Aufgaben:Problem 12

Hamiltonian of a 1D fermionic oscillator $$ H_F = -i\omega\psi_1\psi_2, $$ where the fermionic wave functions anticommute \begin{equation} \{\psi_i,\psi_j\} = \hbar\delta_{ij}. \end{equation} We introduce the lowering and rising operators \begin{equation} \alpha = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 - i\psi_2 \right), \quad \alpha^{\dagger} = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 + i\psi_2 \right). \end{equation}

%% Part a) \begin{bf} a) \end{bf}\emph{ Show that $\{\alpha,\alpha^\dagger\} = 1$, $\{\alpha,\alpha\} = \{\alpha^\dagger,\alpha^\dagger\} = 0$ and $\alpha^2 = \left( \alpha^\dagger \right)^2 = 0$.} We start by using (3) \[ \{\alpha,\alpha^\dagger\} = \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \} \]

\[

= \frac{1}{2\hbar} \left[ (\psi_1\psi_1 + \psi_2\psi_2 + i\psi_1\psi_2 - i\psi_2\psi_1) + (\psi_1\psi_1 + \psi_20\psi_2 - i\psi_1\psi_2 + i\psi_2\psi_1) \right]

\]

\[ = \frac{1}{2\hbar} (2\psi_1\psi_1 + 2\psi_2\psi_2) = \frac{1}{2\hbar} \left( \{\psi_1, \psi_1\} + \{\psi_2, \psi_2\} \right) \] Using (2) we get \begin{equation}

\{\alpha,\alpha^\dagger\} = \frac{2\hbar}{2\hbar} = 1.

\end{equation} Next we see, that \[ \{\alpha,\alpha\} = \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 - i\psi_2 \right) \} \] \[ = \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) \right] \] \[ = \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} -2i\{\psi_1,\psi_2\} \right) = 0. \] Similarly we get \[ \{\alpha^\dagger,\alpha^\dagger\} = \frac{1}{2\hbar} \{\left( \psi_1 + i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \} \] \[ = \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) \right] \] \[ = \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} + 2i\{\psi_1,\psi_2\} \right) = 0. \] Thus it is clear that \begin{equation}

\{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0.

\end{equation}

\begin{equation}

\{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0.

\end{equation}

%% Part b) \begin{bf} b) \end{bf}\emph{ Show that $H_F = \hbar\omega\left(\alpha^\dagger\alpha - \frac{1}{2} \right)$.}

Using (3) we can express $\psi_1$ and $\psi_2$ from $\alpha$ and $\alpha^\dagger$ \begin{equation} \alpha + \alpha^\dagger = \frac{2}{\sqrt{2\hbar}}\psi_1 \Rightarrow \psi_1 = \sqrt{\frac{\hbar}{2}}\left( \alpha + \alpha^\dagger \right) \end{equation}

\begin{equation} \alpha - \alpha^\dagger = \frac{-i2}{\sqrt{2\hbar}}\psi_2 \Rightarrow \psi_2 = i\sqrt{\frac{\hbar}{2}}\left( \alpha - \alpha^\dagger \right) \end{equation} Now we can put these into (1) and express $H_F$ from $\alpha$ and $\alpha^\dagger$ \[ H_F = -i\omega\psi_1\psi_2 = (-i\omega)i\frac{\hbar}{2} \left( \alpha + \alpha^\dagger \right)\left( \alpha - \alpha^\dagger \right) \] \[ = \frac{\hbar\omega}{2} \left( \alpha^2 - \alpha\alpha^\dagger + \alpha^\dagger\alpha - \left(\alpha^\dagger\right)^2 \right) \] Using results (4),(5) and (6) from section a), we get \begin{equation} H_F = \frac{\hbar\omega}{2} \left( -\left( 1 - \alpha^\dagger\alpha \right) + \alpha^\dagger\alpha \right) = \hbar\omega \left( \alpha^\dagger\alpha - \frac{1}{2} \right). \end{equation}