Quantum Mechanics Term 1

1 Elements of Linear Algebra

1.1 First Principle of Quantum Mechanics

A quantum mechanical state is represented by a unit vector $|\psi ⟩$ inside a Hilbert space $\mathscr{H}$.
If $|{\psi }_1⟩,|{\psi }_2⟩\in \mathscr{H}$ and $a,b\in ℂ$, then $|\psi ⟩=a|{\psi }_1⟩+b|{\psi }_2⟩\in \mathscr{H}$.

1.2 Notation

• For vectors in $\mathscr{H}$, we denote them using kets i.e. $|\psi ⟩$.

• For the null vector, we use $|0⟩$, we can also say that $|\psi ⟩=0$

• For two vectors $|V⟩,|W⟩\in \mathscr{H}$, we write the inner product as

  $⟨V|W⟩\equiv (|V⟩,|W⟩)$

The properties of the inner product reformulated in this notation is as follows:

1. $⟨W|V⟩={⟨V|W⟩}^{*}$

2. $⟨V|V⟩\ge 0$

3. $⟨V|V⟩=0$  iff  $|V⟩=0$

4. If $|X⟩=a|W⟩+b|Z⟩$, then $⟨V|X⟩=a⟨V|W⟩+b⟨V|Z⟩$, $⟨X|V⟩=a^{*}⟨W|V⟩+b^{*}⟨Z|V⟩$

1.3 Expansion in Orthonormal basis and Column notation

Suppose that $dim(\mathscr{H})=n$ and given a basis $\{|i⟩,i=1,\mathrm{\dots },n\}$. We write

$|\psi ⟩={\displaystyle \sum _{i=1}^n}{c}_i|i⟩=\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill c_n\hfill \end{array}\right)$

for any $c_i\in ℂ$ and $|\psi ⟩\in \mathscr{H}$.
If the basis is orthornormal i.e. $⟨i|j⟩={\delta }_{ij}$, then we can find the constants using the inner product

$⟨j|\psi ⟩={\displaystyle \sum _{i=1}^n}{c}_i⟨j|i⟩={\displaystyle \sum _{i=1}^n}{c}_i{\delta }_{ij}=c_j$

Hence, for the product of two vectors $|\psi ⟩={\sum }_{i=1}^n{c}_i|i⟩,|\varphi ⟩={\sum }_{j=1}^n{d}_j|j⟩$,

$⟨\varphi |\psi ⟩$ $={\displaystyle \sum _{i=1}^n}{c}_i⟨\varphi |i⟩$

$={\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^n}{c}_i{d}_j^{*}⟨j|i⟩$

$={\displaystyle \sum _{i=1}^n}{\displaystyle \sum _{j=1}^n}{c}_i{d}_j^{*}{\delta }_{ij}$

$={\displaystyle \sum _{i=1}^n}{c}_i{d}_i^{*}$

Relating this to our matrix notation, it is intuitive for us to write $|\varphi ⟩$ as a row vector i.e.

$⟨\varphi |=\left(d_1^{*}{d}_2^{*}\mathrm{\dots }{d}_n^{*}\right)$

so that $⟨\varphi |\psi ⟩$ works exactly like matrix multiplication.

1.4 Dual Space

The dual space ${\mathscr{H}}^{*}$ of $\mathscr{H}$ is the set of linear functionals s.t. $\mathrm{\Phi }:\mathscr{H}\to ℂ$

$\mathrm{\Phi }(a|V⟩+b|W⟩)=a\mathrm{\Phi }(|V⟩)+b\mathrm{\Phi }(|W⟩)$

With this definition, it is clear that ${\mathscr{H}}^{*}$ is a linear space itself since if ${\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2\in {\mathscr{H}}^{*}$ and for any $|\psi ⟩\in \mathscr{H}$

$(a{\mathrm{\Phi }}_1+b{\mathrm{\Phi }}_2)|\psi ⟩=a{\mathrm{\Phi }}_1(|\psi ⟩)+b{\mathrm{\Phi }}_2(|\psi ⟩)\in {\mathscr{H}}^{*}$

The dimension of ${\mathscr{H}}^{*}$ is $n$ since in order to uniquely determine an element $\mathrm{\Phi }\in {\mathscr{H}}^{*}$ we need to know how the basis $\{|i⟩\}$ gets mapped to $ℂ$ under $\mathrm{\Phi }$. Hence, $dim({\mathscr{H}}^{*})=dim(\mathscr{H})=n$.

Given a basis $\{|i⟩\}$ of $\mathscr{H}$, we can always choose a dual basis $\{{\mathrm{\Phi }}_j\}$ of ${\mathscr{H}}^{*}$ s.t. ${\mathrm{\Phi }}_j(|i⟩)={\delta }_{ij}$.
So for $\mathrm{\Phi }={\sum }_{j=1}^n=d_j{\mathrm{\Phi }}_j$.

$\mathrm{\Phi }(|\psi ⟩)={\displaystyle \sum _{j=1}}{d}_j{\mathrm{\Phi }}_j(|\psi ⟩)={\displaystyle \sum _{j=1}^n}{d}_j{\displaystyle \sum _{i=1}^n}{c}_i{\mathrm{\Phi }}_j(|i⟩)={\displaystyle \sum _{j=1}}{c}_j{d}_j$

1.5 Adjoint Vector

Given an inner product, we can define the adjoint $⟨\psi |\in {\mathscr{H}}^{*}$ of a vector $|\psi ⟩\in \mathscr{H}$. The defining property is

$(⟨\psi |)(|V⟩)=⟨\psi |V⟩$

We call the inner product a braket as it is a product between the adjoint (a ’bra’) and the vector (a ket).

If $|K⟩=a|V⟩+b|W⟩$, we find $⟨K|$ using the above definition.

$(⟨K|)(|\psi ⟩)$ $=a^{*}⟨V|\psi ⟩+b^{*}⟨W|\psi ⟩$

$=(a^{*}⟨V|+b^{*}⟨W|)(|\psi ⟩)$

$\Rightarrow ⟨K|$ $=a^{*}⟨V|+b^{*}⟨W|$

It follows that the dual basis of an orthonormal basis $\{|i⟩\}$ is ${\mathrm{\Phi }}_j=⟨j|$.
As a consequence, we have that if $|\psi ⟩={\sum }_{i=1}^n{c}_i|i⟩$

1. $⟨\psi |={\displaystyle \sum _{i=1}^n}{c}_i^{*}⟨i|=(c_1^{*}{c}_2^{*}\mathrm{\dots }{c}_n^{*})$

2. $c_i$ $=⟨i|\psi ⟩$

   $|\psi ⟩$ $={\displaystyle \sum _{i=1}^n}|i⟩⟨i|\psi ⟩$

   $⟨\psi |$ $={\displaystyle \sum _{i=1}^n}{⟨i|\psi ⟩}^{*}⟨i|={\displaystyle \sum _{i=1}^n}⟨\psi |i⟩⟨i|$

1.6 Linear Operators

A linear operator $\mathrm{\Omega }$ is a linear map $\mathrm{\Omega }:\mathscr{H}\to \mathscr{H}$.
Similar to the discussion in dual spaces, the linear operator is completely fixed by its action on a basis $\{|i⟩\}$. Let $\mathrm{\Omega }|i⟩={|i⟩}^{\prime }$. Then $\mathrm{\Omega }|\psi ⟩={\sum }_{i=1}^n{c}_i{|i⟩}^{\prime }$.

Since $|i⟩$ is a basis, there has to be ${\mathrm{\Omega }}_{ji}\in ℂ,i,j=1,\mathrm{\dots },n$ s.t.

$\mathrm{\Omega }|i⟩$ $={\displaystyle \sum _{j=1}^n}{\mathrm{\Omega }}_{ji}|j⟩$

We can also represent linear operators in matrix form. For any $|\psi ⟩\in \mathscr{H}$,

$|\psi ⟩$ $=\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill c_n\hfill \end{array}\right)$

$\mathrm{\Omega }$

$\mathrm{\Omega }|\psi ⟩$

In order to find these ${\mathrm{\Omega }}_{ji}$, we again use the inner product with the basis vectors.

$⟨k|\mathrm{\Omega }|i⟩={\displaystyle \sum _{j=1}^n}{\mathrm{\Omega }}_{ji}⟨k|j⟩={\mathrm{\Omega }}_{ki}$

Example:
Consider the operator $\widehat{R}$ which is a counterclockwise rotation by $\frac{\pi }{2}$ around $|3⟩$ on $\mathscr{H}$ s.t. $|1⟩$ is the x-axis, $|2⟩$ is the y-axis and $|3⟩$ is the z-axis.
First we determine the action $\widehat{R}$ on these basis vectors.

$\widehat{R}|1⟩$ $=|2⟩$

$\widehat{R}|2⟩$ $=-|1⟩$

$\widehat{R}|3⟩$ $=|3⟩$

$\Rightarrow \widehat{R}|v⟩$ $=v_1|2⟩-v_2|1⟩+v_3|3⟩$

whereby $|v⟩$ is any vector in $\mathscr{H}$ with basis $\{|1⟩,|2⟩,|3⟩\}$.
For the matrix form,

$⟨i|\widehat{R}|j⟩$ $=R_{ij}$

$R_{11}$ $=⟨1|\widehat{R}|1⟩=0$

$R_{21}$ $=⟨2|\widehat{R}|1⟩=1$

$R_{12}$ $=⟨1|\widehat{R}|2⟩=-1$

$R_{33}$ $=⟨3|\widehat{R}|3⟩=1$

$\Rightarrow R$

${\mathrm{\Omega }}^{-1}$ is the inverse of $\mathrm{\Omega }$ iff ${\mathrm{\Omega }}^{-1}\mathrm{\Omega }=\mathrm{\Omega }{\mathrm{\Omega }}^{-1}=\widehat{I}$.

Also, for any $|a⟩,|b⟩\in \mathscr{H}$, we can define a linear operator $\widehat{\mathrm{\Lambda }}=|a⟩⟨b|$. Then for any $|\psi ⟩\in \mathscr{H}$,

$\widehat{\mathrm{\Lambda }}|\psi ⟩=|a⟩⟨b|\psi ⟩=⟨b|\psi ⟩|a⟩$

1.6.1 Projection Operators

The defining property of a projection operator $\widehat{P}$ is ${\widehat{P}}^2=\widehat{P}$.
Hence an obvious example is the identity operator.

Consider $\mathscr{H}={\mathscr{H}}_1\oplus {\mathscr{H}}_2$.
We can decompose any $|\psi ⟩\in \mathscr{H}$ s.t. $|\psi ⟩=|{\psi }_1⟩+|{\psi }_2⟩$ whereby $|{\psi }_1⟩\in {\mathscr{H}}_1$ and $|{\psi }_2⟩\in {\mathscr{H}}_2$.
Therefore we define ${\widehat{P}}_1,{\widehat{P}}_2$,

${\widehat{P}}_{{\mathscr{H}}_1}|\psi ⟩$ $=|{\psi }_1⟩$

${\widehat{P}}_{{\mathscr{H}}_2}|\psi ⟩$ $=|{\psi }_2⟩$

${\widehat{P}}_{{\mathscr{H}}_1}{\widehat{P}}_{{\mathscr{H}}_2}$ $=0$

${\widehat{P}}_{{\mathscr{H}}_1}+{\widehat{P}}_{{\mathscr{H}}_2}$ $=\widehat{I}$

We can generalise this to an arbitrary $\mathscr{H}$.
If $\mathscr{H}={\oplus }_{i=1}^n{\mathscr{H}}_i$, we define ${\widehat{P}}_{{\mathscr{H}}_i}$ s.t.

${\widehat{P}}_{{\mathscr{H}}_i}^2$ $={\widehat{P}}_{{\mathscr{H}}_i}$

${\widehat{P}}_{{\mathscr{H}}_i}{\widehat{P}}_{{\mathscr{H}}_j}$ $={\delta }_{ij}{\widehat{P}}_{{\mathscr{H}}_i}$

${\displaystyle \sum _{i=1}^n}{\widehat{P}}_{{\mathscr{H}}_i}$ $=\widehat{I}$

Example:
If the subspace $V\subset \mathscr{H}$ is spanned by the orthonomal basis $\{|i⟩\}$, $i=1,\mathrm{\dots },dim(V)$. Then,

${\widehat{P}}_v={\displaystyle \sum _{j=1}^{dim(V)}}|j⟩⟨j|$

Proof:
Consider the complement $V^c$ i.e. $\mathscr{H}=V\oplus V^c$ whereby $V^c$ is spanned by $\{|\alpha ⟩\}$ with the following relations,

$⟨i|j⟩$ $=⟨\alpha |\beta ⟩={\delta }_{ij}={\delta }_{\alpha \beta }$

$⟨i|\alpha ⟩$ $=0$

i.e. $\{|i⟩,|\alpha ⟩\}$ form an orthonormal basis for $\mathscr{H}$ and for any $|\psi ⟩\in \mathscr{H}$,

$|\psi ⟩$ $={\displaystyle \sum _{j=1}^{dim(V)}}{c}_j|j⟩+{\displaystyle \sum _{\alpha =1}^{dim(V^c)}}{d}_{\alpha }|\alpha ⟩$

$\Rightarrow {\widehat{P}}_V|\psi ⟩$ $={\displaystyle \sum _{j=1}^{dim(V)}}{c}_j|j⟩$

$={\displaystyle \sum _{j=1}^{dim(V)}}⟨j|\psi ⟩|j⟩$

$={\displaystyle \sum _{j=1}^{dim(V)}}|j⟩⟨j|\psi ⟩$

We now need to check that the property for projection operators hold under this definition.

${\widehat{P}}_V{\widehat{P}}_V$ $={\displaystyle \sum _{i,j=1}^{dim(V)}}|i⟩⟨i||j⟩⟨j|$

$={\displaystyle \sum _{i,j=1}^{dim(V)}}{\delta }_{ij}|i⟩⟨j|$

$={\displaystyle \sum _{i=1}^{dim(V)}}|i⟩⟨i|$

$={\widehat{P}}_V$

There is an important special case that we will use often in the future which is the case that $V=\mathscr{H}$, then

${\widehat{P}}_{\mathscr{H}}=\widehat{I}={\displaystyle \sum _{i=1}^{dim(\mathscr{H})}}|i⟩⟨i|$

We can now use the projection operators to find probabilities of observing a specific state.

Example:
Consider a quantum mechanical coin whereby $\mathscr{H}=\{|H⟩,|T⟩\}$. For $|\psi ⟩=a|H⟩+b|T⟩$,

$P(H)$ $=⟨\psi |{\widehat{P}}_H|\psi ⟩=⟨\psi |a|H⟩=a^{*}a={|a|}^2$

Alternatively,

$P(H)$ $=⟨\psi |{\widehat{P}}_H|\psi ⟩=⟨\psi |H⟩⟨H|\psi ⟩={|⟨H|\psi ⟩|}^2={|a|}^2$

1.6.2 Adjoint Operator

The adjoint ${\widehat{O}}^{+}$ of $\widehat{O}$ satisfies the following definition

$⟨V_1|{\widehat{O}}^{+}|V_2⟩={⟨V_2|\widehat{O}|V_1⟩}^{*}$

for $V_1,V_2\in \mathscr{H}$.
If $\widehat{A}=c_1\widehat{B}+c_2\widehat{C}$, then ${\widehat{A}}^{+}=c_1^{*}{\widehat{B}}^{+}+c_2{\widehat{C}}^{+}$
Also, under matrix representation, ${\widehat{O}}^{+}$ is the complex conjugate of the transpose matrix of $\widehat{O}$ since,

${({\widehat{O}}^{+})}_{ij}=⟨i|{\widehat{O}}^{+}|j⟩={⟨j|\widehat{O}|i⟩}^{*}={\widehat{O}}_{ji}^{*}$

1.6.3 Hermitian Operators and the Eigenvalue Problem

An operator $\widehat{O}$ is Hermitian if ${\widehat{O}}^{+}=\widehat{O}$.
The eigenvalue problem for $\widehat{O}$ is the set of all $w\in ℂ$ and $|w⟩\in \mathscr{H}$ s.t. $\widehat{O}|w⟩=w|w⟩$ whereby $w$ is the eigenvalue and $|w⟩\ne 0$.

The eigenvectors of a given value $w$ form a linear subspace $V_w$, the eigenspace corresponding to the eigenvalue $w$.
If $dim(V_w)=1$, then $w$ is non-degenerate.
If $dim(V_w)>1$, then $dim(V_w)$ is called the degeneracy of the eigenvalue $w$

Properties of eigenvalues of Hermitian Operators:

1. The eigenvalues of Hermitian $\widehat{O}$ are all real numbers

2. The different eigenspaces of $w$ are orthogonal, i.e. for $w_1\ne w_2$ and $|w_1⟩\in V_{w_1},|w_2⟩\in V_{w_2}$, then $⟨w_1|w_2⟩=0.$

3. The eigenvectors of $\widehat{O}$ can be used to construct an orthonormal basis of $\mathscr{H}$. If $V_{w_i}$ are all the different eigenspaces of $\widehat{O}$, then

   $\mathscr{H}={\displaystyle \underset{i=1}{\overset{n}{\oplus }}}{V}_{w_i}$

   This is known as the completeness of eigenvectors of Hermitian Operators

Proof:
1) If $\widehat{O}|w⟩=w|w⟩$, we want $w=w^{*}$.
Consider $⟨w|\widehat{O}|w⟩=w⟨w|w⟩$ and take the complex conjugate.

${⟨w|\widehat{O}|w⟩}^{*}$ $=⟨w|{\widehat{O}}^{+}|w⟩=⟨w|\widehat{O}|w⟩=w⟨w|w⟩$

${⟨w|\widehat{O}|w⟩}^{*}$ $=w^{*}⟨w|w⟩$

$\Rightarrow w$ $=w^{*}$

2) By definition, $\widehat{O}|w_1⟩=w_1|w_1⟩$ and $\widehat{O}|w_2⟩=w_2|w_2⟩$. Do the same as the proof above but by considering $⟨w_1|\widehat{O}|w_2⟩$.

$⟨w_1|\widehat{O}|w_2⟩$ $=w_2⟨w_1|w_2⟩$

$\Rightarrow {⟨w_1|\widehat{O}|w_2⟩}^{*}$ $=w_2⟨w_2|w_1⟩$

${⟨w_1|\widehat{O}|w_2⟩}^{*}$ $=⟨w_2|{\widehat{O}}^{+}|w_1⟩$

$=⟨w_2|\widehat{O}|w_1⟩$

$=w_1⟨w_2|w_1⟩$

$\Rightarrow (w_2-w_1)⟨w_2|w_1⟩$ $=0$

$\Rightarrow ⟨w_2|w_1⟩$ $=0$

In order to solve the eigenvalue problem, we need to solve $(\widehat{O}-w\widehat{I})|w⟩=0$.
For $|w⟩$ to be non-trivial, we need $\widehat{O}-w\widehat{I}$ to be non-invertible i.e. having determinant 0.
Hence the roots of the characteristic polynomial $p(w)=det(\widehat{O}-w\widehat{I})$ are the eigenvalues.
Since $p(w)$ has degree $dim(\mathscr{H})$, we can say

$p(w)={\displaystyle \prod _{i=1}^m}{(w-w_i)}^{{\lambda }_i}$

whereby we have $m$ different eigenvalues with corresponding multiplicity ${\lambda }_i$ and ${\sum }_{i=1}^m{\lambda }_i=dim(\mathscr{H})$.
Also $dim(V_{w_i})={\lambda }_i$.

1.7 Commutator

We can define the product between two linear operators $\widehat{A}\widehat{B}$ through composition i.e. $\widehat{A}\widehat{B}|\psi ⟩=\widehat{A}(\widehat{B}|\psi ⟩)$.
However the order matters and the commutator measures the difference,

$[\widehat{A},\widehat{B}]=\widehat{A}\widehat{B}-\widehat{B}\widehat{A}$

Identities:

1. $[\widehat{A},\widehat{B}]=-[\widehat{B},\widehat{A}]$

2. $[\widehat{A},\widehat{B}\widehat{C}]=[\widehat{A},\widehat{B}]\widehat{C}+\widehat{B}[\widehat{A},\widehat{C}]$

3. $[\widehat{A}\widehat{B},\widehat{C}]=\widehat{A}[\widehat{B},\widehat{C}]+[\widehat{A},\widehat{C}]\widehat{B}$

1.8 Simultaneous Diagonalisation of two Hermitian Operators

Theorem:
If $\widehat{O}$ and $\widehat{\mathrm{\Lambda }}$ are two commuting Hermitian Operators i.e. $[\widehat{O},\widehat{\mathrm{\Lambda }}]=0$, then there exists a common eigenbasis or a common eigenspace decomposition.

Suppose that $|w⟩\in V_w$ i.e. $\widehat{O}|w⟩=w|w⟩$. Then,

$\widehat{O}\widehat{\mathrm{\Lambda }}|w⟩=\widehat{\mathrm{\Lambda }}\widehat{O}|w⟩$ $=w\widehat{\mathrm{\Lambda }}|w⟩$

This means that $\widehat{\mathrm{\Lambda }}|w⟩\in V_w$ and that $\widehat{\mathrm{\Lambda }}:V_w\to V_w$. Hence,

$V_{w_i}={\displaystyle \underset{j=1}{\overset{d_i}{\oplus }}}{V}_{w_i,{\lambda }_{i,j}}$

whereby $d_i$ is the number of distinct eigenvalues of $\widehat{\mathrm{\Lambda }}$ appearing in the $i-th$ eigenspace of $w$.
In essence, this means that after we have decomposed $\mathscr{H}$ using the eigenvalues of $\widehat{O}(w_i)$, if we can find a commuting operator $\widehat{\mathrm{\Lambda }}$ we can further decompose $V_{w_i}$ in the eigenspaces of $\widehat{\mathrm{\Lambda }}$.
This helps in reducing the dimension of each individual eigenspace.

Example:
Consider

They are obviously Hermitian since they only have real entries and are symmetric. One can check that they commute by doing matrix multiplication. Also, we are given that $\widehat{O}$ has 2 eigenvalues with the corresponding eigenspaces:

$V_{w=2}$

$V_{w=0}$

Although there is not much point in decomposing the first eigenspace since it is non-degenarate already, we can see that $V_{w=2}=V_{w=2,\lambda =3}$.
So now, the main issue is we want to solve the eigenvalue problem of $\widehat{\mathrm{\Lambda }}$ in the eigenspace $V_{w=0}$ in order to decompose it. I.e. we want to show,

$V_{w=0}$

In order to solve the eigenvalue problem, we need the new matrix form of $\widehat{\mathrm{\Lambda }}$ constrained in this space.

${({\mathrm{\Lambda }}_{2\times 2})}_{ij}$ $=⟨w=0,i|\widehat{\mathrm{\Lambda }}|w=0,j⟩$

$\widehat{\mathrm{\Lambda }}|w=0,1⟩$ $=\widehat{\mathrm{\Lambda }}{\displaystyle \frac{1}{\sqrt{2}}}\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$

$={\displaystyle \frac{1}{\sqrt{2}}}\left(\begin{array}{c}\hfill -1\hfill \\ \hfill -2\hfill \\ \hfill -1\hfill \end{array}\right)$

$=|w=0,1⟩-\sqrt{2}|w=0,2⟩$

$\widehat{\mathrm{\Lambda }}|w=0,2⟩$ $=\widehat{\mathrm{\Lambda }}\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)$

$=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill -1\hfill \end{array}\right)$

$=-\sqrt{2}|w=0,1⟩$

$\Rightarrow {\mathrm{\Lambda }}_{2\times 2}$

Solving for its eigenvalues we have that they are $-1,2$ with corresponding eigenvectors $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill \sqrt{2}\hfill \end{array}\right)x,\left(\begin{array}{c}\hfill -\sqrt{2}\hfill \\ \hfill 1\hfill \end{array}\right)y$.

$\Rightarrow |w=0,\lambda =-1⟩$ $=x(|w=0,1⟩+\sqrt{2}|w=0,2⟩)$

$={\displaystyle \frac{1}{\sqrt{6}}}\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 2\hfill \\ \hfill 1\hfill \end{array}\right)$

$\Rightarrow |w=0,\lambda =-2⟩$ $=y(-\sqrt{2}|w=0,1⟩+|w=0,2⟩)$

$={\displaystyle \frac{1}{\sqrt{3}}}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \\ \hfill -1\hfill \end{array}\right)$

Note that we chose the constants $x,y$ s.t. the eigenvector is a unit vector.
Hence, we have decomposed $V_{w=0}$,

1.9 Spectral Decomposition

If $\widehat{O}$ is Hermitian and $w_i$ are its eigenvalues, then

$\widehat{O}={\displaystyle \sum _{i=1}^m}{w}_i{\widehat{P}}_{w_i}$

${\widehat{P}}_{w_i}$ is the projection operator on the eigenspace $V_{w_i}$.

Proof:
For any $|\psi ⟩\in \mathscr{H}$,

$\widehat{O}|\psi ⟩$ $=\widehat{O}\widehat{I}|\psi ⟩=\widehat{O}{\displaystyle \sum _{i=1}^m}{\widehat{P}}_{w_i}|\psi ⟩={\displaystyle \sum _{i=1}^m}\widehat{O}({\widehat{P}}_{w_i}|\psi ⟩)={\displaystyle \sum _{i=1}^m}{w}_i{\widehat{P}}_{w_i}|\psi ⟩$

$\Rightarrow \widehat{O}$ $={\displaystyle \sum _{i=1}^m}{w}_i{\widehat{P}}_{w_i}$

If $f(x)$ is a function $f:ℂ\to ℂ$ admits a taylor expansion. Then

$\widehat{f}(\widehat{O})={\displaystyle \sum _{i=1}^m}f(w_i){\widehat{P}}_{w_i}$

Proof: We know that since different eigenspaces are orthogonal, we must have

${\widehat{P}}_{w_i}{\widehat{P}}_{w_j}={\delta }_{ij}{\widehat{P}}_{w_i}.$ (1)

From equation (1) we more generally have that

${\left({\widehat{P}}_{w_i}\right)}^m{\left({\widehat{P}}_{w_j}\right)}^n={\delta }_{ij}{\widehat{P}}_{w_i}$ (2)

for any two natural numbers $m$ and $n$.

$\widehat{f}(\widehat{O})$ $={\displaystyle \sum _{i=0}^{\mathrm{\infty }}}{c}_i{\widehat{O}}^i={\displaystyle \sum _{i=1}^{\mathrm{\infty }}}{c}_i{\left({\displaystyle \sum _{j=1}^m}{w}_j{\widehat{P}}_{w_j}\right)}^i={\displaystyle \sum _{i=1}^{\mathrm{\infty }}}{c}_i{\displaystyle \sum _{j=1}^m}{w}_j^i{\widehat{P}}_{w_j}$

$={\displaystyle \sum _{j=1}^m}\left({\displaystyle \sum _{i=1}^{\mathrm{\infty }}}{c}_i{w}_j^i\right){\widehat{P}}_{w_j}={\displaystyle \sum _{j=1}^m}f(w_j){\widehat{P}}_{w_j},$

where we used equation (2) in expanding ${\sum }_{i=1}^{\mathrm{\infty }}{c}_i{\left({\sum }_{j=1}^m{w}_j{\widehat{P}}_{w_j}\right)}^i$.

Properties:
1) $V_{w_i}$ are also eigenspaces of $\widehat{f}(\widehat{O})$
2) The corresponding eigenvalues are $f(w_i)$

Proof:

$\widehat{f}(\widehat{O})|w_i⟩={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n{\widehat{O}}^n|w_i⟩={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n{w}_i^n|w_i⟩=f(w_i)|w_i⟩$

2 Principles of Quantum Mechanics

1. Quantum, states form a Hilbert space $\mathscr{H}$ with physical states having norm 1, i.e. $⟨\psi |\psi ⟩=1$.

2. Physical observables are represented by Hermitian Operators $\widehat{O}:\mathscr{H}\to \mathscr{H}$.

3. A measurement of $\widehat{O}$ given a state $|\psi ⟩$ can only return one of its eigenvalues $w_i$. The probability is $P(w_i)=⟨\psi |{\widehat{P}}_{w_i}|\psi ⟩$.

4. Right after a measurement which yielded the eigenvalue $w_i$, the state collapses to

   $|{\psi }_{after}⟩$ $=c{\widehat{P}}_{w_i}|{\psi }_{before}⟩$

   ${|c|}^2$ $={\displaystyle \frac{1}{⟨{\psi }_{before}|{\widehat{P}}_{w_i}|{\psi }_{before}⟩}}$

   $c$ is the normalising constant

In order for $P(w_i)$ to make sense, we need ${\sum }_{i=1}^{m}P(w_i)=1$.

${\displaystyle \sum _{i=1}^m}P(w_i)={\displaystyle \sum _{i=1}^m}⟨\psi |{\widehat{P}}_{w_i}|\psi ⟩=⟨\psi |{\displaystyle \sum _{i=1}^m}{\widehat{P}}_{w_i}|\psi ⟩=⟨\psi |\psi ⟩=1$

We can also derive the formula for expectation.

$={\displaystyle \sum _{i=1}^m}{w}_{i}P(w_i)={\displaystyle \sum _{i=1}^m}{w}_i⟨\psi |{\widehat{P}}_{w_i}|\psi ⟩=⟨\psi |{\displaystyle \sum _{i=1}^m}{w}_i{\widehat{P}}_{w_i}|\psi ⟩=⟨\psi |\widehat{O}|\psi ⟩$

More generally,

$⟨\widehat{f}(\widehat{O})⟩=⟨\psi |\widehat{f}(\widehat{O})|\psi ⟩$

It is also important to note that measurements cannot completely fix a quantum mechanical state.
Suppose that in the case of the QM coin, $|\psi ⟩=a|H⟩+b|T⟩$.

$P(H)$ $={|a|}^2=h$

$P(T)$ $={|b|}^2=t$

$\Rightarrow a$ $=h^{\frac{1}{2}}{e}^{i{\varphi }_1}$

$b$ $=t^{\frac{1}{2}}{e}^{i{\varphi }_2},{\varphi }_1,{\varphi }_2\in ℝ$

$\Rightarrow |\psi ⟩$ $=e^{i{\varphi }_1}(h^{\frac{1}{2}}|H⟩+t^{\frac{1}{2}}{e}^{i({\varphi }_2-{\varphi }_1)}|T⟩)$

The overall phase is not important but a relative phase remains unfixed.

3 Time Evolution of Quantum Mechanical states

Quantum Mechanical systems have a special Hermitian operator $\widehat{H}$, the Hamiltonian operator which obeys the Schrodinger equation.

$$i\mathrm{\hslash }\frac{d}{dt}|\psi (t)⟩=\widehat{H}(t)|\psi (t)⟩$$ (3)

However, we assume that we are dealing with isolated systems i.e. $\widehat{H}$ is independent of time. Also, we need to find $|\psi (t)⟩$ s.t. it satisfies Schrodinger’s equation and $|\psi (t=0)⟩=|{\psi }_0⟩$. We try the following,

$|\psi (t)⟩$ $=e^{-i\frac{\widehat{H}t}{\mathrm{\hslash }}}|{\psi }_0⟩$

${\displaystyle \frac{d}{dt}}|\psi (t)⟩$ $=-i{\displaystyle \frac{\widehat{H}}{\mathrm{\hslash }}}{e}^{-i\frac{\widehat{H}t}{\mathrm{\hslash }}}|{\psi }_0⟩$

$\Rightarrow i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}|\psi (t)⟩$ $=\widehat{H}|\psi (t)⟩$

Note that in general $[\widehat{H}(t_1),\widehat{H}(t_2)]\ne 0$.

We can represent $|\psi (t)⟩$ in terms of the eigenbasis of $\widehat{H}$.
Method 1:
We use spectral decomposition of $\widehat{H}$ which has eigenvalues $E_i,i=1,\mathrm{\dots },m$ and ${\widehat{P}}_{E_i}$ are the respective projection operations.

$e^{-i\frac{\widehat{H}}{\mathrm{\hslash }}t}={\displaystyle \sum _{i=1}^m}{e}^{-i\frac{E_i}{\mathrm{\hslash }}t}{\widehat{P}}_{E_i}$

If $|E_i,j⟩$ is an orthonormal basis of $V_{E_i}$, then

${\widehat{P}}_{E_i}$ $={\displaystyle \sum _{j=1}^{dim(V_{E_i})}}|E_i,j⟩⟨E_i,j|$

$\Rightarrow |\psi (t)⟩$ $={\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^{dim(V_{E_i})}}{e}^{-i\frac{E_i}{\mathrm{\hslash }}t}|E_i,j⟩⟨E_i,j|{\psi }_0⟩$

Method 2:
The initial state $|\psi (t=0)⟩$ can be written as a linear combination of the eigenbasis,

$|\psi (t=0)⟩$ $={\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^{dim(V_{E_i})}}{c}_{ij}|E_i,j⟩$

$\Rightarrow |\psi (t)⟩$ $=e^{-i\frac{\widehat{H}}{\mathrm{\hslash }}t}{\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^{dim(V_{E_i})}}{c}_{ij}|E_i,j⟩$

$={\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^{dim(V_{E_i})}}{c}_{ij}{e}^{-i\frac{\widehat{H}}{\mathrm{\hslash }}t}|E_i,j⟩$

$={\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^{dim(V_{E_i})}}{c}_{ij}{e}^{-i\frac{E_i}{\mathrm{\hslash }}t}|E_i,j⟩$

Recall that,

$|\psi ⟩$ $={\displaystyle \sum _{i=1}^m}{c}_i|i⟩\Rightarrow c_i=⟨i|\psi ⟩$

$\Rightarrow c_{ij}$ $=⟨E_i,j|\psi (t=0)⟩$

We also observe that if the initial state $|{\psi }_0⟩$ is an eigenvector of $\widehat{H}$ with eigenvalue $E$, then $|\psi (t)⟩=e^{-i\frac{E}{\mathrm{\hslash }}t}|{\psi }_0⟩$.

3.1 Time evolution of expectation values

Ehrenfest’s Theorem:

$$\frac{d}{dt}⟨\widehat{\mathrm{\Lambda }}⟩=\frac{i}{\mathrm{\hslash }}⟨[\widehat{H},\widehat{\mathrm{\Lambda }}]⟩$$ (4)

Proof:

${\displaystyle \frac{d}{dt}}⟨\widehat{\mathrm{\Lambda }}⟩$ $={\displaystyle \frac{d}{dt}}⟨\psi (t)|\widehat{\mathrm{\Lambda }}|\psi (t)⟩$

$=\left[{\displaystyle \frac{d}{dt}}⟨\psi (t)|\right]\widehat{\mathrm{\Lambda }}|\psi (t)⟩+⟨\psi (t)|{\displaystyle \frac{d}{dt}}(\widehat{\mathrm{\Lambda }}|\psi (t)⟩)$

${\left(i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}|\psi (t)⟩\right)}^{+}$ $={(\widehat{H}(t)|\psi (t)⟩)}^{+}$

$\Rightarrow -i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}⟨\psi (t)|$ $=⟨\psi (t)|\widehat{H}(t)$

$\Rightarrow {\displaystyle \frac{d}{dt}}⟨\widehat{\mathrm{\Lambda }}⟩$ $={\displaystyle \frac{i}{\mathrm{\hslash }}}⟨\psi (t)|\widehat{H}(t)\widehat{\mathrm{\Lambda }}|\psi (t)⟩-{\displaystyle \frac{i}{\mathrm{\hslash }}}⟨\psi (t)|\widehat{\mathrm{\Lambda }}\widehat{H}(t)|\psi (t)⟩$

$={\displaystyle \frac{i}{\mathrm{\hslash }}}⟨[\widehat{H},\widehat{\mathrm{\Lambda }}]⟩$

Observations:
We can write $⟨\psi (t)|\psi (t)⟩=⟨\psi (t)|\widehat{I}|\psi (t)⟩=⟨\widehat{I}⟩$.

${\displaystyle \frac{d}{dt}}⟨\widehat{I}⟩={\displaystyle \frac{i}{\mathrm{\hslash }}}⟨[\widehat{H},\widehat{I}]⟩=0$

This means that the norm of a state remains invariant under time evolution.

If $\widehat{\mathrm{\Lambda }}$ commutes with $\widehat{H}$, then $⟨\widehat{\mathrm{\Lambda }}⟩$ is independent of time and is a conserved quantity. Hence, operators that commute with the Hamiltonian has its expectation being a conserved quantity and can be simultanueously diagonalised with $\widehat{H}$.

Example: (Spin in an external magnetic field)
An electron has a two-dimensional ’internal’ spin Hilbert space. It is described by 3-pauli matrices ${\widehat{S}}_i,i=x,y,z$. They satisfy

$[{\widehat{S}}_i,{\widehat{S}}_j]=i\mathrm{\hslash }{ϵ}_{ijk}{\widehat{S}}_k$

We study the time evolution of spin according to

$\widehat{H}$ $=\gamma \widehat{S}𝑩,\gamma \in ℝ\mathit{\hspace{1em}}\gamma ={\displaystyle \frac{e}{m_{e}c}}$

$\widehat{S}𝑩$ $=B_x{\widehat{S}}_x+B_y{\widehat{S}}_y+B_z{\widehat{S}}_z$

We choose a magnetic field s.t. $B_x=B_y=0$, hence $\widehat{H}=\gamma B_z{\widehat{S}}_z$.

${\displaystyle \frac{d}{dt}}⟨{\widehat{S}}_z⟩={\displaystyle \frac{i\gamma B_z}{\mathrm{\hslash }}}⟨[{\widehat{S}}_z,{\widehat{S}}_z]⟩=0$

This means that $⟨{\widehat{S}}_z⟩$ does not change with time i.e. conserved.

$[{\widehat{S}}_x,{\widehat{S}}_z]$ $=i\mathrm{\hslash }{ϵ}_{xzy}{\widehat{S}}_y=-i\mathrm{\hslash }{\widehat{S}}_y$

$=i\mathrm{\hslash }{\widehat{S}}_x$

${\displaystyle \frac{d}{dt}}⟨{\widehat{S}}_x⟩$ $={\displaystyle \frac{i{B}_z\gamma }{\mathrm{\hslash }}}⟨[{\widehat{S}}_z,{\widehat{S}}_x]⟩$

$=-B_z\gamma ⟨{\widehat{S}}_y⟩$

${\displaystyle \frac{d}{dt}}⟨{\widehat{S}}_y⟩$ $={\displaystyle \frac{i{B}_z\gamma }{\mathrm{\hslash }}}⟨[{\widehat{S}}_z,{\widehat{S}}_y]⟩$

$=B_z\gamma ⟨{\widehat{S}}_x⟩$

So we have two couple ODEs which we can solve simultaneously.

${\displaystyle \frac{d^2}{d{t}^2}}⟨{\widehat{S}}_x⟩$ $=-B_z\gamma {\displaystyle \frac{d}{dt}}⟨{\widehat{S}}_y⟩$

$=-B_z^2{\gamma }^2⟨{\widehat{S}}_x⟩$

$\Rightarrow ⟨{\widehat{S}}_x⟩$ $=b\cos (B_z\gamma t)+a\sin (B_z\gamma t)\mathit{\hspace{1em}}a,b\in ℝ$

$\Rightarrow ⟨{\widehat{S}}_y⟩$ $=b\sin (B_z\gamma t)-a\cos (B_z\gamma t)$

$\Rightarrow \left(\begin{array}{c}\hfill ⟨{\widehat{S}}_x⟩\hfill \\ \hfill ⟨{\widehat{S}}_y⟩\hfill \end{array}\right)$

The constants $a,b$ are fixed by initial conditions and we have showed that the expectations transform in time by a rotation of $\frac{\pi }{2}-B_z\gamma t$.

3.2 Unitary operators

Any operator $\widehat{U}$ s.t. $\widehat{U}{\widehat{U}}^{+}=\widehat{I}$ is called unitary. Unitary operators preserve the inner product.

Suppose that ${|{\psi }_1⟩}^{\prime }=\widehat{U}|{\psi }_1⟩,{|{\psi }_2⟩}^{\prime }=\widehat{U}|{\psi }_2⟩$.

$⟨{\psi }_2^{\prime }|{\psi }_1^{\prime }⟩=⟨{\psi }_2|{\widehat{U}}^{+}\widehat{U}|{\psi }_1⟩=⟨{\psi }_2|{\psi }_1⟩$

These operators generalise rotation in Euclidean space.

We saw before that the norm of the state does not change under time evolution.

$\widehat{U}(t)$ $=e^{\frac{-it}{\mathrm{\hslash }}\widehat{H}}$

$\Rightarrow {\widehat{U}}^{+}(t)$ $=e^{\frac{it}{\mathrm{\hslash }}\widehat{H}}$

$\Rightarrow {\widehat{U}}^{+}(t)$ $={\widehat{U}}^{-1}(t)$

We can show that time evolution is unitary in general by writing $|\psi (t)⟩=\widehat{U}(t)|\psi (t=0)⟩$ and demanding that it satisfies Schrodinger’s equation.

$\left(i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}-\widehat{H}(t)\right)\widehat{U}(t)|\psi (t)⟩$ $=0$

$\Rightarrow i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}\widehat{U}(t)$ $=\widehat{H}(t)\widehat{U}(t)$

$\Rightarrow -i\mathrm{\hslash }{\displaystyle \frac{d}{dt}}{\widehat{U}}^{+}(t)$ $={\widehat{U}}^{+}(t)\widehat{H}(t)$

We now consider the derivative of ${\widehat{U}}^{+}(t)\widehat{U}(t)$.

${\displaystyle \frac{d}{dt}}({\widehat{U}}^{+}(t)\widehat{U}(t))$ $=\left({\displaystyle \frac{d}{dt}}{\widehat{U}}^{+}(t)\right)\widehat{U}(t)+{\widehat{U}}^{+}(t){\displaystyle \frac{d}{dt}}\widehat{U}(t)$

$={\displaystyle \frac{i}{\mathrm{\hslash }}}{\widehat{U}}^{+}(t)\widehat{H}(t)\widehat{U}(t)-{\displaystyle \frac{i}{\mathrm{\hslash }}}{\widehat{U}}^{+}(t)\widehat{H}(t)\widehat{U}(t)$

$=0$

This means that ${\widehat{U}}^{+}(t)\widehat{U}(t)$ does not change in time.

${\widehat{U}}^{+}(t)\widehat{U}(t)={\widehat{U}}^{+}(t=0)\widehat{U}(t=0)=\widehat{I}$

Since $\widehat{U}(t=0)=\widehat{I}$.

4 Generalisation to infinite dimensional Hilbert spaces

4.1 Position Operator

So far we are given an observable $\widehat{O}$ and knowing the possible outcomes gives us the spectrum $w_i$ (eigenvalues), there is a complete basis $|w_i⟩$ which is also an eigenbasis of $\widehat{O}$. Hence, for a particle on the real line, we have a Hermitian position operator $\widehat{x}$ with its discrete spectrum $x_i$ and corresponding eigenbasis $|x_i⟩$.

We promote $x_i$ to eigenvalues and $\widehat{x}|x_i⟩=x_i|x_i⟩$. We should also have $⟨x_i|x_j⟩={\delta }_{ij}$. The $|x_i⟩$s are now a complete basis.

$\Rightarrow |\psi ⟩$ $={\displaystyle \sum _{i=0}^N}{c}_i|x_i⟩$

$={\displaystyle \sum _{i=0}^N}⟨x_i|\psi ⟩|x_i⟩$

$={\displaystyle \sum _{i=0}^N}\psi (x_i)|x_i⟩$

whereby we denote $\psi (x_i)$ is the wavefunction which will be useful in expressing the probability of finding the eigenvalue $x_i$ in a measurement of $\widehat{x}$.

$P(x_i)$ $=⟨\psi |{\widehat{P}}_{x_i}|\psi ⟩$

${\widehat{P}}_{x_i}$ $=|x_i⟩⟨x_i|$

$\Rightarrow P(x_i)$ $=⟨\psi |x_i⟩⟨x_i|\psi ⟩={|\psi (x_i)|}^2$

By extension, the probability to find the particle between $x_i$ and $x_{i+h}$ is

$P([x_i,x_{i+h}])={\displaystyle \sum _{j=i}^{i+h}}P(x_j)={\displaystyle \sum _{j=i}^{i+h}}{|\psi (x_j)|}^2$

To discuss the continuum limit we segment the line of length $L$ by taking intervals of length $\mathrm{\Delta }x=\frac{L}{N}\to 0$ i.e. $N\to \mathrm{\infty }$.
This is problematic, we show this by considering an arbitrary vector

$|f⟩$ $={\displaystyle \sum _{i=0}^N}f(x_i)|x_i⟩$

$\Rightarrow ⟨f|f⟩$ $={\displaystyle \sum _{i=0}^N}{|f(x_i)|}^2$

Hence in the $N\to \mathrm{\infty }$ limit, the norm blows up and is not normalisable. So, we modify the inner product to

$⟨f|f⟩$ $={\displaystyle \frac{L}{N}}{\displaystyle \sum _{i=0}^N}{|f(x_i)|}^2=\mathrm{\Delta }x{\displaystyle \sum _{i=0}^N}{|f(x_i)|}^2\to {\displaystyle {\int }_a^b}{|f(x)|}^2𝑑x\mathit{\hspace{1em}}asN\to \mathrm{\infty }$

More generally, for the inner product

$⟨f_1|f_2⟩=\mathrm{\Delta }x{\displaystyle \sum _{i=0}^N}{f}_1^{*}(x_i)f_2(x_i)\to {\displaystyle {\int }_a^b}{f}_1^{*}(x)f_2(x)𝑑x\mathit{\hspace{1em}}asN\to \mathrm{\infty }$

This modifications mean that there is change for the product between basis vectors $\{x_i\}$. When we take $N\to \mathrm{\infty }$, we are replacing the discrete $x_i$ by a continuous $x$ and $|x_i⟩\to |x⟩$.
The $|x⟩$ are still eigenvalues for a Hermitian $\widehat{x}$. Also, $⟨x|x^{\prime }⟩=0$ for $x\ne x^{\prime }$.
We now need to check for the case that $x=x^{\prime }$ by using completeness,

$\widehat{I}$ $={\displaystyle {\int }_a^b}|x⟩⟨x|𝑑x$

$f(x)$ $=⟨x|f⟩$

$=⟨x|\widehat{I}|f⟩$

$={\displaystyle {\int }_a^b}⟨x|x^{\prime }⟩⟨x^{\prime }|f⟩𝑑x^{\prime }$

$={\displaystyle {\int }_a^b}⟨x|x^{\prime }⟩f(x^{\prime })𝑑x^{\prime }$

$\Rightarrow ⟨x|x^{\prime }⟩$ $=\delta (x-x^{\prime })$

The probability to find the particle in the interval $[a,b]$ wiil then be

$P([a,b])$ $=⟨\psi |{\displaystyle {\int }_a^b}|x⟩⟨x|𝑑x|\psi ⟩={\displaystyle {\int }_a^b}⟨\psi |x⟩⟨x|\psi ⟩𝑑x={\displaystyle {\int }_a^b}{|\psi (x)|}^2𝑑x$

$\Rightarrow ⟨\widehat{x}⟩$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}x{|\psi (x)|}^2𝑑x$

Note that we derived the expression for the expectation by considering ${|\psi (x)|}^2$ as the probability density for position.
Alternatively,

$⟨\widehat{x}⟩$ $=⟨\psi |\widehat{x}|\psi ⟩$

$=⟨\psi |\widehat{x}\widehat{I}|\psi ⟩$

$=⟨\psi |\widehat{x}{\displaystyle {\int }_a^b}|x⟩⟨x|𝑑x|\psi ⟩$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}⟨\psi |\widehat{x}|x⟩⟨x|\psi ⟩𝑑x$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}x⟨\psi |x⟩⟨x|\psi ⟩𝑑x$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}x{|\psi (x)|}^2𝑑x$

4.2 Momentum Operator

In classical mechanics, momentum is the generator of translations.
Suppose that ${\widehat{T}}_{l_1}$ translates states by $l_1$ If ${\psi }_{l_1}(x)=⟨x|{\psi }_{l_1}⟩$, ${\psi }_{l_1}(x)=\psi (x+l_1)$. Then

$⟨x|{\widehat{T}}_{l_1}|\psi ⟩=⟨x|{\psi }_{l_1}⟩=⟨x+l_1|\psi ⟩$

We observe that ${\widehat{T}}_l$ forms a group since ${\widehat{T}}_{l_1}{\widehat{T}}_{l_2}={\widehat{T}}_{l_1+l_2}$ and ${\widehat{T}}_{l=0}=\widehat{I}$.
For small translations,

${\widehat{T}}_{ϵ}$ $=\widehat{I}+\widehat{G}ϵ+O({ϵ}^2)$

$\Rightarrow ⟨x|\widehat{I}+ϵ\widehat{G}|\psi ⟩$ $=⟨x+ϵ|\psi ⟩$

$=\psi (x+ϵ)$

$=\psi (x)+ϵ{\psi }^{\prime }(x)+O({ϵ}^2)$

$=⟨x|\psi ⟩+ϵ{\displaystyle \frac{d}{dx}}⟨x|\psi ⟩+O({ϵ}^2)$

$\Rightarrow ⟨x|\widehat{G}|\psi ⟩$ $={\displaystyle \frac{d}{dx}}⟨x|\psi ⟩$

$\widehat{G}$ is defined as the generator of the translation. So we should look for a constant $c$ s.t. $\widehat{p}=c\widehat{G}$ and it should be hermitian while having dimensions of momentum i.e. $m\frac{s}{t}$.
For hermiticity, we check the adjoint ${\widehat{G}}^{+}$ by considering its matrix representation.

$G_{x{x}^{\prime }}$ $=⟨x|\widehat{G}|x^{\prime }⟩$

$={\displaystyle \frac{d}{dx}}⟨x|x^{\prime }⟩$

$={\displaystyle \frac{d}{dx}}\delta (x-x^{\prime })$

$\Rightarrow G_{x{x}^{\prime }}^{*}$ $={\displaystyle \frac{d}{dx}}\delta (x-x^{\prime })$

$G_{x^{\prime }x}$ $={\displaystyle \frac{d}{d{x}^{\prime }}}\delta (x^{\prime }-x)$

$=-{\displaystyle \frac{d}{dx}}\delta (x-x^{\prime })$

$=-G_{x{x}^{\prime }}^{*}$

Hence, $\widehat{G}$ is anti-Hermitian so for $c$ we should have $c^{*}=-c$ so that

${\widehat{p}}^{+}=c^{*}{\widehat{G}}^{+}=-c(-\widehat{G})=c\widehat{G}=\widehat{p}$

Since $\widehat{G}$ has units $\frac{1}{s}$, we need $c$ to have units $\frac{m{s}^2}{t}$. Hence we choose $c=-i\mathrm{\hslash }$.

$$⟨x|\widehat{p}|\psi ⟩=-i\mathrm{\hslash }\frac{d}{dx}⟨x|\psi ⟩$$ (5)

Last year, we had $\widehat{p}\psi (x)=-i\mathrm{\hslash }\frac{d}{dx}\psi (x)$ which is the same expression as the equation above.
Also, it makes sense to write $\frac{d}{dt}|\psi (t)⟩$ but no sense to write $\frac{d}{dx}|\psi ⟩$ but rather $\frac{d}{dx}⟨x|\psi ⟩$.

Now we need to inspect the commutator between the position and momentum operators.

$[\widehat{x},\widehat{p}]|\psi ⟩$ $=\widehat{I}[\widehat{x},\widehat{p}]|\psi ⟩$

$={\displaystyle \int 𝑑x|x⟩⟨x|\widehat{x}\widehat{p}|\psi ⟩}-{\displaystyle \int 𝑑x|x⟩⟨x|\widehat{p}\widehat{x}|\psi ⟩}$

$⟨x|\widehat{x}\widehat{p}|\psi ⟩$ $=x⟨x|\widehat{p}|\psi ⟩$

$=-i\mathrm{\hslash }x{\displaystyle \frac{d}{dx}}⟨x|\psi ⟩$

$⟨x|\widehat{p}\widehat{x}|\psi ⟩$ $=-i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}⟨x|\widehat{x}|\psi ⟩$

$=-i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}x⟨x|\psi ⟩$

$=-i\mathrm{\hslash }⟨x|\psi ⟩-i\mathrm{\hslash }x{\displaystyle \frac{d}{dx}}⟨x|\psi ⟩$

$\Rightarrow [\widehat{x},\widehat{p}]|\psi ⟩$ $=i\mathrm{\hslash }{\displaystyle \int 𝑑x|x⟩⟨x|\psi ⟩}$

$=i\mathrm{\hslash }\left({\displaystyle \int 𝑑x|x⟩⟨x|}\right)|\psi ⟩$

$=i\mathrm{\hslash }|\psi ⟩$

$\Rightarrow [\widehat{x},\widehat{p}]$ $=i\mathrm{\hslash }$

For finite translations ${\widehat{T}}_l$, we break $l$ into $N$ small pieces of $ϵ$ and use the group property.

${\widehat{T}}_l$ $={\widehat{T}}_{ϵ}\mathrm{\dots }{\widehat{T}}_{ϵ}$

$={\left(\widehat{I}+ϵ{\displaystyle \frac{i\widehat{p}}{\mathrm{\hslash }}}\right)}^N$

$={\left(\widehat{I}+{\displaystyle \frac{i\widehat{p}}{\mathrm{\hslash }}}{\displaystyle \frac{l}{N}}\right)}^N$

$\to e^{i\frac{\widehat{p}}{\mathrm{\hslash }}l}\mathit{\hspace{1em}}asN\to \mathrm{\infty }$

$\widehat{p}$ generates translations in space whereas before we have shown that $\widehat{H}$ generates translations in time. Hence similarly, ${\widehat{T}}_l$ is unitary since ${\widehat{T}}_l^{+}{\widehat{T}}_l=\widehat{I}$.

4.2.1 Momentum Eigenbasis

We want to find the eigenbasis $|p⟩$ s.t. it satisfies

$\widehat{p}|p⟩$ $=p|p⟩$

$\widehat{I}$ $={\displaystyle \int 𝑑p|p⟩⟨p|}$

$⟨p|p^{\prime }⟩$ $=\delta (p-p^{\prime })$

In order to find $|p⟩$ we express it in terms of $|x⟩$.

$|p⟩$ $=\widehat{I}|p⟩$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x⟨x|p⟩|x⟩$

So we need to find $⟨x|p⟩$

$⟨x|\widehat{p}|p⟩$ $=p⟨x|p⟩$

$\Rightarrow -i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}⟨x|p⟩$ $=p⟨x|p⟩$

$\Rightarrow {\displaystyle \frac{d}{dx}}⟨x|p⟩$ $={\displaystyle \frac{ip}{\mathrm{\hslash }}}⟨x|p⟩$

$\Rightarrow ⟨x|p⟩$ $=c_p{e}^{\frac{ip}{\mathrm{\hslash }}x}$

We fix the constants $c_p$ by demanding that the eigenbasis has unit norm.

$⟨p|p^{\prime }⟩$ $=⟨p|{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x|x⟩⟨x||p^{\prime }⟩$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x⟨x|p⟩⟨x|p^{\prime }⟩$

$=c_p{c}_{p^{\prime }}^{*}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{\frac{-i}{\mathrm{\hslash }}(p-p^{\prime })x}𝑑x$

$=\mathrm{\hslash }{c}_p{c}_{p^{\prime }}^{*}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{i(p^{\prime }-p)y}𝑑y$

$=2\pi \mathrm{\hslash }{c}_p{c}_{p^{\prime }}^{*}\delta (p^{\prime }-p)$

$\Rightarrow c_p$ $={\displaystyle \frac{1}{\sqrt{2\pi \mathrm{\hslash }}}}$

$\Rightarrow ⟨x|p⟩$ $={\displaystyle \frac{1}{\sqrt{2\pi \mathrm{\hslash }}}}{e}^{\frac{ipx}{\mathrm{\hslash }}}$

4.3 The position and momentum representations

We have the $|x⟩$ and $|p⟩$ eigenbasis. So we can write any vector in these bases,

$|\psi ⟩$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|x⟩⟨x|\psi ⟩𝑑x$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\psi (x)|x⟩𝑑x$

$|\psi ⟩$ $={\displaystyle \int 𝑑p|p⟩⟨p|\psi ⟩}$

$={\displaystyle \int \tilde{\psi }(p)|p⟩𝑑p}$

$\tilde{\psi }(p)$ $=⟨p|\psi ⟩$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x⟨p|x⟩⟨x|\psi ⟩$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x⟨p|x⟩\psi (x)$

$={\displaystyle \frac{1}{\sqrt{2\pi \mathrm{\hslash }}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{ipx}{\mathrm{\hslash }}}\psi (x)𝑑x$

whereby $\tilde{\psi }(p)$ is the analogue of $\psi (x)$ s.t. ${|\tilde{\psi }(p)|}^2$ is the probability density of momentum i.e.

$$P([p_1,p_2])=⟨\psi |{\widehat{P}}_{[p_1,p_2]}|\psi ⟩=⟨\psi |{\int }_{p_1}^{p_2}𝑑p|p⟩⟨p||\psi ⟩={\int }_{p_1}^{p_2}{|\tilde{\psi }(p)|}^2𝑑p$$ (6)

We can also examine how the operators $\widehat{x},\widehat{p}$ act under these eigenbases.

$⟨x|\widehat{x}|\psi ⟩$ $=x⟨x|\psi ⟩=x\psi (x)$

$⟨p|\widehat{p}|\psi ⟩$ $=p\tilde{\psi }(p)$

$⟨x|\widehat{p}|\psi ⟩$ $=-i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}⟨x|\psi ⟩=-i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}\psi (x)$

$⟨p|\widehat{x}|\psi ⟩$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}⟨p|x^{\prime }⟩⟨x^{\prime }|\widehat{x}|\psi ⟩𝑑x^{\prime }$

$={\displaystyle \frac{1}{\sqrt{2\pi \mathrm{\hslash }}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{x}^{\prime }\psi (x^{\prime })e^{-\frac{ip{x}^{\prime }}{\mathrm{\hslash }}}𝑑x^{\prime }$

$={\displaystyle \frac{1}{\sqrt{2\pi \mathrm{\hslash }}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}i\mathrm{\hslash }{\displaystyle \frac{d}{dp}}(e^{-\frac{ip{x}^{\prime }}{\mathrm{\hslash }}})\psi (x^{\prime })𝑑x^{\prime }$

$=i\mathrm{\hslash }{\displaystyle \frac{d}{dp}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}⟨p|x^{\prime }⟩⟨x^{\prime }|\psi ⟩𝑑x^{\prime }$

$=i\mathrm{\hslash }{\displaystyle \frac{d}{dp}}⟨p|\psi ⟩$

$=i\mathrm{\hslash }{\displaystyle \frac{d}{dp}}\tilde{\psi }(p)$

4.4 Canonical Quantisation of Classical Systems

The independent $x$ and $p$ classical variables become $\widehat{x},\widehat{p}$ in Quantum Mechanics.
The classical observables $O(x,p)$ become Hermitian operators which can be constructed by substituting $\widehat{O}(\widehat{x},\widehat{p})=O(x\to \widehat{x},p\to \widehat{p})$.

For 2 dimensions $|x,y⟩$, $\widehat{x}$ and $\widehat{y}$ should be degenerate i.e.

$\widehat{x}|x,y⟩$ $=x|x,y⟩$

$\widehat{y}|x,y⟩$ $=y|x,y⟩$

$=0$

In higher dimensions, we have cartesian coordinates $x_i\to {\widehat{x}}_i$ with $[{\widehat{x}}_i,{\widehat{x}}_j]=0$.

${\widehat{x}}_i|x_1,x_2,\mathrm{\dots },x_n⟩=x_i|x_1,\mathrm{\dots },x_n⟩$

There are also the corresponding momentum operators ${\widehat{p}}_i$ satisfying

$[{\widehat{p}}_i,{\widehat{p}}_j]$ $=0$

$=i\mathrm{\hslash }{\delta }_{ij}$

Examples:
The Hamiltonian of a particle in a 3D potential $V(x,y,)$. The classical Hamiltonian is

$H$ $={\displaystyle \frac{p_x^2+p_y^2+p_z^2}{2m}}+V(x,y,z)$

$\Rightarrow \widehat{H}$ $={\displaystyle \frac{{\widehat{p}}_x^2+{\widehat{p}}_y^2+{\widehat{p}}_z^2}{2m}}+\widehat{V}(\widehat{x},\widehat{y},\widehat{z})$

Promoting $x,y,z$ to operators in the potential term is unambiguous since as operators they all commute with each other. E.g. if $V(x,y)=xy$, then $\widehat{V}=\widehat{x}\widehat{y}=\widehat{y}\widehat{x}$. However, in the following example, the choice matters.

$H={\displaystyle \frac{p^2}{2m}}+px$

The problem arises in the second term $px$ since as operators they do not commute. Do we choose $\widehat{p}\widehat{x}$ or $px=xp\to \widehat{x}\widehat{p}$? Actually we choose $\frac{\widehat{x}\widehat{p}+\widehat{p}\widehat{x}}{2}$ to preserve Hermiticity.

4.5 Schrodinger’s Wave Mechanics

Schrodinger’s wave equation of a 3D particle in $V(x,y,z)$ under Dirac’s formalism is

$$i\mathrm{\hslash }\frac{d}{dt}|\psi (t)⟩=\frac{1}{2m}({\widehat{p}}_x^2+{\widehat{p}}_y^2+{\widehat{p}}_z^2)|\psi (t)⟩+\widehat{V}(\widehat{x},\widehat{y},\widehat{z})|\psi (t)⟩$$ (7)

In Schrodinger’s wave mechanics, we want to describe the wave function $\psi (x,y,z,t)=⟨x,y,z|\psi (t)⟩$. So we multiply the equation by $⟨x,y,z|$.

$$\Rightarrow -i\mathrm{\hslash }\frac{\partial }{\partial t}\psi (x,y,z,t)=-\frac{{\mathrm{\hslash }}^2}{2m}({\partial }_x^2+{\partial }_y^2+{\partial }_z^2)\psi (x,y,z,t)+V(x,y,z)\psi (x,y,z,t)$$ (8)

For the second term, we used the fact that $\widehat{V}(\widehat{x},\widehat{y},\widehat{z})|x,y,z⟩=V(x,y,z)|x,y,z⟩$ and taking the adjoint. Also, we used the lemma that if $f$ is real then ${(\widehat{f}(\widehat{O}))}^{+}=\widehat{f}({\widehat{O}}^{+})$.

5 Simple Harmonic Oscillator

The Hamiltonian for a 1D particle is

$$\widehat{H}=\frac{{\widehat{p}}^2}{2m}+\frac{1}{2}m{\omega }^2{\widehat{x}}^2$$ (9)

We are interested in the energy spectrum of the SHO, so we want to solve the eigenvalue problem $\widehat{H}|E⟩=E|E⟩$. However, before we do that, we can deduce that all the eigenvalues of $\widehat{H}$ are non-negative.

We want to show that $\widehat{H}$ is a positive definite operator i.e. $⟨\psi |\widehat{H}|\psi ⟩>0\forall |\psi ⟩\ne 0$.
First we consider a general Hermitian operator $\widehat{A}$.

$⟨\psi |{\widehat{A}}^2|\psi ⟩=⟨\psi |\widehat{A}\widehat{A}|\psi ⟩=⟨\psi |{\widehat{A}}^{+}\widehat{A}|\psi ⟩={|\widehat{A}|\psi ⟩|}^2\ge 0$

Choose $|\psi ⟩=|E⟩$ and $\widehat{A}=\widehat{H}$

$⟨E|\widehat{H}|E⟩$ $={\displaystyle \frac{1}{2m}}⟨\psi |{\widehat{p}}^2|\psi ⟩+{\displaystyle \frac{1}{2}}mw{\omega }^2⟨\psi |{\widehat{x}}^2|\psi ⟩\ge 0$

$⟨E|\widehat{H}|E⟩$ $=E⟨E|E⟩=E$

$\Rightarrow E$ $\ge 0$

To solve the eigenvalue problem we need the creation and annihilation operators.

$\widehat{a}$ $={\displaystyle \frac{1}{\sqrt{2m\mathrm{\hslash }\omega }}}(m\omega \widehat{x}+i\widehat{p})$

${\widehat{a}}^{+}$ $={\displaystyle \frac{1}{\sqrt{2m\mathrm{\hslash }\omega }}}(m\omega \widehat{x}-i\widehat{p})$

We can then use these to redefine $\widehat{x},\widehat{p}$.

$\widehat{x}$ $=\sqrt{{\displaystyle \frac{\mathrm{\hslash }}{2m\omega }}}(\widehat{a}+{\widehat{a}}^{+})$

$\widehat{p}$ $=-i\sqrt{{\displaystyle \frac{\mathrm{\hslash }m\omega }{2}}}(\widehat{a}-{\widehat{a}}^{+})$

$=1$

For the last equation we used the fact that $[\widehat{x},\widehat{p}]=i\mathrm{\hslash }$. Using these equations,

$$\widehat{H}=\mathrm{\hslash }\omega \left({\widehat{a}}^{+}\widehat{a}+\frac{1}{2}\right)=\mathrm{\hslash }\omega \widehat{k}$$ (10)

So we just need to solve $\widehat{k}|k⟩=k|k⟩$ since $\widehat{H}|k⟩=\mathrm{\hslash }\omega k|k⟩$ which means $E=\mathrm{\hslash }\omega k$.

We can also derive the commutator between $\widehat{a},{\widehat{a}}^{+}$ with $|k⟩$.

$[\widehat{a},\widehat{k}]$ $=[\widehat{a},{\widehat{a}}^{+}\widehat{a}]$

$={\widehat{a}}^{+}[\widehat{a},\widehat{a}]+[\widehat{a},{\widehat{a}}^{+}]\widehat{a}$

$=\widehat{a}$

$\Rightarrow {[\widehat{a},\widehat{k}]}^{+}={\widehat{a}}^{+}$

$\Rightarrow [{\widehat{k}}^{+},{\widehat{a}}^{+}]={\widehat{a}}^{+}$

$\Rightarrow [\widehat{k},{\widehat{a}}^{+}={\widehat{a}}^{+}$

Using these, we can observe an interesting fact.

$\widehat{k}\widehat{a}|k⟩$ $=(\widehat{k}\widehat{a}-\widehat{a}\widehat{k}+\widehat{a}\widehat{k})|k⟩$

$=[\widehat{k},\widehat{a}]|k⟩+\widehat{a}k|k⟩$

$=(k-1)\widehat{a}|k⟩$

This shows that $\widehat{a}|k⟩$ is an eigenvector of $\widehat{k}$ with eigenvalue $k-1$. This implies that $\widehat{a}|k⟩=c_{k-1}|k-1⟩$ as there is no degeneracy in 1D.

$\Rightarrow {\widehat{a}}^n|k⟩={\displaystyle \prod _{i=1}^n}{c}_{k-i}|k-n⟩$

As we can always apply $\widehat{a}$ to lower the eigenvalue, this means we can make the eigenvalues of $\widehat{k}$ negative. However, this contradicts with our previous finding that all the eigenvalues of $\widehat{H}$ are non-negative. Hence there has to be a ground state $|0⟩$ s.t. $\widehat{a}|0⟩=0$.

Opposite of $\widehat{a}$,${\widehat{a}}^{+}$ raises the eigenvalues and all the eigenstates of $\widehat{k}$ can be found using $|n⟩=c_n{({\widehat{a}}^{+})}^n|0⟩$. These $|n⟩$ form the eigenbasis of $\widehat{H}$.
We fix the constants by demanding unit norm as usual.

$⟨n|n⟩$ $={|c_n|}^2⟨0|{\widehat{a}}^n{({\widehat{a}}^{+})}^n|0⟩$

$={|c_n|}^2⟨0|{\widehat{a}}^{n-1}\widehat{a}{({\widehat{a}}^{+})}^n|0⟩$

$={|c_n|}^2⟨0|{\widehat{a}}^{n-1}({({\widehat{a}}^{+})}^n\widehat{a}+[\widehat{a},{({\widehat{a}}^{+})}^n])|0⟩$

$={|c_n|}^2\left(⟨0|{\widehat{a}}^{n-1}{({\widehat{a}}^{+})}^n\widehat{a}|0⟩+⟨0|{\widehat{a}}^{n-1}[\widehat{a},{({\widehat{a}}^{+})}^n]|0⟩\right)$

$={|c_n|}^2\left(⟨0|{\widehat{a}}^{n-1}{({\widehat{a}}^{+})}^n\widehat{a}|0⟩+n⟨0|{\widehat{a}}^{n-1}{({\widehat{a}}^{+})}^{n-1}|0⟩\right)$

$=n{|c_n|}^2⟨0|{\widehat{a}}^{n-1}{({\widehat{a}}^{+})}^{n-1}|0⟩$

$=n!{|c_n|}^2$

$\Rightarrow c_n$ $={\displaystyle \frac{1}{\sqrt{n!}}}$

$\Rightarrow |n⟩$ $={\displaystyle \frac{1}{\sqrt{n!}}}{({\widehat{a}}^{+})}^n|0⟩$

We used the facts that $a|0⟩=0$ and $[\widehat{a},{({\widehat{a}}^{+})}^n]=n{({\widehat{a}}^{+})}^{n-1}$ which is easily proven by induction.

Now we just need to find the eigenvalues of $\widehat{k}$ from these eigenvectors.

$\widehat{k}|n⟩$ $=\left({\widehat{a}}^{+}\widehat{a}+{\displaystyle \frac{1}{2}}\right)c_n{({\widehat{a}}^{+})}^n|0⟩$

$=c_n{\widehat{a}}^{+}({({\widehat{a}}^{+})}^n\widehat{a}+[\widehat{a},{({\widehat{a}}^{+})}^n])|0⟩+{\displaystyle \frac{1}{2}}|n⟩$

$=c_n{\widehat{a}}^{+}n{({\widehat{a}}^{+})}^{n-1}|0⟩+{\displaystyle \frac{1}{2}}|n⟩$

$=\left(n+{\displaystyle \frac{1}{2}}\right)|n⟩$

$\Rightarrow \widehat{H}|n⟩$ $=\mathrm{\hslash }\omega \left(n+{\displaystyle \frac{1}{2}}\right)|n⟩$

5.1 The 2-D Isotropic Simple Harmonic Oscillator

$\widehat{H}$ $={\displaystyle \frac{{\widehat{p}}_x^2+{\widehat{p}}_y^2}{2m}}+{\displaystyle \frac{1}{2}}m{\omega }^2({\widehat{x}}^2+{\widehat{y}}^2)$

$={\displaystyle \frac{{\widehat{p}}_x^2}{2m}}+{\displaystyle \frac{1}{2}}m{\omega }^2{\widehat{x}}^2+{\displaystyle \frac{{\widehat{p}}_y^2}{2m}}+{\displaystyle \frac{1}{2}}m{\omega }^2{\widehat{y}}^2$

$={\widehat{H}}_x+{\widehat{H}}_y$

$[\widehat{x},\widehat{y}]=[{\widehat{p}}_x,{\widehat{p}}_y]=[{\widehat{p}}_x,\widehat{y}]=[{\widehat{p}}_y,\widehat{x}]=0$

$=[\widehat{y},{\widehat{p}}_y]=i\mathrm{\hslash }$

$=0$

The last equation suggests that we can diagonalise simultaneously w.r.t. ${\widehat{H}}_x,{\widehat{H}}_y$, so we label the eigenbasis of $\widehat{H}$ with two integers $n_x,n_y$ and define the corrresponding creation-annihilation operators.

${\widehat{a}}_x$ $={\displaystyle \frac{1}{\sqrt{2\mathrm{\hslash }m\omega }}}(m\omega \widehat{x}+i{p}_x)$

${\widehat{a}}_y$ $={\displaystyle \frac{1}{\sqrt{2\mathrm{\hslash }m\omega }}}(m\omega \widehat{y}+i{p}_y)$

$=[{\widehat{a}}_y,{\widehat{a}}_y^{+}]$ $=1$

$=[{\widehat{a}}_x^{+},{\widehat{a}}_y]$ $=0$

Our new ground state $|0,0⟩$ now satisfies

${\widehat{a}}_x|0,0⟩={\widehat{a}}_y|0,0⟩=0$

Our general eigenvector will have the following form and corresponding eigenvalue,

$|n_x,n_y⟩$ $={\displaystyle \frac{1}{\sqrt{n_x!n_y!}}}{({\widehat{a}}_x^{+})}^{n_x}{({\widehat{a}}_y^{+})}^{n_y}|0,0⟩$

$\widehat{H}|n_x,n_y⟩$ $=\mathrm{\hslash }\omega (n_x+n_y+1)|n_x,n_y⟩$

This shows that the 2D oscillator is degenerate.

5.2 The Fermionic Simple Harmonic Oscillator

The SHO we considered before are Bosonic which means that we can add as many ’quanta’ as we want by acting ${\widehat{a}}^{+}$ on the ground state. This is different for the fermionic case which obeys the following principle:
Pauli’s Exclusion Principle:
No two fermionic quanta can occupy the same state

The Hamiltonian for this system is

$\widehat{H}$ $=\mathrm{\hslash }\omega {\widehat{b}}^{+}\widehat{b}$

$\mathrm{\{}{\widehat{b}}^{+},\widehat{b}\}$ $={\widehat{b}}^{+}\widehat{b}+\widehat{b}{\widehat{b}}^{+}=1$

$\mathrm{\{}{\widehat{b}}^{+},{\widehat{b}}^{+}\}$ $=\{\widehat{b},\widehat{b}\}=0$

There is one thing we can note about the operators $\widehat{b},{\widehat{b}}^{+}$,

$\mathrm{\{}{\widehat{b}}^{+},{\widehat{b}}^{+}\}$ $=0$

$\iff 2{\widehat{b}}^{+}{\widehat{b}}^{+}$ $=0$

$\iff {({\widehat{b}}^{+})}^2$ $=0$

Analagously, ${\widehat{b}}^2=0$
Let $\widehat{H}=\mathrm{\hslash }\omega \widehat{N},\widehat{N},{\widehat{b}}^{+}\widehat{b}$ and consider ${\widehat{N}}^2$.

${\widehat{N}}^2$ $={\widehat{b}}^{+}\widehat{b}{\widehat{b}}^{+}\widehat{b}$

$={\widehat{b}}^{+}(\{\widehat{b},{\widehat{b}}^{+}\}-{\widehat{b}}^{+}\widehat{b})\widehat{b}$

$={\widehat{b}}^{+}(1-{\widehat{b}}^{+}\widehat{b})\widehat{b}$

$={\widehat{b}}^{+}\widehat{b}-{({\widehat{b}}^{+})}^2{\widehat{b}}^2$

$={\widehat{b}}^{+}\widehat{b}$

$=\widehat{N}$

This means that the eigenvalues of $\widehat{N}$ can only be 0 or 1, since

$\widehat{N}|n⟩$ $=n|n⟩$

${\widehat{N}}^2|n⟩$ $=n\widehat{N}|n⟩$

$=n^2|n⟩$

$\Rightarrow n^2$ $=n$

$\Rightarrow n$ $=0or\mathrm{\hspace{0.25em}1}$

So this implies that the eigenvalues of $\widehat{H}$ can only be $0$ or $\mathrm{\hslash }\omega$. This is exactly Pauli’s exclusion principle as it allows for different states to have different quantas.
However, in principle we could have degeneracy i.e. we have more than one states with $\widehat{N}|0,s⟩=0$ with $s=0,1,\mathrm{\dots }$. Yet here we assumed non degeneracy to simplify the system.

$\widehat{N}|0⟩$ $=0$

$\iff {\widehat{b}}^{+}\widehat{b}|0⟩$ $=0$

$\Rightarrow ⟨0|{\widehat{b}}^{+}\widehat{b}|0⟩$ $=0$

$\Rightarrow \widehat{b}|0⟩$ $=0$

So we have that $|0⟩$ is an eigenvector of $\widehat{b}$ with eigenvalue 0.

$\widehat{N}{\widehat{b}}^{+}|0⟩$ $={\widehat{b}}^{+}\widehat{b}{\widehat{b}}^{+}|0⟩$

$={\widehat{b}}^{+}(\{\widehat{b},{\widehat{b}}^{+}\}-{\widehat{b}}^{+}\widehat{b})|0⟩$

$=1.{\widehat{b}}^{+}|0⟩$

$\Rightarrow |1⟩$ $=c{\widehat{b}}^{+}|0⟩$

$⟨1|1⟩$ $={|c|}^2⟨0|\widehat{b}{\widehat{b}}^{+}|0⟩$

$={|c|}^2⟨0|\{\widehat{b},{\widehat{b}}^{+}\}-{\widehat{b}}^{+}\widehat{b}|0⟩$

$={|c|}^2⟨0|0⟩$

$={|c|}^2$

$\Rightarrow c$ $=1$

So it only remains to check $\widehat{b}|1⟩$ and ${|b⟩}^{+}|1⟩$ to complete the system.

$\widehat{b}|1⟩$ $=\widehat{b}{\widehat{b}}^{+}|0⟩$

$=(\{\widehat{b},{\widehat{b}}^{+}\}-{\widehat{b}}^{+}\widehat{b})|0⟩$

$=|0⟩$

${\widehat{b}}^{+}|1⟩$ $={({\widehat{b}}^{+})}^2|0⟩$

$=0$

6 Symmetries in Quantum Mechanics

Suppose ${\widehat{U}}_l$ is a one-parameter family of Unitary operators and ${\widehat{U}}_{l=0}=\widehat{I}$ which acts on our states $|\psi ⟩$. This process is called active transformation.
An example was in the case of translations whereby

${\widehat{U}}_l={\widehat{T}}_l=e^{-\frac{il}{\mathrm{\hslash }}\widehat{p}}$

Close to the identity , we can expand using Taylor’s approximation

${\widehat{U}}_l=\widehat{I}-i\widehat{G}l+O(l^2)$

whereby $\widehat{G}$ is the generator. If we demand that ${\widehat{U}}_l$ is unitary, then

${\widehat{U}}_l^{+}{\widehat{U}}_l$ $=(\widehat{I}+i{\widehat{G}}^{+}l+O(l^2))(\widehat{I}-i\widehat{G}l+O(l^2))$

$=\widehat{I}+il({\widehat{G}}^{+}-\widehat{G})+O(l^2)$

$\Rightarrow \widehat{G}$ $={\widehat{G}}^{+}$

This means that the generator has to be Hermitian.
We want to ask for any observable $\widehat{O}$, how does its expectation transform

${⟨\widehat{O}⟩}^{\prime }$ $=⟨\psi |\widehat{O}|\psi ⟩$

$=⟨\psi |{\widehat{U}}_l^{+}\widehat{O}{\widehat{U}}_l|\psi ⟩$

This is called passive transformation whereby we transform the operator.
The basic operators are $\widehat{x},\widehat{p}$ which transform like

${\widehat{x}}^{\prime }$ $={\widehat{U}}_l^{+}\widehat{x}{\widehat{U}}_l$

${\widehat{p}}^{\prime }$ $={\widehat{U}}_l^{+}\widehat{p}{\widehat{U}}_l$

Hence for any observable, we transform them as follows

${\widehat{O}}^{\prime }$ $={\widehat{U}}_l^{+}\widehat{O}(\widehat{x},\widehat{p}){\widehat{U}}_l$

$=\widehat{O}({\widehat{U}}_l^{+}\widehat{x}{\widehat{U}}_l,{\widehat{U}}_l^{+}\widehat{p}{\widehat{U}}_l)$

$=\widehat{O}({\widehat{x}}^{\prime },{\widehat{p}}^{\prime })$

Example:
Take $\widehat{O}={\widehat{x}}^4$.

${\widehat{O}}^{\prime }$ $={({\widehat{U}}_l^{+}\widehat{x}{\widehat{U}}_l)}^4$

$={\widehat{U}}_l^{+}{\widehat{x}}^4{\widehat{U}}_l$

In fact in general this is true for any series:

${\widehat{x}}^n{\widehat{p}}^m\to {\widehat{U}}_l^{+}{\widehat{x}}^n{\widehat{p}}^m{\widehat{U}}_l={({\widehat{U}}_l^{+}\widehat{x}{\widehat{U}}_l)}^n{({\widehat{U}}_l^{+}\widehat{p}{\widehat{U}}_l)}^m$

How does the position and momentum operator transform under a translation. Before we address this question, we need the following two facts:

$[\widehat{x},{\widehat{p}}^n]$ $=i\mathrm{\hslash }n{\widehat{p}}^{n-1}$

$=i\mathrm{\hslash }{\displaystyle \frac{d\widehat{f}}{d\widehat{p}}}$

The first can be proven through induction whereas the prove of the second can be proven as follows:

$\widehat{f}(\widehat{p})$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n{\widehat{p}}^n$

$\Rightarrow [\widehat{x},\widehat{f}(\widehat{p})]$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n[\widehat{x},{\widehat{p}}^n]$

$=i\mathrm{\hslash }{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}n{c}_n{\widehat{p}}^{n-1}$

$=i\mathrm{\hslash }{\displaystyle \frac{d\widehat{f}}{d\widehat{p}}}$

${\widehat{x}}^{\prime }$ $={\widehat{T}}_l^{+}\widehat{x}{\widehat{T}}_l$

$=e^{\frac{il}{\mathrm{\hslash }}\widehat{p}}\widehat{x}{e}^{-\frac{il}{\mathrm{\hslash }}\widehat{p}}$

$=e^{\frac{il}{\mathrm{\hslash }}\widehat{p}}([\widehat{x},e^{\frac{-il}{\mathrm{\hslash }}\widehat{p}}]+e^{\frac{-il}{\mathrm{\hslash }}\widehat{p}}\widehat{x})$

$=e^{\frac{il}{\mathrm{\hslash }}\widehat{p}}i\mathrm{\hslash }.-{\displaystyle \frac{il}{\mathrm{\hslash }}}{e}^{\frac{-il}{\mathrm{\hslash }}\widehat{p}}+\widehat{x}$

$=l\widehat{I}+\widehat{x}$

And this is what we expect, by translating we are adding $l$ to the position of the particle.

${\widehat{p}}^{\prime }$ $=e^{\frac{il}{\mathrm{\hslash }}\widehat{p}}\widehat{p}{e}^{-\frac{il}{\mathrm{\hslash }}\widehat{p}}$

$=\widehat{p}$

Again, translation does not affect the momentum of the particle.

Definition:
The operator $\widehat{O}(\widehat{x},\widehat{p})$ is invariant under the transformation ${\widehat{U}}_l$ if $\widehat{O}(\widehat{x^{\prime }},{\widehat{p}}^{\prime })=\widehat{O}(\widehat{x},\widehat{p})={\widehat{U}}_l^{+}\widehat{O}{\widehat{U}}_l$.
Under this assumption, if $\widehat{G}$ is the generator of the transformation ${\widehat{U}}_l$, then $[\widehat{O},\widehat{G}]=0$.

Proof:
For ,

$\widehat{O}$ $=(\widehat{I}+i\widehat{G}ϵ+O({ϵ}^2))\widehat{O}(\widehat{I}-i\widehat{G}ϵ+O({ϵ}^2))$

$=\widehat{O}+iϵ(\widehat{G}\widehat{O}-\widehat{O}\widehat{G})+O({ϵ}^2)$

$\Rightarrow [\widehat{O},\widehat{G}]$ $=0$

Noether’s Theorem in Quantum Mechanics
If $\widehat{H}(\widehat{x},\widehat{p})$ is invariant under a family ${\widehat{U}}_l$ and $\widehat{G}$ is the generator, then $\widehat{G}$ defines a conserved quantity.

Proof:

${\displaystyle \frac{d}{dt}}⟨\widehat{G}⟩={\displaystyle \frac{1}{i\mathrm{\hslash }}}⟨[\widehat{G},\widehat{H}]⟩=0$

Since $[\widehat{G},\widehat{H}]=0$, this means that $\widehat{G},\widehat{H}$ have a common eigenbasis and they can be diagonalised simultaneously.
A basis of eigenstates will be labelled by the eigenvalues of $\widehat{G}$ and $\widehat{H}$

$\widehat{H}|E,g⟩$ $=E|E,g⟩$

$\widehat{G}|E,g⟩$ $=g|E,g⟩$

For a third operator, we need $[\widehat{H},{\widehat{G}}_1]=[\widehat{H},{\widehat{G}}_2]=[{\widehat{G}}_1,{\widehat{G}}_2]=0$. Then we can have our vectors labelled as $|E,g_1,g_2⟩$.

6.1 Rotations in 2-D

Classically for an anticlockwise rotation,

$\left(\begin{array}{c}\hfill x^{\prime }\hfill \\ \hfill y^{\prime }\hfill \end{array}\right)$

$=U_{\varphi }\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$

We translate this to Quantum Mechanics by considering the action of rotation on the position eigenstates $|x,y⟩$.

${\widehat{U}}_{\varphi }|x,y⟩=|x\cos (\varphi )-y\sin (\varphi ),y\cos (\varphi )+x\sin (\varphi )⟩$

We are mostly interested in the generator of the transformation ${\widehat{L}}_z$. Under small angle appoximations,

$\varphi$

${\widehat{U}}_{ϵ}$ $=\widehat{I}-{\displaystyle \frac{i}{\mathrm{\hslash }}}{\widehat{L}}_zϵ+O({ϵ}^2)$

$\Rightarrow (\widehat{I}-{\displaystyle \frac{i}{\mathrm{\hslash }}}{\widehat{L}}_zϵ)|x,y⟩$ $=|x-ϵy,y+ϵx⟩$

So we need an expression for ${\widehat{L}}_z$ in terms of $\widehat{x},\widehat{p}$.

$⟨x,y|(\widehat{I}+{\displaystyle \frac{i}{\mathrm{\hslash }}}{\widehat{L}}_z)|\psi ⟩$ $=⟨x-ϵy,y+ϵx|\psi ⟩\mathit{\hspace{1em}}\forall |\psi ⟩$

$=\psi (x-ϵy,y+ϵx)$

$=\psi (x,y)-ϵy{\partial }_x\psi (x,y)+ϵx{\partial }_y\psi (x,y)+O({ϵ}^2)$

$\Rightarrow {\displaystyle \frac{i}{\mathrm{\hslash }}}⟨x,y|{\widehat{L}}_z|\psi ⟩$ $=-ϵy{\partial }_x\psi (x,y)+ϵx{\partial }_y\psi (x,y)$

$⟨x,y|{\widehat{p}}_x|\psi ⟩$ $=-i\mathrm{\hslash }{\partial }_x⟨x,y|\psi ⟩$

$\Rightarrow {\displaystyle \frac{i}{\mathrm{\hslash }}}⟨x,y|{\widehat{L}}_z|\psi ⟩$ $={\displaystyle \frac{i}{\mathrm{\hslash }}}(-y⟨x,y|{\widehat{p}}_x|\psi ⟩+x⟨x,y|{\widehat{p}}_y|\psi ⟩)$

$={\displaystyle \frac{i}{\mathrm{\hslash }}}(-⟨x,y|\widehat{y}{\widehat{p}}_x|\psi ⟩+⟨x,y|\widehat{x}{\widehat{p}}_y|\psi ⟩)$

$\Rightarrow {\widehat{L}}_z$ $=\widehat{x}{\widehat{p}}_y-\widehat{y}{\widehat{p}}_x={\widehat{p}}_y\widehat{x}-{\widehat{p}}_x\widehat{y}$

This corresponds to classical angular momentum which we have shown to be the generator of rotations.

As a sanity check, we can see if ${\widehat{L}}_z$ commutes with rotationally symmetric Hamiltonians.

$\widehat{H}={\displaystyle \frac{{\widehat{p}}_x^2+{\widehat{p}}_y^2}{2m}}+\widehat{V}({\widehat{x}}^2+{\widehat{y}}^2)$

This gets rather complicated so we evaluate the commutator by parts.

$[{\widehat{L}}_z,{\widehat{p}}_x^2+{\widehat{p}}_y^2]$ $={[{\widehat{L}}_z,{\widehat{p}}_x]}^2+[{\widehat{L}}_z,{\widehat{p}}_y^2]$

$=[\widehat{x}{\widehat{p}}_y,{\widehat{p}}_x]+[-{\widehat{p}}_x\widehat{y},{\widehat{p}}_y^2]$

$=0$

$=-i\mathrm{\hslash }{\displaystyle \frac{\partial \widehat{f}}{\partial {\widehat{x}}_i}}$

$\Rightarrow [{\widehat{p}}_x,\widehat{V}({\widehat{x}}^2+{\widehat{y}}^2)]$ $=-i\mathrm{\hslash }{\widehat{V}}^{\prime }({\widehat{x}}^2+{\widehat{y}}^2)2\widehat{x}$

$\Rightarrow [{\widehat{p}}_y,\widehat{V}({\widehat{x}}^2+{\widehat{y}}^2)]$ $=-i\mathrm{\hslash }{\widehat{V}}^{\prime }({\widehat{x}}^2+{\widehat{y}}^2)2\widehat{y}$

$\Rightarrow [{\widehat{L}}_z,\widehat{V}]$ $=\widehat{x}[{\widehat{p}}_y,\widehat{V}]-\widehat{y}[{\widehat{p}}_x,\widehat{V}]$

$=0$

$\Rightarrow [\widehat{H},{\widehat{L}}_z]$ $=0$

Since this a possible measurement, we are also interested in its eigenvalues.

${\widehat{L}}_z|l⟩$ $=l|l⟩$

$\Rightarrow ⟨x,y|{\widehat{L}}_z|l⟩$ $=l⟨x,y|l⟩$

$\iff -i\mathrm{\hslash }(x{\partial }_y-y{\partial }_x){\psi }_l(x,y)$ $=l{\psi }_l(x,y)$

This is a PDE whch we can solve by going to polar coordinates.

$x$ $=r\cos (\varphi )$

$y$ $=r\sin (\varphi )$

$\Rightarrow -i\mathrm{\hslash }{\partial }_{\varphi }{\psi }_l(r,\varphi )$ $=l{\psi }_l(r,\varphi )$

$\Rightarrow {\psi }_l(r,\varphi )$ $=e^{\frac{il}{\mathrm{\hslash }}\varphi }$

We have one constraint on $l$ which is $\varphi \sim \varphi +2\pi$.

$e^{\frac{il}{\mathrm{\hslash }}(\varphi +2\pi )}$ $=e^{\frac{il}{\mathrm{\hslash }}\varphi }$

$\Rightarrow e^{\frac{il}{\mathrm{\hslash }}2\pi }$ $=1$

$\Rightarrow l$ $=\mathrm{\hslash }m,m\in \mathbb{Z}$

This means that the eigenvalues of the angular momentum are quantised.