Discussion
Consider two non-identical spin \(\frac{1}{2}\) particles labelled 1 and 2 in the spin product state \(|\tfrac{1}{2}, \tfrac{1}{2}\rangle |\tfrac{1}{2}, -\tfrac{1}{2}\rangle\). The Hamiltonian of the system is
\[ H = \frac{4\lambda}{\hbar^{2}}\vec{S}_{1}\cdot\vec{S}_{2} , \]
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?
Answer: A
The expectation value is \(-\lambda\).
Write the dot product with raising and lowering operators:
\[ \begin{aligned} \vec{S}_{1}\cdot\vec{S}_{2} &= S_{1z}S_{2z} \\ &\quad + \tfrac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+}) \end{aligned} \]
In the product state \(|\uparrow\rangle|\downarrow\rangle\), the flip terms turn the state into \(|\downarrow\rangle|\uparrow\rangle\) (or zero), which is orthogonal to it, so their expectation vanishes. Only the \(z\) term survives:
\[ \begin{aligned} \langle H \rangle &= \frac{4\lambda}{\hbar^{2}}\left(\frac{\hbar}{2}\right)\left(-\frac{\hbar}{2}\right) \\ &= -\lambda \end{aligned} \]
\(+\lambda\) is the value for parallel spins (the triplet \(|\uparrow\uparrow\rangle\)). \(-3\lambda\) would be the singlet, and \(\pm 2\lambda\) do not follow from the spin algebra.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development