Spin System Expectation Value Calculation

The problem asks for the expectation value of the Hamiltonian $H = \frac{4\lambda}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$ for a system of two spin $\frac{1}{2}$ particles in a specific initial state.

Initial State Description

The initial state is given as a product state:

$ |\psi\rangle = |\frac{1}{2}, \frac{1}{2}\rangle_1 |\frac{1}{2}, -\frac{1}{2}\rangle_2 $

This represents particle 1 in the spin-up state ($m_{s1} = +\frac{1}{2}$) and particle 2 in the spin-down state ($m_{s2} = -\frac{1}{2}$). We can denote these as $|+\rangle_1$ and $|-\rangle_2$ respectively.

$ |\psi\rangle = |+\rangle_1 |-\rangle_2 $

Hamiltonian Operator

The Hamiltonian 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 for particles 1 and 2.

Expectation Value Calculation

The expectation value of the Hamiltonian in the state $|\psi\rangle$ is $\langle H \rangle = \langle \psi | H | \psi \rangle$.

$ \langle H \rangle = \left\langle \psi \left| \frac{4\lambda}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2 \right| \psi \right\rangle = \frac{4\lambda}{\hbar^2} \langle \psi | \vec{S}_1 \cdot \vec{S}_2 | \psi \rangle $

We need to calculate the expectation value $\langle \vec{S}_1 \cdot \vec{S}_2 \rangle$ for the state $|\psi\rangle = |+\rangle_1 |-\rangle_2$. We can evaluate the expectation value of each component:

• Z-component: $\langle S_{1z}S_{2z} \rangle$
  • For spin $\frac{1}{2}$, $S_z |+\frac{1}{2}\rangle = +\frac{\hbar}{2} |+\frac{1}{2}\rangle$ and $S_z |-\frac{1}{2}\rangle = -\frac{\hbar}{2} |-\frac{1}{2}\rangle$.
  • $\langle S_{1z}S_{2z} \rangle = \langle +\frac{1}{2}|_1 \langle -\frac{1}{2}|_2 S_{1z}S_{2z} |+\frac{1}{2}\rangle_1 |-\frac{1}{2}\rangle_2$
  • $= (\langle +\frac{1}{2}|_1 S_{1z} |+\frac{1}{2}\rangle_1) (\langle -\frac{1}{2}|_2 S_{2z} |-\frac{1}{2}\rangle_2)$
  • $= (\frac{\hbar}{2}) (-\frac{\hbar}{2}) = -\frac{\hbar^2}{4}$
• X-component: $\langle S_{1x}S_{2x} \rangle$
  • The expectation value of $S_x$ for a spin $\frac{1}{2}$ particle in the $|+\frac{1}{2}\rangle$ or $|-\frac{1}{2}\rangle$ state is zero. $\langle +\frac{1}{2}| S_x |+\frac{1}{2}\rangle = 0$.
  • $\langle S_{1x}S_{2x} \rangle = (\langle +\frac{1}{2}|_1 S_{1x} |+\frac{1}{2}\rangle_1) (\langle -\frac{1}{2}|_2 S_{2x} |-\frac{1}{2}\rangle_2)$
  • $= (0)(0) = 0$
• Y-component: $\langle S_{1y}S_{2y} \rangle$
  • Similarly, the expectation value of $S_y$ for a spin $\frac{1}{2}$ particle in the $|+\frac{1}{2}\rangle$ or $|-\frac{1}{2}\rangle$ state is zero. $\langle +\frac{1}{2}| S_y |+\frac{1}{2}\rangle = 0$.
  • $\langle S_{1y}S_{2y} \rangle = (\langle +\frac{1}{2}|_1 S_{1y} |+\frac{1}{2}\rangle_1) (\langle -\frac{1}{2}|_2 S_{2y} |-\frac{1}{2}\rangle_2)$
  • $= (0)(0) = 0$

Summing the components:

$ \langle \vec{S}_1 \cdot \vec{S}_2 \rangle = \langle S_{1x}S_{2x} \rangle + \langle S_{1y}S_{2y} \rangle + \langle S_{1z}S_{2z} \rangle = 0 + 0 - \frac{\hbar^2}{4} = -\frac{\hbar^2}{4} $

Final Result

Substitute the expectation value of $\vec{S}_1 \cdot \vec{S}_2$ back into the Hamiltonian expectation value equation:

$ \langle H \rangle = \frac{4\lambda}{\hbar^2} \left( -\frac{\hbar^2}{4} \right) $ $ \langle H \rangle = -\lambda $