Difference between revisions of "Aufgaben:Problem 12"

Latest revision as of 12:43, 9 July 2015

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 \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \).

Part b)

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

Option One

Solution Part a) Writing it out

We start by using (3) $$ \begin{align} \{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \} \\ &= \frac{1}{2\hbar} \left[ (\psi_1\psi_1 + \psi_2\psi_2 + i\psi_1\psi_2 - i\psi_2\psi_1) + (\psi_1\psi_1 + \psi_2\psi_2 - i\psi_1\psi_2 + i\psi_2\psi_1) \right] \\ &= \frac{1}{2\hbar} (2\psi_1\psi_1 + 2\psi_2\psi_2) = \frac{1}{2\hbar} \left( \{\psi_1, \psi_1\} + \{\psi_2, \psi_2\} \right) \\ \end{align} $$

Using (2) we get

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

Next we see, that

$$ \begin{align} \{\alpha,\alpha\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 - i\psi_2 \right) \} \\ &= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) \right] \\ &= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} -2i\{\psi_1,\psi_2\} \right) = 0. \\ \end{align} $$ Similarly we get $$ \begin{align} \{\alpha^\dagger,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 + i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \} \\ &= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) \right] \\ &= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} + 2i\{\psi_1,\psi_2\} \right) = 0. \\ \end{align} $$

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

Solution Part b)

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

Option Two

First of all we show that the anticommutator is bilinear and symmetrical:

$$ \{\mu (A+B),\nu (C+D)\} = \mu (A+B) \nu (C+D) + \nu (C+D) \mu (A+B) = \mu \nu ((AC+AD+BC+BD) + (CA+CB+DA+DB)) = \mu \nu (\{A,C\} + \{A,D\} + \{B,C\} + \{B,D\}) $$ $$ \{A,B\} = AB + BA = BA + AB = \{B,A\} $$

Solution Part a) Use bilinearity of the anticommutator

$$ \begin{align} \\ \{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ &= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ \psi_2, \psi_2 \}) \\ &= \frac{1}{2\hbar} (\hbar + \hbar) \\ &= 1 \\ \end{align} $$

$$ \begin{align} \\ \{\alpha,\alpha\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 - i\psi_2 \} \\ &= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} - i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ -i \psi_2, -i \psi_2 \}) \\ &= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} -0 -0 - \{ \psi_2, \psi_2 \}) \\ &= \frac{1}{2\hbar} (\hbar - \hbar) \\ &= 0 \\ \end{align} $$

$$ \begin{align} \\ \{\alpha^\dagger\,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 + i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ &= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} + i \{ \psi_2, \psi_1 \} + \{ +i \psi_2, +i \psi_2 \}) \\ &= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + 0 + 0 - \{ \psi_2, \psi_2 \}) \\ &= \frac{1}{2\hbar} (\hbar - \hbar) \\ &= 0 \\ \end{align} $$

And thus as above: $$ \{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0 $$ and $$ \{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0 $$

Solution Part b)

$$ \begin{align} \alpha^\dagger \alpha\ &= \frac{1}{2\hbar} (\psi_1 + i \psi_2)(\psi_1 - i\psi_2) \\ &= \frac{1}{2\hbar} (\psi_1^2 + \psi_2^2 - i \psi_1 \psi_2 + i \psi_2 \psi_1) \\ &= \frac{1}{2\hbar} (\frac{1}{2} \{\psi_1, \psi_1 \} + \frac{1}{2} \{\psi_2, \psi_2 \} - i \psi_1 \psi_2 - i \psi_1 \psi_2 + i \psi_1 \psi_2 + i \psi_2 \psi_1) \\ &= \frac{1}{2\hbar} (\hbar - 2i(\psi_1 \psi_2) + i\{ \psi_1, \psi_2 \} \\ &= \frac{1}{2} - \frac{i}{\hbar}(\psi_1 \psi_2) + 0 \\ \end{align} $$ $$ \Rightarrow \psi_1 \psi_2 = -\frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \quad (*) $$

Now, we insert \((*)\) into \( H_F \).

$$ \begin{align} H_F &= -i \omega \psi_1 \psi_2 \\ &\stackrel{(*)}{=} i \omega \frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \\ &=\omega \hbar (\alpha^\dagger \alpha\ - \frac{1}{2} ) \end{align} $$