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\begin{document}
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\noindent {\bf Assignment 1}\\

\noindent DUE: Tuesday, 2/7/2023, work individually\\[-0.2in]

\begin{enumerate}
\item Derive Eq.~(1.38) in two ways: (a) Using matrix multiplication from the
definitions of $\Xb$ and $\Zb$, and (b) Algebraically, that is, not using matrix multiplication,
only the relations $\Zb = \nbt-\nb$, $\Xb\nb = \nbt\Xb$, and $\Xb\nbt = \nb\Xb$.
 
\item Beginning with the form for ${\Sb}_{ij}$ in Eq.~(1.35), prove,
that $\Sb_{ij}^2 = 1$. Do {\it not} use the matrix form for ${\Sb}_{ij}$.
Rather, derive this algebraically using only the properties of $\tilde{\nb}$,
${\nb}$, and ${\Xb}$ as in Eqs.~(1.33) and (1.34), and the 
fact that ${\Xb}^2 = 1$. Remember that
as the ${\nb}$'s and ${\Xb}$ are all single-Cbit operators, and by virtue of
the positional notation (the $i$'s and $j$'s), those with
different indices {\it commute} in tensor products: That is, ${\Xb}_i{\Xb}_j
= {\Xb}_i\otimes{\Xb}_j = {\Xb}_j\otimes{\Xb}_i = {\Xb}_j{\Xb}_i$,
and similarly, ${\Xb}_i{\nb}_j = {\nb}_j{\Xb}_i$, etc. On the other hand,
those with the same index do {\it not} necessarily commute, e.g., ${\nb}_i{\Xb}_i
= {\Xb}_i\tilde{\nb}_i$, as dictated by Eq.~(1.34).

\item Along similar lines to problem (1), prove the identity
${\Sb}_{ij} = {\Cb}_{ij}{\Cb}_{ji}{\Cb}_{ij}$ (Eq.~(1.23)) in two ways:
\begin{enumerate}
\item Straightforwardly, using the matrix representations of ${\Sb}_{ij}$ and
${\Cb}_{ij}$.
\item Using only using the 
identity given in Eq.~(1.36), i.e., ${\Cb}_{ij} = \tilde{\nb}_i+{\Xb}_j{\nb}_i$ and the properties
of the ${\nb}$'s and ${\Xb}$ as in the previous problem. Follow the hint given
by Mermin on page 13: Substitute the formula for ${\Cb}_{ij}$ given
in Eq.~(1.36) in Eq.~(1.23). You should find that of the resulting 8 terms,
4 of the terms vanish, and the remaining form yield the formula for ${\Sb}_{ij}$
given in Eq.~(1.35).
\end{enumerate}
 [Full disclosure: In this case, you will certainly find the technique
of part (a) to be easier! But the technique of part (b) will help you understand
the notation far more deeply, and provides 
valuable algebraic practice with tools that are more easily generalized to larger
matrices. Similar comment for problem (2): deriving the formula directly
from matrix multiplication of the 4$\times$4 matrices is far easier.]

\item
Write down the explicit $4 \times 4$ matrix 
 for the tensor product
of the Hadamard operator with itself, i.e.,
${\Hb}\otimes{\Hb}$.
Tensor products of operators {\it as matrices}
is discussed in lecture, although it is not spelled out in the book. To build a tensor product of two $2\times 2$ matrices, in general, you multiply each element of the first matrix times the {\it entire} second matrix, which gives four $2\times 2$ blocks which, when stacked together as bricks, form a $4 \times 4$ matrix. Here's what I mean (for {\it any} $a, b, c, d, e, f, g, h$):
\begin{eqnarray*}
\left[\begin{array}{cc}
a & b   \\
c & d   \\
\end{array}\right]\;
\otimes
\left[\begin{array}{cc}
e & f   \\
g & h   \\
\end{array}\right]\;
= 
\left[\begin{array}{cc}
a
\left[\begin{array}{cc}
e & f   \\
g & h   \\
\end{array}\right]\;
&
b
\left[\begin{array}{cc}
e & f   \\
g & h   \\
\end{array}\right]\;\\ \\
c
\left[\begin{array}{cc}
e & f   \\
g & h   \\
\end{array}\right]\;
&
d
\left[\begin{array}{cc}
e & f   \\
g & h   \\
\end{array}\right]\;\\
\end{array}\right]\;
= 
 \left[\begin{array}{cccc}
ae & af &be &bf  \\
ag & ah &bg &bh  \\
ce & cf &de &df  \\
cg & ch &dg &dh  \\
\end{array}\right]\;
\end{eqnarray*}
Simply apply this formula to two $\Hb$'s; you should get 4 rows, all consisting only of 1's and $-1$'s. 
Also note there is a normalizing factor of $\frac{1}{\sqrt{2}}$ in $\Hb$. Be sure to work out the corresponding
 normalizing factor in $\Hb\otimes \Hb$.

Once you have an exact form for $\Hb\otimes \Hb$, show 
that each row and column is orthogonal. (This is a nice sanity check that your answer is correct. In fact, observe that is the case for $\Hb$ itself.)


\item Using matrices in a straightforward manner, and
the result of the previous problem, verify the relation ${\Hb}_i{\Hb}_j{\Cb}_{ij}{\Hb}_i{\Hb}_j = {\Cb}_{ji}$. Note that here ${\Hb}_i{\Hb}_j$ does not mean the product of two matrices, but is rather a shorthand for the tensor product ${\Hb}_i\otimes{\Hb}_j$, as worked out in problem 4. So I am asking you to compute the product of three $4 \times 4$ matrices.



\end{enumerate}




\end{document}
