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 that \( \Phi_n(x) \) is an eigenfunction of the Fourier-transform with eigenvalue \((-i)^n \), i.e. \( \hat \Phi_n(k) = (-i)^n\Phi_n(k).\)
+
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
 +
$$
 +
\alpha = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 - i\psi_2 \right), \quad \alpha^{\dagger} = \frac{1}{\sqrt{2\hbar}} \left( \psi_1 + i\psi_2 \right) \quad (3)
 +
$$
  
−
''Proof.''
+
=== Part a) ===
 +
Show that \( \{\alpha,\alpha^\dagger\} = 1 \), \( \{\alpha,\alpha\} = \{\alpha^\dagger,\alpha^\dagger\} = 0 \) and \( \alpha^2 = \left( \alpha^\dagger \right)^2 = 0 \).
  
−
We first check the case \( \hat \Phi_0(k) \) (you don't have to know this calculation, we did that in Series 8, ex. 3 \( \color{red}{see \: discussion} \)) where we complete the square in the exponent by using the substitution \( \eta \equiv \frac{x + ik}{\sqrt{2}} \):
 
  
−
$$ \hat \Phi_0(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R} } e^{-\frac{1}{2}x^2} e^{-ikx} dx = \frac{\sqrt{2}}{\sqrt{2\pi}} \int_{\mathbb{R} } e^{-\eta^2 - \frac{k^2}{2}} d\eta = e^{-\frac{1}{2} k^2} = \Phi_0(k) \tag{3.14} $$
+
=== Part b) ===
 +
Show that \( H_F = \hbar\omega\left(\alpha^\dagger\alpha - \frac{1}{2} \right) \).
  
−
For the general case:
 
  
−
$$ \sqrt{2\pi} \hat \Phi_n(k) = \int_{\mathbb{R} } (-1)^n e^{\frac{1}{2}x^2} \Big( (\frac{d}{dx})^n e^{-x^2} \Big) e^{-ikx} dx = \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}x^2 - ikx} dx $$
+
==Option One==
  
−
where we partially integrated. The first term of each partial integration is zero since:
+
==== Solution  Part a) Writing it out====
  
−
$$ \left| \left( \frac{d}{dx} \right)^m e^{\frac{1}{2}x^2 - ikx} \left( \left(\frac{d}{dx}\right)^l e^{-x^2} \right) \right| = \left| p(x) e^{-\frac{1}{2}x^2-ikx}\right| = \left| p(x)\right| e^{-\frac{1}{2}x^2} $$
+
We start by using (3)
 +
$$
 +
\begin{align}
 +
\{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \} \\   
 +
&= \frac{1}{2\hbar} \left[ (\psi_1\psi_1 + \psi_2\psi_2 + i\psi_1\psi_2 - i\psi_2\psi_1) + (\psi_1\psi_1 + \psi_2\psi_2 - i\psi_1\psi_2 + i\psi_2\psi_1) \right] \\
 +
&= \frac{1}{2\hbar} (2\psi_1\psi_1 + 2\psi_2\psi_2) = \frac{1}{2\hbar} \left( \{\psi_1, \psi_1\} + \{\psi_2, \psi_2\} \right) \\
 +
\end{align}
 +
$$
  
−
where \( p(x) = \mathcal{O}(x^j) \) (using the Landau notation) is a (complex) polynomial in \( x \) and \(l, m, j \in \mathbb{N}, \) leading to
+
Using (2) we get
  
−
$$ \left( \frac{d}{dx} \right)^m e^{\frac{1}{2}x^2 - ikx} \left( \left(\frac{d}{dx}\right)^l e^{-x^2} \right)\bigg \vert_{-\infty}^{\infty} = 0 $$
+
$$
 +
\{\alpha,\alpha^\dagger\} = \frac{2\hbar}{2\hbar} = 1 \quad (4)
 +
$$
  
−
since the exponential goes to \( 0 \) faster than any polynomial goes to \( \infty \) for \( x \rightarrow \pm \infty: \) $$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty} = \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{\infty} = 0. $$
+
Next we see, that
  
−
We then go on like this:
+
$$
 +
\begin{align}
 +
\{\alpha,\alpha\} &= \frac{1}{2\hbar} \{\left( \psi_1 - i\psi_2 \right), \left( \psi_1 - i\psi_2 \right) \} \\
 +
&= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 - i\psi_1\psi_2 - i\psi_2\psi_1 \right) \right] \\
 +
&= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} -2i\{\psi_1,\psi_2\} \right) = 0. \\
 +
\end{align}
 +
$$
 +
Similarly we get
 +
$$
 +
\begin{align}
 +
\{\alpha^\dagger,\alpha^\dagger\} &= \frac{1}{2\hbar} \{\left( \psi_1 + i\psi_2 \right), \left( \psi_1 + i\psi_2 \right) \}  \\
 +
&= \frac{1}{2\hbar} \left[ \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) + \left( \psi_1\psi_1 - \psi_2\psi_2 + i\psi_1\psi_2 + i\psi_2\psi_1 \right) \right] \\
 +
&= \frac{1}{2\hbar} \left( \{\psi_1,\psi_1\} - \{\psi_2,\psi_2\} + 2i\{\psi_1,\psi_2\} \right) = 0. \\
 +
\end{align}
 +
$$
  
−
$$ \sqrt{2\pi} \hat \Phi_n(k) = e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx = i^n e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx $$
+
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 completed the square in the exponent and then used the following lemma:
+
==== 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 Schwarz's theorem. With the induction assumption everything follows. \( \square \)
+
$$
 +
\{\mu (A+B),\nu (C+D)\} = \mu (A+B) \nu (C+D) + \nu (C+D) \mu (A+B) = \mu \nu ((AC+AD+BC+BD) + (CA+CB+DA+DB)) = \mu \nu (\{A,C\} + \{A,D\} + \{B,C\} + \{B,D\})
 +
$$
 +
$$
 +
\{A,B\} = AB + BA = BA + AB = \{B,A\}
 +
$$
  
−
''Okay, ladies and gentlemen, listen carefully. Lemma 2 is incredibly boring to prove. We've found a shorter and more general way to do this. I've marked the subparts that divide in two versions separately.''
+
==== Solution Part a) Use bilinearity of the anticommutator====
  
−
'''Lemma 2:''' Let \( f \in \mathcal{S}(\mathbb{R}) \) and \( \lambda (k) \in C^{\infty} \) be a function only depending on k. Then: \( \left( \frac{d}{dk} \right)^{n} \lambda (k) \hat f(k) = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n \lambda (k) f(x) e^{-ikx} dx \).
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar + \hbar) \\
 +
&= 1 \\
 +
\end{align}
 +
$$
  
−
''Proof.''
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 - i\psi_2 \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} - i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ -i \psi_2, -i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} -0 -0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
\( \vdash \) : \( \mathcal{S}(\mathbb{R}) \subset L^1( \mathbb{R} ) \)
+
$$
 +
\begin{align} \\
 +
\{\alpha^\dagger\,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 + i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} + i \{ \psi_2, \psi_1 \} + \{ +i \psi_2, +i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + 0 + 0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
Indeed, from Series 11, Ex. 1 we know:
+
And thus as above:
 +
$$
 +
\{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0
 +
$$
 +
and
 +
$$
 +
\{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0
 +
$$
  
−
$$ \left| f(x) \right| \leq \frac{C}{(1 + \vert x \vert^2)} $$
 
  
−
for some \( C \in \mathbb{R} \) and thus:
+
===Solution Part b)===
  
−
$$ \int_{\mathbb{R}} \left| f(x) \right| dx \leq \int_{\mathbb{R}}  \left| \frac{C}{(1 + \vert x \vert^2)} \right| dx < \infty $$
+
$$
 +
\begin{align}
 +
\alpha^\dagger \alpha\ &= \frac{1}{2\hbar} (\psi_1 + i \psi_2)(\psi_1 - i\psi_2) \\
 +
&= \frac{1}{2\hbar} (\psi_1^2 + \psi_2^2 - i \psi_1 \psi_2 + i \psi_2 \psi_1) \\
 +
&= \frac{1}{2\hbar} (\frac{1}{2} \{\psi_1, \psi_1 \} + \frac{1}{2} \{\psi_2, \psi_2 \} - i \psi_1 \psi_2 - i \psi_1 \psi_2 + i \psi_1 \psi_2 + i \psi_2 \psi_1) \\
 +
&= \frac{1}{2\hbar} (\hbar - 2i(\psi_1 \psi_2) + i\{ \psi_1, \psi_2 \} \\
 +
&= \frac{1}{2} - \frac{i}{\hbar}(\psi_1 \psi_2) + 0 \\
 +
\end{align}
 +
$$
 +
$$
 +
\Rightarrow \psi_1 \psi_2 = -\frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \quad (*)
 +
$$
  
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
Now, we insert \((*)\) into \(  H_F  \).
  
−
''Version 1''
+
$$
−
 
+
−
Using the property
+
−
 
+
−
Now, we use an induction argument for \( \left( \frac{d}{dk} \right)^n \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n f(x) e^{-ikx} dx \)
+
−
 
+
−
(well-defined since \( f \in \mathcal{S}(\mathbb{R}) \Rightarrow \hat f(k) \in \mathcal{S}(\mathbb{R}) \subset C^{\infty} \) from a lemma in the script):
+
−
 
+
−
\( n = 0 \) is the trivial case.
+
−
 
+
−
\( n - 1 \rightarrow n: \)
+
−
 
+
−
Let \( h_m \) be an arbitrary zero-sequence.
+
−
 
+
−
$$ \sqrt{2\pi} \left( \frac{d}{dk} \right)^{n} \hat f(k) =  \sqrt{2\pi} \frac{d}{dk} \left( \frac{d}{dk} \right)^{n-1} \hat f(k) =^{\color{green}{*}} \lim_{m \rightarrow \infty} \int_{\mathbb{R}} \frac{1}{h_m} \left( \frac{d}{dk} \right)^{n-1} f(x) e^{ - ikx} \left( e^{-ixh_m} - 1 \right) dx $$
+
−
 
+
−
$$ \color{green}{*} \left( \left( \frac{d}{dk} \right)^{j} e^{ - ikx} \right) \bigg \vert_{k + h_m} = \left( p(x)e^{ - ikx} \right) \bigg \vert_{k + h_m} = p(x)e^{ - i(k+h_m)x} = \left( \frac{d}{dk} \right)^{j} e^{ - ikx} e^{ - ih_mx} $$
+
−
 
+
−
We then proceed:
+
−
+
−
$$ \begin{align} \left| \frac{1}{h_m} \left( \frac{d}{dk} \right)^{n-1} f(x) e^{-ikx} \left( e^{-i h_m x} - 1 \right) \right| &= \left| \frac{2}{h_m} f(x) \sin \left( \frac{xh_m}{2} \right) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \leq \left| \frac{2}{h_m} f(x) \frac{xh_m}{2} \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \\
+
−
&= \left| x \cdot f(x)  \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| = \left| p(x) \cdot f(x) \right| \in L^1
+
−
\end{align}$$
+
−
 
+
−
where we used \( \vert \sin (x) \vert \leq \vert x \vert \) and \( p(x) \in \mathbb{C}[x] \). Using Lebesgue dominated convergence theorem, the induction proof is concluded.
+
−
 
+
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
−
 
+
−
''Version 2:''
+
−
 
+
−
Using the property from the script: \( \frac{d}{dk} \widehat{f} (k) = (-i) \cdot \widehat{\left( xf \right) }(k) \) we see easily that \( \left( \frac{d}{dk} \right)^n \widehat{f} (k) = \left( -i \right)^n \widehat{ \left( x^n \cdot f \right) } (k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n f(x) e^{-ikx} dx \) since \( \left( \frac{d}{dk} \right)^n e^{-ikx} = (-ix)^n \cdot e^{-ikx} \). ''If you like you may of course do an induction here but it seems clear enough to me.''
+
−
 
+
−
Note that \( x^{\alpha} \cdot f \in L^1 \) for any \( \alpha \in \{ 0, ..., n \} \) follows from \( f \in \mathcal{S}(\mathbb{R}) \).
+
−
 
+
−
<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
+
−
 
+
−
''Here the fans of version 1 and version 2 have to work together again.''
+
−
 
+
−
We now use the generalized Leibniz-rule to show:
+
−
 
+
−
$$\begin{align}
+
−
\left(\frac{d}{dk}\right)^{n} \lambda (k)\int_\mathbb{R} f(x) e^{-ikx}dx &= \sum_{j=0}^{n} \binom{n}{j}\left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} \int_\mathbb{R} f(x) e^{-ikx} dx \\
+
−
&= \sum_{j=0}^{n} \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \int_\mathbb{R} \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\
+
−
&= \sum_{j=0}^{n} \int_\mathbb{R} \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\
+
−
&= \int_\mathbb{R} \sum_{j=0}^{n}  \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\
+
−
&= \int_\mathbb{R} \left( \frac{d}{dk} \right)^n \lambda (k) f(x) e^{-ikx} dx
+
−
\end{align}$$
+
−
 
+
−
And the Lemma is proven. \( \square \)
+
−
 
+
−
From this Lemma follows directly for \( f(x) = e^{- \frac{x^2}{2} } \in \mathcal{S}(\mathbb{R}) \) and \( e^{-\frac{k^2}{2}} \in C^{\infty} \)
+
−
 
+
−
$$ \left( \frac{d}{dk} \right)^n e^{\frac{-k^2}{2}} \hat f(k) = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n e^{- \frac{k^2}{2} - x^2 + \frac{1}{2}x^2 - ikx} dx = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n e^{- x^2 + \frac{1}{2} (x - ik)^2 }  dx $$
+
−
 
+
−
We then finish the proof as follows:
+
−
 
+
−
$$ \sqrt{2\pi} \hat \Phi_n(k) = i^n e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n \int_{\mathbb{R} } e^{-x^2 + \frac{1}{2}(x - ik)^2} dx = i^n \sqrt{2\pi} e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n e^{-k^2} = (-i)^n \sqrt{2\pi} \Phi_n(k) $$
+
−
 
+
−
Where for the first equality we used Lemma 2 and for the second equation (3.14). \( \square \)
+
−
 
+
−
===Aternative Solution (not proof-read yet) ===
+
−
 
+
−
Claim: \( \hat \Phi_n(k) = (-i)^n\Phi_n(k)\)
+
−
 
+
−
proof by induction:
+
−
 
+
−
n = 0, using the Fundamental identity in the script (Fourier-SchwartzAdded 95):
+
−
 
+
−
$$ \hat \Phi_0(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R} } e^{-\frac{1}{2}x^2} e^{-ikx} dx = e^{-\frac{1}{2} k^2} = \Phi_0(k)$$
+
−
 
+
−
induction step: $$  \hat \Phi_{n+1}(k) =\frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R} } (-1)^{n+1} e^{\frac{1}{2}x^2} \Big( (\frac{d}{dx})^{n+1} e^{-x^2} \Big) e^{-ikx} dx $$
+
−
 
+
−
integrating by parts
+
−
 
+
−
$$= \frac{1}{\sqrt{2\pi}}(-1)^{n+1} \left( \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) e^{\frac{1}{2}x^2} e^{-ikx} \right) \bigg \vert_{-\infty}^{\infty} - \frac{1}{\sqrt{2\pi}}(-1)^{n+1} \int_{\mathbb{R} } \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) \frac{d}{dx}\Big(e^{\frac{1}{2}x^2} e^{-ikx}\Big) dx $$
+
−
 
+
−
since
+
−
 
+
−
$$\bigg \vert \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) e^{\frac{1}{2}x^2} e^{-ikx} \bigg \vert = \vert p(x) e^{-\frac{1}{2}x^2} \vert$$
+
−
 
+
−
goes to \(0\) as x goes to \( \pm \infty \) , the left term is \(0\)
+
−
 
+
−
performing the differential in the right term we get:
+
−
 
+
−
$$ \frac{1}{\sqrt{2\pi}} (-1)^n \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) x e^{-ikx} dx \ + \ \frac{1}{\sqrt{2\pi}} (-1)^{n+1}ik\int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) e^{-ikx} dx$$
+
−
 
+
−
$$ = (x\Phi(x))\hat(k) \ - \ ik\hat\Phi_n(k)$$
+
−
 
+
−
using the propertiy of the fouriertransform we proved in the script (FourierSchwartz-Added page 96), we are allowed to use them because \(x\Phi(x)\) is in Schwartzspace and thus is integrable and also goes to zero as the absolute value of x goes to infinity:
+
−
 
+
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$$ = i \frac{d}{dk}\hat\Phi_n(k) \ - \ ik\hat\Phi_n(k)$$
+
−
 
+
−
using the induction assumtion:
+
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+
−
$$= i \frac{d}{dk}\Big( (-i)^n (-1)^n e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^n
+
−
e^{-k^2} \Big)\Big) \ - \ ik\hat\Phi_n(k)$$
+
−
 
+
−
with the produkt rule:
+
−
 
+
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$$= i (-i)^n (-1)^n \Big( k e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^n
+
−
e^{-k^2} \Big) +  e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^{n+1}
+
−
e^{-k^2} \Big)\Big)\ - \ ik\hat\Phi_n(k)$$
+
−
 
+
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$$=  ik\hat\Phi_n(k)\ +\ \ (-i)^{n+1} (-1)^{n+1}e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^{n+1} e^{-k^2} \Big)\ \ +\ ik\hat\Phi_n(k) = (-i)^{n+1} \Phi_{n+1}(k) $$
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+
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<p style="text-align:right;">\(\square\)</p>
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==Part b) ==
+
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+
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Let \( \chi_{[a,b]} \) be the characteristic function, \( a, b \in \mathbb{R}, a < b \). Show explicitly, i.e. without using Plancherel, that
+
−
 
+
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$$ \Vert \hat \chi_{[a,b]} \Vert_2^2 = b - a. $$
+
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''Hint:'' You may use without proof that
+
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+
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$$ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} $$
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+
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+
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===Solution===
+
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+
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''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 \( \Vert f \Vert_{2} = \left( \int_{\mathbb{R}} \vert f(x) \vert^2 dx \right)^{\frac{1}{2}} \)
+
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+
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But we show the identity like badasses by direct calculation.
+
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+
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We have: $$ \hat \chi_{[a,b]}(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$
+
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+
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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:
+
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+
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$$  
+
 
\begin{align}
 
\begin{align}
−
\hat \chi_{[a,b]}(k) &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy \\
+
H_F &= -i \omega \psi_1 \psi_2 \\
−
&= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \left( \frac{-2}{ik(b-a)} e^{-\frac{ik(b-a)}{2}y} \bigg \vert_{-1}^{1} \right) \\
+
&\stackrel{(*)}{=} i \omega \frac{\hbar}{i}(\alpha^\dagger \alpha\ - \frac{1}{2} ) \\
−
&= \frac{1}{\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \frac{1}{ik} \left( - e^{-ik\frac{(b-a)}{2}} + e^{ik \frac{b-a}{2}} \right) \\
+
&=\omega \hbar (\alpha^\dagger \alpha\ - \frac{1}{2} )
−
&= \frac{\sqrt{2}}{k\sqrt{\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \sin \left( \frac{b-a}{2} k \right)  
+
 
\end{align}
 
\end{align}
 
$$
 
$$
−
 
−
 
−
 
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And so:
 
−
 
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$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \int_{\mathbb{R} } |\hat \chi_{[a,b]}(k)|^2 dk = \frac{2}{\pi} \int_{\mathbb{R} }  \frac{\sin^2(\frac{b-a}{2}k)}{k^2} dk $$
 
−
 
−
With the substitution \( \eta \equiv \frac{b-a}{2}k \) we get:
 
−
 
−
$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \frac{(b-a)}{\pi} \int_{\mathbb{R} }  \frac{\sin^2(\eta)}{\eta^2} d\eta $$
 
−
 
−
It remains to evaluate the integral \( \int_{\mathbb{R} }  \frac{\sin^2\left(\eta\right)}{\eta^2} d\eta \).
 
−
 
−
By using the identity \( \sin^2 \left( x \right) = \frac{1}{2} \left( 1 - \cos \left( 2x \right) \right) \) and partial integration we have
 
−
 
−
$$ \begin{align}
 
−
\int_{\mathbb{R} }  \frac{\sin^2\left(\eta\right)}{\eta^2} d\eta &= - \frac{1}{2\eta} \left( 1 - \cos \left( 2\eta \right) \right) \Bigg \vert_{- \infty}^{\infty} + \int_{\mathbb{R} } \frac{\sin \left(2\eta \right) }{\eta} d\eta \\
 
−
&= 2 \int_{0}^{\infty} \frac{\sin \left(2\eta \right) }{\eta} d\eta \\
 
−
&= 2 \int_{0}^{\infty} \frac{\sin\left( \xi \right)}{\xi} d\xi = \pi
 
−
\end{align}$$
 
−
 
−
where obviously \( \left| \frac{1}{2\eta} \left( 1 - \cos \left( 2\eta \right) \right) \right| \leq \left| \frac{1}{\eta} \right| \rightarrow 0 \) for \( \eta \rightarrow \pm \infty \).
 
−
 
−
The result therefore 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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