Update src/4_vector_spaces_QM.md

src/4_vector_spaces_QM.md

@@ -135,12 +135,9 @@ We can also express the basis vectors in this manner. Given that the basis vecto
 the basis state $|\phi_i\rangle$ will have as component in the $j$ direction
 $$ (\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,$$ 
 and thus the vector column expression of the basis vectors will be very simple
-$$ 
-|\phi_1\rangle= \begin{pmatrix} 1\\0\\0 \\\vdots\end{pmatrix} \;, \quad
-|\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots
-$$
+$$ |\phi_1\rangle= \begin{pmatrix} 1\\0\\0 \\\vdots\end{pmatrix} \;, \quad |\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots $$
 
-!!! example "Evaluating an inner product"
+!!! note "Evaluating the 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
@@ -168,7 +165,7 @@ $$
      $$\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
+## 4.4. A two-dimensional Hilbert space
 
 As a practical example to illustrate the basic ideas of vector spaces applied to quantum physics presented above, we will consider a quantum system which is fundamental for quantum mechanics and its applications. This system corresponds to the possible states that the intrinsic angular momentum of an electron, known as *spin*, can occupy. As you will see in following courses, the Hilbert space for the electron spin has dimension $n=2$, meaning that we can found an electron *pointing* either in the up direction, denoted by $|+\rangle$, or the down direction, denoted by $|-\rangle$.
 
@@ -207,11 +204,13 @@ $$
 $$
 The values of the coefficients $c_+$ and $c_-$ for these examples above are, respectively,
 $$
-(c_+,c_) = (3,-2i) \, ,\qquad
+(c_+,c_-) = (3,-2i) \, ,\qquad
 (c_+,c_-) = (i,-4) \, ,\qquad
 (c_+,c_-) = (2,5) \, .\qquad 
 $$
-Note however that many other bases are possible, and that the physics of a quantum system do not depend on the basis that we choose.
+!!! danger ""
+     Note however that many other bases are possible, and that the physics of a quantum system do not depend on the basis that we choose.
+
 The bra vectors associated to these ket vectors will be given by
 $$ |{\Psi}\rangle=   \left( \begin{array}{c}3 \\ -2i \end{array} \right) \, , \qquad
 \langle{\Psi}|=\left( 3, 2i \right) \, ,$$
@@ -222,27 +221,28 @@ $$|{\Psi}\rangle=  \left( \begin{array}{c}2 \\ 5 \end{array} \right) \, , \qquad
 Note however that the above vectors are not normalised (the inner product with themselves is different from unity), and thus
 cannot represent physical states. We show below an explicit example of a normalised state vector belonging to this Hilbert space.
 
-We also know how we can evaluate the inner product between any two state vectors belonging to this Hilbert space. If we have two state vectors given by
-$$
-|\psi\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
-|\phi \rangle = \left( \begin{array}{c}0 \\1 \end{array} \right)
-$$
-then their inner product is
-$$
-\langle \psi | \phi \rangle =
-\frac{1}{\sqrt{2}} \left( 1 , i  \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
-\frac{i}{\sqrt{2}} 
-$$
-and the associated probability will be given by
-$$
-|\langle \psi|\phi\rangle|^2 = \frac{1}{2}
-$$
-meaning that if I measure the state $| \phi \rangle $, I will have a 50% probability
-of finding it in the state $| \psi \rangle$. Recall that probabilities must always be smaller than 1 to make physical sense. Note also that I am using normalised vectors, you can check yourselves that
-$$
-\langle \psi |\psi \rangle =  \langle \phi | \phi\rangle = 1 \, ,
-$$
-as required to ensure the probabilistic description of the state vector.
+!!! done "Example: Evaluating the inner product"
+     We also know how we can evaluate the inner product between any two state vectors belonging to this Hilbert space. If we have two state vectors given by
+     $$
+     |\psi\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
+     |\phi \rangle = \left( \begin{array}{c}0 \\1 \end{array} \right)
+     $$
+     then their inner product is
+     $$
+     \langle \psi | \phi \rangle =
+     \frac{1}{\sqrt{2}} \left( 1 , i  \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
+     \frac{i}{\sqrt{2}} 
+     $$
+     and the associated probability will be given by
+     $$
+     |\langle \psi|\phi\rangle|^2 = \frac{1}{2}
+     $$
+     meaning that if I measure the state $| \phi \rangle $, I will have a 50% probability
+     of finding it in the state $| \psi \rangle$. Recall that probabilities must always be smaller than 1 to make physical sense. Note also that I am using normalised vectors, you can check yourselves that
+     $$
+     \langle \psi |\psi \rangle =  \langle \phi | \phi\rangle = 1 \, ,
+     $$
+     as required to ensure the probabilistic description of the state vector.
 
 ## Problems