%xxx ``states'' vs. [???] --- make the pre-observation and post-observation
%vocabulary clear and consistent.

%xxx change $(.,.)$ to bra-ket notation throughout.

%% ================================================================
\section{Introduction}

This is a reference sheet, vocabulary sheet, and to-do list for my spring 2008
independent study course under Janek Wehr on quantum networking theory.  For
notation and terminology, please see \cite{bib:kerlprob} (probability) and
\cite{bib:kerlqm} (basic quantum mechanics).

%% ================================================================
\section{Single qubits}

%% ----------------------------------------------------------------
\subsection{Single-qubit state space}
$$
	\ket 0 = \colvectwo{1}{0}, \qquad
	\ket 1 = \colvectwo{0}{1}
$$

\bfidx{two-level quantum system}.  Make the distinction between physical
implementations (which vary) and the mathematical abstraction (which is
clean and simple).

\bfidx{Bloch sphere}.  Include figure ...

\bfidx{measurements}.  Two uses:  input preparation, and output assessment.

Need algebra of matrices to invent ``really good measurements''.

%% ----------------------------------------------------------------
\subsection{Single-qubit operators}

Observable: self-adjoint operator on the state space.

Measurement:  the observable's spectral projector.

Work out some examples.

%% ----------------------------------------------------------------
\subsection{Lack of time evolution for quantum computation}
\label{sec:lack_of_time_evo_for_QC}

\bfidxtwo{Time evolution}{time evolution, lack of} is generated by the
\bfidx{Hamiltonian}.  Here the Hamiltonian is zero: $e^{iHt} = I$.  Or
eigenstate of the Hamiltonian.

Cite \cite{bib:kerlqm}.

\bfidx{physics kitchen} \ldots type up notes.

%% ================================================================
\newpage
\section{Multiple qubits}

%% ----------------------------------------------------------------
\subsection{Multiple-qubit states}

linear combinations

binary/unary diagram --- a key concept (for me anyway).

\bfidx{entanglement} as indecomposability of tensor.

\bfidx{EPR pair ...}

measurements of entanglement

\bfidx{maximally entangled}

\bfidx{Bell state}

\bfidx{singlet conversion}

\bfidx{singlet conversion probability}

\bfidx{pure states} and
\bfidx{mixed states}.  \bfidx{density operator}.

emph distinction between pure state, mixed state, and entangled state.

$$
	\rho = \sum_j p_j \ket{\phi_j} \bra{\phi_j}
$$
with $sum_j p_j = 1$.
Note that $\rho$ is said to be a \bfidx{trace-class operator}. [xxx def
and xref.  This is a trivial distinction in finite dimensions.]
[xxx $p_j = \braket{\psi_j}{\psi_j} = \|\psi_j\|^2$?]

Pure state:  rank-one projection.  Iff $\rho^2=\rho$.

$$
	\expn{A} = \sum_j p_j \braCket{\phi_j}{A}{\phi_j} = \tr[\rho A].
$$

\bfidx{spectral resolution} (when does it exist?)
\begin{align*}
	P_i &:= \ket{a_i}\bra{a_i} \\
	A   & = \sum_i a_i P_i = \sum_i a_i \ket{a_i}\bra{a_i}
\end{align*}

%% ----------------------------------------------------------------
\subsection{Multiple-qubit operators}

%% ================================================================
\newpage
\section{Entanglement}

\begin{defn}
A \bfidx{separable state} is a decomposable tensor; a \bfidx{entangled state}
is an indecomposable tensor.
\end{defn}

[xxx rank notion, and examples]

[xxx triviality of determining entanglement in the finite-dimensional case.
Only for rank-two tensors?  More difficult for higher-rank tensor products?]

%% ================================================================
\newpage
\section{Quantum teleportation}

\bfidx{teleportation}.  Alice and Bob.  Draw up the figure.

Cite result \cite{bib:nc}:  this can be done with maximum entanglement.
It cannot be done ($p<1$) without maximum entanglement.  Include the proof.

%% ================================================================
\newpage
\section{Entanglement swapping}

Open question:  how to do entanglement swapping when input states are mixed.

%% ----------------------------------------------------------------
\subsection{Lattices}

\bfidx{lattice}

%% ----------------------------------------------------------------
\subsection{Triangles}

draw the figure.

Such a thing can be constructed.
\bfidx{star-triangle formulation}.  Cite.

Explicitly write down the preparation matrix.

Open question:  to what use can such a thing be put?

%% ================================================================
\appendix
\newpage
\section{Tensor products and Kronecker notation}
\label{appx:tensors}

Keep it brief and practical.  State (with don't-be-scared caveat) the
abstract-algebra definition.  Then immediately give examples showing that these
are just pairs (or $n$-tuples) with the manipulation rules (which is the
practical meaning of the equivalence relations) of scalar-through and
multilinearity.  These give different ways to write the same thing, and give us
some flexibility for computations.

%Tensor product of functions in general:  given
%$$
%	f_1, \ldots, f_n : X \to k
%$$
%where $k$ is a field,
%$$
%	(f_1 \tct f_n)(\vecx_1, \ldots, \vecx_n) :=
%	f_1(\vecx_1) \cdot \cdots \cdot f_n(\vecx_n).
%$$
%
%Maybe talk about vectors and covectors (and how they consume one another
%by duality) but \emph{only} if necessary for the discussion in this paper.

Lift stuff from prolrev.

\bfidx{decomposable} tensors.  xref both ways between this and entangled
states in QM.

%% ----------------------------------------------------------------
\subsection{Tensor products and array notation}

Tableaux for tensor product of vectors.  Lift from prolrev.

%% ----------------------------------------------------------------
\subsection{Kronecker notation}

Work out a quick $2 \times 2$ example of why this is the right thing to do.

%% ----------------------------------------------------------------
\subsection{Linear operators on tensor products}

$$
	(A \tensor B)(\vecu \tensor \vecv) := (A\vecu) \tensor (B\vecv)
$$

work it out by expansion of eigenbases.  Use $\vecu$ and $\vecv$ for
this appx; also be sure to use $u_i$ and $v_j$ for coeffs.

\bfidx{Kronecker product}

xxx:
$$
	\begin{matrix}
		  &   &  \\
		  &   &  \\
		i & \Bigg\downarrow &
		\begin{pmatrix}
			(A\vecu)_1 \\
			(A\vecu)_2 \\
			(A\vecu)_3 \\
		\end{pmatrix}
	\end{matrix}
	\quad
	=
	\quad
	\begin{matrix}
		  &   & k \\
		  &   & \longrightarrow \\
		i & \Bigg\downarrow &
		\begin{pmatrix}
			A_{11} & A_{12} & A_{13} \\
			A_{21} & A_{22} & A_{23} \\
			A_{31} & A_{32} & A_{33} \\
		\end{pmatrix}
	\end{matrix}
	\quad
	\begin{matrix}
		  &   &  \\
		  &   &  \\
		k & \Bigg\downarrow &
		\begin{pmatrix}
			u_1 \\
			u_2 \\
			u_3 \\
		\end{pmatrix}
	\end{matrix}
$$

Mnemonic: adjacency of coefficients in 2nd argument of vkron, so B is blocked
tighter.

$$
\begin{pmatrix}
	A_{11}B_{11} & A_{11}B_{12} & \vline &A_{12}B_{11} & A_{12}B_{12} \\
	A_{11}B_{21} & A_{11}B_{22} & \vline &A_{12}B_{21} & A_{12}B_{22} \\
	\hline
	A_{21}B_{11} & A_{21}B_{12} & \vline &A_{22}B_{11} & A_{22}B_{12} \\
	A_{21}B_{21} & A_{21}B_{22} & \vline &A_{22}B_{21} & A_{22}B_{22} \\
\end{pmatrix}
\begin{pmatrix}
	u_1 v_1 \\
	u_1 v_2 \\
	\hline
	u_2 v_1 \\
	u_2 v_2 \\
\end{pmatrix}
$$

$$
\begin{pmatrix}
	A_{11} B & \vline & A_{12} B \\
	\hline
	A_{21} B & \vline & A_{22} B \\
\end{pmatrix}
\begin{pmatrix}
	u_1 \vecv \\
	\hline
	u_2 \vecv \\
\end{pmatrix}
$$

%% ----------------------------------------------------------------
\subsection{Inner products on tensor products}

xxx present this as a special case of the above:
$$
	\langle (\veca\tensor\vecb), (\vecc\tensor\vecd) \rangle
	=
	\langle \veca,\vecc \rangle
	\langle \vecb,\vecd \rangle.
$$

%% ----------------------------------------------------------------
\subsection{Norms on tensor products}

Prove (easily)
$$
	\|\vecu\tensor\vecv\| = \|\vecu\| \; \|\vecv\|
$$
as a consequence of the above.

%%% ----------------------------------------------------------------
%\subsection{To be reorganized}
%
%% Tensor products of finite-dimensional vector spaces
%
%Do this just as a warm-up and intuition builder for xxx.
%
%Oddly and only to ease the transition below, think of
%$$
%	V = L^2(\{1, 2, \ldots, N\}).
%$$
%For each index $j=1,2, \ldots, N$, we have a complex number at that index.  The
%square-summability condition removes nothing, since all the sums are finite.
%
%Then as is well-known (one hopes!) from abstract algebra, if $V$
%has basis $\{\vece_j : j = 1, 2, \ldots, N\}$, then
%$V \tensor V$
%has basis
%$$
%	\{\vece_j \tensor \vece_k : j, k = 1, 2, \ldots, N\}.
%$$
%This basis has $N^2$ elements, and the $n$th tensor power of $V$ with
%itself has dimension $N^n$.  I hope this makes it palatable that
%we can think of
%$$
%	V\tensor V = L^2(\{1, 2, \ldots, N\}^2).
%$$
%and in general
%$$
%	V^{\tensor n} = L^2(\{1, 2, \ldots, N\}^n).
%$$
%
%For $F\in V\tensor V$,
%$$
%	F = \sum_{j=1}^N \sum_{k=1}^N c_{jk} \vece_j \tensor \vece_k;
%$$
%for $F\in V^{\tensor n}$,
%$$
%	F = \sum_{r_1=1}^N \cdots \sum_{r_n=1}^N
%		c_{r_1, \ldots, r_n}
%		\vece_{r_1} \tensor \cdots \tensor \vece_{r_n}.
%$$
%
%Something about selection of coordinates \ldots what is the appropriate notion
%of inner product?  It's done by product of inner products on slots.  Write down
%the formula and maybe an example; prove what is reasonable to prove.


%% ================================================================
\section{Trace and partial trace}

do it in the kronecker rep too.  $2 \times 2$ on $A$ and $B$.

type up the handwritten notes.

I think left-partial is trace within blocks; right partial is sum of blocks.

Is this just
$$
	\tr_{\textrm{left}}
	\pcmatrix{
		A & \vline & B \\
		\hline
		C & \vline & D \\
	}
	\quad = \quad
	\pcmatrix{
		\tr(A) & 0 \\
		0 & \tr(D) \\
	}
$$
and
$$
	\tr_{\textrm{right}}
	\pcmatrix{
		A & \vline & B \\
		\hline
		C & \vline & D \\
	}
	\quad = \quad
	A+D?
$$
Work out some computational examples.  And of course code it up. \texttt{:)}

% [ A | B ]
% [---+---]
% [ C | D ]

% left partial trace: 
%   [ trA |  0  ]
%   [-----+-----]
%   [  0  | trD ]
% right partial trace: A+D

%% ================================================================
\section{Density operators}

\begin{defn}
An \bfidx{ensemble} is a list of $n$ state vectors $\{\phi_1, \ldots, \phi_n\}$
along with respective probabilities $\{p_1, \ldots, p_n\}$ with $0 \le p_i \le
1$ and $\sum_{i=1}^n p_i = 1$.
\end{defn}

One may think of an ensemble as a probability mass function.

\begin{defn}
A \bfidx{density matrix} is a positive-definite matrix with trace 1.
\end{defn}

\begin{prop}
\label{prop:rho_from_ens}
An $n \times n$ density matrix may always be obtained from an ensemble by
$$
	\rho = \sum_{i=1}^n p_i \ket{\phi_i}\bra{\phi_i}.
$$
\begin{proof}
xxx type it up.
\end{proof}
\end{prop}

\begin{prop}
\label{prop:rho_from_psi}
An $n \times n$ density matrix may be obtained from a state $\psi$
and a basis $\{\phi_1, \ldots, \phi_n\}$ by
$$
	\rho = \sum_{i=1}^n p_i \ket{\phi_i}\bra{\phi_i}
$$
where
$$
	p_i = |\braket{\psi}{\phi_i}|^2.
$$
\begin{proof}
xxx type it up.
\end{proof}
\end{prop}

\begin{rem}
Two different ensembles can give the same density matrix.  N\&C give
a theorem characterizing the conditions under which this can happen.
[xxx type this up and include examples.]
\end{rem}

\begin{defn}
A state $\psi$ is said to be a \bfidx{pure state} with respect to an ensemble
$\{\phi_1, \ldots, \phi_n\}$ if it is equal to one of the $\phi_i$'s.
Otherwise it is said to be a \bfidx{mixed state}.
\end{defn}

\begin{rem}
Pure and mixed states have no meaning except with respect to a specified
ensemble.  A state that is pure with respect to one ensemble may be mixed with
respect to another.
\end{rem}

\begin{prop}
A density matrix $\rho$ has $\tr(\rho)=1$.  Furthermore, if $\rho$ is a density
matrix for a state $\psi$ and a basis $\{\phi_1, \ldots, \phi_n\}$ (as in the
statement of proposition \ref{prop:rho_from_psi}), then $\tr(\rho^2) \le 1$,
with equality if and only if $\psi$ is pure with respect to the basis.
\begin{proof}
For the duration of this proof, index the basis vectors as
$$
	\{\phi^{(1)}, \ldots, \phi^{(n)}\}.
$$
Then subscripts will denote elements of a vector.  For example, if
$\phi^{(1)}=(0.6, 0.8)$, then $\phi^{(1)}_2 = 0.8$.

By proposition \ref{prop:rho_from_ens}, there is an ensemble such that
$$
	\rho = \sum_{k=1}^n
	p_k \ket{\phi^{(k)}} \bra{\phi^{(k)}}.
$$
Now, each $\ket{\phi^{(k)}} \bra{\phi^{(k)}}$ is an outer-product matrix
with $ij$th entry equal to
$$
	(\ket{\phi^{(k)}} \bra{\phi^{(k)}})_{ij} = \sum_{k=1}^n
		\phi^{(k)}_i
		\phi^{(k)*}_j.
$$
Since the trace of an $n \times n$ matrix $A$ is $\tr(A) = \sum_{i=1}^n
A_{ii}$, we have
\begin{align*}
	\tr(\rho) &= \sum_{i=1}^n \rho_{ii} \\
	&=
		\sum_{i=1}^n
		\sum_{k=1}^n
		p_k
		\phi^{(k)}_i
		\phi^{(k)*}_i
		\\
	&=
		\sum_{k=1}^n
		p_k
		\sum_{i=1}^n
		|\phi^{(k)}_i|^2
		\\
	&=
		\sum_{k=1}^n
		p_k
		\|\phi^{(k)}\|
		\\
	&=
		\sum_{k=1}^n
		p_k
		\\
	&= 1.
\end{align*}

For the second claim, first note that for an $n \times n$
matrix $A$, we have
$$
	\tr(A^2) =
		\sum_{i=1}^n
		\sum_{j=1}^n
		A_{ij}
		A_{ji}.
$$
Then
\begin{align*}
	\tr(\rho^2) &=
		\sum_{i=1}^n
		\sum_{j=1}^n
		\rho_{ij}
		\rho_{ji} \\
	&=
		\sum_{i=1}^n
		\sum_{j=1}^n
		\left(
		\sum_{k=1}^n p_k
			\phi^{(k)}_i
			\phi^{(k)*}_j
		\right)
		\left(
		\sum_{\ell=1}^n p_\ell
			\phi^{(\ell)}_j
			\phi^{(\ell)*}_i
		\right)
		\\
	&=
		\sum_{i=1}^n
		\sum_{j=1}^n
		\sum_{k=1}^n
		\sum_{\ell=1}^n
		p_k
		p_\ell
		\phi^{(k)}_i
		\phi^{(k)*}_j
		\phi^{(\ell)}_j
		\phi^{(\ell)*}_i \\
	&=
		\sum_{k=1}^n
		p_k
		\sum_{\ell=1}^n
		p_\ell
		\sum_{i=1}^n
		\phi^{(k)}_i
		\phi^{(\ell)*}_i
		\sum_{j=1}^n
		\phi^{(\ell)}_j
		\phi^{(k)*}_j \\
	&=
		\sum_{k=1}^n
		p_k
		\sum_{\ell=1}^n
		p_\ell
		\braket{\phi^{(k)}}{\phi^{(\ell)}}
		\braket{\phi^{(\ell)}}{\phi^{(k)}} \\
	&=
		\sum_{k=1}^n
		p_k
		\sum_{\ell=1}^n
		p_\ell
		|\braket{\phi^{(k)}}{\phi^{(\ell)}}|^2.
\end{align*}
Now, since I assume the basis is orthonormal, we have
\begin{align*}
	\tr(\rho^2) &= 
		\sum_{k=1}^n
		p_k
		\sum_{\ell=1}^n
		p_\ell
		\delta_{k\ell}
	= \sum_{k=1}^n p_k^2.
\end{align*}

To finish the proof, note that $\sum_{k=1}^n p_k^2$ is the diagonal part of
$(\sum_{k=1}^n p_k)^2$.  We have
\begin{align*}
	1 = (1)^2 &= \left(\sum_{k=1}^n p_k\right)^2 \\
	&=
		\sum_{k=1}^n
		\sum_{\ell=1}^n
		p_k
		p_\ell \\
	&=
		\sum_{k=1}^n
		\sum_{\ell \ne k}
		p_k
		p_\ell
		+
		\sum_{k=1}^n
		p_k^2.
\end{align*}
If only one $p_k$ is 1, then we clearly have equality.  Note that all the terms
in the sum are non-negative.  If two $p_k$'s are non-zero (say, $p_1$ and
$p_2$) then the off-diagonal term is non-zero, so the diagonal sum is less than
1.
\end{proof}
\end{prop}

\begin{rem}
The trace $\tr(\rho)$ is basis-independent; $\tr(\rho^2)$ is basis-dependent.
For example, let
$$
	\psi=\colvectwo{1}{0}.
$$
If
$$
	\phi_1 = \colvectwo{1}{0} \qand
	\phi_2 = \colvectwo{0}{1},
$$
then
$$
	\rho=\pcmatrix{
		1 & 0 \\
		0 & 0
	},
	\quad
	\rho^2=\pcmatrix{
		1 & 0 \\
		0 & 0
	},
	\quad
	\tr(\rho) = 1,
	\qand
	\tr(\rho^2) = 1.
$$
On the other hand, if
$$
	\phi_1 = \colvectwo{1/\sqrt{2}}{ 1/\sqrt{2}} \qand
	\phi_2 = \colvectwo{1/\sqrt{2}}{-1/\sqrt{2}},
$$
then
$$
	\rho=\pcmatrix{
		1/2 & 0 \\
		0 & 1/2
	},
	\quad
	\rho^2=\pcmatrix{
		1/4 & 0 \\
		0 & 1/4
	},
	\quad
	\tr(\rho) = 1,
	\qand
	\tr(\rho^2) = 0.5.
$$
\end{rem}

\begin{rem}
In the finite-dimensional case, determining purity of a state is trivial: form
the density matrix $\rho$ and compute $\tr(\rho^2)$.  This is 1 iff $\psi$ is
pure.
\end{rem}
