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\begin{document}

       \begin{center}
      {\large Quantum Mechanics Homework\\}  
      {\small November 19, 2$\infty$8 \\}  
      {\small John Baez} 
      {\small homework by C.Pro} 
      \end{center}


1. If $T$ is a normal operator, then the spectrum of $T$ consists of approximate eigenvalues.
\vskip 1em
\textit{\underline{Solution:} } Let $\varepsilon >0$. If $$\lambda \in\sigma(T) = \overline{\bigcup_{\psi\in\textbf{H}} \supp\mu_{T,\psi}},$$ then for some $\psi(\neq 0) \in \textbf{H}$ we have $B\cap \supp\mu_{T,\psi}\neq\emptyset$  where $ B=\left\{ z\in\C\,|\, |z-\lambda|<\varepsilon \right\}. $ 
Consider the characteristic function $ \chi_B\in\mathcal{B}(\sigma(T)))$ on $B$. Applying the functional calculus for the normal operator $T$ to $\chi_B$, we get a bounded linear functional with the same name.  Set $\psi_\varepsilon = \chi_B(T)\psi$ and note $\psi_\varepsilon\neq 0$ for
\ban 
     ||\psi_\varepsilon||^2                 &=& \langle \psi_\varepsilon, \psi_\varepsilon \rangle\\
                                            &=& \langle \psi, \chi_B(T)^*\chi_B(T) \psi\rangle \\
     																				&=& \int_{\supp\mu_{T,\psi}}(\chi_B(z))^2\, d\mu_{T,\psi}\\ 
                                            &=& \int_{B\cap\supp\mu_{T,\psi}}d\mu_{T,\psi}\\
                                            &=& \mu_{T,\psi}(B\cap\supp\mu_{T,\psi})\\
                                            &>& 0
\ean
where the last inequality follows from the fact we can find an open set contained in $B\cap \supp\mu_{T,\psi}\neq\emptyset$ and every open set contained in $\supp \mu_{T,\psi}$ has positive measure.

Now define the function $f(z)= (\overline{z}-\overline{\lambda})(z-\lambda) = |z-\lambda|^2$ and note $f\in\mathcal{B}(\sigma(T))$. We can apply the functional calculus to $f$ to obtain
\ban
||(T-\lambda) \psi_{\varepsilon}||^2 &=& \langle \psi_\varepsilon,(T^*-\overline\lambda)(T-\lambda)\psi_\varepsilon \rangle\\
&=& \langle \psi, \chi_B(T)^*f(T)\chi_B(T)\psi\rangle\\
&=& \int_{\supp\mu{T,\psi}} |z-\lambda|^2(\chi_B(z))^2\,d\mu{T,\psi}\\
&=& \int_{B\cap\supp\mu_{T,\psi}}|z-\lambda|^2\,d\mu_{T,\psi}\\
&\leq& \varepsilon^2\int_{B\cap\supp\mu_{T,\psi}}d\mu_{T,\psi}\\
&=& \varepsilon^2||\psi_\varepsilon||^2.
\ean
Hence $||(T-\lambda) \psi_{\varepsilon}||\leq \varepsilon ||\psi_\varepsilon||$ and after normalizing we see $\lambda$ is an approximate eigenvalue as desired.
\vskip 2em
2. If $U$ is a unitary operator, then spectrum of $U$ is a subset of the unit circle in $\C$.
\vskip 1em
\textit{\underline{Solution:} } Let $\lambda \in \sigma(U)$. Since $U$ is normal, $\lambda$ is an approximate eigenvalue. Also, for all $\psi\in \mathbf{H}$, $||U\psi||^2 = \langle U\psi,U\psi\rangle = \langle \psi, U^*U\psi\rangle = ||\psi||^2$. So for any $\varepsilon>0$ there is a $\psi\in\mathbf{H}$ with $||\psi||=1$ and $||(U-\lambda)\psi||<\sqrt\varepsilon$. By the triangle inequality, 
\ban
|1-|\lambda|^2 | &=& \left|||\psi||^2 -|\lambda|^2||\psi||^2\right|\\
                 &=& \left| ||U\psi||^2 -\lambda\overline{\lambda}||\psi||^2\right|\\
                 &=& \left| ||U\psi||^2 -||\lambda\psi||^2 \right|\\
                 &\leq& ||(U-\lambda)\psi||^2\\
                 &<& \varepsilon.
\ean
and we have no choice but to say $|\lambda|=1$.
\vskip 2em
0. If $T$ is a normal operator, then $$\sigma(T) \subset \overline{\bigcup_{\psi\in\textbf{H}} \supp\mu_{T,\psi}}.$$
\textit{\underline{Solution:} } If $\lambda \notin$ the closure of $\bigcup \supp\mu_{T,\psi},$ then we can find a $\varepsilon_0>0$ such that $B = \left\{z\in\C\,|\,|z-\lambda|<\varepsilon_0 \right\}$ is entirely contained outside of $\bigcup \supp\mu_{T,\psi}.$ Therefore the function $$f(z) = \frac{1}{z-\lambda}$$ is bounded by $1/\varepsilon_0$ on $\bigcup \supp\mu_{T,\psi}$. If $g(z) = z-\lambda$, then $f(z)g(z)=g(z)f(z)=1$. Applying the spectral theorem for normal operators, we see that $g(T) = T-\lambda$ and that $f(T)(T-\lambda)=(T-\lambda)f(T) = id_{\mathbf{H}}$. Hence $f(T)=(T-\lambda)^{-1}$ and thus $\lambda\notin\sigma(T)$.
\vskip 2em


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