, \(S=0\): \(S(S+1)=0\), so \(\vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left(0-\frac{3}{2}\right) = -\frac{3\hbar^2}{4}\).
Triplet, \(S=1\): \(S(S+1)=2\), so \(\vec{S}_1\cdot\vec{S}_2 = \frac{\hbar^2}{2}\left(2-\frac{3}{2}\right) = \frac{\hbar^2}{4}\).
Step 4: Get the energy levels.
\[ E_{singlet} = \frac{A}{\hbar^2}\left(-\frac{3\hbar^2}{4}\right) = -\frac{3A}{4}, \qquad E_{triplet} = \frac{A}{\hbar^2}\left(\frac{\hbar^2}{4}\right) = \frac{A}{4} \]
Since \(A = 10.56\) eV is positive, \(E_{singlet} < E_{triplet}\), so the singlet is the ground state and the triplet is the excited state.
Step 5: Compute the gap.
\[ \Delta E = E_{triplet} - E_{singlet} = \frac{A}{4} - \left(-\frac{3A}{4}\right) = A \]
So the excitation energy is simply \(A\) itself:
\[ \Delta E = 10.56\text{ eV} \]
Final Answer:
Rounded to two decimal places, \[ \boxed{\Delta E = 10.56\text{ eV}} \]
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