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\noindent\bf
Chapter 2  Quantum Mechanics
\rm

\medskip
Quantum Information Science uses quantum properties to help store,
process, and transmit information. In this chapter, we describe some
basic background on quantum mechanics. We first use vector states to
describe quantum systems. Then we demonstrate the formulation using
density matrices.

\medskip\noindent
\bf Copenhagen interpretation \rm

\begin{itemize}
\item[A1]
A vector state $|x\rangle$
is a unit vector in a Hilbert space $\cH$ (usually 
$\IC^n$). Linear combinations (superposition) of the physical states
are allowed in the state space.

\item[A2]
Every physical quantity (observable) corresponds to a 
Hermitian operator (matrix) $A$.
Suppose a state $|x\rangle = c_1|u_1\rangle + c_2 |u_2\rangle$ 
such that $A|u_i\ra = a_i |u_i\ra$ for $i \in \{1,2\}$.
Then applying a measurement of $|x\ra$
corresponding to $A$ will cause
the {\bf wave function} (that describes the quantum state)
to {\bf collapse} to $|u_1\ra$ or $|u_2\ra$ 
with probability of $|c_1|^2$ and $|c_2|^2$, respectively.
Here $c_1, c_2$ are called the probability amplitude of the state $|x\ra$.

\item[A3] The time dependence of a state is governed by the
Schr\"{o}dinger equation
$$i \ph {\partial |x\ra \over \partial t} = H |x\ra,$$
where $\ph$ is the Planck constant with 
$$\ph = 6.62607004 × 10^{-34} m^2 kg/s,$$ 
and $H$ is a Hermitian operator (matrix)
corresponding to the energy of the system known as the Hamiltonian.
\end{itemize}

\newpage
\medskip\noindent
\bf Remarks \rm
\begin{enumerate}
\item The phase of the state does not matter, i.e.,
$|x\ra$ and $e^{i\alpha}|x\ra$ represents the same states.

\item In the finite dimensional case, 
if the state and the observable are represented by
$$|x\ra = \sum_{j=1}^n c_j |u_j\ra \in \IC^n \quad \hbox{ and } \quad
A = \sum_{j=1}^n \lambda_j |u_j\ra \la u_j| = \sum_{j=1}^n \lambda_j P_j,$$
then the projective measurement of the state is
$$\la x|A| x\ra = \sum_{j=1}^n \lambda_j |c_j|^2.$$
Once the measurement is applied, the state becomes (collapses to) 
$${P_i|x\ra \over \sqrt{\la x| P_i |x\ra}} = {P_i |x\ra \over |c_i|}.$$
%with a probabilty of $|c_i|^2$.

\item In the Schr\"{o}dinger equation, if 
$H(t)$ does not depend on $t$, then 
$$|x(t)\ra = e^{-iHt/\ph}|x(0)\ra.$$
Otherwise,  
$$|x(t)\ra = \exp\left({-i\over \ph} \int_0^t H(s)ds\right)|x(0)\ra.$$

\end{enumerate}

\noindent
{\bf Example} 
If 
$$H = \frac{-\ph}{2}w \sigma_x \quad  \hbox{ and } \quad |\psi(0)\ra = 
\begin{pmatrix}1 \cr 0\cr\end{pmatrix} 
\quad \hbox{ so that } \quad  i\ph \frac{\partial |\psi\ra}{\partial t} = H|\psi\ra ,$$ 
then
$|\psi(t)\ra = \exp(i\frac{w}{2}\sigma_x t) |\psi(0)\ra$. Hence,
$$|\psi(t)\ra = ((\cos wt/2) I_2 + (i\sin wt/2) \sigma_x) |\psi(0)\ra 
=  \begin{pmatrix} \cos wt/2 \cr i\sin wt/2\cr\end{pmatrix}.$$


\newpage
\noindent
{\bf Two ways to compute} 
$$\exp(B) \quad \hbox{ with } \quad B = (iwt/2)\sigma_x
 = (iwt/2)\begin{pmatrix}0 & 1 \cr 1 & 0\cr\end{pmatrix}.$$

\medskip\noindent
Method 1. Use the fact that 
$$B = (iwt/2)\sigma_x = (iwt/2)P_1 + (-iwt/2)P_2 \quad 
\hbox{ with } P_1 = \frac{1}{2}\begin{pmatrix}1 & 1 \cr 1 & 1 \cr\end{pmatrix},
P_2 = \frac{1}{2}\begin{pmatrix}1 & -1 \cr -1 & 1 \cr\end{pmatrix}.$$
Thus, 
$$\exp(B) = e^{iwt/2} P_1 + e^{-iwt/2}P_2 = 
\frac{1}{2}\begin{pmatrix}e^{iwt/2} + e^{-iwt/2} & e^{iwt/2} - e^{-iwt/2} \cr
e^{iwt/2} - e^{-iwt/2} & e^{iwt/2} + e^{-iwt/2}\cr\end{pmatrix} = 
\begin{pmatrix}\cos(wt/2) & i\sin(wt/2)\cr i\sin(wt/2) & \cos(wt/2)\cr\end{pmatrix}.$$

\medskip\noindent
Method 2. Use the fact that $\sigma_x^n = I_2$ if $n$ is even, and
$\sigma_x^n = \sigma_x$ if $n$ is odd.
Then 
$$\exp(B) =  \sum_n B^n/n! = \sum_{2j} B^{2j}/(2j)! + \sum_{2j-1} B^{2j-1}/(2j-1)!
= \cos(wt/2) I_2 + i\sin(wt/2) \sigma_x.$$



\newpage
\noindent
{\bf The uncertainty principle}

\medskip
Let
${\rm Exp}_x(A) = \la x|A|x\ra = \mu$
and 
$${\rm Var}_x(A) = {\rm Exp}_x((A-\mu I)^2) = \la x|(A-\mu I)^2|x\ra
= \|(A-\mu I)|x\ra\|^2.$$

\medskip
In an deterministic model, the variance of measurements should go to zero as
the apparatus is made very accurate.

\medskip\noindent
{\bf Theorem}
For any observable $A$ and $B$ and for any state $|x\ra$, we have
$${\rm Var}_x(A){\rm Var}_x(B) \ge {1\over 4} \la x| [A,B] |x\ra,$$
where $[A,B] = AB-BA$ is the commutator of $A$ and $B$.

\medskip
\it Proof. \rm Exercise 2.1. Use the notation  $[A,B] = AB-BA, \{A,B\} = AB+BA\}$, 

the Lie braket and the Poisson bracket, respectively.

\newpage
\medskip\noindent
{\bf Bipartite systems}

A system may have two components described by
two Hilbert spaces $\cH_1$ and $\cH_2$. Then
the bipartite
system is represented by $\cH = \cH_1 \otimes \cH_2$.
A general state in $\cH$ has the form
$$|x\ra = \sum_{i,j} c_{ij} |e_{1,i}\ra \otimes 
|e_{2,j}\ra\quad \hbox{ with }
\sum_{i,j} |c_{ij}|^2 = 1,$$
where $\{|e_{r,1}\ra, |e_{r,2}\ra, \dots \}$ is an orthonormal basis
for $\cH_r$ with $r \in \{1, 2\}$.

\smallskip
A state of the form $|x\ra = |x_1 \ra \otimes |x_2\ra$
is a separable state or a tensor product state.
Otherwise, it is an entangled state.

\medskip\noindent
{\bf Theorem} Every state $|x\ra$
in $\cH_1 \otimes \cH_2$ admits a Schmidt 
decomposition
$$|x\ra = \sum_{j=1}^r \sqrt{s_j} |u_j\ra \otimes |v_j\ra,$$
where $s_j > 0$ are the Schmidt coefficients satisfying 
$\sum_{j=1}^r s_j = 1$, $r$ is the Schmidt number of $|x\ra$,
$\{|u_1\ra, \dots, |u_r\ra\}$ is an orthonormal set of $\cH_1$ and
$\{|v_1\ra, \dots, |v_r\ra\}$ is an orthonormal set of $\cH_2$.


\medskip\noindent
{\bf Remark} Extending the results to $\cH_1 \otimes \cdots \otimes \cH_k$
for $k \ge 3$ is an open problem.

\newpage
\noindent
{\bf Mixed States and Density Matrices}

\smallskip
A system is in a mixed state if there is a probability $p_i$ that
the system is in state $|x_i\ra$ for $i = 1,  \dots, N$. 
If there is only one possible state, i.e., $p_1 = 1$, then the system
is in pure state. The mean value of the
measurement of the system corresponding to
the observable described by the Hermitian matrix $A$ is
$$\la A \ra = \sum_{j=1}^N p_j \la x_j | A | x_j \ra 
= \tr(A\rho),$$
where
$$\rho = \sum_{j=1}^N  p_j |x_j\ra \la x_j|$$
is a density operator (matrix).


\newpage\noindent
{\bf Description of a quantum system in mixed states.}
\begin{itemize}
\item[A1'] A physical state is specified by a density matrix
$\rho: \cH\rightarrow \cH$, which is positive semidefinite
with trace equal to one.

\item[A2'] The mean value of an observable associate with the
Hermitian matrix $A$ is $\la A \ra = \tr(\rho A)$.

\item[A3'] The temporal evolution of the density matrix is given by
the Liouville-von Neumann equation
$$i \ph {d\over dt} \rho = [H, \rho] = H\rho - \rho H,$$
where $H$ is the system Hamiltonian.
\end{itemize}

\medskip\noindent
{\bf Theorem 2.1} The following conditions are equivalent for a given state 
(density matrix) $\rho$.

\medskip
\centerline{
(a) $\rho$ is pure. \qquad  (b) $\rho^2 = \rho$. \qquad (c) $\tr(\rho^2) = 1$.}


\medskip
\noindent
{\bf Remarks}
\begin{itemize}
\item A mixed state $\rho$ is a Hermitian matrix with nonnegative eigenvalues
(Exercise 2.3) summing up to one.
Equivalently, $\la v |\rho|v \ra \ge 0$ for all vectors $|v\ra$ and $\tr \rho = 1$.

Note
$\rho$ is a nonnegative combination of Hermitian matrices so that it is Hermitian,
and has real eigenvalues.
If $\rho$ has a negative eigenvalue $\lambda_1$, then ...

\item A density matrix 
$\rho$ is a pure state $\rho = |\psi \ra \la \psi|$, 
i.e., $\rho$ has eigenvalues $1, 0, \dots, 0$.
This is the same as saying that $\tr \rho = 1$ (Exercise 2.4).

Suppose $\rho$ has eigenvalues $\lambda_1, \dots, \lambda_n$. Then 
$\tr \rho^2 = \sum_{j=1}^n \lambda_j^2$. 

\end{itemize}

\medskip\noindent
{\bf Definition 2.1} 
Suppose $\cH = \cH_1 \otimes \cH_2$.
A state $\rho$ is {\bf uncorrelated} if $\rho = \rho_1\otimes \rho_2$;
it is {\bf separable} if it is a convex combination of {uncorrelated 
states},
i.e., 
$$\rho = \sum_{j=1}^r p_j \rho_{1,j} \otimes \rho_{2,j}.$$
Otherwise, it is {\bf inseparable}.


\newpage\noindent
{\bf Partial transpose}


\medskip
Let $\rho = \sum_{j=1}^r c_j \rho_{1,j}\otimes \rho_{2,j} 
\in \cH = \cH_1 \otimes \cH_2$.
The {\bf partial transpose} of $\rho$ with respect to $\cH_2$  is
$$ \rho^\pt = \sum_{j=1}^r \rho_{1,j}\otimes \rho_{2,j}^t.$$

\medskip
In matrix form, if $\rho = (\rho_{ij}) \in M_m(M_n)$, then $\rho^{\pt} = (\rho_{ij}^t)$.
 


\medskip\noindent
{\bf Remark} If $\rho$ is separable, then so is $\rho^\pt$.
If $\rho^\pt$ has negative eigenvalues, which can be detected by
$N(A) = (\sum_{j} |\lambda_i(A)| - 1)/2$, then it is not physical
and $\rho$ is not separable. Converse holds for $\IC^r \otimes \IC^s$
for $r + s \le 5$.

\medskip\noindent 
{\bf See Example 2.5 and 2.6.} 

\newpage\noindent
{\bf Partial Trace and Purification}

\medskip
Let $A \in \cH_1 \otimes \cH_2$. The partial trace of $A$ over $\cH_2$
is an operator acting on $\cH_1$ defined by
$$A_1 = \tr_2 A = \sum_{u} (I_m\otimes \la u|)A(I_m\otimes |u\ra),$$
where $m,n$ are the dimension of $\cH_1$ and $\cH_2$.

\medskip
In matrix form, if $\rho = (\rho_{ij}) \in M_m(M_n)$, then $\tr_2(\rho) = (\tr \rho_{ij})
\in M_n$. One can define $\tr_1(\rho_{ij}) = \rho_{11} + \cdots + \rho_{mm}$,
which corresponds to 
$$A_2 = \tr_1 A = \sum_{u} (\la u| \otimes I_n) A (|u\ra \otimes I_n).$$

\medskip\noindent
{\bf Theorem (Purification)} Let $\rho_1 = \sum_{j=1}^n p_j|x_j\ra \la x_j|$.
Suppose $|\psi\ra = \sum_{j=1}^n \sqrt p_j|x_j\ra \otimes |y_j\ra$.
Then $\tr_2(|\psi \ra \la \psi|) = \rho_1$.


\newpage\noindent
{\bf Fidelity}

\medskip\noindent
{\bf Definition 2.2} The fidelity of two density matrices 
$\rho_1$ and $\rho_2$ is defined as
$$F(\rho_1, \rho_2) = \tr((\rho_1^{1/2}\rho_2\rho_1^{1/2})^{1/2}) \ge 0.$$

\medskip\noindent
{\bf Remarks}
\begin{enumerate}
\item If $A = \sum_{j} \lambda_j P_j$ with 
$\lambda_j \ge 0$, then $A^{1/2} = \sum_{j} \sqrt \lambda_j P_j.$

\item If $A, B$ are $m\times n$ and $n\times m$, then $AB$ and $BA$ have the same
nonzero eigenvalues.

\it Proof. \rm Note that $$\begin{pmatrix} AB & 0 \cr B & 0_n \end{pmatrix} 
\begin{pmatrix} I_m & A \cr 0 & I_n\cr\end{pmatrix}
= \begin{pmatrix} AB & ABA \cr B & BA\cr \end{pmatrix} 
= \begin{pmatrix} I_m & A \cr 0 & I_n\cr\end{pmatrix}
\begin{pmatrix} 0_m & 0 \cr B & BA \end{pmatrix}.$$ 
So, $\begin{pmatrix} AB & 0 \cr B & 0 \cr \end{pmatrix}$ and 
$\begin{pmatrix} 0_m & 0 \cr B & BA \end{pmatrix}$ are similar. The result follows.

\item We have $F(\rho_1,\rho_2) = F(\rho_2,\rho_1)$
as $\rho_1^{1/2} \rho_2\rho_1^{1/2}$ and 
$\rho_2^{1/2}\rho_1\rho_2^{1/2}$ have the same eigenvalues.

\item For any unitary $U$, $F(U\rho_1U^\dag, U\rho_2 U^\dag) = F(\rho_1, \rho_2)$.
(Exercise 2.10).

\item Suppose $\rho_1, \rho_2$ have eigenvalues $a_1 \ge\cdots \ge a_n$
and $b_1 \ge \cdots \ge b_n$. Then
$$F(\rho_1, \rho_2) =  \max\{
|\tr(\rho_1^{1/2}\rho_2^{1/2}U)|: U \ \hbox{unitary}\}
\le \sum_{j=1}^n \sqrt{a_jb_j} \le 1.$$

\item For any two density matrices $\rho_1$ and $\rho_2$, we have
$$0 \le F(\rho_1,\rho_2) \le 1.$$

The first equality holds if and only if $\rho_1\rho_2 = 0$; the second equality holds
if and only if $\rho_1 = \rho_2$.
\end{enumerate}


\end{document}
