updates

latex/chapters/chap0.tex

@@ -309,7 +309,6 @@ $$
 \operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i
 $$
 
-
 \end{defn}
 Or we can define the map $L_v: \mathscr{A}\to \mathscr{A}\otimes \mathscr{B}$ by
 
@@ -509,7 +508,7 @@ Recall from classical probability theory, we call the initial probability distri
 
     Given a non-commutative probability space $(\mathscr{B}(\mathscr{H}),\mathscr{P})$,
 
-    A state is a unit vector $\bra{\psi}$ in the Hilbert space $\mathscr{H}$, such that $\bra{\psi}\ket{\psi}=1$. 
+    A state is a unit vector $\ket{\psi}$ in the Hilbert space $\mathscr{H}$, such that $\bra{\psi}\ket{\psi}=1$. 
     
     Every state uniquely defines a map $\rho:\mathscr{P}\to[0,1]$, $\rho(P)=\bra{\psi}P\ket{\psi}$ (commonly named as density operator) such that:
     \begin{itemize}
@@ -518,7 +517,7 @@ Recall from classical probability theory, we call the initial probability distri
     \end{itemize}
 \end{defn}
 
-Note that the pure states are the density operators that can be represented by a unit vector $\bra{\psi}$ in the Hilbert space $\mathscr{H}$, whereas mixed states are the density operators that cannot be represented by a unit vector in the Hilbert space $\mathscr{H}$.
+Note that the pure states are the density operators that can be represented by a unit vector $\ket{\psi}$ in the Hilbert space $\mathscr{H}$, whereas mixed states are the density operators that cannot be represented by a unit vector in the Hilbert space $\mathscr{H}$.
 
 If $(|\psi_1\rangle,|\psi_2\rangle,\cdots,|\psi_n\rangle)$ is an orthonormal basis of $\mathscr{H}$ consisting of eigenvectors of $\rho$, for the eigenvalues $p_1,p_2,\cdots,p_n$, then $p_j\geq 0$ and $\sum_{j=1}^n p_j=1$.