Exercise 6.10

The Hamiltonian is given as

\[ \mathbf{H} = \frac{\omega}{2} \vec{\sigma} \cdot \vec{\tau} \]

The Hamiltonian should have units of energy. Therefore we will use the following form of the Hamiltonian for this exercise:

\[ \mathbf{H} = \frac{\hbar \omega}{2} \vec{\sigma} \cdot \vec{\tau} \]

The possible energies of the system are the eigenvalues of the Hamiltonian, specified by the time-independent Schrödinger Equation:

\[ \mathbf{H}\ket{E_i} = E_i \ket{E_i} \]

Using the results from Exercise 6.9, we already know the eigenvectors of the operator \(\vec{\sigma} \cdot \vec{\tau}\), which is the Hamiltonian (apart from a scalar):

\[ \mathbf{H} \ket{sing} = -3 \frac{\hbar \omega}{2} \ket{sing} \] \[ \mathbf{H} \ket{T_1} = \frac{\hbar \omega}{2} \ket{T_1} \] \[ \mathbf{H} \ket{T_2} = \frac{\hbar \omega}{2} \ket{T_2} \] \[ \mathbf{H} \ket{T_3} = \frac{\hbar \omega}{2} \ket{T_3} \]

Hence there are two possible energies: \(-\frac{3}{2} \hbar \omega\) and \(\frac{1}{2}\hbar \omega\).

The state of space of the two spin system has four dimensions, as basis vectors we can use \(\ket{sing}, \ket{T_1}, \ket{T_2}, \ket{T_3}\), since these vectors are eigenvectors of the Hermitian operator \(\mathbf{H}\) and thus form a complete set of basis vectors. They also form an orthonormal set, which is easy to show.

Regarding the time evolution of a state, we make use of the recipe for a Schrödinger ket, which is given in the textbook in chapter 4.13.

Initial state \(\ket{uu}\)

1. The Hamiltonian is \[ \mathbf{H} = \frac{\hbar \omega}{2} \vec{\sigma} \cdot \vec{\tau} \]

2. The initial state is given as \(\ket{uu}\).

3. The eigenvalues and eigenvectors are already known (see above).

4. The initial coefficients \(\alpha_j\) are: \[ \begin{flalign*} \alpha_0 (0) & = \braket{sing | uu }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} - \bra{ud} )(\ket{uu})&\\ & = 0 \end{flalign*} \] \[ \begin{flalign*} \alpha_1 (0) & = \braket{T_1 | uu }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} + \bra{du} )(\ket{uu})&\\ & = 0 \end{flalign*} \] \[ \begin{flalign*} \alpha_2 (0) & = \braket{T_2 | uu }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} + \bra{dd} )(\ket{uu})&\\ & = \frac{1}{\sqrt{2}} \end{flalign*} \] \[ \begin{flalign*} \alpha_3 (0) & = \braket{T_3 | uu }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} - \bra{dd} )(\ket{uu})&\\ & = \frac{1}{\sqrt{2}} \end{flalign*} \]

5. Rewrite \(\Psi(0) = \ket{uu}\) in terms of the eigenvectors \(\ket{E_j}\): \[ \Psi(0) = \frac{1}{\sqrt{2}} \ket{T_2} + \frac{1}{\sqrt{2}} \ket{T_3} \]

6. Capture time dependence: \[ \Psi(t) = \alpha_2(t) \ket{T_2} + \alpha_3(t) \ket{T_3} \]

7. Use Eq. 4.30 from the textbook: \[ \begin{flalign*} \alpha_2(t) & = \alpha_2(0) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} \frac{\hbar \omega}{2}t} = \alpha_2(0) \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t}\\ \alpha_3(t) & = \alpha_3(0) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} \frac{\hbar \omega}{2}t} = \alpha_3(0) \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \end{flalign*} \] \[ \Psi(t) = \frac{1}{\sqrt{2}} \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} (\ket{T_2} + \ket{T_3}) \]

\[ \begin{flalign*} \ket{T_2} + \ket{T_3} & = \frac{1}{\sqrt{2}} \ket{uu} + \frac{1}{\sqrt{2}} \ket{dd} + \frac{1}{\sqrt{2}} \ket{uu} - \frac{1}{\sqrt{2}} \ket{dd}&\\ & = \frac{1}{\sqrt{2}} \ket{uu} + \frac{1}{\sqrt{2}} \ket{uu}\\ & = \sqrt{2} \ket{uu} \end{flalign*} \]

\[ \Rightarrow \Psi(t) = \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \ket{uu} \]

Initial state \(\ket{ud}\)

\[ \begin{flalign*} \alpha_0 (0) & = \braket{sing | ud }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} - \bra{du} )(\ket{ud})&\\ & = \frac{1}{\sqrt{2}} \end{flalign*} \] \[ \begin{flalign*} \alpha_1 (0) & = \braket{T_1 | ud }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} + \bra{du} )(\ket{ud})&\\ & = \frac{1}{\sqrt{2}} \end{flalign*} \] \[ \begin{flalign*} \alpha_2 (0) & = \braket{T_2 | ud }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} + \bra{dd} )(\ket{ud})&\\ & = 0 \end{flalign*} \] \[ \begin{flalign*} \alpha_3 (0) & = \braket{T_3 | ud }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} - \bra{dd} )(\ket{ud})&\\ & = 0 \end{flalign*} \]

\[ \Psi(0) = \frac{1}{\sqrt{2}} \ket{sing} + \frac{1}{\sqrt{2}} \ket{T_1} \] \[ \Psi(t) = \alpha_0 (t) \ket{sing} + \alpha_1 (t) \ket{T_1} \] \[ \begin{flalign*} \alpha_0(t) & = \alpha_0(0) \mathrm{e}^{\mathrm{i}\frac{3 \omega}{2} t}\\ \alpha_1(t) & = \alpha_1(0) \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \end{flalign*} \] \[ \Psi(t)= \frac{1}{\sqrt{2}} (\mathrm{e}^{\mathrm{i}\frac{3 \omega}{2} t} \ket{sing} + \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \ket{T_1}) \]

Initial state \(\ket{du}\)

\[ \begin{flalign*} \alpha_0 (0) & = \braket{sing | du }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} - \bra{du} )(\ket{du})&\\ & = - \frac{1}{\sqrt{2}} \end{flalign*} \] \[ \begin{flalign*} \alpha_1 (0) & = \braket{T_1 | du }&\\ & = \frac{1}{\sqrt{2}}(\bra{ud} + \bra{du} )(\ket{du})&\\ & = \frac{1}{\sqrt{2}} \end{flalign*} \] \[ \begin{flalign*} \alpha_2 (0) & = \braket{T_2 | du }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} + \bra{dd} )(\ket{du})&\\ & = 0 \end{flalign*} \] \[ \begin{flalign*} \alpha_3 (0) & = \braket{T_3 | du }&\\ & = \frac{1}{\sqrt{2}}(\bra{uu} - \bra{dd} )(\ket{du})&\\ & = 0 \end{flalign*} \]

If we compare this with the initial state \(\ket{ud}\), the only difference is the sign of \(\alpha_0\). Hence, \[ \Psi(t)= -\frac{1}{\sqrt{2}} (\mathrm{e}^{\mathrm{i}\frac{3 \omega}{2} t} \ket{sing} - \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \ket{T_1}) \]

Initial state \(\ket{dd}\)

\[ \alpha_0 (0) = \braket{sing | dd } = 0 \] \[ \alpha_1 (0) = \braket{T_1 | dd } = 0 \] \[ \alpha_2 (0) = \braket{T_2 | dd } = \frac{1}{\sqrt{2}} \] \[ \alpha_3 (0) = \braket{T_3 | dd } = -\frac{1}{\sqrt{2}} \]

If we compare this with the initial state \(\ket{uu}\), the only difference is the sign of \(\alpha_3\). Hence,

\[ \Psi(t) = \frac{1}{\sqrt{2}} \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} (\ket{T_2} - \ket{T_3}) \]

\[ \Rightarrow \Psi(t) = \mathrm{e}^{-\mathrm{i}\frac{\omega}{2} t} \ket{dd} \]