Difference between revisions of "Aufgaben:Problem 12"

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−
==Part a)==
 
−
Let \( \Phi_0(x) = e^{-\frac{1}{2}x^2} \) and define \( \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2}\left(\left(\frac{d}{dx}\right)^n e^{-x^2}\right) \). Show that \( \hat \Phi_n(k) = C\Phi_n(k)\), with \( C = C(n) \in \mathbb{C} \).
 
  
−
===Solution===
+
== Problem 12 ==
 +
(by Madiso)
  
−
We show this by proving the following claim:
+
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
 +
$$
 +
\alpha = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 - i\psi_2 \right), \quad \alpha^{\dagger} = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 + i\psi_2 \right) \quad (3)
 +
$$
  
−
\( \vdash : \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2} \Big(\frac{d}{dx}\Big)^n e^{-x^2}\) is an eigenfunction of the Fourier-transform with the eigenvalue \((-i)^n \).
+
=== Part a) ===
 +
Show that \( \{\alpha,\alpha^\dagger\} = 1 \), \( \{\alpha,\alpha\} = \{\alpha^\dagger,\alpha^\dagger\} = 0 \) and \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \).
  
−
''Proof.''
 
  
−
We first check the case \( \hat \Phi_0(k) \) where we complete the square in the exponent by using the substitution \( \eta \equiv \frac{x + ik}{\sqrt{2}} \):
+
=== Part b) ===
 +
Show that \( H_F = \hbar\omega\left(\alpha^\dagger\alpha - \frac{1}{2} \right) \).
  
−
$$ \hat \Phi_0(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R} } e^{-\frac{1}{2}x^2} e^{-ikx} dx = \frac{\sqrt{2}}{\sqrt{2\pi}} \int_{\mathbb{R} } e^{-\eta^2 - \frac{k^2}{2}} d\eta = e^{-\frac{1}{2} k^2} = \Phi_0(k) \tag{3.14} $$
 
  
−
For the general case:
+
==Option One==
  
−
$$ \sqrt{2\pi} \hat \Phi_n(k) = \int_{\mathbb{R} } (-1)^n e^{\frac{1}{2}x^2} \Big( (\frac{d}{dx})^n e^{-x^2} \Big) e^{-ikx} dx = \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}x^2 - ikx} dx $$
+
==== Solution  Part a) Writing it out====
  
−
where we partially integrated. The first term of each partial integration is zero since:
+
We start by using (3)
 +
$$
 +
\begin{align}
 +
\{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 + i\psi_2 \right)  \} \\ 
 +
&= \frac{1}{2\hbar} \left[ (\psi_1\psi_1 + \psi_2\psi_2 + i\psi_1\psi_2 - i\psi_2\psi_1) + (\psi_1\psi_1 + \psi_2\psi_2 - i\psi_1\psi_2 + i\psi_2\psi_1) \right] \\
 +
&= \frac{1}{2\hbar} (2\psi_1\psi_1 + 2\psi_2\psi_2) = \frac{1}{2\hbar} \left( \{\psi_1, \psi_1\} + \{\psi_2, \psi_2\} \right) \\
 +
\end{align}
 +
$$
  
−
$$ \left| \left( \frac{d}{dx} \right)^m e^{\frac{1}{2}x^2 - ikx} \left( \left(\frac{d}{dx}\right)^l e^{-x^2} \right) \right|  = \left| p(x) \right| e^{-\frac{1}{2}x^2} $$
+
Using (2) we get
  
−
where \( p(x) = \mathcal{O}(n^j) \) (using the Landau notation) and \(l, m, j \in \mathbb{N} \) leading to
+
$$
 +
\{\alpha,\alpha^\dagger\} = \frac{2\hbar}{2\hbar} = 1 \quad (4)
 +
$$
  
−
$$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty}^{\infty} = 0 $$
+
Next we see, that
  
−
Furthermore
+
$$
 +
\begin{align}
 +
\{\alpha,\alpha\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 - i\psi_2 \right) \} \\
 +
&= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) \right] \\
 +
&= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} -2i\{\psi_1,\psi_2\} \right) = 0. \\
 +
\end{align}
 +
$$
 +
Similarly we get
 +
$$
 +
\begin{align}
 +
\{\alpha^\dagger,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 + i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \}  \\
 +
&= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) \right] \\
 +
&= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} + 2i\{\psi_1,\psi_2\} \right) = 0. \\
 +
\end{align}
 +
$$
  
−
$$ \sqrt{2\pi} \Phi_n(k) = e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx = i^n e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx $$
+
Thus it is clear that
 +
$$
 +
\{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0 \quad (5)
 +
$$
 +
and
 +
$$
 +
\{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0 \quad (6)
 +
$$
  
−
where we used the following
+
==== Solution Part b)====
  
−
'''Lemma 1:''' \( \forall n \in \mathbb{Z}_{\geqslant 0}: i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} \)  
+
Using (3) we can express  \(\psi_1\) and \(\psi_2\) from \(\alpha\) and \(\alpha^\dagger\)
 +
$$
 +
\alpha + \alpha^\dagger = \frac{2}{\sqrt{2\hbar}}\psi_1 \Rightarrow \psi_1 = \sqrt{\frac{\hbar}{2}}\left( \alpha + \alpha^\dagger \right)
 +
$$
  
−
''Proof of Lemma 1:''
+
$$
 +
\alpha - \alpha^\dagger = \frac{-i2}{\sqrt{2\hbar}}\psi_2 \Rightarrow \psi_2 = i\sqrt{\frac{\hbar}{2}}\left( \alpha - \alpha^\dagger \right)
 +
$$
 +
Now we can put these into (1) and express \(H_F\) from \(\alpha\) and \(\alpha^\dagger\)
 +
$$
 +
H_F = -i\omega\psi_1\psi_2 = (-i\omega)i\frac{\hbar}{2} \left( \alpha + \alpha^\dagger \right)\left( \alpha - \alpha^\dagger \right)
 +
$$
  
−
By induction.  
+
$$
 +
= \frac{\hbar\omega}{2} \left( \alpha^2 - \alpha\alpha^\dagger + \alpha^\dagger\alpha - \left(\alpha^\dagger\right)^2 \right)
 +
$$
 +
Using results (4),(5) and (6) from section a), we get
 +
$$
 +
H_F = \frac{\hbar\omega}{2} \left( -\left( 1 - \alpha^\dagger\alpha \right) + \alpha^\dagger\alpha  \right) = \hbar\omega \left( \alpha^\dagger\alpha - \frac{1}{2} \right).
 +
$$
  
−
\( n = 0 \) : \( i^0 e^{\frac{1}{2}(x - ik)^2} = e^{\frac{1}{2}(x - ik)^2} \) is the trivial case.
 
  
−
\( n-1 \rightarrow n \) :
+
==Option Two==
  
−
$$ i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} i (-i) (x-ik) e^{\frac{1}{2}(x - ik)^2} =  i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} (x-ik) e^{\frac{1}{2}(x - ik)^2} =  i^{n-1} \Big( \frac{d}{dk} \Big)^{n - 1}  \frac{d}{dx} e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \frac{d}{dx} \Big( \frac{d}{dk} \Big)^{n - 1} e^{\frac{1}{2}(x - ik)^2} $$
+
First of all we show that the anticommutator is bilinear and symmetrical:
  
−
where for the last equality we used the Schwarz's theorem. With the induction assumption everything follows. \( \square \)
+
$$
 +
\{\mu (A+B),\nu (C+D)\} = \mu (A+B) \nu (C+D) + \nu (C+D) \mu (A+B) = \mu \nu ((AC+AD+BC+BD) + (CA+CB+DA+DB)) = \mu \nu (\{A,C\} + \{A,D\} + \{B,C\} + \{B,D\})
 +
$$
 +
$$
 +
\{A,B\} = AB + BA = BA + AB = \{B,A\}
 +
$$
  
−
'''Lemma 2:''' \( \frac{d}{dk} \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \frac{d}{dk} f(x) e^{-ikx} dx \) where \( f(x) = e^{- \frac{1}{2} x^2 - ikx} \).
+
==== Solution Part a) Use bilinearity of the anticommutator====
  
−
''Proof of Lemma 2:''
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 + i\psi_2 )  \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar + \hbar) \\
 +
&= 1 \\
 +
\end{align}
 +
$$
  
−
$$ \sqrt{2\pi} \frac{d}{dk} \hat f(k) = \lim_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh} - 1 \right) dx $$
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 - i\psi_2 \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} - i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ -i \psi_2, -i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} -0 -0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
We use \( \vert \sin(x) \vert =  \vert x - \frac{x^3}{6} + \frac{x^5}{120} \pm\cdots \vert \leq \vert x \vert \) for
+
$$
 +
\begin{align} \\
 +
\{\alpha^\dagger\,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 + i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} + i \{ \psi_2, \psi_1 \} + \{ +i \psi_2, +i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + 0 + 0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
$$ \begin{align}
+
And thus as above:
−
\left| \frac{1}{h} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh} - 1 \right) \right| &= e^{- \frac{x^2}{2}} \frac{1}{h} \left| e^{-ihx} e^{-\frac{ixh}{2}} \left( e^{- \frac{ixh}{2} } - e^{\frac{ixh}{2}} \right) \right| \\
+
$$
−
&= e^{- \frac{x^2}{2}} \frac{2}{h} \left| \sin \left( \frac{xh}{2} \right) \right| \leq \frac{2}{h} \left| \frac{xh}{2} \right| e^{-\frac{x^2}{2}} \\
+
\{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0
−
&= \vert x \vert e^{-\frac{x^2}{2}}
+
$$
−
\end{align} $$
+
and
−
 
+
$$
−
and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)
+
\{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0
−
 
+
$$
−
 
+
−
 
+
−
The rest is a piece of cake:
+
−
 
+
−
$$ \sqrt{2\pi} \hat \Phi_n(k) = i^n e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n \int_{\mathbb{R} } e^{-x^2 + \frac{1}{2}(x - ik)^2} dx = i^n \sqrt{2\pi} e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n e^{-k^2} = (-i)^n \sqrt{2\pi} \Phi_n(k) $$  
+
  
−
(for the second last equality we've used equation (3.14)). \( \square \)
 
  
−
==Part b) ==  
+
===Solution Part b)===
  
−
Let \( \chi_{[a,b]} \) be the characteristic function, \( a, b \in \mathbb{R}, a < b \). Show explicitly, i.e. without using Plancherel, that
+
$$
−
 
+
−
$$ \Vert \hat \chi_{[a,b]} \Vert_2^2 = b - a. $$
+
−
 
+
−
''Hint:'' You may use without proof that
+
−
 
+
−
$$ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} $$
+
−
 
+
−
 
+
−
===Solution===
+
−
 
+
−
''Nice to know:'' Plancherel's theorem states that for any function \( f \in L^2: \Vert \hat f \Vert_{2} = \Vert f \Vert_{2} \) where \( \Vert f \Vert_{2} = \left( \int_{\mathbb{R}} \vert f(x) \vert^2 dx \right)^{\frac{1}{2}} \)
+
−
 
+
−
But we show the identity like badasses by direct calculation.
+
−
 
+
−
We have: $$ \hat \chi_{[a,b]} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$
+
−
 
+
−
Now we use the substitution \( y \equiv \frac{2(x-a)}{b-a} -1 \) to scale the integral to the interval \( \left[ -1, 1 \right] \). Thus:
+
−
 
+
−
$$  
+
 
\begin{align}
 
\begin{align}
−
\hat \chi_{[a,b]} &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy \\
+
\alpha^\dagger \alpha\ &= \frac{1}{2\hbar} (\psi_1 + i \psi_2)(\psi_1 - i\psi_2) \\
−
&= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \left( \frac{-2}{ik(b-a)} e^{-\frac{ik(b-a)}{2}y} \bigg \vert_{-1}^{1} \right) \\
+
&= \frac{1}{2\hbar} (\psi_1^2 + \psi_2^2 - i \psi_1 \psi_2 + i \psi_2 \psi_1) \\
−
&= \frac{1}{\sqrt{2\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \frac{1}{ik} \left( - e^{-ik\frac{(b-a)}{2}} + e^{ik \frac{b-a}{2}} \right) \\
+
&= \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{\sqrt{2}}{k\sqrt{\pi}} e^{-ik \left( \frac{1}{2}b + \frac{3}{2}a \right)} \sin \left( \frac{b-a}{2} k \right)
+
&= \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}
 
\end{align}
 +
$$
 +
$$
 +
\Rightarrow \psi_1 \psi_2 = -\frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \quad (*)
 
$$
 
$$
  
 +
Now, we insert \((*)\) into \(  H_F  \).
  
−
 
+
$$
−
And so:
+
\begin{align}
−
 
+
H_F &= -i \omega \psi_1 \psi_2 \\
−
$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \int_{\mathbb{R} } |\hat \chi_{[a,b]}(k)|^2 dk = \frac{2}{\pi} \int_{\mathbb{R} }  \frac{\sin^2(\frac{b-a}{2}k)}{k^2} dk $$
+
&\stackrel{(*)}{=} i \omega \frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \\
−
 
+
&=\omega \hbar (\alpha^\dagger \alpha\ - \frac{1}{2} )
−
With the substitution \( \eta \equiv \frac{b-a}{2}k \) we get:
+
\end{align}
−
 
+
$$
−
$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \frac{(b-a)}{\pi} \int_{\mathbb{R} }  \frac{\sin^2(\eta)}{\eta^2} d\eta $$
+
−
 
+
−
It remains to evaluate the integral \( I(\alpha) = \int_{\mathbb{R} }  \frac{\sin^2\left(\alpha \eta\right)}{\eta^2} d\eta \) at \( \alpha = 1 \).
+
−
 
+
−
'''Amrein's Lemma:''' \( \frac{dI}{d\alpha} = \int_{\mathbb{R} } \frac{d}{d\alpha} \frac{\sin^2\left(\alpha \eta\right)}{\eta^2} d\eta \).
+
−
 
+
−
''Proof of Lemma:''
+
−
 
+
−
$$ \frac{dI}{d\alpha} = \lim_{h \rightarrow 0 } \frac{1}{h} \int_{\mathbb{R} } \frac{\sin^2 \left( (\alpha + h) \eta \right) - \sin^2 \left( \alpha \eta \right) }{\eta^2} d\eta $$
+
−
 
+
−
Since \( \sin(x) \leq 1 \Rightarrow \sin^2(x) \leq \vert \sin(x) \vert \) and \(\left| \sin(x) \right| \leq |x| \) (see section a), Lemma 2) we have
+
−
 
+
−
$$ \left| \frac{1}{h} \frac{\left| \sin \left( ( \alpha + h) \eta \right) \right| - \left| \sin \left(  \alpha \eta \right) \right|}{\eta^2} \right| \leq \left| \frac{1}{h} \frac{(\alpha + h)\eta - \alpha\eta}{\eta^2} \right| = \left| \frac{1}{\eta} \right| $$
+
−
 
+
−
and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)
+
−
 
+
−
$$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty}  \frac{2 \eta \sin\left( \alpha \eta \right) \cos\left( \alpha \eta \right) }{\eta^2} $$ and with the identity $$ \sin(x) \cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that:
+
−
 
+
−
$$ \begin{align}
+
−
\frac{dI}{d\alpha} &= 2 \int_{0}^{\infty}  \frac{\sin\left( 2 \alpha \eta \right)}{\eta} d\eta \\
+
−
&= 2 \int_{0}^{\infty} \frac{\sin\left( \xi \right)}{\xi} d\xi = \pi
+
−
\end{align} $$
+
−
 
+
−
 
+
−
Solving this equation is a thing we learned a long time ago:
+
−
 
+
−
\( I(\alpha) = \pi \alpha + c \) and since \( I(0) = 0: I(\alpha) = \pi \alpha \).
+
−
 
+
−
As \( I(1) = \pi \) the result is:
+
−
 
+
−
$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = b-a $$
+

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 \( H_F \).

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

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