Question

Consider two non-identical spin-$\dfrac{1}{2}$ particles labelled 1 and 2 in the spin product state $\left[\left|\dfrac{1}{2}, -\dfrac{1}{2}\right\rangle \left|\dfrac{1}{2}, \dfrac{1}{2}\right\rangle\right]$. The Hamiltonian of the system is $H = \dfrac{4\lambda}{\hbar^2} \, \vec{S_1} \cdot \vec{S_2}$, where $\vec{S_1}$ and $\vec{S_2}$ are the spin operators of particles 1 and 2, respectively, and $\lambda$ is a constant with appropriate dimensions. What is the expectation value of $H$ in the above state?

• A. $-\lambda$
• B. $-2\lambda$
• C. $\lambda$
• D. $2\lambda$

Hint:

For two spin-$\dfrac{1}{2}$ particles in the singlet state, the expectation value of the spin dot product is $-\dfrac{3}{4} \hbar^2$.

Answer 1

The Correct Option is A

- The expectation value of the Hamiltonian is given by the inner product of the wavefunction with the Hamiltonian operator. In this case, we use the fact that for spin-$\dfrac{1}{2}$ particles, the spin dot product has a known result: \[ \langle \vec{S_1} . \vec{S_2} \rangle = -\dfrac{3}{4}\hbar^2, \] for the singlet state of two spin-$\dfrac{1}{2}$ particles.
- Using this, the expectation value of $H$ is: \[ \langle H \rangle = \dfrac{4\lambda}{\hbar^2} \left(-\dfrac{3}{4}\hbar^2\right) = -\lambda. \] Thus, the correct answer is (A).