Update src/4_vector_spaces_QM.md

src/4_vector_spaces_QM.md

@@ -140,35 +140,33 @@ $$
 |\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots
 $$
 
-Let us show how we can use the matrix representation to evaluate the inner product (bracket) between two state vectors when expanded in terms of their components in the same basis:
-$$
-|\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle \,, \qquad
-|\chi\rangle=\sum_{i=1}^n \chi_i |\phi_i\rangle\, .
-$$
-First of all, we note that we can write the above expansions in the following way
-$$
-|\psi\rangle=\sum_{i=1}^n |\phi_i \rangle \langle \phi_i | \psi \rangle \, ,
-$$
-and thus we see that the basis vectors provide a very useful representation of the *identity operator*:
-$$
-\hat{I} = \sum_i |\phi_i\rangle \langle\phi_i| \, ,
-$$
-We can insert this  identify operator within the bracket to evaluate the inner
-product $\langle \chi|\psi\rangle$ between the two state vectors to evaluate the inner product $\langle \chi|\psi\rangle$:
-$$
-\langle \chi|\psi\rangle=
-\langle\chi|\hat{I} |\psi\rangle=\sum_{i=1}^n \langle\chi| \phi_i \rangle \langle\phi_i|\psi\rangle \, .
-$$
-Next, using that $\chi_i = \langle \phi_i|\chi \rangle$ are the components of the
-state vector $|\chi\rangle$ and that  $\langle \chi| \phi_n \rangle=(\langle\phi_i|\chi\rangle)^*$,
-we have that $\langle \chi |\phi_i\rangle =\chi_i^*$
-and therefore the inner product of the two state vectors $|\psi\rangle$
-and  $|\chi\rangle$ can be expressed in terms of their components
-as follows
-$$\langle\chi|\psi\rangle=\sum_{i=1}^n\chi_i^*\psi_i.$$
-which in the matrix representation of state vectors can also be written as
-$$\langle \chi|\psi\rangle=\begin{pmatrix} \chi^*_1 , \chi^*_2 &,\ldots \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix} \, .$$ Therefore, we can present  bra vector $\langle \chi|$ as row vectors and ket vectors as column vector. 
-The row vector can thus be treated as the *complex conjugate* of the corresponding column vector.
+!!! example "Evaluating an inner product"
+     Let us show how we can use the matrix representation to evaluate the inner product (bracket) between two state vectors when expanded in terms of their components in the same basis:
+     $$ |\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle \,, \qquad |\chi\rangle=\sum_{i=1}^n \chi_i |\phi_i\rangle\, .$$
+     First of all, we note that we can write the above expansions in the following way
+     $$
+     |\psi\rangle=\sum_{i=1}^n |\phi_i \rangle \langle \phi_i | \psi \rangle \, ,
+     $$
+     and thus we see that the basis vectors provide a very useful representation of the *identity operator*:
+     $$
+     \hat{I} = \sum_i |\phi_i\rangle \langle\phi_i| \, ,
+     $$
+     We can insert this  identify operator within the bracket to evaluate the inner
+     product $\langle \chi|\psi\rangle$ between the two state vectors to evaluate the inner product $\langle \chi|\psi\rangle$:
+     $$
+     \langle \chi|\psi\rangle=
+     \langle\chi|\hat{I} |\psi\rangle=\sum_{i=1}^n \langle\chi| \phi_i \rangle \langle\phi_i|\psi\rangle \, .
+     $$
+     Next, using that $\chi_i = \langle \phi_i|\chi \rangle$ are the components of the
+     state vector $|\chi\rangle$ and that  $\langle \chi| \phi_n \rangle=(\langle\phi_i|\chi\rangle)^*$,
+     we have that $\langle \chi |\phi_i\rangle =\chi_i^*$
+     and therefore the inner product of the two state vectors $|\psi\rangle$
+     and  $|\chi\rangle$ can be expressed in terms of their components
+     as follows
+     $$\langle\chi|\psi\rangle=\sum_{i=1}^n\chi_i^*\psi_i.$$
+     which in the matrix representation of state vectors can also be written as
+     $$\langle \chi|\psi\rangle=\begin{pmatrix} \chi^*_1 , \chi^*_2 &,\ldots \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix} \, .$$ Therefore, we can present  bra vector $\langle \chi|$ as row vectors and ket vectors as column vector. 
+     The row vector can thus be treated as the *complex conjugate* of the corresponding column vector.
 
 ## A two-dimensional Hilbert space