Quantum States and Measurement

In Chapter 3, we wrote the state of a register as a weighted sum of its configurations, where each configuration \(i\) carries a complex amplitude \(v_i\) with \(\sum_i |v_i|^2 = 1\). Naming the complete state \(\ket{\psi}\) (read “ket psi”), we write

\[\ket{\psi} = \sum_i v_i\ket{i}.\]

Our two rules for combining ket sums—adding the weights of matching configurations and multiplying all weights by a common factor—are the defining operations of a vector space. We therefore view \(\ket{\psi}\) as a state vector. For a system with two distinguishable configurations, \(0\) and \(1\), any state takes the form \(\ket{\psi} = v_0\ket{0} + v_1\ket{1}\). A weighted combination of configurations is called a superposition, and a two-state quantum system is called a qubit (short for quantum bit).

Fixing the natural order of configurations—\(0\) first, \(1\) second—lets us write the amplitudes as a vertical column of numbers. Each configuration ket becomes the column with \(1\) in its own row and \(0\) elsewhere:

\[\ket{0} = \begin{pmatrix}1\\0\end{pmatrix},\qquad \ket{1} = \begin{pmatrix}0\\1\end{pmatrix},\qquad \ket{\psi} = v_0\ket{0} + v_1\ket{1} = v_0\begin{pmatrix}1\\0\end{pmatrix} + v_1\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}v_0\\v_1\end{pmatrix}.\]

An ordered column of numbers is a vector, its entries \(v_0, v_1\) are its coordinates, and the reference kets \(\ket{0}\) and \(\ket{1}\) form the computational basis.

In column form, adding two vectors and multiplying a vector by a number \(c\) act coordinate by coordinate:

\[\begin{pmatrix}a_0\\a_1\end{pmatrix} + \begin{pmatrix}b_0\\b_1\end{pmatrix} = \begin{pmatrix}a_0+b_0\\a_1+b_1\end{pmatrix},\qquad c\begin{pmatrix}a_0\\a_1\end{pmatrix} = \begin{pmatrix}ca_0\\ca_1\end{pmatrix}.\]

These two operations are vector addition and scalar multiplication. Note that while any two-entry column is a vector, only columns whose coordinates satisfy the normalization condition \(|v_0|^2 + |v_1|^2 = 1\) represent physical states of a qubit. For instance, in the decomposition \(\ket{\psi} = v_0\ket{0} + v_1\ket{1}\), the scaled basis vectors \(v_0\ket{0} = \binom{v_0}{0}\) and \(v_1\ket{1} = \binom{0}{v_1}\) are unnormalized component vectors whose sum is the normalized state vector \(\ket{\psi}\).

In the real-amplitude example above, the coordinates \(3/5\) and \(4/5\) yield outcome probabilities \((3/5)^2 = 9/25\) for \(0\) and \((4/5)^2 = 16/25\) for \(1\), which sum to \(1\). The two coordinates belong to a single state vector \(\ket{\psi}\) rather than describing two separate systems.

For a system with \(N\) configurations ordered from \(0\) to \(N-1\), a state vector is a column of \(N\) complex coordinates, while ket notation writes only the nonzero terms explicitly.

To keep track of several state vectors at once, we label the coordinates of \(\ket{\psi}\) as \(\psi_i\) and those of \(\ket{\phi}\) as \(\phi_i\). Just as \(\bra{j}S\ket{i}\) denotes the amplitude to reach configuration \(j\) from \(i\) in one step \(S\), we write \(\braket{j}{\psi}\) for the amplitude of finding state \(\ket{\psi}\) in basis configuration \(j\):

\[\braket{j}{\psi} = \psi_j.\]

When \(\ket{\psi}\) is itself a basis state \(\ket{i}\), its only nonzero coordinate is \(1\) at position \(i\), so

\[\braket{j}{i} = \begin{cases}1, & j=i,\\ 0, & j\ne i.\end{cases}\]

Now suppose we ask for the transition amplitude \(\braket{\phi}{\psi}\) from an initial state \(\ket{\psi} = \sum_j \psi_j\ket{j}\) to a target state \(\ket{\phi} = \sum_j \phi_j\ket{j}\). Every route from \(\ket{\psi}\) to \(\ket{\phi}\) passes through some basis configuration \(j\), leaving \(\ket{\psi}\) into \(j\) with amplitude \(\braket{j}{\psi}\) and entering \(\ket{\phi}\) from \(j\) with amplitude \(\braket{\phi}{j}\). By the product and sum rules,

\[\braket{\phi}{\psi} = \sum_j \braket{\phi}{j}\,\braket{j}{\psi}.\]

How is the incoming amplitude \(\braket{\phi}{j}\) related to the coordinate \(\phi_j = \braket{j}{\phi}\)? A normalized state \(\ket{\psi}\) must have total transition amplitude \(1\) to remain in itself, \(\braket{\psi}{\psi} = 1\). Since normalization requires \(\sum_j |\psi_j|^2 = \sum_j \psi_j^*\psi_j = 1\), matching \(\sum_j \braket{\psi}{j}\braket{j}{\psi}\) with \(\sum_j \psi_j^*\psi_j\) fixes the incoming amplitude to be the complex conjugate of the outgoing amplitude:

\[\braket{\phi}{j} = \braket{j}{\phi}^* = \phi_j^*.\]

Substituting \(\braket{\phi}{j} = \phi_j^*\) gives the general formula for the inner product of \(\ket{\phi}\) and \(\ket{\psi}\) (read “bra phi ket psi”):

\[\braket{\phi}{\psi} = \sum_j \phi_j^*\psi_j.\]

Without complex conjugation, the normalized state \((\ket{0}+i\ket{1})/\sqrt{2}\) would yield \(\tfrac{1}{2} + \tfrac{i^2}{2} = 0\) with itself instead of \(1\).

In matrix notation, the conjugated row vector paired with \(\ket{\phi}\) is called its bra, written \(\bra{\phi}\):

\[\ket{\phi} = \begin{pmatrix}\phi_0\\\phi_1\end{pmatrix},\qquad \bra{\phi} = \begin{pmatrix}\phi_0^* & \phi_1^*\end{pmatrix},\qquad \braket{\phi}{\psi} = \begin{pmatrix}\phi_0^* & \phi_1^*\end{pmatrix}\begin{pmatrix}\psi_0\\\psi_1\end{pmatrix} = \phi_0^*\psi_0 + \phi_1^*\psi_1.\]

Swapping the two states conjugates every term in the sum, giving conjugate symmetry:

\[\braket{\psi}{\phi} = \bigl(\braket{\phi}{\psi}\bigr)^*.\]

Reversing the direction of a transition therefore preserves the magnitude \(|\braket{\phi}{\psi}|\) and reverses the phase; for real vectors, \(\braket{\psi}{\phi} = \braket{\phi}{\psi}\).

The inner product equips our space of state vectors with geometric notions of length, angle, and perpendicular projection.

The norm (or length) of a vector \(\ket{\psi}\) is defined as

\[\|\psi\| = \sqrt{\braket{\psi}{\psi}} = \sqrt{\sum_j |\psi_j|^2}.\]

Normalized states satisfy \(\|\psi\| = 1\) and are therefore unit vectors. Two vectors \(\ket{\phi}\) and \(\ket{\psi}\) are orthogonal (perpendicular) when their inner product vanishes, \(\braket{\phi}{\psi} = 0\). A set of unit vectors that are mutually orthogonal is called orthonormal; the computational basis \(\{\ket{j}\}\) is orthonormal because \(\braket{j}{i}\) equals \(1\) for \(j=i\) and \(0\) for \(j\ne i\).

Given any unit vector \(\ket{\phi}\), we can split a state \(\ket{\psi}\) into a component along \(\ket{\phi}\) and a remainder \(\ket{\psi_\perp}\) orthogonal to \(\ket{\phi}\):

\[\ket{\psi} = \braket{\phi}{\psi}\,\ket{\phi} + \ket{\psi_\perp},\qquad \braket{\phi}{\psi_\perp} = 0.\]

Because the two pieces are orthogonal, their squared lengths obey the Pythagorean relation

\[\|\psi\|^2 = |\braket{\phi}{\psi}|^2 + \|\psi_\perp\|^2.\]

For two unit vectors (\(\|\psi\|=\|\phi\|=1\)), \(|\braket{\phi}{\psi}|^2 = 1 - \|\psi_\perp\|^2\) always lies between \(0\) and \(1\): it equals \(1\) when \(\ket{\psi_\perp}=0\) (so \(\ket{\psi}\) and \(\ket{\phi}\) differ only by a global phase) and \(0\) when \(\ket{\psi}\) and \(\ket{\phi}\) are orthogonal. When both unit vectors have real coordinates separated by an angle \(\theta\) in the plane, \(\braket{\phi}{\psi} = \cos\theta\).

For example, if \(\ket{\psi} = (\sqrt{3}/2)\ket{0} + (1/2)\ket{1}\) and \(\ket{\phi} = \ket{0}\), then \(\ket{\psi}\) makes a \(30^\circ\) angle with \(\ket{0}\), its projection along \(\ket{0}\) has amplitude \(\braket{0}{\psi} = \cos 30^\circ = \sqrt{3}/2\) (squared length \(3/4\)), and the orthogonal remainder \((1/2)\ket{1}\) has squared length \(1/4\).

An amplitude as a component

What to observe

At separation angle \(\theta\), the transition probability is \(\cos^2\theta\): \(3/4\) at \(30^\circ\), \(1/2\) at \(45^\circ\), and \(0\) at \(90^\circ\) where the states are orthogonal. At \(180^\circ\), the inner product is \(-1\) and the probability returns to \(1\).

With states written as column vectors, we can represent any single-step transition rule \(U\) as a rectangular array of numbers—a transition matrix (or simply a matrix)—whose entry in row \(j\) and column \(i\) is the arrow weight from initial configuration \(i\) to final configuration \(j\):

\[U_{ji} = \bra{j}U\ket{i}.\]

The column index \(i\) on the right names the starting configuration, and the row index \(j\) on the left names the destination configuration. Consequently, column \(i\) of \(U\) is the column vector \(U\ket{i} = \sum_j \bra{j}U\ket{i}\,\ket{j}\) formed by the outgoing arrow weights from configuration \(i\).

When \(U\) acts on a general state \(\ket{\psi} = \sum_i v_i\ket{i}\), the updated coordinate \(v'_j = \sum_i U_{ji}v_i\) is given by multiplying row \(j\) of \(U\) into the column vector \(\ket{\psi}\):

\[\ket{\psi'} = U\ket{\psi} = \sum_i v_i\,U\ket{i}.\]

Because matrix–vector multiplication distributes over vector addition and scalar multiplication, every matrix operation obeys linearity: for any states \(\ket{\phi}, \ket{\psi}\) and complex scalars \(a, b\),

\[U(a\ket{\phi} + b\ket{\psi}) = aU\ket{\phi} + bU\ket{\psi}.\]

A linear operation is therefore completely specified by its columns \(U\ket{i}\). For example, the single-qubit operations from Chapter 3—bit-flip \(X\), phase-flip \(Z\), and Hadamard \(H\)—have the transition matrices

\[X = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\qquad Z = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix},\qquad H = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}.\]

Geometrically, \(U\ket{\psi} = v_0 U\ket{0} + v_1 U\ket{1}\) combines the transformed basis vectors \(U\ket{0}\) and \(U\ket{1}\) using the original coordinates \(v_0\) and \(v_1\). For the real matrix

\[U = \frac{1}{5}\begin{pmatrix}4 & -3\\ 3 & 4\end{pmatrix},\]

acting on \(\ket{\psi} = (3/5)\ket{0} + (4/5)\ket{1}\) gives \((3/5)U\ket{0} + (4/5)U\ket{1} = \ket{1}\):

When a step \(U\) is followed by a second step \(V\), the two-step sum rule across intermediate configurations \(j\),

\[\bra{k}VU\ket{i} = \sum_j \bra{k}V\ket{j}\,\bra{j}U\ket{i},\qquad (VU)_{ki} = \sum_j V_{kj}U_{ji},\]

is the matrix product \(VU\), with the rightmost matrix \(U\) acting first. Two Hadamard steps multiply to the identity matrix \(I = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\), which leaves every vector unchanged: \(HH = I\).

Which matrices preserve total probability for every input state, and how can such a transition be reversed? In Section 3.4.2, we found that a step \(U\) preserves normalization across all superpositions if and only if its outgoing weights from each basis configuration are normalized (\(\sum_j |U_{ji}|^2 = 1\)) and the cross sums between distinct inputs vanish (\(\sum_j U_{ji}^* U_{jk} = 0\) for \(i\ne k\)). In the geometric language of Section 4.1.3, these two conditions state simply that the columns of \(U\) are orthonormal:

\[\sum_j U_{ji}^* U_{jk} = \delta_{ik},\]

where \(\delta_{ik}\) is \(1\) when \(i=k\) and \(0\) when \(i\ne k\).

As we saw with return and cross paths in Section 3.4.2, this condition also tells us immediately how to invert \(U\): reverse every transition arrow and take the complex conjugate of its weight. For any matrix \(A\), this operation defines the adjoint (or conjugate transpose), written \(A^\dagger\) (read “A dagger”):

\[\bra{i}A^\dagger\ket{j} = \bigl(\bra{j}A\ket{i}\bigr)^*,\qquad (A^\dagger)_{ij} = A_{ji}^*.\]

In matrix form, column \(i\) of \(A\) becomes the conjugated row \(i\) of \(A^\dagger\).

Now multiply \(U^\dagger\) by \(U\). The \((i,k)\) entry of the product \(U^\dagger U\) is the inner product of column \(i\) and column \(k\) of \(U\):

\[(U^\dagger U)_{ik} = \sum_j (U^\dagger)_{ij}U_{jk} = \sum_j U_{ji}^* U_{jk}.\]

Thus the columns of a square matrix \(U\) are orthonormal if and only if

\[U^\dagger U = I \qquad\bigl(\text{and hence } UU^\dagger = I\bigr).\]

A square matrix satisfying \(U^\dagger U = I\) is called unitary. Every unitary matrix is reversible, with its inverse given directly by its adjoint, \(U^{-1} = U^\dagger\).

For instance, the square root of NOT from Section 3.4.1 has \(a = (1+i)/2\) and \(b = (1-i)/2\); reversing its arrows without conjugation leaves the matrix unchanged and gives \(\sqrt{X}\sqrt{X} = X\), whereas conjugating the weights gives \((\sqrt{X})^\dagger\) with \(a^* = (1-i)/2\) and \(b^* = (1+i)/2\), so return paths add to \(|a|^2+|b|^2 = 1\) and switch paths cancel. Gates such as \(X\), \(Z\), and \(H\) satisfy \(U^\dagger = U\), so they are their own inverses.

For any matrix \(A\), the definition of \(A^\dagger\) implies \(\bra{\phi}A\ket{\psi} = \braket{A^\dagger\phi}{\psi}\). When \(U\) is unitary, applying \(U\) to any two states \(\ket{\chi}\) and \(\ket{\psi}\) therefore preserves their inner product:

\[\braket{U\chi}{U\psi} = \bra{\chi}U^\dagger U\ket{\psi} = \braket{\chi}{\psi}.\]

By contrast, when a matrix \(A\) is not unitary, \(A^\dagger\) does not equal \(A^{-1}\). For the horizontal shear

\[A = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix},\qquad A^\dagger = \begin{pmatrix}1 & 0\\ 1 & 1\end{pmatrix},\]

the columns of \(A\) are not orthogonal, neither \(A\) nor \(A^\dagger\) preserves vector lengths, and neither \(A^\dagger A\) nor \(AA^\dagger\) restores \(\ket{\psi} = (3/5)\ket{0} + (4/5)\ket{1}\):

State postulate. The state of an isolated finite-dimensional quantum system is represented by a unit vector \(\ket{\psi}\) in a complex vector space with inner product \(\braket{\phi}{\psi} = \sum_j \phi_j^*\psi_j\). Two unit vectors that differ only by a global phase factor \(e^{i\theta}\) represent the same physical state.

For a single qubit, \(\ket{\psi} = v_0\ket{0} + v_1\ket{1}\), normalization (\(|v_0|^2 + |v_1|^2 = 1\)) lets us write \(|v_0| = \cos(\theta/2)\) and \(|v_1| = \sin(\theta/2)\) for an angle \(0 \le \theta \le \pi\). Factoring out the global phase of \(v_0\) leaves a single relative phase \(\phi\) on \(\ket{1}\), so every pure one-qubit state can be written in terms of two angles \((\theta,\phi)\):

\[\ket{\psi} = \cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1},\qquad 0 \le \theta \le \pi.\]

Evolution postulate. Over any interval during which a quantum system remains isolated from outside interaction or measurement, its state vector evolves by a unitary operator \(U\):

\[\ket{\psi} \longmapsto U\ket{\psi},\qquad U^\dagger U = I.\]

As established in Section 4.1.5, unitary evolution is reversible (\(U^{-1} = U^\dagger\)) and preserves the inner product \(\braket{U\chi}{U\psi} = \braket{\chi}{\psi}\) between every pair of states. In the real state plane, unitary steps act as rigid rotations \(U(\delta) = \begin{pmatrix}\cos\delta & -\sin\delta\\ \sin\delta & \cos\delta\end{pmatrix}\) or reflections on the unit circle:

Whereas closed evolution transforms a superposition reversibly, observing a quantum system—performing a measurement—returns a single classical outcome and updates the state to match the observed outcome.

Measurement postulate. Given a system in the normalized state \(\ket{\psi}\) and any orthonormal basis \(\{\ket{b_j}\}\), an ideal measurement in that basis yields outcome \(j\) with probability

\[p(j) = |\braket{b_j}{\psi}|^2.\]

Conditional on observing outcome \(j\), the state of the system immediately becomes \(\ket{b_j}\).

When the chosen basis is the computational basis \(\{\ket{0},\ket{1}\}\) of a qubit in state \(\ket{\psi} = v_0\ket{0} + v_1\ket{1}\), outcome \(j\) occurs with probability \(|\braket{j}{\psi}|^2 = |v_j|^2\) and leaves the system in \(\ket{j}\). For example, measuring \(\ket{\psi} = (\sqrt{3}/2)\ket{0} + (1/2)\ket{1}\) yields \(0\) with probability \(3/4\) (leaving \(\ket{0}\)) and \(1\) with probability \(1/4\) (leaving \(\ket{1}\)). Because the state after outcome \(1\) is \(\ket{1}\), repeating the same measurement immediately afterward yields \(1\) with certainty.

This postulate also gives the operational meaning of the general transition amplitude \(\braket{\phi}{\psi}\) from Section 4.1.2: when \(\ket{\psi}\) is measured in any orthonormal basis containing \(\ket{\phi}\), the outcome corresponding to \(\ket{\phi}\) appears with probability \(|\braket{\phi}{\psi}|^2\). Geometrically, measurement projects \(\ket{\psi}\) onto the chosen basis directions and renormalizes the selected component. Writing the projector onto \(\ket{b_j}\) as the outer product \(P_j = \ketbra{b_j}{b_j}\), which acts by \(P_j\ket{\psi} = \ket{b_j}\braket{b_j}{\psi}\) (so \(\ketbra{0}{0} = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}\)), we have

\[p(j) = \|P_j\ket{\psi}\|^2,\qquad \ket{\psi_j} = \frac{P_j\ket{\psi}}{\sqrt{p(j)}}\quad\bigl(p(j) > 0\bigr).\]

In the real state plane, measuring \(\ket{\psi} = \cos\alpha\ket{0} + \sin\alpha\ket{1}\) in an orthonormal basis rotated by angle \(\beta\),

\[\ket{b_0} = \cos\beta\ket{0} + \sin\beta\ket{1},\qquad \ket{b_1} = -\sin\beta\ket{0} + \cos\beta\ket{1},\]

gives probabilities \(|\braket{b_0}{\psi}|^2 = \cos^2(\alpha-\beta)\) and \(|\braket{b_1}{\psi}|^2 = \sin^2(\alpha-\beta)\). Setting \(\beta = 0^\circ\) measures in the computational (\(Z\)) basis \(\{\ket{0},\ket{1}\}\); setting \(\beta = 45^\circ\) measures in the diagonal (\(X\)) basis \(\{\ket{+},\ket{-}\}\), where \(\ket{\pm} = (\ket{0}\pm\ket{1})/\sqrt{2}\).

One state, different measurements

For \(\ket+\), the Z-basis probabilities are \((1/2,1/2)\); the X-basis probabilities are \((1,0)\). For \(\ket-\), the Z-basis probabilities are also \((1/2,1/2)\), while the X-basis probabilities are \((0,1)\).

What to observe

Across 100 fresh copies, outcome \(1\) appears about \(25\pm 4\) times in accordance with \(|1/2|^2=1/4\). Re-measuring the same qubit after outcome \(1\) yields \(1\) deterministically because the first measurement projected the state onto \(\ket{1}\).

Unlike unitary evolution, a projective measurement is irreversible: both \(\ket{0}\) and \(\ket{+}\) can yield outcome \(0\) in the \(Z\) basis and collapse to the same post-measurement state \(\ket{0}\), erasing the distinction between them. If a qubit is measured in the \(Z\) basis between two Hadamard steps starting from \(\ket{0}\), the measurement projects \(\ket{+}\) onto \(\ket{0}\) or \(\ket{1}\) with probability \(1/2\) each, eliminating the second branch so the final Hadamard produces \(0\) or \(1\) at random instead of returning to \(\ket{0}\).

How do we form the state vector of a system made of multiple qubits? As we saw with two-bit configurations in Section 3.4.4, a configuration of a pair of qubits \(A\) and \(B\) specifies a bit for each qubit simultaneously, giving four joint basis configurations: \(00, 01, 10, 11\). Ordering them lexicographically turns a two-qubit state into a four-component column vector:

\[\ket{\Psi} = c_{00}\ket{00} + c_{01}\ket{01} + c_{10}\ket{10} + c_{11}\ket{11} = \begin{pmatrix}c_{00}\\c_{01}\\c_{10}\\c_{11}\end{pmatrix},\qquad \sum_{i,j\in\{0,1\}} |c_{ij}|^2 = 1.\]

When qubit \(A\) is prepared independently in \(\ket{\psi} = a\ket{0} + b\ket{1}\) and qubit \(B\) in \(\ket{\phi} = c\ket{0} + d\ket{1}\), the product rule assigns amplitude \(c_{ij} = \psi_i\phi_j\) to each joint configuration \(\ket{ij}\). This operation is the tensor product, written \(\ket{\psi}\otimes\ket{\phi}\):

\[\begin{pmatrix}a\\b\end{pmatrix}\otimes\begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}ac\\ad\\bc\\bd\end{pmatrix},\qquad \ket{i}\otimes\ket{j} = \ket{ij}.\]

Any composite state that factors into a tensor product \(\ket{\psi}\otimes\ket{\phi}\) of individual subsystem states is called a product state.

Composite-state postulate. The state space of a composite quantum system is the tensor product of the state spaces of its subsystems. If the subsystems have orthonormal bases \(\{\ket{i}\}\) and \(\{\ket{j}\}\), the composite system has the orthonormal basis \(\{\ket{ij} = \ket{i}\otimes\ket{j}\}\), and every normalized vector in this space is a valid physical state.

Combining subsystems of dimensions \(m\) and \(n\) produces a state space of dimension \(mn\); for \(n\) qubits, the state vector has \(2^n\) coordinates. Because any normalized superposition of the joint basis states is permitted, a composite state need not be a product state. For example, the state

\[\frac{\ket{00}+\ket{11}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\\0\\1\end{pmatrix},\]

which we produced in Section 3.4.4 by applying \(H\) on the first qubit followed by CNOT, cannot be written as \((a\ket{0}+b\ket{1})\otimes(c\ket{0}+d\ket{1})\): matching \(ac = 1/\sqrt{2}\) and \(bd = 1/\sqrt{2}\) forces \(a,b,c,d\) to be nonzero, which makes \(ad = 0\) and \(bc = 0\) impossible. A composite state that cannot be factored into a product state is entangled, and this property is called entanglement. In Chapter 5, we combine these four postulates into the circuit model of quantum computation, and in Chapter 6 we study multi-qubit gates and entangled states in detail.