Difference between revisions of "Aufgaben:Problem 12"
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== Problem 12 == | == Problem 12 == | ||
| + | (by Madiso) | ||
| − | Hamiltonian of a 1D fermionic oscillator | + | We consider the Hamiltonian of a 1D fermionic oscillator |
$$ | $$ | ||
| − | H_F = -i\omega\psi_1\psi_2, | + | H_F = -i\omega\psi_1\psi_2 \quad (1), |
$$ | $$ | ||
| − | where the fermionic wave functions | + | where the anti-commutator of the fermionic wave functions is given by |
$$ | $$ | ||
| − | \{\psi_i,\psi_j\} = \hbar\delta_{ij} | + | \{\psi_i,\psi_j\} = \hbar\delta_{ij} \quad (2) |
$$ | $$ | ||
We introduce the lowering and rising operators | 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) | + | \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) === | === Part a) === | ||
Show that \( \{\alpha,\alpha^\dagger\} = 1 \), \( \{\alpha,\alpha\} = \{\alpha^\dagger,\alpha^\dagger\} = 0 \) and \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \). | 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) === | === 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 | ||
| + | $$ | ||
| + | H_F = \frac{\hbar\omega}{2} \left( -\left( 1 - \alpha^\dagger\alpha \right) + \alpha^\dagger\alpha \right) = \hbar\omega \left( \alpha^\dagger\alpha - \frac{1}{2} \right). | ||
| + | $$ | ||
Revision as of 13:59, 9 June 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 \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \).
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 $$ H_F = \frac{\hbar\omega}{2} \left( -\left( 1 - \alpha^\dagger\alpha \right) + \alpha^\dagger\alpha \right) = \hbar\omega \left( \alpha^\dagger\alpha - \frac{1}{2} \right). $$
