Qiskit Lesson 2: Multiple Systems

Quantum Computing

Qiskit

Quantum Information

Tensor Products

Entanglement

Lesson 2 notes on compound classical and quantum systems: Cartesian products, tensor products, marginal and conditional states, CNOT, Bell states, partial measurement, and local unitary operations.

Author

Chao Ma

Published

May 23, 2026

Lesson 1 described a single system.

Lesson 2 asks what changes when we combine systems. The main answer is that state spaces multiply: classical states use Cartesian products, while vector representations use tensor products.

That same idea appears in several forms:

classical multi-system states use Cartesian products and probability vectors
quantum state multi-systems use tensor products and complex amplitudes

Classical Multi-System States

Classical Compound Systems

Suppose $X$ has classical state set $\Sigma$ and $Y$ has classical state set $\Gamma$.

The compound system $(X,Y)$ has classical state set:

\[ \Sigma \times \Gamma = \{(a,b): a\in\Sigma,\ b\in\Gamma\}. \]

For example, if:

\[ \Sigma=\{0,1\}, \qquad \Gamma=\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}, \]

then:

\[ \Sigma\times\Gamma = \{ (0,\clubsuit),(0,\diamondsuit),(0,\heartsuit),(0,\spadesuit), (1,\clubsuit),(1,\diamondsuit),(1,\heartsuit),(1,\spadesuit) \}. \]

For $n$ systems $X_1,\dots,X_n$ with classical state sets $\Sigma_1,\dots,\Sigma_n$, the compound state set is:

\[ \Sigma_1\times\cdots\times\Sigma_n = \{(a_1,\dots,a_n): a_1\in\Sigma_1,\dots,a_n\in\Sigma_n\}. \]

If $X_1,X_2,X_3$ are bits, then:

\[ \{0,1\}^3 = \{ 000,001,010,011,100,101,110,111 \}. \]

The convention is lexicographic order, with significance decreasing from left to right.

For ten bits:

\[ \Sigma_1=\cdots=\Sigma_{10}=\{0,1\}, \qquad \Sigma_1\times\cdots\times\Sigma_{10}=\{0,1\}^{10}. \]

The ordered states begin:

\[ 0000000000,\ 0000000001,\ 0000000010,\ 0000000011,\ \dots,\ 1111111111. \]

Probabilistic States of Multiple Systems

A probabilistic state of a compound system assigns probabilities to the Cartesian product of the individual state sets.

For two bits $(X,Y)$, one possible state is:

\[ \begin{aligned} \Pr((X,Y)=(0,0)) &= \frac{1}{2},\\ \Pr((X,Y)=(0,1)) &= 0,\\ \Pr((X,Y)=(1,0)) &= 0,\\ \Pr((X,Y)=(1,1)) &= \frac{1}{2}. \end{aligned} \]

Using the lexicographic basis $00,01,10,11$, the probability vector is:

\[ \begin{pmatrix} 1/2\\ 0\\ 0\\ 1/2 \end{pmatrix}. \]

Equivalently:

\[ |\pi\rangle = \frac{1}{2}|00\rangle+\frac{1}{2}|11\rangle. \]

The basis vectors are:

\[ |00\rangle= \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \quad |01\rangle= \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \quad |10\rangle= \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \quad |11\rangle= \begin{pmatrix}0\\0\\0\\1\end{pmatrix}. \]

In general:

\[ |\pi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma} p_{ab}|ab\rangle. \]

Independence

For a probabilistic state of $(X,Y)$, the systems $X$ and $Y$ are independent if:

\[ \Pr((X,Y)=(a,b))=\Pr(X=a)\Pr(Y=b) \]

for every $a\in\Sigma$ and $b\in\Gamma$.

In vector notation, this means there are probability vectors:

\[ |\phi\rangle=\sum_{a\in\Sigma}q_a|a\rangle, \qquad |\psi\rangle=\sum_{b\in\Gamma}r_b|b\rangle \]

such that:

\[ p_{ab}=q_ar_b. \]

This is exactly what the tensor product encodes.

Tensor Products of Vectors

Let:

\[ |\phi\rangle=\sum_{a\in\Sigma}\alpha_a|a\rangle, \qquad |\psi\rangle=\sum_{b\in\Gamma}\beta_b|b\rangle. \]

Their tensor product is:

\[ |\phi\rangle\otimes|\psi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma}\alpha_a\beta_b|ab\rangle. \]

Equivalently:

For example:

\[ |\phi\rangle=\frac{1}{4}|0\rangle+\frac{3}{4}|1\rangle, \qquad |\psi\rangle=\frac{2}{3}|0\rangle+\frac{1}{3}|1\rangle. \]

Then:

\[ \begin{aligned} |\phi\rangle\otimes|\psi\rangle &= \frac{1}{4}\frac{2}{3}|00\rangle + \frac{1}{4}\frac{1}{3}|01\rangle + \frac{3}{4}\frac{2}{3}|10\rangle + \frac{3}{4}\frac{1}{3}|11\rangle\\ &= \frac{1}{6}|00\rangle + \frac{1}{12}|01\rangle + \frac{1}{2}|10\rangle + \frac{1}{4}|11\rangle. \end{aligned} \]

For column vectors:

\[ \begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_m \end{pmatrix} \otimes \begin{pmatrix} \beta_1\\ \vdots\\ \beta_k \end{pmatrix} = \begin{pmatrix} \alpha_1\beta_1\\ \vdots\\ \alpha_1\beta_k\\ \alpha_2\beta_1\\ \vdots\\ \alpha_m\beta_k \end{pmatrix}. \]

The tensor product is bilinear:

and:

\[ (\alpha|\phi\rangle)\otimes|\psi\rangle = \alpha(|\phi\rangle\otimes|\psi\rangle). \]

For $n$ systems:

\[ |\psi\rangle = |\phi_1\rangle\otimes\cdots\otimes|\phi_n\rangle \]

is defined by:

\[ \langle a_1\cdots a_n|\psi\rangle = \langle a_1|\phi_1\rangle\cdots\langle a_n|\phi_n\rangle. \]

Measuring Probabilistic States

If two bits are in the probabilistic state:

\[ \frac{1}{2}|00\rangle+\frac{1}{2}|11\rangle, \]

then measuring both bits gives outcome $00$ with probability $1/2$ and outcome $11$ with probability $1/2$.

If only system $X$ is measured, the probability of observing $a\in\Sigma$ is the marginal probability:

\[ \Pr(X=a) = \sum_{b\in\Gamma}\Pr((X,Y)=(a,b)). \]

The conditional probability is:

\[ \Pr(Y=b\mid X=a) = \frac{\Pr((X,Y)=(a,b))}{\Pr(X=a)}. \]

For:

\[ \sum_{(a,b)\in\Sigma\times\Gamma}p_{ab}|ab\rangle = \sum_{a\in\Sigma}|a\rangle\otimes \left(\sum_{b\in\Gamma}p_{ab}|b\rangle\right), \]

the marginal probability is:

\[ \Pr(X=a)=\sum_{b\in\Gamma}p_{ab}, \]

and the conditional state of $Y$ after observing $X=a$ is:

\[ \frac{\sum_{b\in\Gamma}p_{ab}|b\rangle} {\sum_{c\in\Gamma}p_{ac}}. \]

For example:

\[ \frac{1}{12}|00\rangle + \frac{1}{4}|01\rangle + \frac{1}{3}|10\rangle + \frac{1}{3}|11\rangle \]

can be grouped by the first system as:

\[ |0\rangle\otimes \left(\frac{1}{12}|0\rangle+\frac{1}{4}|1\rangle\right) + |1\rangle\otimes \left(\frac{1}{3}|0\rangle+\frac{1}{3}|1\rangle\right). \]

Operations on Probabilistic States

The controlled-NOT operation, or CNOT, acts on a two-bit system $(X,Y)$.

$X$ is the control bit.
$Y$ is the target bit.
if $X=1$, flip $Y$
if $X=0$, leave $Y$ unchanged

Its action on the standard basis is:

\[ \begin{aligned} |00\rangle &\mapsto |00\rangle,\\ |01\rangle &\mapsto |01\rangle,\\ |10\rangle &\mapsto |11\rangle,\\ |11\rangle &\mapsto |10\rangle. \end{aligned} \]

So the matrix is:

\[ \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}. \]

Another example is a randomized operation:

with probability $1/2$, set $Y=X$
with probability $1/2$, set $X=Y$

The resulting stochastic matrix is an average of two deterministic matrices:

\[ \begin{pmatrix} 1 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 \end{pmatrix}. \]

Tensor Products of Matrices

Let:

\[ M=\sum_{a,b\in\Sigma}\alpha_{ab}|a\rangle\langle b|, \qquad N=\sum_{c,d\in\Gamma}\beta_{cd}|c\rangle\langle d|. \]

Their tensor product is:

\[ M\otimes N = \sum_{a,b\in\Sigma} \sum_{c,d\in\Gamma} \alpha_{ab}\beta_{cd}|ac\rangle\langle bd|. \]

Equivalently:

For block matrices, the rule is:

\[ A\otimes B = \begin{pmatrix} a_{11}B & a_{12}B\\ a_{21}B & a_{22}B \end{pmatrix} \]

for a $2\times 2$ matrix $A$.

For $n$ matrices:

The mixed-product property is:

\[ (M_1\otimes\cdots\otimes M_n) (N_1\otimes\cdots\otimes N_n) = (M_1N_1)\otimes\cdots\otimes(M_nN_n). \]

This says that independent local operations can be multiplied locally before tensoring globally.

Quantum State Multi-Systems

Quantum States of Multiple Systems

A quantum state of a compound system is a column vector whose indices correspond to the Cartesian product of the individual classical state sets.

For two qubits, the standard basis is:

\[ |00\rangle,\ |01\rangle,\ |10\rangle,\ |11\rangle. \]

A general two-qubit state has the form:

\[ |\psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle, \]

with normalization:

\[ |\alpha_{00}|^2+|\alpha_{01}|^2+|\alpha_{10}|^2+|\alpha_{11}|^2=1. \]

Examples:

\[ \frac{1}{2}|00\rangle + \frac{i}{2}|01\rangle - \frac{1}{2}|10\rangle - \frac{i}{2}|11\rangle \]

has equal measurement probability $1/4$ for all four outcomes.

The state:

\[ \frac{3}{5}|00\rangle-\frac{4}{5}|11\rangle \]

has probability $9/25$ of measuring $00$ and $16/25$ of measuring $11$.

The basis state:

\[ |01\rangle \]

has probability 1 of measuring $01$.

Product States and Entanglement

If $|\phi\rangle$ is a state of system $X$ and $|\psi\rangle$ is a state of system $Y$, then:

\[ |\phi\rangle\otimes|\psi\rangle \]

is a state of the compound system $(X,Y)$.

States of this form are product states. They represent systems that are independent in the quantum-state description.

More generally:

\[ |\psi_1\rangle\otimes\cdots\otimes|\psi_n\rangle \]

is a product state of $(X_1,\dots,X_n)$.

Not every compound quantum state is a product state. States that cannot be factored into tensor products are entangled.

Bell States

The Bell states are four maximally entangled two-qubit states. They form an orthonormal basis for the four-dimensional two-qubit state space.

The first Bell state is:

\[ |\Phi^+\rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle. \]

Measuring both qubits gives $00$ with probability $1/2$ and $11$ with probability $1/2$. The outcomes are perfectly correlated.

The second Bell state is:

\[ |\Phi^-\rangle = \frac{1}{\sqrt{2}}|00\rangle - \frac{1}{\sqrt{2}}|11\rangle. \]

It has the same measurement probabilities as $|\Phi^+\rangle$, but the relative phase is different.

The third Bell state is:

\[ |\Psi^+\rangle = \frac{1}{\sqrt{2}}|01\rangle + \frac{1}{\sqrt{2}}|10\rangle. \]

Measuring gives $01$ or $10$ with equal probability. The outcomes are perfectly anti-correlated.

The fourth Bell state is:

\[ |\Psi^-\rangle = \frac{1}{\sqrt{2}}|01\rangle - \frac{1}{\sqrt{2}}|10\rangle. \]

This is also anti-correlated, with a different relative phase.

GHZ and W States

The GHZ state is:

\[ \frac{1}{\sqrt{2}}|000\rangle + \frac{1}{\sqrt{2}}|111\rangle. \]

All three qubits are perfectly correlated. Measuring one qubit determines the others.

The W state is:

\[ \frac{1}{\sqrt{3}}|001\rangle + \frac{1}{\sqrt{3}}|010\rangle + \frac{1}{\sqrt{3}}|100\rangle. \]

Exactly one qubit is in state $1$. Each of the three positions has probability $1/3$.

The W state is important because its entanglement is more robust to losing or measuring one subsystem than GHZ-style entanglement.

Measuring Quantum Compound Systems

If a compound system is measured completely, the outcome belongs to the Cartesian product:

\[ (a_1,\dots,a_n)\in \Sigma_1\times\cdots\times\Sigma_n. \]

For a joint quantum state $|\psi\rangle$, the probability of observing $a_1\cdots a_n$ is:

\[ \left|\langle a_1\cdots a_n|\psi\rangle\right|^2. \]

For example:

\[ |\psi\rangle = \frac{3}{5}|0\rangle|\heartsuit\rangle - \frac{4i}{5}|1\rangle|\spadesuit\rangle. \]

The probability of outcome $(0,\heartsuit)$ is:

\[ \left|\frac{3}{5}\right|^2=\frac{9}{25}. \]

The probability of outcome $(1,\spadesuit)$ is:

\[ \left|-\frac{4i}{5}\right|^2=\frac{16}{25}. \]

For a two-system state:

\[ |\psi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma}\alpha_{ab}|ab\rangle, \]

if only $X$ is measured, then:

\[ \Pr(\text{outcome is }a) = \sum_{b\in\Gamma}|\langle ab|\psi\rangle|^2 = \sum_{b\in\Gamma}|\alpha_{ab}|^2. \]

Now group the state by outcomes of $X$:

\[ |\psi\rangle = \sum_{a\in\Sigma}|a\rangle\otimes|\Phi_a\rangle. \]

Then:

\[ \Pr(\text{outcome is }a)=\||\Phi_a\rangle\|^2. \]

If the measurement of $X$ gives outcome $a$, the post-measurement state is:

\[ |a\rangle\otimes \frac{|\Phi_a\rangle}{\||\Phi_a\rangle\|}. \]

The division by the norm re-normalizes the remaining state.

Unitary Operations on Multiple Systems

When independent unitary operations are applied to separate systems, the combined global operation is the tensor product of the local unitary matrices.

If $U_1,\dots,U_n$ act on systems $X_1,\dots,X_n$, then the global operation is:

\[ U_1\otimes\cdots\otimes U_n. \]

If we apply a unitary $U$ to system $X$ and do nothing to system $Y$, the global operation on $(X,Y)$ is:

\[ U\otimes I. \]

The identity matrix $I$ represents doing nothing to the unaddressed subsystem. This is a recurring pattern: even a local physical operation must be represented as an operation on the full compound state space.

--- title: "Qiskit Lesson 2: Multiple Systems" author: "Chao Ma" date: "2026-05-23" categories: [Quantum Computing, Qiskit, Quantum Information, Tensor Products, Entanglement] description: "Lesson 2 notes on compound classical and quantum systems: Cartesian products, tensor products, marginal and conditional states, CNOT, Bell states, partial measurement, and local unitary operations." toc: true --- <iframe width="100%" height="500" src="https://www.youtube.com/embed/DfZZS8Spe7U?list=PLOFEBzvs-VvqKKMXX4vbi4EB1uaErFMSO&index=4" title="Qiskit Lesson 2" frameborder="0" allowfullscreen></iframe> Lesson 1 described a single system. Lesson 2 asks what changes when we combine systems. The main answer is that state spaces multiply: classical states use Cartesian products, while vector representations use tensor products. That same idea appears in several forms: - classical multi-system states use Cartesian products and probability vectors - quantum state multi-systems use tensor products and complex amplitudes ## Classical Multi-System States ### Classical Compound Systems Suppose $X$ has classical state set $\Sigma$ and $Y$ has classical state set $\Gamma$. The compound system $(X,Y)$ has classical state set: $$ \Sigma \times \Gamma = \{(a,b): a\in\Sigma,\ b\in\Gamma\}. $$ For example, if: $$ \Sigma=\{0,1\}, \qquad \Gamma=\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}, $$ then: $$ \Sigma\times\Gamma = \{ (0,\clubsuit),(0,\diamondsuit),(0,\heartsuit),(0,\spadesuit), (1,\clubsuit),(1,\diamondsuit),(1,\heartsuit),(1,\spadesuit) \}. $$ For $n$ systems $X_1,\dots,X_n$ with classical state sets $\Sigma_1,\dots,\Sigma_n$, the compound state set is: $$ \Sigma_1\times\cdots\times\Sigma_n = \{(a_1,\dots,a_n): a_1\in\Sigma_1,\dots,a_n\in\Sigma_n\}. $$ If $X_1,X_2,X_3$ are bits, then: $$ \{0,1\}^3 = \{ 000,001,010,011,100,101,110,111 \}. $$ The convention is lexicographic order, with significance decreasing from left to right. For ten bits: $$ \Sigma_1=\cdots=\Sigma_{10}=\{0,1\}, \qquad \Sigma_1\times\cdots\times\Sigma_{10}=\{0,1\}^{10}. $$ The ordered states begin: $$ 0000000000,\ 0000000001,\ 0000000010,\ 0000000011,\ \dots,\ 1111111111. $$ ### Probabilistic States of Multiple Systems A probabilistic state of a compound system assigns probabilities to the Cartesian product of the individual state sets. For two bits $(X,Y)$, one possible state is: $$ \begin{aligned} \Pr((X,Y)=(0,0)) &= \frac{1}{2},\\ \Pr((X,Y)=(0,1)) &= 0,\\ \Pr((X,Y)=(1,0)) &= 0,\\ \Pr((X,Y)=(1,1)) &= \frac{1}{2}. \end{aligned} $$ Using the lexicographic basis $00,01,10,11$, the probability vector is: $$ \begin{pmatrix} 1/2\\ 0\\ 0\\ 1/2 \end{pmatrix}. $$ Equivalently: $$ |\pi\rangle = \frac{1}{2}|00\rangle+\frac{1}{2}|11\rangle. $$ The basis vectors are: $$ |00\rangle= \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \quad |01\rangle= \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \quad |10\rangle= \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \quad |11\rangle= \begin{pmatrix}0\\0\\0\\1\end{pmatrix}. $$ In general: $$ |\pi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma} p_{ab}|ab\rangle. $$ ### Independence For a probabilistic state of $(X,Y)$, the systems $X$ and $Y$ are independent if: $$ \Pr((X,Y)=(a,b))=\Pr(X=a)\Pr(Y=b) $$ for every $a\in\Sigma$ and $b\in\Gamma$. In vector notation, this means there are probability vectors: $$ |\phi\rangle=\sum_{a\in\Sigma}q_a|a\rangle, \qquad |\psi\rangle=\sum_{b\in\Gamma}r_b|b\rangle $$ such that: $$ p_{ab}=q_ar_b. $$ This is exactly what the tensor product encodes. ### Tensor Products of Vectors Let: $$ |\phi\rangle=\sum_{a\in\Sigma}\alpha_a|a\rangle, \qquad |\psi\rangle=\sum_{b\in\Gamma}\beta_b|b\rangle. $$ Their tensor product is: $$ |\phi\rangle\otimes|\psi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma}\alpha_a\beta_b|ab\rangle. $$ Equivalently: $$ \langle ab|\left(|\phi\rangle\otimes|\psi\rangle\right) = \langle a|\phi\rangle\langle b|\psi\rangle. $$ For example: $$ |\phi\rangle=\frac{1}{4}|0\rangle+\frac{3}{4}|1\rangle, \qquad |\psi\rangle=\frac{2}{3}|0\rangle+\frac{1}{3}|1\rangle. $$ Then: $$ \begin{aligned} |\phi\rangle\otimes|\psi\rangle &= \frac{1}{4}\frac{2}{3}|00\rangle + \frac{1}{4}\frac{1}{3}|01\rangle + \frac{3}{4}\frac{2}{3}|10\rangle + \frac{3}{4}\frac{1}{3}|11\rangle\\ &= \frac{1}{6}|00\rangle + \frac{1}{12}|01\rangle + \frac{1}{2}|10\rangle + \frac{1}{4}|11\rangle. \end{aligned} $$ For column vectors: $$ \begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_m \end{pmatrix} \otimes \begin{pmatrix} \beta_1\\ \vdots\\ \beta_k \end{pmatrix} = \begin{pmatrix} \alpha_1\beta_1\\ \vdots\\ \alpha_1\beta_k\\ \alpha_2\beta_1\\ \vdots\\ \alpha_m\beta_k \end{pmatrix}. $$ The tensor product is bilinear: $$ (|\phi_1\rangle+|\phi_2\rangle)\otimes|\psi\rangle = |\phi_1\rangle\otimes|\psi\rangle + |\phi_2\rangle\otimes|\psi\rangle, $$ and: $$ (\alpha|\phi\rangle)\otimes|\psi\rangle = \alpha(|\phi\rangle\otimes|\psi\rangle). $$ For $n$ systems: $$ |\psi\rangle = |\phi_1\rangle\otimes\cdots\otimes|\phi_n\rangle $$ is defined by: $$ \langle a_1\cdots a_n|\psi\rangle = \langle a_1|\phi_1\rangle\cdots\langle a_n|\phi_n\rangle. $$ ### Measuring Probabilistic States If two bits are in the probabilistic state: $$ \frac{1}{2}|00\rangle+\frac{1}{2}|11\rangle, $$ then measuring both bits gives outcome $00$ with probability $1/2$ and outcome $11$ with probability $1/2$. If only system $X$ is measured, the probability of observing $a\in\Sigma$ is the marginal probability: $$ \Pr(X=a) = \sum_{b\in\Gamma}\Pr((X,Y)=(a,b)). $$ The conditional probability is: $$ \Pr(Y=b\mid X=a) = \frac{\Pr((X,Y)=(a,b))}{\Pr(X=a)}. $$ For: $$ \sum_{(a,b)\in\Sigma\times\Gamma}p_{ab}|ab\rangle = \sum_{a\in\Sigma}|a\rangle\otimes \left(\sum_{b\in\Gamma}p_{ab}|b\rangle\right), $$ the marginal probability is: $$ \Pr(X=a)=\sum_{b\in\Gamma}p_{ab}, $$ and the conditional state of $Y$ after observing $X=a$ is: $$ \frac{\sum_{b\in\Gamma}p_{ab}|b\rangle} {\sum_{c\in\Gamma}p_{ac}}. $$ For example: $$ \frac{1}{12}|00\rangle + \frac{1}{4}|01\rangle + \frac{1}{3}|10\rangle + \frac{1}{3}|11\rangle $$ can be grouped by the first system as: $$ |0\rangle\otimes \left(\frac{1}{12}|0\rangle+\frac{1}{4}|1\rangle\right) + |1\rangle\otimes \left(\frac{1}{3}|0\rangle+\frac{1}{3}|1\rangle\right). $$ ### Operations on Probabilistic States The controlled-NOT operation, or CNOT, acts on a two-bit system $(X,Y)$. - $X$ is the control bit. - $Y$ is the target bit. - if $X=1$, flip $Y$ - if $X=0$, leave $Y$ unchanged Its action on the standard basis is: $$ \begin{aligned} |00\rangle &\mapsto |00\rangle,\\ |01\rangle &\mapsto |01\rangle,\\ |10\rangle &\mapsto |11\rangle,\\ |11\rangle &\mapsto |10\rangle. \end{aligned} $$ So the matrix is: $$ \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}. $$ Another example is a randomized operation: - with probability $1/2$, set $Y=X$ - with probability $1/2$, set $X=Y$ The resulting stochastic matrix is an average of two deterministic matrices: $$ \begin{pmatrix} 1 & 1/2 & 1/2 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1/2 & 1/2 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 \end{pmatrix}. $$ ### Tensor Products of Matrices Let: $$ M=\sum_{a,b\in\Sigma}\alpha_{ab}|a\rangle\langle b|, \qquad N=\sum_{c,d\in\Gamma}\beta_{cd}|c\rangle\langle d|. $$ Their tensor product is: $$ M\otimes N = \sum_{a,b\in\Sigma} \sum_{c,d\in\Gamma} \alpha_{ab}\beta_{cd}|ac\rangle\langle bd|. $$ Equivalently: $$ \langle ac|M\otimes N|bd\rangle = \langle a|M|b\rangle \langle c|N|d\rangle. $$ For block matrices, the rule is: $$ A\otimes B = \begin{pmatrix} a_{11}B & a_{12}B\\ a_{21}B & a_{22}B \end{pmatrix} $$ for a $2\times 2$ matrix $A$. For $n$ matrices: $$ \langle a_1\cdots a_n| M_1\otimes\cdots\otimes M_n |b_1\cdots b_n\rangle = \langle a_1|M_1|b_1\rangle \cdots \langle a_n|M_n|b_n\rangle. $$ The mixed-product property is: $$ (M_1\otimes\cdots\otimes M_n) (N_1\otimes\cdots\otimes N_n) = (M_1N_1)\otimes\cdots\otimes(M_nN_n). $$ This says that independent local operations can be multiplied locally before tensoring globally. ## Quantum State Multi-Systems ### Quantum States of Multiple Systems A quantum state of a compound system is a column vector whose indices correspond to the Cartesian product of the individual classical state sets. For two qubits, the standard basis is: $$ |00\rangle,\ |01\rangle,\ |10\rangle,\ |11\rangle. $$ A general two-qubit state has the form: $$ |\psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle, $$ with normalization: $$ |\alpha_{00}|^2+|\alpha_{01}|^2+|\alpha_{10}|^2+|\alpha_{11}|^2=1. $$ Examples: $$ \frac{1}{2}|00\rangle + \frac{i}{2}|01\rangle - \frac{1}{2}|10\rangle - \frac{i}{2}|11\rangle $$ has equal measurement probability $1/4$ for all four outcomes. The state: $$ \frac{3}{5}|00\rangle-\frac{4}{5}|11\rangle $$ has probability $9/25$ of measuring $00$ and $16/25$ of measuring $11$. The basis state: $$ |01\rangle $$ has probability 1 of measuring $01$. ### Product States and Entanglement If $|\phi\rangle$ is a state of system $X$ and $|\psi\rangle$ is a state of system $Y$, then: $$ |\phi\rangle\otimes|\psi\rangle $$ is a state of the compound system $(X,Y)$. States of this form are product states. They represent systems that are independent in the quantum-state description. More generally: $$ |\psi_1\rangle\otimes\cdots\otimes|\psi_n\rangle $$ is a product state of $(X_1,\dots,X_n)$. Not every compound quantum state is a product state. States that cannot be factored into tensor products are entangled. ### Bell States The Bell states are four maximally entangled two-qubit states. They form an orthonormal basis for the four-dimensional two-qubit state space. The first Bell state is: $$ |\Phi^+\rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle. $$ Measuring both qubits gives $00$ with probability $1/2$ and $11$ with probability $1/2$. The outcomes are perfectly correlated. The second Bell state is: $$ |\Phi^-\rangle = \frac{1}{\sqrt{2}}|00\rangle - \frac{1}{\sqrt{2}}|11\rangle. $$ It has the same measurement probabilities as $|\Phi^+\rangle$, but the relative phase is different. The third Bell state is: $$ |\Psi^+\rangle = \frac{1}{\sqrt{2}}|01\rangle + \frac{1}{\sqrt{2}}|10\rangle. $$ Measuring gives $01$ or $10$ with equal probability. The outcomes are perfectly anti-correlated. The fourth Bell state is: $$ |\Psi^-\rangle = \frac{1}{\sqrt{2}}|01\rangle - \frac{1}{\sqrt{2}}|10\rangle. $$ This is also anti-correlated, with a different relative phase. ### GHZ and W States The GHZ state is: $$ \frac{1}{\sqrt{2}}|000\rangle + \frac{1}{\sqrt{2}}|111\rangle. $$ All three qubits are perfectly correlated. Measuring one qubit determines the others. The W state is: $$ \frac{1}{\sqrt{3}}|001\rangle + \frac{1}{\sqrt{3}}|010\rangle + \frac{1}{\sqrt{3}}|100\rangle. $$ Exactly one qubit is in state $1$. Each of the three positions has probability $1/3$. The W state is important because its entanglement is more robust to losing or measuring one subsystem than GHZ-style entanglement. ### Measuring Quantum Compound Systems If a compound system is measured completely, the outcome belongs to the Cartesian product: $$ (a_1,\dots,a_n)\in \Sigma_1\times\cdots\times\Sigma_n. $$ For a joint quantum state $|\psi\rangle$, the probability of observing $a_1\cdots a_n$ is: $$ \left|\langle a_1\cdots a_n|\psi\rangle\right|^2. $$ For example: $$ |\psi\rangle = \frac{3}{5}|0\rangle|\heartsuit\rangle - \frac{4i}{5}|1\rangle|\spadesuit\rangle. $$ The probability of outcome $(0,\heartsuit)$ is: $$ \left|\frac{3}{5}\right|^2=\frac{9}{25}. $$ The probability of outcome $(1,\spadesuit)$ is: $$ \left|-\frac{4i}{5}\right|^2=\frac{16}{25}. $$ For a two-system state: $$ |\psi\rangle = \sum_{(a,b)\in\Sigma\times\Gamma}\alpha_{ab}|ab\rangle, $$ if only $X$ is measured, then: $$ \Pr(\text{outcome is }a) = \sum_{b\in\Gamma}|\langle ab|\psi\rangle|^2 = \sum_{b\in\Gamma}|\alpha_{ab}|^2. $$ Now group the state by outcomes of $X$: $$ |\psi\rangle = \sum_{a\in\Sigma}|a\rangle\otimes|\Phi_a\rangle. $$ Then: $$ \Pr(\text{outcome is }a)=\||\Phi_a\rangle\|^2. $$ If the measurement of $X$ gives outcome $a$, the post-measurement state is: $$ |a\rangle\otimes \frac{|\Phi_a\rangle}{\||\Phi_a\rangle\|}. $$ The division by the norm re-normalizes the remaining state. ### Unitary Operations on Multiple Systems When independent unitary operations are applied to separate systems, the combined global operation is the tensor product of the local unitary matrices. If $U_1,\dots,U_n$ act on systems $X_1,\dots,X_n$, then the global operation is: $$ U_1\otimes\cdots\otimes U_n. $$ If we apply a unitary $U$ to system $X$ and do nothing to system $Y$, the global operation on $(X,Y)$ is: $$ U\otimes I. $$ The identity matrix $I$ represents doing nothing to the unaddressed subsystem. This is a recurring pattern: even a local physical operation must be represented as an operation on the full compound state space.