Notes › PHYS 3571: Quantum Computing Lecture 5

2-Qubit States

calendar_today February 5, 2026 article 442 words access_time 2 min replay Modified February 10, 2026

Table of Contents

2-Qubit Basis

Subscript tells you which qubit each ket is apart of
- Qubit 0: $\left\{ \ket{0}_{0}, \ket{1}_{0} \right\} \to \left\{ \left( \begin{matrix}1 \\ 0\end{matrix} \right)_{0}, \left( \begin{matrix}0 \\ 1\end{matrix} \right)_{0} \right\}$
- Qubit 1: $\left\{ \ket{0}_{1}, \ket{1}_{1} \right\}\to \left\{ \left( \begin{matrix}1 \\ 0\end{matrix} \right)_{1}, \left( \begin{matrix}0 \\ 1\end{matrix} \right)_{1} \right\}$
To form the 2-qubit basis, take the tensor product of the single-qubit bases:
The topmost line of the quantum circuit is the rightmost bit of the bitstring
- “Little endian” => LSB is the rightmost

$$ \begin{align} \ket{0}_{1} \otimes \ket{0}_{0} &= \ket{0 \ 0} = \left( \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right) \\ \ket{0}_{1} \otimes \ket{1}_{0} &= \ket{0 \ 1} = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right) \\ \ket{1}_{1} \otimes \ket{0}_{0} &= \ket{1 \ 0} = \left( \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right) \\ \ket{1}_{1} \otimes \ket{1}_{0} &= \ket{1 \ 1} = \left( \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right) \end{align} $$

Constructing the matrices for tensor product basis:

$$\ket{0}_{1} \otimes \ket{0}_{0} = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \otimes \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 1 \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)_{0} \\ 0 \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)_{0} \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right) = \ket{0 \ 0}$$

Inner Product

Take the single-qubit inner product, grouping by the qubit index

$$\braket{ 0 \ 0 | 0 \ 0 } = (\bra{0}_{1} \otimes \bra{0}_{0})(\ket{0}_{1} \otimes \ket{0}_{0}) = \braket{ 0 | 0 }_{1} \braket{ 0 | 0 }_{0} = 1 $$

CNOT Example

CNOT With Haddamard Example

Measuring Individual Qubits from 2-Qubit State

Projection operator onto state $\ket{0}_{1}$:

$$\hat{\mathbb{P}}(\ket{0}_{1} ) = (\ket{0}_{1}\bra{0}_{1} ) \otimes \hat{I}_{0}$$

The probability is:

$$P(\ket{0}_{1} ) = \braket{ \psi | \hat{\mathbb{P}}(\ket{0}_{1} ) | \psi }$$

Shortcut: it’s the sum of the squares of the coefficients for any 2-qubit whose first is 0/1
Q: If we measured $q_1$ and got $\ket{0}_{1}$, what is the new wave function?
A:

$$\ket{\psi_{\text{new}}} = \frac{\ket{\hat{\mathbb{P}}}}{\sqrt{ \braket{ \psi | \hat{\mathbb{P}}(\ket{0}_{1} ) | \psi } }}$$

Entangled States

A 2-qubit state is entangled if it can’t be written as the product of 2 single-qubit states
General unentangled state:

$$\begin{align} \ket{\psi} &= (\alpha_{1}\ket{0}_{1} + \beta_{1} \ket{1}_{1} ) \otimes (\alpha_{0}\ket{0}_{0} + \beta_{0 \ket{1}_{0} } ) \\ &= \alpha_{1} \alpha_{0} \ket{0 \ 0} + \alpha_{1} \beta_{0}\ket{0 \ 1} + \beta_{1} \alpha_{0} \ket{1 \ 0} + \beta_{1} \beta_{0} \ket{1 \ 1} \end{align}$$

Entangled state
- Entangled because there’s no solution to get it into the unentangled state

$$\ket{\psi_{+}} = \frac{1}{\sqrt{ 2 }} \ket{0 \ 0} + \frac{1}{\sqrt{ 2 }} \ket{1 \ 1} $$

Graph

mindmap ↗