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\begin{document}
\title{Assignment Two}
{\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} l r }
Advanced Quantum Theory & Due at the beginning of class \\
AMATH 473, 673 / PHYS 454 & on September 28, 2011
\end{tabular*}}
\maketitle
\begin{problem}
\emph{(Adjoint Operators)} (This problem was originally assigned to the first assignment, but was postponed to this one. Please refer to the first assignment for the definition of $\fket{v}$). In class and in the notes you were given a nebulous ``definition'' of the adjoint to an operator on a Hilbert space: ``The adjoint of an operator $\hat{A}$ acting on a Hilbert space $\Hilbert$, denoted $\hat{A}^\dag$, is defined implicitly as the unique linear operator satisfying $ \bra{v}\hat{A}^{\dag}\ket{u} = \conj{\bra{u}\hat{A}\ket{v}}$
for all $\ket u$ and $\ket v$ in $\Hilbert$.'' This problem will justify this pre-definition. Let $\Hilbert$ be a Hilbert space and let $\LHilbert:=\mathcal{L}(\Hilbert,\Hilbert)$ denote the set of linear operators from $\Hilbert$ to $\Hilbert$. We do not assume that $\Hilbert$ must be finite dimensional, however you may assume throughout this question that the results of the previous problem hold for Hilbert spaces of all dimensions.
\begin{itemize}
\item[a)] Suppose $\hat{A}\in\LHilbert$. Prove that for any $\ket{v}\in\Hilbert$ we have that $\fket{v}\circ \hat{A}$ is a dual vector, where $\circ$ denotes the composition of functions. We use the notation $\bra{v}\hat{A}=\fket{v}\circ \hat{A}$.
\item[b)] Applying the result of (Assignment 1) 2(b) to (Assignment 2) 1(a), for each vector $\ket{v}\in \Hilbert$ there exists a vector $\ket{\phi_v}\in\Hilbert$ such that $\bra{\phi_v}=\bra{v}\hat{A}$. We now define the function $\hat{A}^\dag:\Hilbert\rightarrow\Hilbert$ by $\hat{A}^\dag\ket{v}:=\ket{\phi_v}$ for each vector $\ket{v}\in\Hilbert$. Show that $\hat{A}^\dag\in\mathcal{L}(\Hilbert)$.
\item[c)] As a sanity check, use the definition in (b) to find the adjoint of the following matrix acting on the Hilbert space of columns, $\Complex^3$: $$A=\begin{pmatrix}4&i&3 \\ 0&2i&1 \\ 3&0&1 \end{pmatrix}$$ You should of course end up with the conjugate transpose. Use the fact that the dual vector to column vector is its conjugate transpose, and a dual vector acting on a vector , or a dual vector composed with a matrix, is just matrix multiplication.
\item[d)] In general, the kernel of a linear operator $\hat{A}:\vsV\rightarrow\vsW$, denoted by $\ker{\hat{A}}$, is defined as those vectors in $\vsV$ which $\hat{A}$ takes to the zero vector $\ket{\o_W}$, i.e., $\ker{\hat{A}}:=\{\ket{v}\in\vsV~:~\hat{A}\ket{v}=\ket{\o_W}\}$. Prove that the intersection of the kernels of all dual vectors in $\Hilbert^\dag$ is the zero subspace, i.e., that $\cap_{\bra{v}\in\Hilbert^\dag}\ker{\bra{v}} = \{\ket{\o}\}$.
\item[e)] Prove that the linear operator $\hat{A}^\dag$ as defined in part (b) is the unique linear operator such that $\bra{v}\hat{A}^{\dag}\ket{u} = \conj{\bra{u}\hat{A}\ket{v}}$ for all $\ket u$ and $\ket v$ in $\Hilbert$. (\emph{Hint:} First prove that $\hat{A}^\dag$ has the correct property, and then part (c) may be useful for uniqueness.)
\end{itemize}
\end{problem}
\begin{problem} \emph{(Properties of Adjoints)} For $\hat{A},\hat{B}\in\LHilbert$ demonstrate the following properties \footnote{Hint: In Problem 1 (d) and (e) you implicitly proved that operators $\hat{A},\hat{B}\in\LHilbert$ are equal if and only if $\bra{v}\hat{A}\ket{w} = \bra{v}\hat{B}\ket{w}$ for all $\ket{v},\ket{w}\in\Hilbert$. Feel free to use this as much as needed.}:
\begin{itemize}
\item[a)] $(\hat A \ket v)^\dag = \bra v \hat A^\dag$;
\item[b)] $(\hat{A}^{\dag})^{\dag} = \hat{A}$;
\item[c)] $(\alpha\hat{A})^{\dag} = \conj{\alpha}\hat{A}^{\dag}$;
\item[d)] $(\hat{A} + \hat{B})^{\dag} = \hat{A}^{\dag} + \hat{B}^{\dag}$;
\item[e)] $(\hat{A}\hat{B})^{\dag} = \hat{B}^{\dag}\hat{A}^{\dag}$.
\end{itemize}
\end{problem}
\begin{problem}
\emph{(The Trace is Independent of Basis)} Given a finite dimensional Hilbert space $\Hilbert$ and a linear operator $\hat{A}\in\LHilbert$ and an orthonormal basis $\beta=\{\ket{k}\}_{k=1}^{d}$ for $\Hilbert$, define the trace of $\hat{A}$ with respect to this basis as $Tr(A;\beta):=\sum_{k=1}^{d} A_{kk}^{\beta}$ where $A_{ij}^{\beta}=\bra{i}\hat{A}\ket{j}$ are the matrix elements of $\hat{A}$ with respect to $\beta$.
\begin{itemize}
\item[a)] Show that for any $\hat{A},\hat{B}\in\LHilbert$ and any orthonormal basis $\beta=\{\ket{k}\}_{k=1}^{d}$ for $\Hilbert$ we get that $Tr(\hat{A}\hat{B}, \beta) = Tr(\hat{B}\hat{A}, \beta)$. Use this to quickly show that if $\hat{A},\hat{B},\hat{C},\hat{D}\in\LHilbert$, then $$Tr(\hat{A}\hat{B}\hat{C}\hat{D},\beta)= Tr(\hat{D}\hat{A}\hat{B}\hat{C},\beta)=Tr(\hat{B}\hat{C}\hat{D}\hat{A},\beta)$$ This is called the cyclic property of the trace; we cannot permute the products in the trace arbitrarily, but we can shift the ones on the end to the opposite side.
\item[b)] If $\alpha=\{\ket{k_\alpha}\}_{k=1}^{d}$ and $\beta=\{\ket{k_\beta}\}_{k=1}^{d}$ are two orthonormal bases for $\Hilbert$, define the change of basis operator $\hat{E}_{\alpha}^{\beta}\in\LHilbert$ as $\hat{E}_{\alpha}^{\beta} = \sum_{k=1}^{d} \ket{k_\alpha}\bra{k_\beta}$. Show that the inverse of $\hat{E}_{\alpha}^{\beta}$ is $\hat{E}_{\beta}^{\alpha}$. Also show that this operator indeed changes basis, i.e., show that $A_{ij}^{\beta} = (\hat{E}_{\alpha}^{\beta} \hat{A} \hat{E}_{\beta}^{\alpha})_{ij}^{\alpha}$ for all $1\leq i,j \leq d$ and any $\hat{A}\in\LHilbert$.
\item[c)] Suppose $\alpha=\{\ket{k_\alpha}\}_{k=1}^{d}$ and $\beta=\{\ket{k_\beta}\}_{k=1}^{d}$ are two orthonormal bases for $\Hilbert$ and $\hat{A}\in\LHilbert$, then show that $Tr(\hat{A}, \alpha) = Tr(\hat{A}, \beta)$. This means that the trace function is independent of the orthonormal basis we choose, and hence we reduce the notation to $Tr(\hat{A})$.
\end{itemize}
\end{problem}
\end{document}