Update src/4_vector_spaces_QM.md

src/4_vector_spaces_QM.md

@@ -14,7 +14,7 @@ The lecture on vector spaces in quantum mechanics consists of the following part
 
 - [4.4. A two-dimensional Hilbert space](#44-two-dimensional-hilbert-space)
 
-and at the end of the lecture one can find the exercises 
+and at the end of the lecture there is a set of exercises 
 
 - [4.5. Problems](#45-problems)
 
@@ -76,37 +76,57 @@ The set of all possible state vectors describing a given physical system forms a
 
 ### Bra vectors
 
-We need now to extend a bit the Dirac notation for elements of this vector space. We need to introduce a quantity $\langle{\Psi}|$, known as a  *bra vector*,
-which represents the  *complex conjugates* of the corresponding ket vector. Bra vectors are elements of the vector space $\mathcal{H}^{*}$, which is called the *dual space* of the original Hilbert space $\mathcal{H}$.
+We need now to extend the Dirac notation to describe other elements of this vector space. We need to introduce a quantity $\langle{\Psi}|$, known as a  *bra vector*, which represents the  *complex conjugates* of the corresponding ket vector. Bra vectors are elements of the vector space $\mathcal{H}^{*}$, called the *dual space* of the original Hilbert space $\mathcal{H}$.
 
-If a ket vector is given by $| \Psi\rangle= c_1 |\psi_1\rangle+c_2|\psi_2\rangle$, then the corresponding bra vector will be given by 
-$$\langle{\Psi}|= c_1^*\langle{\psi_1}|+c_2^*\langle{\psi_2}| \, .$$
-As mentioned above, the vector space spanned by all bra vectors $\langle{\Psi}|$ is referred to as the  dual space and is represented by $\mathcal{H}^*$. For each ket vector belonging to $\mathcal{H}$, there will exist an associated bra vector belonging to the dual space $\mathcal{H}^*$.
+!!! info "Bra vector"
+     If a ket vector is given by $$| \Psi\rangle= c_1 |\psi_1\rangle+c_2|\psi_2\rangle \, ,$$ 
+     then the corresponding bra vector will be given by 
+     $$\langle{\Psi}|= c_1^*\langle{\psi_1}|+c_2^*\langle{\psi_2}| \, .$$
 
-Below we will further discuss the concept of bra vectors when presenting the matrix representation of elements of the Hilbert space.
+As mentioned above, the vector space spanned by all bra vectors $\langle{\Psi}|$ is referred to as the dual space and is represented by $\mathcal{H}^*$. For each ket vector belonging to $\mathcal{H}$, there will exist an associated bra vector belonging to the dual space $\mathcal{H}^*$.
+
+Below, we will further discuss the concept of bra vectors when presenting the matrix representation of elements of the Hilbert space.
       
-##  Inner product of state vectors
+## 4.2. Inner product of state vectors
 
 Assume that $|{\psi}\rangle$ and $|{\phi}\rangle$ are any two state vectors belonging to the
 state (Hilbert) space $\mathcal{H}$, then we can define the  *inner product*
-between them, $\langle{\psi}|{\phi}\rangle$, as follows. This inner product in quantum mechanics is the analog of the
-usual scalar product that one encounters in vector spaces and that we reviewed in the previous lecture. As in usual vector spaces, the inner product of two state vectors is an  *scalar*, in this case a complex number in general. 
+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 $|{\psi}\rangle$. 
 
-The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the  *probability amplitude* (not the probability) of measuring a  system characterised by the state $|{\phi}\rangle$ to be in the state $|{\psi}\rangle$. This inner product can also be understood as measuring the *overlap* between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. 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 $0 \le |\langle \psi | \phi \rangle|^2 \le 1$.
+!!! info ""
+     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 $$|\langle \psi | \phi \rangle|^2$$. 
+     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:
       
-- *Complex conjugate*: $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
-        
-- *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 "Complex conjugate:" 
+     $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
         
-- *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 "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:" 
+     $$\langle \psi | \psi \rangle\geq0$. If $\langle \psi | \psi \rangle = 0$$ 
+     then this implies that the state vector $|\psi\rangle=0$ is the null element of the Hilbert space.
 
-- *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.
+!!! 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.
         
-The quantity $\sqrt{\langle \psi | \psi \rangle}$ is known as the  *length* or the *norm* of the state vector $|\psi\rangle$. You can see from the properties of complex algebra that this length must be a real number.
+!!! info "Norm:"
+     The quantity $\sqrt{\langle \psi | \psi \rangle}$ is known as the  *length* or the *norm* of the state vector $|\psi\rangle$. You can see from the properties of complex algebra that this length must be a real number.
 
-A physically valid state $|\psi \rangle$ must be normalized to unity, that is $\langle \psi | \psi \rangle=1$. Note that a state that cannot be normalized to unity cannot represent a physically acceptable state.
+A physically valid state $|\psi \rangle$ must be normalized to unity, that is $\langle \psi | \psi \rangle=1$. Note that a state that cannot be normalized to unity does not represent a physically acceptable state.
        
 A set of orthonormal basis vectors $\{|\psi_i\rangle\text{;}\; i=1,2,3,...,n\}$ will have the property $\langle \psi_i |\psi_j \rangle=\delta_{ij}$ where $\delta_{ij}$ is a mathematical symbol known as the *Kronecker delta*, which equals unity if $i=j$ and zero if $i\neq j$.