math-primer

#quantum-computing

In Quantum Computing, one is interested in a finite dimensional complex linear vector space. The elements of these vector space are represented in the dirac notion using $| α ⟩$ (ket)

Two kets can be added to give a new vector $| γ ⟩ \in V$ $$ \begin{array}{ll} | \gamma\rangle &= | \alpha \rangle + | \beta \rangle \ \gamma_i &= \alpha_i + \beta_i\end{array}$$vector addition has the following properties:$$\begin{array}{ll}|\alpha \rangle + | \beta\rangle &= |\beta \rangle + |\alpha\rangle \ | \alpha \rangle + (|\beta \rangle + | \gamma \rangle ) &= (|\alpha\rangle + |\beta\rangle) + | \gamma \rangle \end{array}$$
We can also multiply a vector $| α ⟩ \in V$ with a complex number $c \in C$ to obtain a new vector $c | α ⟩$ . The following properties hold for $c, d \in C$ and $| α ⟩, | β ⟩ \in V$ :$$\begin{array}{ll}c(|\alpha\rangle + |\beta\rangle) &= c|\alpha\rangle + c | \beta \rangle \ (c+ d) |\alpha\rangle &= c |\alpha \rangle + d(|\alpha\rangle) \ (cd) | \alpha \rangle &= c(d|\alpha\rangle)\end{array}$$
A vector space contains the zero vector with the following properties:$$\begin{array}{ll}|\alpha\rangle + 0 &= | \alpha\rangle \ 0|\alpha\rangle &= 0 \ 1 |\alpha \rangle &= |\alpha\rangle \ |\alpha\rangle - |\alpha \rangle &= 0\end{array}$$

We do not use the ket notation for the zero vector

Linear Independence

c_{1} | α_{1} ⟩ + c_{2} | α_{2} ⟩ + \dots | c_{m} | α_{m} ⟩ = 0

iff $c_{1}, \dots c_{m} = 0$

Inner Product

denoted $⟨ α | β ⟩$
Prop: $⟨ α | β ⟩ = ⟨ β | α ⟩^{*}$ (skew-symmetric)
Prop: $⟨ α | c β + d γ ⟩ = c ⟨ α | β ⟩ + d ⟨ α | γ ⟩$ (linearity)
Prop: $⟨ α | α ⟩ \geq 0$ (positivity)
$⟨ α |$ is called the dual vector/bra. The dual vector is a linear operator from the vector space $V$ to the complex numbers $C$ , defined by $⟨ α | (| β ⟩) = ⟨ α | β ⟩$ . For example is $| α ⟩ = (α_{1} \dots α_{n})$ and $| β ⟩ = (β_{1} \dots β_{n})$ then $⟨ α | β ⟩ = \sum_{i = 1}^{n} α_{i}^{*} β_{i}$
norm of a ket is defined as $| | | α ⟩ | | = \sqrt{⟨ α | α ⟩}$ . The normalised unit vector $| α ⟩ / | | (| α ⟩) | |$ has unit norm$$|| ( | \alpha \rangle )|| = \sqrt{\sum_{i=1}^{n} | \alpha_i|^2}$$
Hilbert Space $\equiv$ Complex vector space equipped with an inner product

Cauchy-Schwartz Inequality

| \langle \alpha | \beta \rangle | {#2} \leq \langle \alpha | \alpha \rangle \langle \beta | \beta \rangle

Proof: For any $| α ⟩, | β ⟩ \in V$ and $c \in C$ we have$$\begin{array}{ll}\langle \alpha - c \beta | \alpha - c \beta \rangle \geq 0 \ \langle \alpha | \alpha \rangle - c \langle \alpha | \beta \rangle - c^* \langle \alpha | \beta \rangle - c c^* \langle \beta | \beta \rangle \geq 0\end{array}$$ we set $c = \frac{⟨ β | α ⟩}{⟨ β | β ⟩}$ $$\begin{array}{ll}\langle \alpha | \alpha \rangle - \frac{\langle \alpha | \beta \rangle \langle \beta | \alpha \rangle}{\langle \beta | \beta \rangle} - \frac{\langle \alpha | \beta \rangle \langle \beta | \alpha \rangle}{\langle \beta | \beta \rangle} - \frac{\langle \alpha | \beta \rangle \langle \beta | \alpha \rangle}{\langle \beta | \beta \rangle} &\geq 0 \ \langle \alpha | \alpha \rangle \langle \beta | \beta \rangle - 3 | \langle \alpha | \beta \rangle |
{ #2}
&\geq 0 \ \langle \alpha | \alpha \rangle \langle \beta | \beta \rangle &\geq | \langle \alpha | \beta \rangle |^2 \end{array}$$

Orthonormality Condition

$| α ⟩, | β ⟩$ are said to be orthogonal if their product is zero$$\langle \alpha | \beta \rangle = 0$$A set of vectors $| α_{1} ⟩, | α_{2} ⟩ \dots | α_{n} ⟩$ is said to be orthogonal if:$$\langle \alpha_i | \alpha_j\rangle = \delta_{ij}$$

Orthonormal vectors are linearly independent
Dimension $n$ of vector spaces is given by the maximum number of linearly independent vectors
A set of linearly independent vectors $| α_{1} ⟩ \dots | α_{n} ⟩$ in an $n$ -dimensional vector space is said to be a basis for for the vector space. Any vector $| α ⟩$ can be expanded over a basis ${| α_{1} ⟩, \dots, | α_{n} ⟩}$ $$|\alpha\rangle = \displaystyle\sum_{i=1}^{n} a_i |\alpha_i\rangle$$
$| α_{i} ⟩$ are known as the complete set of vectors
$a_{i} \in C$ are the components of the vector $| α ⟩$ w.r.t. basis ${| α_{1} ⟩, \dots, | α_{n} ⟩}$
$a_{i}$ 's are uniquely determined for an orthonormal basis$$a_i =\langle \alpha_i | \alpha\rangle$$
The ordered collection of components ${a_{1} \dots a_{n}}$ constitutes a representation of the vector $| α ⟩$

Linear Operators

Completeness Relation

We know that $a_{i} = ⟨ α_{i} | α ⟩$ $$\begin{array}{ll} |\alpha\rangle &=\sum a_i |\alpha_i\rangle \ &= \sum \langle\alpha_i | \alpha\rangle | \alpha_i \rangle \ &= \sum (|\alpha_i \rangle \langle \alpha_i |) |\alpha\rangle \end{array}$$Thus $\sum | α_{i} ⟩ ⟨ α_{i} | = I$

Matrix Representation of an Operator

Consider a linear operator$$A |\alpha\rangle = |\beta\rangle$$Expanding over an orthonormal basis ${| γ_{1} ⟩ \dots | γ_{n} ⟩}$ $$|\alpha \rangle = \sum a_i | \gamma_i \rangle \hspace{2em} |\beta\rangle = \sum b_i | \gamma_i \rangle$$Using $b_{i} = ⟨ γ_{i} | β ⟩$ $$\begin{array}{ll}&= \langle \gamma_i | A \alpha\rangle \ &= \sum_j \langle \gamma_i | A | \gamma_i \rangle a_j \ &= \sum_j \langle \gamma_i | A \gamma_j \rangle a_j \ &= A_{ij}a_j \end{array}$$

Pauli Matrices

σ_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) $ $ $ $ σ_{y} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) $ $ $ $ σ_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})

Prop: $σ_{x}^{2} = σ_{y}^{2} = σ_{z}^{2} = I$
Prop: $$\begin{array}{ll}\sigma_x\sigma_y &= i\sigma_z \ \sigma_y\sigma_z &= i\sigma_x \ \sigma_z\sigma_x &= i\sigma_y\end{array}$$

Pauli Matrices	Eigenvectors	Eigenvalues
$σ_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix})$ , $\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - 1 \end{matrix})$	$\pm$ 1
$σ_{y} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ i \end{matrix})$ , $\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - i \end{matrix})$	$\pm$ 1
$σ_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$	$(\begin{matrix} 1 \\ 0 \end{matrix})$ , $(\begin{matrix} 0 \\ 1 \end{matrix})$	$\pm$ 1

Projectors

If $| α ⟩ \in V$ is a unit vector, the uni-dimensional projector $P_{α}$ is defined for some $| β ⟩$ as$$\begin{array}{ll}P_{\alpha} | \beta \rangle &= |\alpha\rangle\langle \alpha | \beta \rangle \ &= \langle \alpha | \beta \rangle |\alpha\rangle\end{array}$$This operator is called a projector since it projects a generic vector $| β ⟩$ along the dimension $| α ⟩$

Prop: $P_{α} | α ⟩ = | α ⟩$
Prop: $P_{α} | γ ⟩ = 0$ when $| γ ⟩$ is orthogonal to $| α ⟩$
Prop: $P_{α}^{2} = P_{α}$

Eigenvalues and Eigenvectors

An eigenvector of a linear operator $A$ is a non-zero vector $| α ⟩$ s.t.$$A |\alpha\rangle = \alpha | \alpha\rangle$$$$\begin{array}{ll}|\alpha\rangle &= \displaystyle \sum_{i=1}^{n} \alpha_i | \gamma_i\rangle \ A |\alpha\rangle &= \displaystyle \sum_{i=1}^{n} (\sum_j A_{ij} a_j) | \gamma_i \rangle \ A | \alpha\rangle &= \alpha | \alpha \rangle \ A | \alpha \rangle - \alpha | \alpha \rangle &= 0 \ \displaystyle \sum_{i=1}^{n} (\sum_{j=1}^{n} A_{ij}a_j - \alpha a_i) | \gamma_i \rangle &= 0\end{array}$$

Hermitian Operators

For any linear operator A on a Hilbert Space $H \exists$ a unique linear operator $A^{†}$ on $H$ called the adjoint or Hermitian conjugate s.t.$$\langle \alpha | A \beta \rangle = \langle A^{\dagger} \alpha | \beta \rangle $$$$\begin{array}{ll}\langle A \alpha | \beta \rangle &= \langle \beta | A \alpha\rangle^* \ &= \langle A^{\dagger}\beta | \alpha \rangle \ &= \langle \alpha | A^{\dagger} \beta \rangle\end{array}$$

The eigenvectors of an Hermitian operator form an orthonormal set in the Hilbert Space $H$
Any vector in the Hilbert Space $H$ can be expressed as a linear superposition of vectors of said basis
We know $A_{i j} = ⟨ γ_{i} | A γ_{j} ⟩$ (from #Matrix Representation of an Operator). Then $A_{j i} = ⟨ γ_{j} | A γ_{i} ⟩ ⟹ A_{j i}^{*} = ⟨ A γ_{i} | γ_{j} ⟩ = ⟨ γ_{j} | A^{†} γ_{i} ⟩$ $$A_{ji}^{*} = A^{\dagger}_{ij}$$
Matrix elements of $A^{†}$ are the complex conjugates of the matrix elements of $A^{⊤}$

Inverse Operators

For an operator $A$ , an operator $B$ is said to be the inverse if $A B = B A = I$

The inverse of an operator

A

exists iff

det A \neq 0

Unitary Operator

An operator $U$ is said to be unitary if $$\mathcal{U}\mathcal{U}^{\dagger} = \mathcal{U}^{\dagger}\mathcal{U} = I$$

this definition implies $U^{†} = U^{- 1}$
The product $U V$ of two unitary operators is unitary$$(\mathcal{U}\mathcal{V}) = (\mathcal{U}\mathcal{V})^{\dagger} = \mathcal{U}\mathcal{V}\mathcal{V}^{\dagger}\mathcal{U} = I$$
Unitary operators preserve inner products$$\langle \mathcal{U} \alpha | \mathcal{U}\beta \rangle = \langle \alpha | \mathcal{U}^{\dagger}\mathcal{U} | \beta\rangle$$
#Pauli Matrices are both Hermitian and Unitary

Change of Basis

We can convert from one basis $| γ_{i} ⟩$ to another basis $| γ_{i}^{^{'}} ⟩$ using unitary transformation $S$ $$|\gamma_i^{'} \rangle = \sum_j S_{ji} | \gamma_i \rangle$$
Thus if $| α ⟩ = \sum_{i} a_{i} | γ_{i} ⟩$ then we can write$$\begin{array}{ll}|\alpha\rangle &= \sum_j a_j^{'} | \gamma_j^{'} \rangle \ &= \sum_{j} a_j^{'} S_{ji} | \gamma_i\rangle \end{array}$$where $a_{j}^{^{'}} = ⟨ γ_{j}^{^{'}} | α ⟩$

Diagonal Representation

An operator can be represented using its eigenvectors as basis:$$A = \displaystyle \sum_{i=1}^{n} \lambda_i | i \rangle \langle i | $$

An operator is said to be diagonalisable if it has a diagonal representation
Hermitian and Unitary operators are diagonalisable

defn: Normal Operators

$A A^{†} = A^{†} A$

Commutators and Anti-Commutators

We say two operators commute if $A B = B A$ . The commutator of two operators $A$ and $B$ is defined as$$[A, B] = AB - BA$$
- Prop: $[A, B] = - [B, A]$
- Prop: $[A B, C] = A [B, C] + [A, C] B$
The anti-commutator of two operators is defined by:$${ A, B} = AB + BA$$Two operators anti-commute is ${A, B} = 0$

#Pauli Matrices anti-commute

Trace

Trace of a matrix is the sum of its diagonal elements$$\text{Tr(A)} = \displaystyle \sum_{i=1}^{n} A_{ii}$$

Prop: $Tr(A+B) = Tr(A) + Tr(B)$
Prop: $Tr(cA) = c Tr(A)$
Prop: $ \text{Tr(AB)} = \text{Tr(BA)}$
For a unitary operator $U$ $$\text{Tr}(\mathcal{U}^{\dagger}A\mathcal{U}) = \text{Tr}(\mathcal{U}\mathcal{U}^{\dagger}A) = \text{Tr}(A)$$i.e. Trace is invariant under Unitary Transformations