give answer in 2 step with explanation at the end of each step and final answer at the end:b) Find the eigen energies in terms of the variables @ and 2, and the corresponding
cigen states.
Note: Kee

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:b) Find the eigen energies in terms of the variables @ and 2, and the corresponding
cigen states.
Note: Keep in mind the simplifying ideas which have been studied so that you are not solving
for the eigen values of a 4-dimensional matrix.

¢) Assume that the initial state is given by

1
W(0) = HB =12)= |= 0)®|+:2)

| W(0)) Ni (I+:0®[ =: 2) )
Calculate the matrix representation of this state and use it to determine the time dependent
general state W(¢)) of the system.

d) Find 8, |'¥(1)), where $, is the z-component of the total spin operator.
Note: The eigenfunctions for this matrix are well known from our previous work and are related
to the states of total spin.

VIPSolutions · Accepted Answer

Let's tackle part **(b)** of the problem step by step:

## Step : Understanding the Hamiltonian Structure

Usually, such quantum spin problems, the Hamiltonian is of the form:
$$
\hat{H} = \alpha \hat{S}_{1z} \hat{S}_{2z} + \beta (\hat{S}_{1x} \hat{S}_{2x} + \hat{S}_{1y} \hat{S}_{2y})
$$
where $\alpha$ and $\beta$ are constants, and $\hat{S}_{1}$ and $\hat{S}_{2}$ are spin operators for two spin-$\frac{1}{2}$ particles.

**Explanation:**  
This Hamiltonian describes two coupled spins. The first term ($\alpha$) involves the z-components, while the second ($\beta$) couples the x and y components. The total spin states (singlet and triplet) are eigenstates of this Hamiltonian, letting us avoid diagonalizing a 4x4 matrix.

---

## Step 2: Finding the Eigenvalues (Eigen Energies)

The total spin can be either singlet ($S=$) or triplet ($S=1$):
- **Singlet state:** $\left| S=, M= \right\rangle = \frac{1}{\sqrt{2}} (|+\rangle|-\rangle - |-\rangle|+\rangle)$
- **Triplet states:** 
  - $\left|1,1\right\rangle = |+\rangle|+\rangle$
  - $\left|1,\right\rangle = \frac{1}{\sqrt{2}} (|+\rangle|-\rangle + |-\rangle|+\rangle)$
  - $\left|1,-1\right\rangle = |-\rangle|-\rangle$

The Hamiltonian can be rewritten in terms of total spin $\vec{S} = \vec{S}_1 + \vec{S}_2$:
$$
\vec{S}_1 \cdot \vec{S}_2 = \frac{1}{2} (S_{\text{tot}}^2 - S_1^2 - S_2^2)
$$
For spin-$\frac{1}{2}$:
- $S_1^2 = S_2^2 = \frac{3}{4}$
- $S_{\text{tot}}^2 = S(S+1)$, with $S=$ (singlet), $S=1$ (triplet)

The eigenvalues are:
- For **triplet** ($S=1$): $S_{\text{tot}}^2 = 2$
- For **singlet** ($S=$): $S_{\text{tot}}^2 = $

Thus,
$$
\vec{S}_1 \cdot \vec{S}_2 = \begin{cases}
\frac{1}{2} (2 - \frac{3}{2}) = \frac{1}{4} & \text{(triplet)} \
\frac{1}{2} ( - \frac{3}{2}) = -\frac{3}{4} & \text{(singlet)}
\end{cases}
$$

If the Hamiltonian is of the form $H = \alpha S_{1z} S_{2z} + \beta (S_{1x} S_{2x} + S_{1y} S_{2y})$, then:
$$
H = \alpha S_{1z} S_{2z} + \beta \left[ \frac{1}{2}(S_1^+ S_2^- + S_1^- S_2^+) \right]
$$
But in terms of the singlet and triplet states, the eigenenergies are:
- **Triplet:** $E_T = \frac{\alpha}{4} + \frac{\beta}{4}$
- **Singlet:** $E_S = -\frac{\alpha}{4} - \frac{3\beta}{4}$

**Explanation:**  
By using the symmetry of the problem and the known properties of spin-1/2 systems, we avoid diagonalizing the full matrix. The eigenstates are simply the singlet and triplet states, and the eigenenergies are functions of $\alpha$ and $\beta$.

---

## **Final Answer**

**Eigen energies in terms of $\alpha$ and $\beta$:**
- **Triplet states ($S=1$):** $E_T = \frac{\alpha}{4} + \frac{\beta}{4}$
- **Singlet state ($S=$):** $E_S = -\frac{\alpha}{4} - \frac{3\beta}{4}$

The corresponding eigenstates are, respectively, the triplet and singlet total spin states.

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