1st major update to lecture note 4

src/4_vector_spaces_QM.md

@@ -95,31 +95,31 @@ between them, $\langle{\psi}|{\phi}\rangle$, as follows.
 
 The inner product in quantum mechanics is the analog of the usual scalar product that one encounters in vector spaces, and which we reviewed in the previous lecture. As in usual vector spaces, the inner product of two state vectors is a  *scalar* and in this case a complex number in general. 
 
-!!! info ""
-     The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the **probability amplitude** (not the probability) of measuring a system, which characterised by the state $|{\phi}\rangle$, to be in the state $|{\psi}\rangle$. 
+!!! tip ""
+    1.  The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the **probability amplitude** (not the probability) of measuring a system, which characterised by the state $|{\phi}\rangle$, to be in the state $|{\psi}\rangle$. 
 
-!!! info ""
-     This inner product can also be understood as measuring the **overlap** between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. 
+!!! tip ""
+    2.  This inner product can also be understood as measuring the **overlap** between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. 
 
-!!! info "" 
-     Then the  **probability of observing the system to be in the state $|\psi\rangle$** given that it is in the state $|\phi\rangle$ will be given by $$|\langle \psi | \phi \rangle|^2$$. 
+!!! tip "" 
+     3. Then the  **probability of observing the system to be in the state $|\psi\rangle$** given that it is in the state $|\phi\rangle$ will be given by $$|\langle \psi | \phi \rangle|^2$$. 
      Since the latter quantity is a probability, we know that it should satisfy the condition that 
      $$0 \le |\langle \psi | \phi \rangle|^2 \le 1 \, .$$
 
 The inner product (probability amplitude) $\langle \psi | \phi \rangle$ exhibits the following properties:
       
 !!! info "Complex conjugate:" 
-     $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
+     $$\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$$
         
-!!! info "Distributivity and associativity:" 
+!!! info "Distributivity and associativity" 
      $$\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$$
 
-!!! info "Positivity:" 
-     $$\langle \psi | \psi \rangle\geq0$. If $\langle \psi | \psi \rangle = 0$$ 
+!!! info "Positivity" 
+     $$\langle \psi | \psi \rangle\geq0$$. If $\langle \psi | \psi \rangle = 0$$ 
      then this implies that the state vector $|\psi\rangle=0$ is the null element of the Hilbert space.
 
-!!! info "Orthogonality:" 
-     two states $|\psi \rangle$ and $|\phi \rangle$ are said to be *orthogonal* if 
+!!! info "Orthogonality" 
+     Two states $|\psi \rangle$ and $|\phi \rangle$ are said to be *orthogonal* if 
      $$\langle \psi | \phi\rangle=0 \, .$$ 
      By analogy with regular vector spaces, we can think of these two state vectors $|\psi \rangle$ and $|\phi \rangle$ as being *perpendicular* to each other. Note that for a quantum system occupying a certain state, there is a vanishing probability of it being observed in a state orthogonal to it.