Aufgaben:Problem 12

Problem 12

(by Madiso)

We consider the Hamiltonian of a 1D fermionic oscillator $$ H_F = -i\omega\psi_1\psi_2 \quad (1), $$ where the anti-commutator of the fermionic wave functions is given by $$ \{\psi_i,\psi_j\} = \hbar\delta_{ij} \quad (2) $$ We introduce the lowering and rising operators $$ \alpha = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 - i\psi_2 \right), \quad \alpha^{\dagger} = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 + i\psi_2 \right) \quad (3) $$

Part a)

Show that \( \{\alpha,\alpha^\dagger\} = 1 \), \( \{\alpha,\alpha\} = \{\alpha^\dagger,\alpha^\dagger\} = 0 \) and \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \).

Solution

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

$$ \{\alpha,\alpha^\dagger\} = \frac{2\hbar}{2\hbar} = 1 \quad (4) $$

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 $$ \{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0 \quad (5) $$ and $$ \{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0 \quad (6) $$

Part b)

Show that \( H_F = \hbar\omega\left(\alpha^\dagger\alpha - \frac{1}{2} \right) \).

Solution

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

$$ \alpha - \alpha^\dagger = \frac{-i2}{\sqrt{2\hbar}}\psi_2 \Rightarrow \psi_2 = i\sqrt{\frac{\hbar}{2}}\left( \alpha - \alpha^\dagger \right) $$ 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 $$ H_F = \frac{\hbar\omega}{2} \left( -\left( 1 - \alpha^\dagger\alpha \right) + \alpha^\dagger\alpha \right) = \hbar\omega \left( \alpha^\dagger\alpha - \frac{1}{2} \right). $$