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) 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 $$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty} = \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{\infty} = 0. $$
+
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 the Schwarz's theorem. With the induction assumption everything follows. \( \square \)
+
$$
 +
\{\mu (A+B),\nu (C+D)\} = \mu (A+B) \nu (C+D) + \nu (C+D) \mu (A+B) = \mu \nu ((AC+AD+BC+BD) + (CA+CB+DA+DB)) = \mu \nu (\{A,C\} + \{A,D\} + \{B,C\} + \{B,D\})
 +
$$
 +
$$
 +
\{A,B\} = AB + BA = BA + AB = \{B,A\}
 +
$$
  
−
We then finish the proof as follows:
+
==== Solution Part a) Use bilinearity of the anticommutator====
  
−
$$ \sqrt{2\pi} \hat \Phi_n(k) =^{\color{red}{*}} i^n e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n \int_{\mathbb{R} } e^{-x^2 + \frac{1}{2}(x - ik)^2} dx = i^n \sqrt{2\pi} e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n e^{-k^2} = (-i)^n \sqrt{2\pi} \Phi_n(k) $$  
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 + i\psi_2 )  \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar + \hbar) \\
 +
&= 1 \\
 +
\end{align}
 +
$$
  
−
Where for the first equality we used Lemma 2 and for the second equation (3.14). \( \square \)
+
$$
 +
\begin{align} \\
 +
\{\alpha,\alpha\} &= \frac{1}{2\hbar} \{( \psi_1 - i\psi_2 ), ( \psi_1 - i\psi_2 \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} - i \{ \psi_1, \psi_2 \} - i \{ \psi_2, \psi_1 \} + \{ -i \psi_2, -i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} -0 -0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
\( \color{red}{* \: see \: discussion}\)
+
$$
 +
\begin{align} \\
 +
\{\alpha^\dagger\,\alpha^\dagger\} &= \frac{1}{2\hbar} \{( \psi_1 + i\psi_2 ), ( \psi_1 + i\psi_2 ) \} \\ 
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + i \{ \psi_1, \psi_2 \} + i \{ \psi_2, \psi_1 \} + \{ +i \psi_2, +i \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\{ \psi_1, \psi_1 \} + 0 + 0 - \{ \psi_2, \psi_2 \}) \\
 +
&= \frac{1}{2\hbar} (\hbar - \hbar) \\
 +
&= 0 \\
 +
\end{align}
 +
$$
  
−
'''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 \( f(x) = e^{- \frac{1}{2} x^2} \).
+
And thus as above:
 +
$$
 +
\{\alpha,\alpha\} = 2\alpha^2 = 0 \Rightarrow \alpha^2 = 0
 +
$$
 +
and
 +
$$
 +
\{\alpha^\dagger,\alpha^\dagger\} = 2\left(\alpha^\dagger\right)^2 = 0 \Rightarrow \left(\alpha^\dagger\right)^2 = 0
 +
$$
  
−
''Proof of Lemma 2:''
 
  
−
''We've already shown this in Series 10, Exercise 3.''
+
===Solution Part b)===
  
−
$$ \sqrt{2\pi} \frac{d}{dk} \hat f(k) = \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{1}{h_n} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh_n} - 1 \right) dx $$
+
$$
−
 
+
−
where \(h_n\) is an arbitrary zero-sequence and we used the definition of \( \frac{d}{dk} \) as limit of the difference quotient, the definition of \( \hat f(k) \) and linearity of the integral.
+
−
 
+
−
We then use \( \vert \sin(x) \vert \leq \vert x \vert \) (this follows directly from \( \left| \operatorname{sinc}(x) \right| \leq 1 \)) and get:
+
−
 
+
−
$$ \begin{align}
+
−
\left| \frac{1}{h_n} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh_n} - 1 \right) \right| &= e^{- \frac{x^2}{2}} \frac{1}{h_n} \left| e^{-ikx} e^{-\frac{ixh_n}{2}} \left( e^{- \frac{ixh_n}{2} } - e^{\frac{ixh_n}{2}} \right) \right| \\
+
−
&= e^{- \frac{x^2}{2}} \frac{2}{h_n} \left| \sin \left( \frac{xh_n}{2} \right) \right| \leq \frac{2}{h_n} \left| \frac{xh_n}{2} \right| e^{-\frac{x^2}{2}} \\
+
−
&= \vert x \vert e^{-\frac{x^2}{2}}
+
−
\end{align} $$
+
−
 
+
−
and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)
+
−
 
+
−
'''Lemma 3:''' Let \( f(x) = e^{- \frac{1}{2} x^2} \) and \( \lambda (k) \) be an arbitrary function just depending on k. Then: \( \left( \frac{d}{dk} \right)^{n} \lambda (k) \hat f(k) = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n \lambda (k) f(x) e^{-ikx} dx \).
+
−
 
+
−
''Proof of Lemma 3: ''
+
−
 
+
−
$$ \begin{align}
+
−
\frac{d}{dk} \lambda (k) \hat f(k) &= \frac{1}{\sqrt{2\pi}} \frac{d}{dk} \lambda (k) \int_{\mathbb{R}}  f(x) e^{-ikx} dx \\
+
−
&= \frac{1}{\sqrt{2\pi}} \left( \frac{d}{dk} \lambda (k) \right) \int_{\mathbb{R}}  f(x) e^{-ikx} dx + \frac{1}{\sqrt{2\pi}} \lambda (k) \int_{\mathbb{R}}  \frac{d}{dk} f(x) e^{-ikx} dx \\
+
−
&=  \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \left( \frac{d}{dk} \lambda (k) \right) f(x) e^{-ikx} + \lambda (k) \frac{d}{dk} f(x) e^{-ikx} dx \\
+
−
&= \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \frac{d}{dk} \lambda (k) f(x) e^{-ikx} dx
+
−
\end{align} $$
+
−
 
+
−
concluding the prove. \( \square \)
+
−
 
+
−
From this follows directly
+
−
 
+
−
$$ \frac{d}{dk} e^{\frac{-k^2}{2}} \hat f(k) = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \frac{d}{dk} e^{- \frac{k^2}{2} - x^2 + \frac{1}{2}x^2 - ikx} dx = \frac{1}{\sqrt{2\pi}}  \int_{\mathbb{R}} \frac{d}{dk} e^{- x^2 + \frac{1}{2} (x - ik)^2 }  dx $$
+
−
 
+
−
''Generalization of Lemma 2:''
+
−
 
+
−
$$ \left| \frac{1}{h} \left( \frac{d}{dk} \right)^{n-1} f(x) e^{-ikx} \left( e^{-ihx} - 1 \right) \right| = \left| \frac{2}{h} f(x) \sin (\frac{xh}{2}) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \leq \left| x f(x)  \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| = \left| p(x) f(x) \right| \in L^1 $$
+
−
 
+
−
where \( p(x) \in \mathbb{C}[x] \). This holds since \( f \in \mathcal{S}(\mathbb{R}) \). \( \square \)
+
−
 
+
−
''Generalization of Lemma 3:'' Use generalized Leibniz rule.
+
−
 
+
−
''Nice to know, to be checked if necessary to know by heart:'' Gaussian-function is a Schwatz-fct., Polynomial times Schwartz-function is a Schwartz-function, Schwatz-space subspace of L1 (to be found in Felder-Script).
+
−
 
+
−
==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: $$ \hat \chi_{[a,b]}(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$
+
−
 
+
−
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]}(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 \\
+
\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{a+b}{2} \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{a+b}{2} \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{a+b}{2} \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 \( \int_{\mathbb{R} }  \frac{\sin^2\left(\eta\right)}{\eta^2} d\eta \).
+
−
 
+
−
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 \( \eta \rightarrow \infty \).
+
−
 
+
−
The result therefore is:
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−
 
+
−
$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = b-a $$
+

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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