Exercise 6

Note: In my edition of the textbook, the final oberservable is denoted by \(\sigma_x\). I think this is a typo and I’ll assume \(\sigma_y\).

We’ll follow the steps given in the textbook (Chapter 4.13: Recipe for a Schrödinger Ket).

  1. The Hamiltonian is \[ \mathbf H = \frac{\hbar \omega}{2} \sigma_z \]

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

  3. The eigenvalues and eigenvectors of the time-independant Schrödinger Equation \[ \frac{\hbar \omega}{2} \sigma_z \ket{E_i} = E_i \ket{E_i} \] are already known: \[ \begin{flalign*} E_1 &= \frac{\hbar \omega}{2} \\ E_2 &= -\frac{\hbar \omega}{2} \\ \ket{E_1} &= \ket{u} \\ \ket{E_2} &= \ket{d} \end{flalign*} \]

  4. The initial coefficients are \[ \begin{flalign*} \alpha_1(0) = \braket{u|u} = 1 \\ \alpha_2(0) = \braket{d|u} = 0 \end{flalign*} \]

  5. Write \(\Psi(0)\) in terms of the eigenvectors: \[ \Psi(0) = \ket{u} = \alpha_1(0) \ket {u} + \alpha_2(0) \ket{d} \]

  6. Expand \(\Psi(t)\): \[ \Psi(t) = \alpha_1(t) \ket{u} + \alpha_2(t) \ket{d} \]

  7. Use Eq. 4.30: \(\alpha_i(t) = \alpha_i(0) \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} E_i t}\) \[ \Psi(t) = \mathrm{e}^{-\frac{\mathrm{i}}{\hbar} \frac{\hbar \omega}{2} t} \ket{u} + 0 \cdot \ket{d} = \mathrm{e}^{-\frac{\mathrm{i}\omega}{2}t} \ket{u} \]

After time \(t\), \(\sigma_y\) is measured. The possible outcomes are the eigenvalues of \(\sigma_y\), which are \(\pm 1\). The eigenvectors are \[ \begin{flalign*} \ket{i} &= \frac{1}{\sqrt{2}} \ket{u} + \frac{\mathrm{i}}{\sqrt{2}} \ket{d} \\ \ket{o} &= \frac{1}{\sqrt{2}} \ket{u} - \frac{\mathrm{i}}{\sqrt{2}} \ket{d} \end{flalign*} \]

The probability to measure \(\sigma_y = 1\) is \[ \begin{flalign*} P_1(t) &= | \braket{i|\Psi(t)} |^2 \\ &= \braket{i|\Psi(t)} \braket{\Psi(t)|i} \\ &= \left[ \begin{pmatrix} \frac{1}{\sqrt{2}} & - \frac{\mathrm{i}}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \mathrm{e}^{-\frac{\mathrm{i} \omega}{2}t} \\ 0 \end{pmatrix} \right] \cdot \left[ \begin{pmatrix} \mathrm{e}^{\frac{\mathrm{i} \omega}{2}t} \\ 0 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{\mathrm{i}}{\sqrt{2}} \end{pmatrix} \right] = \frac{1}{2} \end{flalign*} \]

The probability to measure \(\sigma_y = -1\) is \[ P_{-1}(t) = | \braket{o|\Psi(t)} |^2 = \frac{1}{2} \]