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\begin{document}
\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#5, Rev. B, due Monday, 2022-02-21 }
\end{center}

\vspace{.3 in}

{\bf 1. The good, \ldots}.

{\bf \quad a)} The \href{https://en.wikipedia.org/wiki/Commutator#Ring_theory}{\it commutator} $[A,B]$ of two matrices $A$ and $B$ is defined as the matrix $[A,B]\equiv AB-BA$.  Calculate $[\sigma_{j},\sigma_{k}]$ for the \href{https://en.wikipedia.org/wiki/pauli_matrices}{\it Pauli matrices} $\sigma_{i}$, $i=x,y,z$.  Do they commute? Compare with H03\#0: $\sigma_{j}\sigma_{k}=I\sigma_{j}\cdot\sigma_{k} + i\sigma_{j}\times\sigma_{k}$, treating the Pauli matrices $\sigma_{k}\sim \bh e_{k}$ as unit vectors.

{\bf \quad b)} Show that the commutator of $A$ and $B$ is antisymmetric: $[A,B]=-[B,A]$, and bilinear: $[\alpha_{i}A_{i},B_{j}\beta_{j}]=\alpha_{i}[A_{i},B_{j}]\beta_{j}$. Thus it is a `matrix version' of the vector cross product, so $[A,A]=0$, and two matrices which commute are in the `same direction' (both are diagonal in some basis).

{\bf \quad c)} Verify that the two matrices %
$A=\scriptsize{\pmatrix{ %
    \phantom{-}9 & -2 & -6 \cr %
    -2 & \phantom{-}9 & -6 \cr %
    -6 & -6 & -7 }}$ %
and %
$B=\scriptsize{\pmatrix{ %
    54 & \phantom{-}10 & -3 \cr %
    10 & -45 & 30 \cr %
    -3 & \phantom{-}30 & 46 }}$.\\ %
commute.  Simultaneously diagonalize both matrices.  Are the eigenvectors
orthogonal?\\ {\it Hint:} \href{https://www.gnu.org/software/octave}{Octave}
or \href{http://www.wolfram.com/solutions/education/students}{Mathematica} is
your friend!

{\bf \quad d)} By adding and subtracting $M=H+iK$ and its adjoint $M\ad=H-iK$, where $H\ad=H$ and $K\ad=K$ are Hermitian (note that $A=iK$ is antisymmetric: $A\ad=-A$), show that any matrix $M$ can be decomposed into its symmetric and antisymmetric parts, similar to how $z=x+iy$ projects into its real and imaginary parts.  

{\bf \quad e)} A \href{https://en.wikipedia.org/wiki/Normal_matrix}{\it normal} matrix is one that commutes with its adjoint: $[N,N\ad]=0$.  Show that $[N,N\ad]=2i[H,K]$, where $N=H+iK$.  This shows that a normal matrix is one whose symmetric and antisymmetric parts commute.  Thus normal matrices are the most general matrices that can be diagonalized with orthogonal eigenvectors.  The eigenvalues of $N$ are $n_{j}=h_{j}+ik_{j}$ where $h_{j}$ and $k_{j}$ are the eigenvalues of $H$ and $K$.  Thus $N$ acts like independent complex numbers!

{\bf \quad f)} [bonus: In analogy with polar complex numbers $z=e^{\sigma+i\phi}=\rho u$, where $\rho=e^{\sigma}>0$ and $u=e^{i\phi}$ is a $unit$: $|u|^{2}=u^{*}u=1$, any matrix $A$ has a \href{https://en.wikipedia.org/wiki/Polar_decomposition#Matrix_polar_decomposition}{\it polar decomposition} $A=RS$ into a rotation $R$ and a stretch $S$, which are the building blocks of all linear operators.  By taking the exponential $M=e^{N}$ of $N=H+iK$, show that any normal matrix can be decomposed into $M=UP$, where $U$ is unitary ($U\ad U=I$) and $P$ is positive definite ($P\ad =P$ with non-negative eigenvalues).]

{\bf \quad g)} [bonus: Using the polar decomposition $A=RS$, where $R\ad R=I$ and $S\ad=S$, and diagonalizing $S=VWV\ad$, show that any matrix may be decomposed $A=UWV\ad$, where $U=RV$ and $V$ are unitary, and $W$ is diagonal.  This \href{https://en.wikipedia.org/wiki/Singular_value_decomposition}{\it singular value decomposition} of matrices is the generalization the spectral theorem for all matrices.  Because it involves both stretches and rotations, every {\it singular value} (eigenvalue) $w_{i}$ has two eigenvectors: a {\it left singular vector} $\vec u_{i}$ and a {\it right singular vector} $\vec v_{i}$.  This is useful for transformations from one vector space into another, not just operators.]


{\bf 2.  The bad, \ldots} 

{\bf \quad a)} The Pauli \href{https://en.wikipedia.org/wiki/Ladder_operator}{\it ladder operators} $\sigma_{\pm}=(\sigma_{x}\pm i\sigma_{y})/2$ transform the unit vectors $(1,0)^{T}$ and $(0,1)^{T}$ from one to the next along a chain, {\it annihilating} the last one.  Calculate $\sigma_{\pm}$, $\sigma_{\pm}^{2}$, $\sigma_{\pm}^{3}, \ldots, \sigma^{n}$.  How does this generalize to higher dimension?

{\bf \quad b)}Calculate the doubly {\it degenerate} eigenvalue $\lambda$ of $\sigma_{\pm}$ and show that they have only one eigenvector $\vec v_{1}$.  The other vectors associated with $\lambda$ are called {\it generalized eigenvectors}, defined by $M\vec v_{k}=\vec v_{k}\lambda_{k}+\vec v_{k-1}$. Note that $\vec v_{1}$ follows the pattern if we define $\vec v_{0}=\vec 0$.  Find the matrix of $M$ in the basis $\{\vec v_{k}\}$ (called a \href{https://en.wikipedia.org/wiki/Jordan_matrix}{Jordan block}).

{\bf \quad c)} Calculate the adjoint $\sigma_{\pm}^{\ad}$, the \href{https://en.wikipedia.org/wiki/Projection_(linear_algebra)}{\it projections} $P_{+}=\sigma_{+}\sigma_{-}$ and $P_{-}=\sigma_{-}\sigma_{+}$, and the commutators $[\sigma_{+},\sigma_{-}]$, $[I,\sigma_{\pm}]$, and $[\sigma_{z},\sigma_{\pm}]$.  Note that $\sigma_{\pm}$ commutes with $I$ because its eigenvalues are all the same, but not with $\sigma_{z}$ with distinct eigenvalues.  This is true in general.  [bonus: use the last commutation relation to show that if $\vec v$ is an eigenvector of $\sigma_{z}$, the $\sigma_{\pm}\vec v$ is also an eigenvector.  Thus $\sigma_{\pm}$ raise an lower one one eigenvector of $\sigma_{z}$ into another, like climbing up or down the rungs of a ladder, which is where their name comes from. ]

%NOT TRUE! {\bf \quad b)} Let the operators $Z_{1}$ and $Z_{2}$ be {\it nilpotent}: $Z_{i}^{n_{i}}$ for some $n_{i}\in \blackboard{N}$.  Use the binomial expansion to show that the sum $Z=Z_{1}+Z_{2}$, is also nilpotent: $Z^{n}=0$ where $n=n_{1}+n_{2}$.  Thus the set of nilpotent operators form a subspace.

{\bf \quad d)} Show that $\sigma_{x}\sigma_{\pm}\sigma_{x}^{-1}=\sigma_{\mp}$  Find the most general $2\times2$ {\it nilpotent} operator via the similarity transform $V\sigma_{+}V^{-1}$ of $\sigma_{+}$ into a generic basis $V=\pmatrix{a & b \cr c & d}$.  Represent it as an outer product.  What is its geometrical interpretation?  Do the same for a generic {\it diagonalizable} operator to represent it as the sum of two outer products (its {\it spectral decomposition}).

{\bf \quad e)} Find the matrix of the derivative operator $D=d/dx$ in the linear space of polynomials $A(x)=a_{0}+a_{1}x+a_{x}x^{2}+\ldots +a_{n}x^{n}$, using the monomials $\bh e_{k} = x^{k}$ as basis functions.  Show that $D^{n}=0$ and, thus the derivative is nilpotent.

% {\bf 3. \ldots and the gnarly.}

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