The indivisible unit of classical information is the bit: an object that can take either one of the two possible values: 0 or 1.


The qubit (short for ‘quantum bit’) is a fundamental information carrying unit used in quantum computers. It describe a state in the simplest possible quantum system. A qubit is a two dimensional quantum system. It is a corresponding unit of quantum information and it is a vector in a two dimensional complex vector space with inner product. The quantum state of $N$ qubits can be expressed as a vector in a space of dimension $2^{N}$. The state of a qubit can be expressed as,

$$ \begin{align*} \ket{\psi} = \alpha \ket{0} + \beta \ket{1} \end{align*} $$

Above is the Dirac notation and where

$$ \ket{0} = \begin{pmatrix}1 \\ 0\end{pmatrix}, \ket{1} = \begin{pmatrix}0 \\ 1\end{pmatrix} $$

are shorthands for the vectors encoding the two basis states of a two dimensional vector space.

System of qubits

A quantum computer contain many number of qubits. So it is necessary to know how to construct the combined state of a system of qubits given the states of the individual qubits.

The joint state of a system of qubits is described using an operation known as tensor product $\otimes$.

Mathematically, taking the tensor product of two states is the same as taking the Kronecker product of their corresponding vectors. Say we have two single qubit states

$$\ket{\phi} =\begin{pmatrix}\alpha \\ \beta \end{pmatrix}$$


$$\ket{\phi'} =\begin{pmatrix}\alpha' \\ \beta' \end{pmatrix}$$

then, the full state of a system composed of two independent qubits is given by

$$ \ket{\phi} \otimes \ket{\phi'} = \begin{pmatrix}\alpha \\ \beta \end{pmatrix} \otimes \begin{pmatrix}\alpha' \\ \beta' \end{pmatrix} = \begin{pmatrix}\alpha\alpha' \\ \alpha\beta' \\ \beta\alpha' \\ \beta\beta' \end{pmatrix} $$.

The shortened form of $\ket{\phi} \otimes \ket{\phi’}$ is $\ket{\phi\phi’}$.

Superposition and entanglement

Superposition is any linear combination of two quantum states, once normalized will also be a valid quantum state. This means any quantum state can be expressed as a linear combination of a few basis states.

States of a system of which cannot be expressed as a tensor product of states of its individual subsystems are called entangled states.

Inner product and Outer product

The inner product or overlap between two quantum states i.e. $\braket{\psi\vert\phi}$. Note that $\braket{\psi\vert \phi} = \braket{\psi\vert\phi} ^{*}$.

Mathematically, take two single qubit states, $\ket{\phi} = \alpha\ket{0} + \beta\ket{1}$ and $\ket{\psi} =\gamma\ket{0} + \delta\ket{1}$ then, $\braket{\psi\vert\phi} = \gamma ^* \alpha + \delta ^* \beta$ where $^{*}$ denotes the complex conjugate.

The outer product of two states outputs a matrix given two states and defined as $\ket{\psi}\bra{\phi}$. Mathematically,

$$ \left|\psi\right>\left<\phi\right| = \begin{pmatrix}\alpha \\ \beta\end{pmatrix} \begin{pmatrix}\gamma^{*} & \delta^{*}\end{pmatrix} = \begin{pmatrix} \alpha\gamma^{*} & \alpha\delta^{*} \\ \beta\gamma^{*} & \beta\delta^{*} \end{pmatrix} $$

Singular Value Decomposition

Given a matrix $A$, you want to decompose into orthogonal $U, V$ and diagonal matrix $D$ i.e. $A = UDV^{T}$.

  1. Take matrix product of $AA^{T}$ and find its eigen values $EV(AA^{T})$
  2. Find the singular values, $SV = \sqrt{EV(AA^{T})} := \sigma_{i}$
  3. Find the eigen vector (as a column vector) then make it orthronormal vectors, and put all into a matrix form, so this is $U$
  4. Now, we need to find $V$, you find it by repeating above steps but instead of $AA^{T}$, use $A^{T}A$. But we have easiest way to get $V$ i.e. the orthonormal eigen vectors for matrix $V$ is $V_{i} = \frac{1}{\sigma_{i}} A^{T} U_{i}$

Density matrix

Density matrix (or operator) $\rho_{A}$ has the following properties:

  1. $\rho_{A}$ is self-adjoint: $\rho_{A} = \rho_{A}^{\dagger}$.
  2. $\rho_{A}$ is positive. For any $\ket{\psi}_A$,

    $$_{A}\braket{\psi\vert\rho_{A}\vert\psi}_{A} = \sum_{\mu} |\sum_{i} a_{i\mu} {}_{A}\braket{\psi\vert i}_{A}|^{2} \geq 0$$.

  3. $\text{Tr}(\rho_{A}) = 1$. We have $\text{Tr}(\rho_{A}) = \sum_{i, \mu} |a_{i\mu}|^{2} = 1$, since $\ket{\psi}_{AB}$.

Partial trace

The partial trace $\text{Tr} _{B}$ is a mapping from the density matrices $\rho _{AB}$ on a composite space $\mathcal{H} _{A} \otimes \mathcal{H} _{B}$ onto density matrices $\rho _{A}$ on $\mathcal{H} _{A}$. It is defined as the linear extension of the mapping

$$ \begin{align*} \text{Tr}_{B} : S \otimes T \mapsto \text{Tr}(T)S. \end{align*} $$

for any matrix $S$ on $\mathcal{H}_A$ and $T$ on $\mathcal{H}_{B}$.

The precise definition is, let {$\ket{a_i}$} be a basis of $\mathcal{H}_{A}$, and {$\ket{b_i}$} be a basis of $\mathcal{H}_B$.

Any density matrix $\rho_{AB}$ on $\mathcal{H}_A \otimes \mathcal{H}_B$ can the be decomposed as

$$ \begin{align*} \rho_{AB} = \sum_{ijkl} c_{ijkl} \ket{a_{i}}\bra{a_{j}} \otimes \ket{b_{k}}\bra{b_{l}} \end{align*} $$

and the partial trace reads as

$$ \begin{align*} \text{Tr}_{B}\rho_{AB} = \sum_{ijkl} c_{ijkl} \ket{a_{i}}\bra{a_{j}} \braket{b_{l}\vert b_{k}} = \rho_{A} \end{align*} $$

which is a density matrix $\rho_A$ on $\mathcal{H} _A$. Note that $\text{Tr} \ket{b_k}\bra{b_l} = \braket{b_l\vert b_k}$ is a complex number and note the exchange of indices compared to the operator.


The two-qubit spin singlet $\ket{\psi^{-}} = \frac{1}{\sqrt{2}}(\ket{01} - \ket{10})$ corresponding density matrix is

$$ \begin{align*} \rho_{AB} = \ket{\psi^{-}}\bra{\psi^{-}} &= \frac{1}{\sqrt{2}}(\ket{01} - \ket{10}) \times \frac{1}{\sqrt{2}}(\bra{01} - \bra{10}) \\ &= \frac{1}{2} (\ket{01}\bra{01} - \ket{01}\bra{10} - \ket{10}\bra{01} \\&\quad+ \ket{10}\bra{10}) \\ &= \frac{1}{2} (\ket{0}\otimes \ket{1} \bra{0}\otimes \bra{1} - \ket{0}\otimes \ket{1} \bra{1}\otimes \bra{0}\\&\quad - \ket{1}\otimes \ket{0} \bra{0}\otimes \bra{1} + \ket{1}\otimes \ket{0} \bra{1}\otimes \bra{0}) \\ &= \frac{1}{2} (\ket{0}\bra{0} \otimes \ket{1}\bra{1} - \ket{0}\bra{1} \otimes \ket{1}\bra{0}\\&\quad - \ket{1}\bra{0} \otimes \ket{0}\bra{1} + \ket{1}\bra{1} \otimes \ket{0}\bra{0}) \end{align*} $$

where we have used two things: one is $\ket{01} = \ket{0}\otimes \ket{1}$ and another one is $\ket{0}\otimes \ket{1} \bra{0}\otimes \bra{1} = \ket{0}\bra{0}\otimes \ket{1}\bra{1}$. Thus we act with partial trace give density matrix $\rho_{A}$ i.e.

$$ \begin{align*} \rho_{A} = \text{Tr}_{B}\ket{\psi^{-}}\bra{\psi^{-}} &= \frac{1}{2} (\ket{0}\bra{0}\braket{1\vert 1} - \ket{0}\bra{1}\braket{0\vert 1}\\&\quad - \ket{1}\bra{0}\braket{1\vert 0} + \ket{1}\bra{1}\braket{0\vert 0}) \\ &= \frac{1}{2} (\ket{0}\bra{0} + \ket{1}\bra{1}) \\ & = \frac{1}{2} \mathbb{I}. \end{align*} $$

where we have used one thing: $\braket{i\vert j} = \delta_{ij}$. Hence, we obtain the reduced state $\rho_{A} = \frac{1}{2} \mathbb{I}$.

Quantum States

Pauli matrices (Spinors)

$$ \begin{align*} \sigma_{1} = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_{2} = \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}, \sigma_{3} = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix} \end{align*} $$

This is the unique two-dimensional irreducible representation (See Lie groups ), up-to a unitary change of basis. The Pauli matrices have the properties of being mutually anti-commuting and squaring to the identity,

$$ \begin{align*} \{ \sigma_{k}, \sigma_{l} \} = 2\delta_{kl}\mathbf{1}, \end{align*} $$

and $(\hat{n}.\vec{\sigma})^{2} = n_{k}n_{l}\sigma_{k}\sigma_{l} = n_{k}n_{k}\mathbf{1} = 1$.

Photon polarizations

A Photon have two independent polarizations. i.e traversal and longitudinal which can taken as quantum states for qubit system.

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Published on Aug 8, 2021

Last revised on Jan 8, 2023