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\subsubsection*{Notice}
{\footnotesize In this assignment I'll refer to theorems, definitions and so forth. When I do this, I don't make any references to where they're coming from -- but unless I state anything else -- they're from the notes we use in the course, \cite{notes}. }


\setcounter{subsection}{1}
\subsection*{Problem 1}
\ramme{ Consider a quantum system described on the finite dimensional Hilbert space $\mathbb{C}^2$. Let the Hamiltonian be given by
        \begin{equation*}
         \H = \left[ \begin{array}{cc}
                      0& 1\\
                      1& 0
                     \end{array} \right] .
        \end{equation*}\vspace{-0.2cm} }

\subsubsection*{Question a}
\ramme{ Determine an orthonormal basis of eigenvectors of $\H$. }
First of all, in order to find eigenvectors for $\H$, I need to determine the eigenvalues. So I find the charactheristic polynomium $p(E)$, given by
\begin{align*}
 p(E) = \det{(\H - E I)} 
      = \left| \begin{array}{cc}
                -E & 1 \\
                1 & -E
               \end{array} \right| 
      = E^2 - 1 ,
\end{align*}
which has the two roots $E_1 = 1$ and $E_2 = -1$. The corresponding eigenvectors are 
\begin{equation*}
 e_1' = \alpha \left[ \begin{array}{c} 1 \\ -1 \end{array} \right]   \qquad\text{and}\qquad
 e_2' = \beta \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] ,
\end{equation*}
for $\alpha,\beta \in \mathbb{R}$. So for the two to be normalized (they are already orthogonal), I simply have to let the constants $\alpha$ and $\beta$ be the reciprocal of their norm, which for both $e_1'$ and $e_2'$ is $\sqrt{2}$. Hence the orthonormal basis for $\H$ is
\begin{equation}
 \{ e_1, e_2 \} = 
 \left\{ \frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ -1 \end{array} \right], 
         \frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \right\} .
 \label{p1:qa:basis}
\end{equation}


\clearpage
\subsubsection*{Question b}
\ramme{ Determine the time evolution of the initial state represented by the vector
        \begin{equation*}
         \left[ \begin{array}{c} 0\\ 1 \end{array} \right] .
        \end{equation*}\vspace{-0.2cm} }
The vector, call it $\psi_0$, can be written as a linear combination of the vectors in the orthonormal basis for $\H$ in \eqref{p1:qa:basis}, by
\begin{equation*}
 \psi_0 = \frac{\sqrt2}{2} e_2 - \frac{\sqrt2}{2} e_1 
        = \frac{\sqrt2}{2} \frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]
          - \frac{\sqrt2}{2} \frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ -1 \end{array} \right]
        = \left[ \begin{array}{c} 0\\ 1 \end{array} \right] ,
\end{equation*}
so since the time evolution for the first basis-vector, according to the discussion on page 5 in chapter 6, must be
\begin{equation*}
 e_1(t) = e_1 \exp{\left( - \frac{i E_1 t}{\hbar} \right)} 
        = e_1 \exp{\left( - \frac{i t}{\hbar} \right)} ,
\end{equation*}
and corresponding, the time evolution for the second basis-vector must be
\begin{equation*}
 e_2(t) = e_2 \exp{\left( - \frac{i E_2 t}{\hbar} \right)} 
        = e_2 \exp{\left( \frac{i t}{\hbar} \right)} ,
\end{equation*}
I have that the time evolution of the state $\psi(t)$, where $\psi(0) = \psi_0$, is the linear combination
\begin{equation*}
 \psi(t) = \frac{\sqrt2}{2} e_2(t) - \frac{\sqrt2}{2} e_1(t)
         = \frac{\sqrt2}{2} e_2 \exp{\left( \frac{i t}{\hbar} \right)} - \frac{\sqrt2}{2} e_1 \exp{\left( - \frac{i t}{\hbar} \right)} .
\end{equation*}



\clearpage
\setcounter{subsection}{2}
\subsection*{Problem 2}
\ramme{ {\bfseries\sffamily Exercise 6.24 in the notes} }

\subsubsection*{Question a}
\ramme{ Show that for all $\psi \in \S(\mathcal{R})$ normalized we have
        \begin{equation}
         \braket{\psi}{\hbar^2 D^2}{\psi} + mk \braket{\psi}{x^2}{\psi} \geq \hbar \sqrt{mk} 
         \label{p2:qa:inequality}
        \end{equation}\vspace{-0.5cm} }
For all $\psi \in \S(\mathbb{R})$ I know that $\H_0 \psi = \H \psi$, when I have the two Hamiltonians $\H_0$ and $\H$ defined by
\begin{equation*}
 \H_0 = \frac{\hbar^2}{2m} D^2 + \frac{1}{2} k x^2, \qquad\qquad
 \H = \hbar \omega \sum_{n=0}^\infty \left( n + \frac{1}{2} \right) \ket{\Omega_n} \bra{\Omega_n} ,
\end{equation*}
where $(\Omega_n)_{n=0}^\infty$ is an orthonormal basis and $\omega = \sqrt{k/m}$. From $\H_0 \psi = \H \psi$ I get that $\H_0 \ket{\psi} = \H \ket{\psi}$, and hence that $\braket{\psi}{\H_0}{\psi} = \braket{\psi}{\H}{\psi}$. So since
\begin{equation*}
 \braket{\psi}{\H_0}{\psi} = \frac{1}{2m} \braket{\psi}{\hbar^2 D^2}{\psi} + \frac{1}{2} k \braket{\psi}{x^2}{\psi} ,
\end{equation*}
and (by Parsevals identity, the fact that $\psi$ is normalized and that $(\Omega_n)_{n=0}^\infty$ is an orthonormal basis)
\begin{equation*}
 \braket{\psi}{\H}{\psi} = \hbar \omega \sum_{n=0}^\infty \left( n + \frac{1}{2} \right) \twobraket{\psi}{\Omega_n} \twobraket{\Omega_n}{\psi}
                         \geq \frac{1}{2} \hbar \omega \sum_{n=0}^\infty \left| \twobraket{\psi}{\Omega_n} \right|^2
                         = \frac{1}{2} \hbar \omega ,
\end{equation*}
I see that
\begin{equation*}
 \frac{1}{m} \braket{\psi}{\hbar^2 D^2}{\psi} + k \braket{\psi}{x^2}{\psi} \geq \hbar \omega .
\end{equation*}
Multiplying on both sides by $m$ and using that $\omega = \sqrt{k/m}$, I then get that
\begin{equation*}
 \braket{\psi}{\hbar^2 D^2}{\psi} + mk \braket{\psi}{x^2}{\psi} \geq \hbar \sqrt{mk} ,
\end{equation*}
just as wanted. 


\clearpage
\subsubsection*{Question b}
\ramme{ Derive from the fact that the inequality in question a holds for all $m,k\geq 0$ the Heisenberg uncertainty relation
        \begin{equation*}
         \braket{\psi}{\hbar^2 D^2}{\psi} \braket{\psi}{x^2}{\psi} \geq \frac{1}{4} \hbar^2 ,
        \end{equation*}
        for all normalized functions $\psi \in \S(\mathbb{R})$. }
First of all, I define $t = \sqrt{mk}$ (which makes sence for $m,k \geq 0$), such that my inequality in \eqref{p2:qa:inequality} becomes a polynoium of second degree in $t$, 
\begin{equation*}
 t^2 \braket{\psi}{x^2}{\psi} - \hbar t + \braket{\psi}{\hbar^2 D^2}{\psi} \geq 0 .
\end{equation*}
Since this polynomium is greater than or equal to zero, it must have either only one or simply no roots (this is easily seen on figure \vref{fig:p2:graph}). This means that the determinant of the polynomium must be less than or equal to zero, so
\begin{equation*}
 \hbar^2 - 4 \braket{\psi}{x^2}{\psi} \braket{\psi}{\hbar^2 D^2}{\psi} \leq 0 ,
\end{equation*}
which implies that
\begin{equation*}
 \frac{1}{4} \hbar^2 \leq \braket{\psi}{x^2}{\psi} \braket{\psi}{\hbar^2 D^2}{\psi} ,
\end{equation*}
just as I wanted to show. 

\begin{figure}[tb]
 \centering
  %\begin{center}
  \psset{xunit=10mm,yunit=5mm}
  \begin{pspicture*}(-5,-1)(5.5,7) %\psgrid
   %Koordinatsystem:
   \psline[arrows=->,linewidth=0.5pt](-5,0)(5,0)          \uput[-90](5,0){$x$}
   \psline[arrows=->,linewidth=0.5pt](-1.5,-0.2)(-1.5,7) %\uput[180](){}
   %Plots:
   \psplot[linecolor=Periwinkle,linewidth=1.5pt]{-5}{5}{x x mul}                       \rput[90](0,-0.3){\color{Periwinkle}One root}
   \psplot[linecolor=SpringGreen,linewidth=1.5pt]{-5}{5}{8 x x mul add  5 x mul sub}   \rput[90](2.5,1.5){\color{SpringGreen}No roots}
   \psplot[linecolor=Peach,linewidth=1.5pt]{-5}{5}{6 x x mul add  5 x mul add}         \rput[90](-2.5,-0.5){\color{Peach}Two roots}
  \end{pspicture*}
  %\end{center}
 \caption{Illustration showing the logic of why a second degree polynomial, taking values greater than or equal to zero, must have one or no roots.}
 \label{fig:p2:graph}
\end{figure}



\clearpage
\setcounter{subsection}{3}
\subsection*{Problem 3}
\ramme{ {\bfseries\sffamily Exercise 6.30 in the notes} \\
        Let $SU(2)$ be the set of $2\times2$ complex unitary matrices $V$ with $\det V = 1$. }

\subsubsection*{Question a}
\ramme{ Show that $SU(2)$ is a matrix group, i.e., show that if $V_1,V_2 \in SU(2)$, then $V_1V_2\in SU(2)$ and $V_1^{-1}\in SU(2)$. }
To show that $SU(2)$ is a matrix group, I first have to show that for $V_1,V_2 \in SU(2)$ then $V_1V_2\in SU(2)$. This is simply done by seing that
\begin{equation}
 \det{(V_1 V_2)} = \det{V_1} \det{V_2} 
                 = 1 \cdot 1 
                 = 1 ,
 \label{p3:qa:det1}
\end{equation}
according to Theorem 2.2.3 in \citet{linalg}, and that
\begin{equation*}
 (V_1 V_2)^\ast (V_1 V_2) = V_2^\ast V_1^\ast V_1 V_2 
                          = V_2^\ast V_2 
                          = I .
\end{equation*}
So $V_1V_2\in SU(2)$. 

Secondly I have to show that for $V_1 \in SU(2)$ then $V_1^{-1}\in SU(2)$. According to the fact that $\det{I} = \det{(V_1 V_1^{-1})} = \det{V_1} \det{V_1^{-1}} = 1$, following from the same argument as used in \eqref{p3:qa:det1}, I see that
\begin{equation*}
 \det{V_1^{-1}} = \frac{1}{\det{V_1}}
                = \frac{1}{1}
                = 1 ,
\end{equation*}
and, since $V_1 \in SU(2)$ implies that $V_1$ is unitary, and hence that $V_1^{-1} = V_1^\ast$, I see that
\begin{equation*}
 (V_1^{-1})^\ast V_1^{-1} = (V_1^{-1})^{-1} V_1^{-1}
                          = V_1 V_1^{-1}
                          = I .
\end{equation*}
Hence it is also true that $V_1^{-1}\in SU(2)$, concluding that $SU(2)$ really is a matrix group. 


\subsubsection*{Question b}
\ramme{ Argue that if $V \in SU(2)$ and if $A = (a_{ij})$ is a $2\times2$ complex matrix, which is Hermitian ($A^\ast = A$) and traceless 
        ($0=\Tr A = a_{11} + a_{22}$) then $VAV^\ast$ is Hermitian and traceless. }
First of all, I see that since $A^\ast = A$ ,
\begin{equation*}
 (V A V^\ast)^\ast = \big( (V A) V^\ast \big)^\ast
                   = (V^\ast)^\ast (V A)^\ast
                   = V A^\ast V^\ast
                   = V A V^\ast ,
\end{equation*}
so $V A V^\ast$ is hermitian. Secondly, since the trace of a matrix is invariant under conjugations, $\Tr{(B^{-1} A B)} = \Tr{A}$ according to Theorem 6.1.1 in \citet{linalg}, and that $V \in SU(2)$ implies that $V$ is unitary, so $V^{-1} = V^\ast$, I see that
\begin{equation*}
 \Tr{(V A V^\ast)} = \Tr{(V A V^{-1})}
                   = \Tr{A}
                   = 0 ,
\end{equation*}
concluding that $V A V^\ast$ is also traceless.


\subsubsection*{Question c}
\ramme{ Use the result of question b to show that given $V\in SU(2)$ there is a linear map $R_V : \mathbb{R}^3 \rightarrow \mathbb{R}^3$, such that
        \begin{equation}
         A_{R_V x} = V A_x V^\ast ,
         \label{p3:qc:RVdef}
        \end{equation}
        where 
        \begin{equation}
         A_z = \left[ \begin{array}{cc}
                       z_3 & z_1 - i z_2 \\
                       z_1 + iz_2 & - z_3
                      \end{array} \right] ,
         \label{p3:qc:matrixform}
        \end{equation}
        when $z = (z_1,z_2,z_3) \in \mathbb{R}^3$. }
First of all, I note that a matrix of the form of $A_z$ in \eqref{p3:qc:matrixform} is the most general form of a hermitian traceless $2\times 2$ matrix. I will call the set of hermitian traceless $2\times 2$ matrices $\mathcal{M}$.

The mapping $\Phi : \mathbb{R}^3 \rightarrow \mathcal{M}$ mapping a vector $x = (x_1,x_2,x_3) \in \mathbb{R}^3$ to the hermitian traceless $2\times 2$ matrix
\begin{equation*}
 \Phi(x) = A_x 
         = \left[ \begin{array}{cc}
                   x_3 & x_1 - i x_2 \\
                   x_1 + ix_2 & - x_3
                  \end{array} \right]
\end{equation*}
is then bijective because such a matrix can only be described by one vector, and all matrices in $\mathcal{M}$ can be described by vectors from $\mathbb{R}^3$. The mapping $\Phi$ is also linear, simply because of the normal calculation rules of matrices,
\begin{align*}
 \Phi(\alpha x + \beta y) &= \left[ \begin{array}{cc}
                                     \alpha x_3 + \beta y_3 & \alpha x_1 + \beta y_1 - i \alpha x_2 - i \beta y_2 \\
                                     \alpha x_1 + \beta y_1 + i \alpha x_2 + i \beta y_2 & - \alpha x_3 - \beta y_3
                                    \end{array} \right] \\
                          &= \left[ \begin{array}{cc}
                                     \alpha x_3 & \alpha x_1 - i \alpha x_2 \\
                                     \alpha x_1 + i \alpha x_2 & - \alpha x_3
                                    \end{array} \right]
                                +
                             \left[ \begin{array}{cc}
                                     \beta y_3 & \beta y_1 - i \beta y_2 \\
                                     \beta y_1 + i \beta y_2 & - \beta y_3
                                    \end{array} \right] \\
                          &= \alpha \left[ \begin{array}{cc}
                                     x_3 & x_1 - i x_2 \\
                                     x_1 + i x_2 & - x_3
                                    \end{array} \right]
                                +
                             \beta \left[ \begin{array}{cc}
                                     y_3 & y_1 - i y_2 \\
                                     y_1 + i y_2 & - y_3
                                    \end{array} \right] \\
                          &= \alpha \Phi(x) + \beta \Phi(y) ,
\end{align*}
where $x = (x_1,x_2,x_3), y = (y_1,y_2,y_3) \in \mathbb{R}^3$ and $\alpha,\beta \in \mathbb{R}$. 

I also define the mapping $\sigma_V : \mathcal{M} \rightarrow \mathcal{M}$ by mapping a matrix $A \in \mathcal{M}$ to the matrix $V A V^\ast$ (where $V\in SU(2)$), which is also in $\mathcal{M}$ because of the result in question b. This mapping is also linear because 
\begin{align*}
 \sigma_V(\alpha A + \beta B) &= V(\alpha A + \beta B)V^\ast \\
                              &= V(\alpha A)V^\ast + V(\beta B)V^\ast \\
                              &= \alpha V A V^\ast + \beta V B V^\ast \\
                              &= \alpha \sigma_V(A) + \beta \sigma_V(B) 
\end{align*}
for $A,B \in \mathcal{M}$ and $\alpha,\beta \in \mathbb{R}$. 

Then the mapping $R_V$ defined as in \eqref{p3:qc:RVdef} must, with the two introduced mappings, then be given by
\begin{equation*}
 R_V x = \Phi^{-1} \big[ \sigma_V \big( \Phi(x) \big) \big]
       = \Phi^{-1} \circ \sigma_V \circ \Phi (x) ,
\end{equation*}
where I note that the inverse, $\Phi^{-1}$, of $\Phi$ is well-defined because $\Phi$ is bijective. The mapping $R_V$ above must be linear, simply because it is a composition of three linear mappings, namely $\Phi^{-1}$, $\sigma_V$ and $\Phi$. 

\begin{figure}[tb]
 \centering
  %\begin{center}
  \psset{nodesep=3pt}
  \psset{xunit=5mm,yunit=4mm}
  \begin{pspicture}(0,-2.5)(15,2.5) %\psgrid
   %Punkter:
   \pnode(0,0){x}
   \pnode(5,0){Ax}
   \pnode(10,0){VAxV}
   \pnode(15,0){RVx}
   %Pile:
   \parabola[linewidth=0.7pt,linecolor=Periwinkle]{->}(0.2,0.8)(2.4,1.8)   \rput[b](2.4,2){$\Phi$}
   \parabola[linewidth=0.7pt,linecolor=Periwinkle]{->}(5.5,0.8)(7.5,1.8)   \rput[b](7.5,2){$\sigma_V$}
   \parabola[linewidth=0.7pt,linecolor=Periwinkle]{->}(10.5,0.8)(12.5,1.8) \rput[b](12.5,2){$\Phi^{-1}$}
   \parabola[linewidth=0.7pt,linecolor=Periwinkle]{->}(0.2,-0.3)(7.5,-2)   \rput[t](7.5,-2.2){$R_V$}
   %Labels:
   \rput[B](x){$x$}
   \rput[B](Ax){$A_x$}
   \rput[B](VAxV){$VA_xV^\ast$}
   \rput[B](RVx){$R_Vx$}
  \end{pspicture}
  %\end{center}
 \caption{The mapping $R_V : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $R_V x = \Phi^{-1} \circ \sigma_V \circ \Phi (x)$, or by the 
          expression in \eqref{p3:qc:RVdef}.}
 \label{fig:p3:mapping}
\end{figure}




\subsubsection*{Question d}
\ramme{ Show that for $V,W \in SU(2)$ then $R_V R_W = R_{VW}$. }
By using the notation from the first definition of $R_V$ in \eqref{p3:qc:RVdef}, I simply calculate that for $V,W \in SU(2)$, 
\begin{align*}
 A_{R_V R_W x} = V A_{R_W x} V^\ast 
               = V W A_x W^\ast V^\ast 
               = V W A_x (V W)^\ast 
               = A_{R_{VW} x} ,
\end{align*}
so $R_V R_W x = R_{VW} x$, and hence $R_V R_W = R_{VW}$ as wanted. 


\clearpage
\subsubsection*{Question e}
\ramme{ Show that the map $R_V$ is an isometry for all $V \in SU(2)$, i.e., $R_V \in O(3)$. }
First of all, I calculate the determinant of a matrix $A_x$ in the form showed in \eqref{p3:qc:matrixform}, where I get that
\begin{align*}
 \det{A_x} &= \left| \begin{array}{cc}
                      x_3& x_1 - i x_2 \\
                      x_1 + i x_2& -x_3
                     \end{array} \right| \\
           &= x_3 (-x_3) - (x_1 - i x_2)(x_1 + i x_2) \\
           &= - (x_1^2 + x_2^2 + x_3^2) \\
           &= - ||x||^2 ,
\end{align*}
where $x = (x_1,x_2,x_3) \in \mathbb{R}^3$ and $||\cdot||$ is the Euclidean norm in $\mathbb{R}^3$. Now I see that the determinant of $A_{R_V x}$, by the definition in \eqref{p3:qc:RVdef}, is
\begin{align*}
 \det{A_{R_V x}} &= \det{(V A_x V^\ast)} \\
                 &= \det{V} \det{A_x} \det{V^\ast} \\
                 &= \det{A_x} \det{V^{-1}} \\
                 &= \det{A_x} \frac{1}{\det{V}} \\
                 &= \det{A_x} ,
\end{align*}
where I used that $V\in SU(2)$ and hence is unitary with $\det V = 1$ and $\det V^{-1} = 1/\det V = 1$, and that $\det{AB} = \det A \det B$ for square matrices (according to Theorem 2.2.3 in \citet{linalg}). 

So $\det A_{R_V x} = \det{A_x} = - ||x||^2$, and hence $R_V$ preserve the norm and is hence an isometry. 





\subsubsection*{Question f}
\ramme{ Let $L^2(\mathbb{R}^3, \mathbb{C}^2) = L^2(\mathbb{R}^3) \times L^2(\mathbb{R}^3)$ denote the Hilbert space of functions 
        $\psi : \mathbb{R}^3 \rightarrow \mathbb{C}^2$, such that $\int || \psi ||^2_{\mathbb{C}^2} \,\di x < \infty$. Show that the map 
        $\rho : SU(2) \rightarrow \mathcal{B}\big(L^2(\mathbb{R}^3, \mathbb{C}^2)\big)$ defined by
        \begin{equation*}
         \big( \rho(V)\psi \big)(x) = V\psi(R_V^{-1} x)
        \end{equation*}
        defines a unitary representation of $SU(2)$ on $L^2(\mathbb{R}^3, \mathbb{C}^2)$. }
To show that the mapping $\rho : SU(2) \rightarrow \mathcal{B}\big(L^2(\mathbb{R}^3, \mathbb{C}^2)\big)$ is a unitary representation, I first have to show that $\rho(V)$ is a unitary operator, which means that it has to be linear, norm preserving and bijective, and then that it complies with $\rho(VW)=\rho(V)\rho(W)$ for all $V,W \in SU(2)$. All this is according to Definition 6.5 and Theorem 5.21. 

First, I need to show that $\rho(V)$ is linear, which is shown by the calculation
\begin{align*}
 \big( \rho(V)(\alpha \psi_1 + \beta \psi_2) \big)(x) &= V(\alpha \psi_1 + \beta \psi_2)(R_V^{-1} x) \\
                                                      &= V \big( \alpha \psi_1 (R_V^{-1} x) + \beta \psi_2 (R_V^{-1} x) \big) \\
                                                      &= \alpha V \psi_1 (R_V^{-1} x) + \beta V \psi_2 (R_V^{-1} x) \\
                                                      &= \alpha \big( \rho(V) \psi_1 \big)(x) + \beta \big( \rho(V) \psi_2 \big)(x) ,
\end{align*}
by the linearity of $V$ (because $V\in SU(2)$ it is unitary, and hence linear according to Theorem 5.1). To show that $\rho(V)$ is norm preserving, I simply calculate
\begin{align*}
 || \rho(V) \psi ||^2 = \int |V \psi (R_V^{-1} x)|^2 \,\di x
                      = \int |\psi (R_V^{-1} x)|^2 \,\di x
                      = \int |\psi (x)|^2 \,\di x
                      = || \psi ||^2 ,
\end{align*}
because $V$ is unitary (and hence norm preserving according to Theorem 5.1) and $R_V$ is an isometry (and hence does not change the integral because it is simply a rotation or mirroring). 

Before I show that $\rho(V)$ is bijective, I want to show that it is a representation, which means that $\rho(VW)=\rho(V)\rho(W)$ for all $V,W \in SU(2)$. This is seen by a simple calculation,
\begin{align*}
 \big( \rho(V) \rho(W) \psi \big)(x) &= \rho(V) \big( W \psi (R_W^{-1} x) \big) \\
                                     &= V W \psi(R_W^{-1} R_V^{-1} x) \\
                                     &= V W \psi \big( (R_V R_W)^{-1} x \big) \\
                                     &= V W \psi (R_{VW}^{-1} x) \\
                                     &= \big( \rho(VW) \psi \big)(x) ,
\end{align*}
where I used the result from question d. From this it follows that $\rho(V)$ i bijective: The inverse must simply be $\rho(V)^{-1} = \rho(V^{-1})$, which is easily proven by calculation,
\begin{equation*}
 \rho(V) \rho(V^{-1}) = \rho(V V^{-1}) 
                      = \rho(\text{id}_{SU(2)})
                      = \text{id}_{\mathcal{B}(L^2(\mathbb{R}^3, \mathbb{C}^2))} ,
\end{equation*}
where the last step follows from the fact that representations map the identity element from the domain to the identity element in the codomain. 

So $\rho(V)$ is indeed a unitary representation.




%First, I want to show that the mapping is a representation, which means that $\rho(VW)=\rho(V)\rho(W)$ for all $V,W \in SU(2)$ (from Definition 6.5). This is seen by a simple calculation,
%\begin{align*}
% \big( \rho(V) \rho(W) \psi \big)(x) &= \rho(V) \big( W \psi (R_W^{-1} x) \big) \\
%                                     &= V W \psi(R_W^{-1} R_V^{-1} x) \\
%                                     &= V W \psi \big( (R_V R_W)^{-1} x \big) \\
%                                     &= V W \psi (R_{VW}^{-1} x) \\
%                                     &= \big( \rho(VW) \psi \big)(x) ,
%\end{align*}
%where I used the result from question d. 
%
%For the mapping to be unitary, it has to be bijective, linear and norm preserving. First of all, that the mapping is bijective follows directly from the fact that it is a representation: The inverse must simply be $\rho(V)^{-1} = \rho(V^{-1})$, which is easily proven by calculation,
%\begin{equation*}
% \rho(V) \rho(V^{-1}) = \rho(V V^{-1}) 
%                      = \rho(\text{id}_{SU(2)})
%                      = \text{id}_{\mathcal{B}(L^2(\mathbb{R}^3, \mathbb{C}^2))} ,
%\end{equation*}
%where the last step follows from the fact that representations map the identity element from the domain to the identity element in the codomain.\fixme{Reference?}
%
%The mapping is seen to be linear by the calculation
%\begin{align*}
% \big( \rho(V)(\alpha \psi_1 + \beta \psi_2) \big)(x) &= V(\alpha \psi_1 + \beta \psi_2)(R_V^{-1} x) \\
%                                                      &= V \big( \alpha \psi_1 (R_V^{-1} x) + \beta \psi_2 (R_V^{-1} x) \big) \\
%                                                      &= \alpha V \psi_1 (R_V^{-1} x) + \beta V \psi_2 (R_V^{-1} x) \\
%                                                      &= \alpha \big( \rho(V) \psi_1 \big)(x) + \beta \big( \rho(V) \psi_2 \big)(x) ,
%\end{align*}
%by the linearity of $V$ (because $V\in SU(2)$ it is unitary, and hence linear according to Theorem 5.1). 
%
%Lastly, the mapping $\rho(V)$ is seen to be norm preserving by
%\begin{align*}
% || \rho(V) \psi ||^2 = \int |V \psi (R_V^{-1} x)|^2 \,\di x
%                      = \int |\psi (R_V^{-1} x)|^2 \,\di x
%                      = \int |\psi (x)|^2 \,\di x
%                      = || \psi ||^2 ,
%\end{align*}
%because $V$ is unitary (and hence norm preserving according to Theorem 5.1) and $R_V$ is an isometry (and hence does not change the integral because it is simply a rotation or mirroring), concluding that the mapping $\rho(V)$ is a unitary representation. 






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