1st major update to lecture note 4

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)
 
@@ -76,105 +76,97 @@ 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. 
+
+!!! tip "Meaning of the inner product in quantum mechanics"
+     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 $|{\psi}\rangle$. 
+     2.   This inner product can also be understood as measuring the **overlap** between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. 
+     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 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 $0 \le |\langle \psi | \phi \rangle|^2 \le 1$.
+### Properties of the inner product
 
 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*: $\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$
-        
-- *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.
-        
-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 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 $i\neq j$.
+!!! info "Properties of the inner product"
+     1. **Complex conjugate:** $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
+     2. **Distributivity and associativity:** $\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$
+     3. **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.
+     4. **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.
+     5. **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 does not represent a physically acceptable state.
+     6. **Orthonormality:** 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 $i\neq j$.
    
-From all the above conditions we see that a Hilbert space is a so-called *complex inner product space*, which is nothing but a complex vector space equipped with a inner product. All the vectors belonging to a Hilbert space $\mathcal{H}$ have a finite norm, that is they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic interpretation of the state vectors described above.
+From all the above conditions, we see that a Hilbert space is a so-called *complex inner product space*, which is nothing else but a complex vector space equipped with a inner product. All the vectors belonging to a Hilbert space $\mathcal{H}$ have a finite norm, which means that they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic interpretation of the state vectors described above.
 
-## Matrix representation of ket and bra vectors
+## 4.3. Matrix representation of ket and bra vectors
  
-As we have discussed, in quantum mechanics a general state vector $|\psi\rangle$ can be represented in terms of the basis vectors, $\{|\phi_i\rangle;i=1,2,...,n\}$, as follows
-$$
-|\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle \, ,
-$$
+As we have discussed, in quantum mechanics a general state vector $|\psi\rangle$ can be represented in terms of the basis vectors, $\{|\phi_i\rangle;i=1,2,...,n\}$, as 
+$$ |\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle $$
 for some values of the complex coefficients $\{ c_i\}$. To determine the values of these coefficients, we can take the inner product between the bra basis vector $\langle \phi_j|$ and the ket state vector $|\psi\rangle$ and use the orthogonality properties of the basis vectors:
-$$
-\langle \phi_j|\psi\rangle = \langle \phi_j|\sum_{i=1}^n c_i |\phi_i\rangle = \sum_{i=1}^n c_i\langle \phi_j|\phi_i\rangle = \sum_{i=1}^n c_i\delta_{ij} = c_j \, .
-$$
+$$ \langle \phi_j|\psi\rangle = \langle \phi_j|\sum_{i=1}^n c_i |\phi_i\rangle = \sum_{i=1}^n c_i\langle \phi_j|\phi_i\rangle = \sum_{i=1}^n c_i\delta_{ij} = c_j \, .$$
 Therefore, if we now denote the coefficients $\{ c_i\}$ of the state vector $|\psi\rangle$ by $\{ \psi_i\}$, we have the expansion
-$$
-|\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle= \sum_{i=1}^n \left( \langle \phi_i|\psi\rangle \right) |\phi_i\rangle \, .
-$$
-By analogy with the Euclidean case, we can understand the coefficients $\psi_i$
-as the *components* of the state vector $ |\psi\rangle$ along the $n$ directions
-spanned by the basis vectors. Here note that in this notation $\psi_i$ is an *scalar* (just a number) and not a vector. Further note that, as opposed to the Euclidean space, the coefficients $\psi_i$ will be in general complex numbers.
+$$ |\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle= \sum_{i=1}^n \left( \langle \phi_i|\psi\rangle \right) |\phi_i\rangle \, .$$
+By analogy with the Euclidean case, we can understand the coefficients $\psi_i$ as the *components* of the state vector $ |\psi\rangle$ along the $n$ directions spanned by the basis vectors. Here, note also that in this notation $\psi_i$ is an *scalar* (just a number) and not a vector. Furthermore, note that, as opposed to the Euclidean space, the coefficients $\psi_i$ will generally be complex numbers.
 
 This analogy with the case of ordinary vectors allows us to write the state $|\psi\rangle$ as a *column vector* with respect to the set of basis vectors $\{|\phi_i\rangle;i=1,2,...,n\}$, which are kept implicit: 
-$$
-|\psi\rangle= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\vdots\\\psi_n\end{pmatrix}.
-$$
-We can also express the basis vectors in this manner. Given that the basis vectors are *orthonormal* among them,
+$$ |\psi\rangle= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\vdots\\\psi_n\end{pmatrix} \, . $$
+
+We can also express the basis vectors in this manner. Given that the basis vectors are *orthonormal* among themselves,
 the basis state $|\phi_i\rangle$ will have as component in the $j$ direction
-$$
-(\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,
-$$
+$$ (\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,$$ 
 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
 $$
 
-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