← Chapter 04: Quantum States and Measurement

Quantum Computing · Chapter 04

Quantum States and Measurement

How do path calculations become a geometry of states, and what happens when we observe a quantum system?

Mithilesh Kumar

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4.1

Vectors and Matrices

States become vectors, amplitudes between states become inner products, and steps become matrices, each built from the path rules.

State vectors

A state \(\ket\psi=\sum_iv_i\ket i\) gives each configuration an amplitude. Its bookkeeping rules are those of vector arithmetic, so we treat \(\ket\psi\) as a vector. A two-configuration quantum system is a qubit; a weighted sum of kets is a superposition.

Order the configurations (0 first) and list the amplitudes:

\[\ket0=\begin{pmatrix}1\\0\end{pmatrix},\quad\ket1=\begin{pmatrix}0\\1\end{pmatrix},\quad v_0\ket0+v_1\ket1=\begin{pmatrix}v_0\\v_1\end{pmatrix}.\]

\(\ket0,\ket1\) form the computational basis. Adding states adds coordinates; multiplying by a number multiplies every coordinate.

Example: \(\tfrac35\ket0+\tfrac45\ket1\) has probabilities \(\tfrac9{25}\) and \(\tfrac{16}{25}\). The coordinates describe one state, not two outcomes at once.

A ket as a vector in the standard basisThe standard basis vectors ket zero and ket one are perpendicular unit vectors. A blue arrow represents ket psi equals three fifths ket zero plus four fifths ket one. Dashed lines mark its coordinates v zero equals three fifths horizontally and v one equals four fifths vertically. The same ket is the column vector with entries three fifths and four fifths. This real-amplitude drawing represents state space, not physical space.A ket and its standard-basis coordinates|1⟩|0⟩|ψ⟩v₁=4/5v₀=3/50Standard basis|0⟩=beginpmatrix1\0endpmatrix|1⟩=beginpmatrix0\1endpmatrix|ψ⟩=3/5|0⟩+4/5|1⟩|ψ⟩=beginpmatrix3/5\4/5endpmatrixA real-amplitude example.|ψ⟩=v₀|0⟩+v₁|1⟩: the coefficients are coordinates.

The amplitude between two states

Write \(\braket j\psi=\psi_j\) for the amplitude of \(j\) in \(\ket\psi\). Each path to a final state \(\ket\phi\) passes through one \(j\), so

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

What is \(\braket\phi j\)? Normalization, \(\sum_j\braket j\psi^*\braket j\psi=1\), matches the path formula for \(\braket\psi\psi\) if we choose

\[\braket\phi j=\braket j\phi^*=\phi_j^* .\]
Paths from an initial state to a final stateThe arrow into configuration j carries ⟨j|ψ⟩, and the arrow from j into the final state carries ⟨φ|j⟩. Each path multiplies the two and the paths add, giving ⟨φ|ψ⟩. For the final state |0⟩, only the path through 0 contributes, because ⟨0|0⟩=1 and ⟨0|1⟩=0. |ψ⟩ → 0 (weight ⟨0|ψ⟩); |ψ⟩ → 1 (weight ⟨1|ψ⟩); 0 → |φ⟩ (weight ⟨φ|0⟩); 1 → |φ⟩ (weight ⟨φ|1⟩). |ψ⟩ → 0 (weight ⟨0|ψ⟩); |ψ⟩ → 1 (weight ⟨1|ψ⟩); 0 → |0⟩ (weight ⟨0|0⟩=1); 1 → |0⟩ (weight ⟨0|1⟩=0).Paths from an initial state to a final stateinitialconfigurationfinal⟨0|ψ⟩⟨1|ψ⟩⟨φ|0⟩⟨φ|1⟩|ψ⟩01|φ⟩initialconfigurationfinal⟨0|ψ⟩⟨1|ψ⟩⟨0|0⟩=1⟨0|1⟩=0|ψ⟩01|0⟩

Inner products

Definition

The inner product (transition amplitude from \(\ket\psi\) to \(\ket\phi\)) is \(\displaystyle\braket\phi\psi=\sum_j\phi_j^*\psi_j\).

The bra \(\bra\phi\) is the conjugated row of \(\ket\phi\). Row times column performs the path sum:

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

Exchanging initial and final states conjugates the amplitude (conjugate symmetry):

\[\braket\psi\phi=\sum_j\psi_j^*\phi_j=\Bigl(\sum_j\phi_j^*\psi_j\Bigr)^*=\braket\phi\psi^* .\]

Example: for \(\ket\psi=(\ket0+i\ket1)/\sqrt2\), \(\braket1\psi=i/\sqrt2\) but \(\braket\psi1=-i/\sqrt2\). The magnitude is unchanged; for real vectors the order does not matter.

This definition is chosen for consistency with the probability rule; its physical meaning for a general final state comes with the measurement postulate.

Length, orthogonality, and projection

The norm is \(\|\psi\|=\sqrt{\braket\psi\psi}\); normalized states are unit vectors. Vectors with \(\braket\phi\psi=0\) are orthogonal; mutually orthogonal unit vectors are orthonormal.

Split a unit vector \(\ket\psi\) into a part along a unit vector \(\ket\phi\) and a part orthogonal to it:

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

The parts are perpendicular, so their squared lengths add:

\[1=|\braket\phi\psi|^2+\|\psi_\perp\|^2 .\]

Hence \(0\le|\braket\phi\psi|^2\le1\): equal to 1 when the states differ by a global phase, 0 when orthogonal. For real unit vectors \(\braket\phi\psi=\cos\theta\).

Splitting a state along a directionIn a real state-vector plane, the unit vector phi points at 15 degrees and the unit vector psi at 75 degrees, so the angle between them is 60 degrees. The component of psi along phi, the inner product of phi and psi times phi, has length one half and lies along phi. The remainder psi perp runs from the end of that component to the tip of psi, at a right angle to phi, with length root three over two. The squared lengths one quarter and three quarters add to one.Splitting a state along a direction|φ⟩|ψ⟩θ⟨φ|ψ⟩ |φ⟩|ψ_(perp)⟩|ψ⟩=⟨φ|ψ⟩ |φ⟩+|ψ_(perp)⟩⟨φ|ψ_(perp)⟩=01=|⟨φ|ψ⟩|²+|ψ_(perp)|²Here θ=60°: ⟨φ|ψ⟩=1/21=1/4+3/4

Coordinates are components

Doing the split along every basis direction gives the coordinates: \(\braket j\psi\) is the component of \(\ket\psi\) along \(\ket j\). Because the basis is orthonormal,

\[\ket\psi=\sum_j\ket j\braket j\psi,\qquad \braket\psi\psi=\sum_j|\braket j\psi|^2 .\]

The probabilities of the configurations are the squared lengths of these perpendicular components.

Example: \(\ket\psi=\tfrac{\sqrt3}2\ket0+\tfrac12\ket1\) and \(\ket\phi=\ket0\). Then \(\braket0\psi=\tfrac{\sqrt3}2=\cos30^\circ\), with probability \(\tfrac34\); the remainder \(\tfrac12\ket1\) is orthogonal to \(\ket0\).

The picture is a plane in state space, not physical space. Complex vectors keep this geometry while the inner product also carries phase.

Inner product: amplitude along the final directionA real state-vector plane shows the final unit vector phi equals ket zero and the initial unit vector psi equals square root three over two ket zero plus one half ket one. Their angle is pi over six. The component of psi along phi has coefficient bra phi ket psi equals square root three over two, so the probability of the final-state outcome is three quarters. The dashed line selects the projection. These are state-vector directions, not directions in physical space.Amplitude as a component of the initial vector|1⟩θ0initial |ψ⟩final |φ⟩=|0⟩⟨φ|ψ⟩=√3/2⟨φ|φ⟩=⟨ψ|ψ⟩=1θ=π/6|⟨φ|ψ⟩|²=3/4For real unit vectors: ⟨φ|ψ⟩=cosθ.

Linearity and transition matrices

A step sends each configuration to the weighted sum of its successors and a state to the same combination of these images:

\[U\ket\psi=v_0U\ket0+v_1U\ket1,\]
\[U(a\ket\phi+b\ket\psi)=aU\ket\phi+bU\ket\psi .\]

This is linearity. The transition matrix stores the arrow weights: \(U_{ji}=\bra jU\ket i\), with column \(i\) for the input and row \(j\) for the output.

Matrix–vector multiplication performs the path sums:

\[v_j^\prime=\sum_iU_{ji}v_i,\qquad \ket{\psi^\prime}=U\ket\psi .\]
Transition amplitudes and matrix entriesTwo input nodes and two output nodes show one step. From input 0, the arrows to outputs 0 and 1 have weights a and b; these form matrix column 0. From input 1, the arrows to outputs 0 and 1 have weights c and d; these form column 1. Output 0 selects row 0, containing a and c. Output 1 selects row 1, containing b and d. Thus U equals the matrix with rows a,c and b,d, and U ji is the weight of i to j.One-step transition graphMatrix of the same transitionInput iOutput jabcd0101from 0from 1to 0to 1U=abcdRows: destinations. Columns: sources.Uⱼᵢ=⟨j|U|i⟩, U=beginpmatrixa&c\b&dendpmatrix

Unitary matrices

Every normalized input must give a normalized output. The two conditions of Chapter 3 say:

  • each column has unit length (outgoing normalization);
  • distinct columns are orthogonal (cross-sum condition).
\[\sum_jU_{ji}^*U_{jk}=\delta_{ik}\quad(\delta_{ik}=1\text{ if }i=k,\ 0\text{ otherwise}).\]

A square matrix with orthonormal columns is unitary. Example:

\[U=\frac15\begin{pmatrix}4&-3\\3&4\end{pmatrix},\qquad U\Bigl(\tfrac35\ket0+\tfrac45\ket1\Bigr)=\ket1 .\]
Geometric action of a transition matrixLeft: the initial ket psi has coordinates three fifths and four fifths in the standard basis. Right: U equals one fifth times the matrix with rows four,minus three and three,four. It rotates ket zero to coordinates four fifths,three fifths and ket one to minus three fifths,four fifths. The transformed components three fifths U ket zero and four fifths U ket one add to U ket psi equals ket one. The coordinate axes stay fixed. This is a real state-vector plane.Initial vectorTransformed vector and basis directions|0⟩|1⟩|ψ⟩4/53/5|0⟩|1⟩U|0⟩U|1⟩U|ψ⟩3/5U|0⟩4/5U|1⟩The state changes; the coordinate axes stay fixed.U=1/5beginpmatrix4&−3\3&4endpmatrix, |ψ⟩=beginpmatrix3/5\4/5endpmatrix, U|ψ⟩=beginpmatrix0\1endpmatrixU|ψ⟩=3/5U|0⟩+4/5U|1⟩=|1⟩

Gate matrices and composition

\(X\) exchanges the basis directions, \(Z\) flips the sign of the \(\ket1\) component, \(H\) sends them to the two diagonals:

\[X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\ H=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.\]

After a step, comparing with \(\ket\phi\) gives the amplitude

\[\bra\phi U\ket\psi=\sum_{i,j}\braket\phi j\,\bra jU\ket i\,\braket i\psi .\]

Two steps, \(U\) first and \(V\) second, give \(VU\ket\psi\). The two-step sum rule is the matrix product:

\[(VU)_{ki}=\sum_jV_{kj}U_{ji}.\]

The rightmost matrix acts first. For example \(HH=I\), the identity matrix:

\[\tfrac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.\]

Undoing a step: the adjoint

Reversing arrows undoes CNOT. Try it on \(\sqrt X\) (stay \(a=\tfrac{1+i}2\), switch \(b=\tfrac{1-i}2\)): its arrows look the same both ways, so reversing gives \(\sqrt X\) again and \(\sqrt X\sqrt X=X\), not \(I\). The return paths give \(a^2+b^2=0\).

Reverse and also conjugate the weights. Return paths: \(a^*a+b^*b=1\); crossing paths: \(b^*a+a^*b=0\). The step is undone.

Adjoint

\(U^\dagger\) reverses every arrow of \(U\) and conjugates its weight: \(\bra iU^\dagger\ket j=\bra jU\ket i^*\). As a matrix, \(U^\dagger\) is the conjugate transpose.

This is the rule \(\braket\phi j=\braket j\phi^*\) applied to the arrows of a step: going backward conjugates.

Adjoint: reverse transitions and conjugate amplitudesThe arrows of the preceding U graph are reversed. The original transition i to j with amplitude U ji becomes a transition j to i with conjugate amplitude U ji star. The reversed arrows have weights a star from 0 to 0, b star from 1 to 0, c star from 0 to 1, and d star from 1 to 1. Thus U dagger has rows a star,b star and c star,d star. The original columns become conjugated rows. For a unitary U this is the inverse transition; the adjoint is defined for any matrix.Reversed transition graphMatrix U†Output iInput ja*b*c*d*0101from 0from 1to 0to 1U†=a*b*c*d*Original columns become conjugated rows.U=beginpmatrixa&c\b&dendpmatrix longmapsto U†=beginpmatrixa*&b*\c*&d*endpmatrix(U†)ᵢⱼ=Uⱼᵢ*=bigl(⟨j|U|i⟩bigr)*

Unitary steps are undone by their adjoint

Apply \(U\), then \(U^\dagger\), and use the sum rule:

\[\bra iU^\dagger U\ket k=\sum_j\bra jU\ket i^*\,\bra jU\ket k .\]

This is 1 for \(i=k\) (normalization) and 0 for \(i\ne k\) (cross sum). Hence

\[U^\dagger U=I,\qquad U^{-1}=U^\dagger .\]

Moving an operator across an inner product: \(\bra\phi U\ket\psi=\braket{U^\dagger\phi}\psi\). With \(\ket\phi=U\ket\chi\):

\[\braket{U\chi}{U\psi}=\braket\chi\psi .\]

A unitary step preserves every length and angle: it moves state space rigidly, like a rotation. Example: \(H^\dagger=H\), so \(H\) undoes itself.

Composition of U and its adjointA three-layer graph shows inputs k, intermediate configurations j, and outputs i. The first step U has matrix rows a,c and b,d. The second step U dagger reverses the original arrows and conjugates their weights. Each entry of U dagger U is the sum of the two path products to that output. Normalized columns give diagonal entries one; orthogonal columns give off-diagonal entries zero. Under these conditions the composite is the identity.One step, then its adjointSum products along the two-step pathsInput kIntermediate jOutput iUU†abcda*c*b*d*010101U† U=beginpmatrixa*a+b*b&a*c+b*d\c*a+d*b&c*c+d*dendpmatrix=beginpmatrix1&0\0&1endpmatrix=IUnit output vectors give the diagonal ones.Orthogonal outputs give the off-diagonal zeros.(U† U)ᵢₖ=Σⱼ Uⱼᵢ*Uⱼₖ=begincases1&i=k\0&i≠ kendcases

The adjoint is not always the inverse

Every matrix has an adjoint, but it undoes the matrix only when the matrix is unitary. Take the shear

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

\(A\) slides points sideways by their height; \(A^\dagger\) slides them upward by their horizontal position. Neither \(AA^\dagger\) nor \(A^\dagger A\) returns \(\tfrac35\ket0+\tfrac45\ket1\): they give \((2,\tfrac75)\) and \((\tfrac75,\tfrac{11}5)\).

Undoing \(A\) needs \(A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}\). The identity \(\bra\phi A\ket\psi=\braket{A^\dagger\phi}\psi\) still holds; the adjoint always moves an operator across an inner product. A shear stretches lengths, so it cannot be a step of a closed quantum system.

A matrix and its adjoint acting on the same vectorOne set of axes shows the initial vector psi in blue at coordinates three fifths,four fifths, A psi in orange at seven fifths,four fifths, and A dagger psi in green at three fifths,seven fifths. A is the horizontal shear with rows one,one and zero,one; its adjoint is the vertical shear with rows one,zero and one,one. An orange arc alpha measures the angle from A psi to psi, about 23.4 degrees. A green arc beta measures the angle from psi to A dagger psi, about 13.7 degrees. The arcs compare directions, not lengths. The adjoint does not undo A.Same initial vector, two different images|0⟩|1⟩|ψ⟩A|ψ⟩A†|ψ⟩αβA=beginpmatrix1&1\0&1endpmatrixA†=beginpmatrix1&0\1&1endpmatrixα≈23.4°β≈13.7°Angles measured from the initial vector.|ψ⟩=beginpmatrix3/5\4/5endpmatrix, A|ψ⟩=beginpmatrix7/5\4/5endpmatrix, A†|ψ⟩=beginpmatrix3/5\7/5endpmatrixA† A=beginpmatrix1&1\1&2endpmatrix≠ I

4.2

Postulates

Which of these calculations describe nature: states, closed evolution, measurement, and composite systems.

The state postulate

Multiplying every amplitude by the same \(e^{i\theta}\) changes no probability, now or after any later step, because every path sum gains the same factor. Such vectors cannot be told apart.

State postulate

A state of a finite quantum system is a unit vector of complex amplitudes, with the inner product above. Vectors differing only by a global phase represent the same physical state.

For a qubit, normalization ties the two magnitudes and the global phase can make the first amplitude real and nonnegative. One angle and one relative phase remain:

\[\ket\psi=\cos\tfrac\theta2\ket0+e^{i\phi}\sin\tfrac\theta2\ket1,\qquad 0\le\theta\le\pi .\]

At \(\theta=0\) or \(\pi\) one amplitude vanishes and no relative phase remains.

The evolution postulate

The mathematics shows that probability-preserving steps are unitary. The postulate asserts that nature uses them.

Evolution postulate

The evolution of a closed system is unitary: \(\ket\psi\mapsto U\ket\psi\) with \(U^\dagger U=I\).

With real amplitudes, unit vectors end on a circle and a unitary is a rotation:

\[U(\delta)=\begin{pmatrix}\cos\delta&-\sin\delta\\\sin\delta&\cos\delta\end{pmatrix}.\]

Lengths and angles between states are kept; in general \(\braket{U\chi}{U\psi}=\braket\chi\psi\).

Unitary evolution preserves state geometryEvery arrow ends on the unit circle: both states have length one before and after the operation. The same rotation carries |ψ⟩ to U|ψ⟩ and |chi⟩ to U|chi⟩, preserving the angle between them. This real-amplitude example shows a rotation in state space; general unitaries preserve inner products in complex state space. Two unit circles show two states before and after a common rotation of 40 degrees. Both radii remain one and the angle between the states is unchanged.Unitary evolution preserves state geometryBeforev₀v₁|ψ⟩|chi⟩γUnit radius · same angle γAfter the same unitary Uv₀v₁U|ψ⟩U|chi⟩γUnit radius · same angle γU|Uψ|=|ψ|=1, ⟨Uchi|Uψ⟩=⟨chi|ψ⟩

The measurement postulate

Measurement postulate

For a normalized \(\ket\psi=\sum_jv_j\ket j\), an ideal measurement in the computational basis gives outcome \(j\) with probability \(|\braket j\psi|^2=|v_j|^2\). Conditional on that outcome, the state becomes \(\ket j\).

Both the probability and the update are physical assumptions, not consequences of path arithmetic.

Example: \(\tfrac{\sqrt3}2\ket0+\tfrac12\ket1\) gives 0 with probability \(\tfrac34\), 1 with \(\tfrac14\). After outcome 1 the state is \(\ket1\), so an immediate repeat gives 1.

Measurement and its conditional stateFor |ψ⟩=(√3/2)|0⟩+(1/2)|1⟩, the first measurement gives the indicated probabilities. Each branch leaves a definite state, so an immediate repeat gives the same result. start → zero: p=3/4; start → one: p=1/4; zero → end0: p=1; one → end1: p=1Measurement and its conditional statep=3/4p=1/4p=1p=1|ψ⟩|0⟩|1⟩|0⟩|1⟩First measurementImmediate repeatArrow labels are probabilities; nodes are conditional states.

Measurement as projection

A state is the sum of its components along the basis directions, \(\ket\psi=\sum_j\ket j\braket j\psi\). Measurement picks one component, with probability equal to its squared length, keeps it, and rescales it to unit length.

The projector \(P_j=\ketbra jj\) keeps the component along \(\ket j\): \(P_j\ket\psi=\ket j\braket j\psi\). As a matrix it is a column times a row, e.g.

\[\ketbra00=\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}.\]

In this language,

\[p(j)=\|P_j\ket\psi\|^2,\qquad \ket{\psi_j}=\frac{P_j\ket\psi}{\sqrt{p(j)}}\quad(p(j)>0).\]
Measurement: project, square the length, then normalize A unit state vector with coordinates square root of three over two and one half has orthogonal projections onto the zero and one axes. The squared lengths give probabilities three quarters and one quarter. Conditional on the observed outcome, the corresponding projection is normalized to the unit basis vector. Only one outcome occurs per run. 1. Resolve the state into projections |ψ⟩ |0⟩ |1⟩ P₀|ψ⟩ = (√3/2)|0⟩ P₁|ψ⟩ = ½|1⟩ Squared projection length = outcome probability 2. Normalize the observed component Outcome 0 · probability 3/4 P₀|ψ⟩ / √(3/4) = |0⟩ Outcome 1 · probability 1/4 P₁|ψ⟩ / √(1/4) = |1⟩ One outcome per run; the retained state has unit length.

Measuring in other bases

Let a unitary \(U\) carry an orthonormal basis \(\ket{b_j}\) to the computational basis. Measuring in the \(b\) basis means applying \(U\) and then measuring:

\[p(j)=|\bra jU\ket\psi|^2=|\braket{U^\dagger j}\psi|^2=|\braket{b_j}\psi|^2 .\]

So \(|\braket\phi\psi|^2\) is the probability of outcome \(\phi\) in any orthonormal basis containing \(\ket\phi\); afterwards the state is \(\ket\phi\).

Real example: \(\ket\psi=\cos\alpha\ket0+\sin\alpha\ket1\), \(\ket{b_0}=\cos\beta\ket0+\sin\beta\ket1\), \(\ket{b_1}=-\sin\beta\ket0+\cos\beta\ket1\) give amplitudes \(\cos(\alpha-\beta)\) and \(\sin(\alpha-\beta)\).

\(\beta=0\) is the Z measurement; \(\beta=45^\circ\) is the X measurement, with \(\ket\pm=(\ket0\pm\ket1)/\sqrt2\). Preparing \(\ket+\): X gives \(+\) with certainty, Z gives each outcome half the time. The measurement asks a different question; the state is the same.

With \(\beta=45^\circ\), \(\ket{b_1}=-\ket-\); an overall sign does not change the projector.

Evolution and measurement together

A history alternates the two kinds of change: unitary evolution between measurements, and at each measurement the observed outcome selects a component, which is rescaled to unit length.

Example: from \(\ket0\), \(H\) gives \((\ket0+\ket1)/\sqrt2\). Uninterrupted, a second \(H\) returns \(\ket0\) with certainty. Measured in between, the state becomes \(\ket0\) or \(\ket1\), and the second \(H\) gives equal probabilities.

The update cannot be undone: after outcome 0, an initial \(\ket0\) and an initial \((\ket0+\ket1)/\sqrt2\) both leave \(\ket0\).

The larger system, including the device and its record, may still evolve unitarily; the conditional description keeps only the reported branch. In the figure, vertical drops are schematic and do not mean a loss of norm or energy.

Continuous evolution interrupted by measurement A schematic state history follows four smooth blue curves separated by three vertical orange drops at measurement times. Each drop represents projection and normalization conditional on the observed outcome. Smooth unitary evolution resumes from each updated state. The vertical position is illustrative, not a probability or state norm. One observed history: evolve → measure → evolve → measure … State trajectory (schematic) Time Unitary evolution Unitary evolution Unitary evolution Evolution resumes |ψ₁⟩ |ψ₂⟩ |ψ₃⟩ Measurement 1 Measurement 2 Measurement 3 At each drop: project onto the observed outcome and normalize. Vertical drops mark conditional updates; their height is not a loss of probability or norm.

Composite systems

A complete configuration of two qubits names one configuration of each: \(00,01,10,11\). A joint state assigns an amplitude to each:

\[\ket\Psi=c_{00}\ket{00}+c_{01}\ket{01}+c_{10}\ket{10}+c_{11}\ket{11},\]

with \(\sum_{i,j}|c_{ij}|^2=1\).

Independent preparations \(a\ket0+b\ket1\) and \(c\ket0+d\ket1\): reaching 01 means 0 in \(A\) and 1 in \(B\), so its amplitude is \(ad\). This is the tensor product:

\[\begin{pmatrix}a\\b\end{pmatrix}\otimes\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}ac\\ad\\bc\\bd\end{pmatrix},\qquad \ket i\otimes\ket j=\ket{ij}.\]
Two-qubit amplitudes on a complete bipartite graph Two nodes for qubit A, zero and one, are each connected to both basis nodes for qubit B. The four edge weights are c00, c01, c10, and c11, one amplitude for each joint configuration. For a product state these weights are ac, ad, bc, and bd. Edges label joint configurations, not time evolution. Four joint configurations, four amplitude weights Qubit A Qubit B c₀₀ = ac c₁₁ = bd c₀₁ = ad c₁₀ = bc |0⟩ₐ|1⟩ₐ |0⟩ᵦ|1⟩ᵦ Each edge joins a pair of local labels to specify one joint basis state. |Ψ⟩ = Σᵢⱼ cᵢⱼ|ij⟩, Σᵢⱼ |cᵢⱼ|² = 1 The product weights ac, ad, bc, bd apply to independent pure-state preparations.

The composite-state postulate and entanglement

Composite-state postulate

The state space of a composite system is the tensor product of the state spaces of its parts, with basis \(\ket{ij}=\ket i\otimes\ket j\). Every unit vector in it is an allowed state.

Dimensions multiply: parts with \(m\) and \(n\) basis states give \(mn\). Are there allowed states that are not products? Take

\[\frac{\ket{00}+\ket{11}}{\sqrt2}.\]

A product would need \(ac\ne0\) and \(bd\ne0\), so all of \(a,b,c,d\ne0\); then \(ad\) and \(bc\) could not vanish. But this state has no 01 or 10 term. A pure joint state that is not a product is entangled.

Measuring both qubits gives 00 or 11, each half the time. A classical coin choosing 00 or 11 gives the same counts, so correlation alone does not reveal entanglement; the difference lies in the amplitudes of the whole state.

Summary

  1. A state is a vector of amplitudes; the inner product \(\braket\phi\psi=\sum_j\phi_j^*\psi_j\) is the transition amplitude, with the conjugate forced by the probability rule.
  2. A step is a matrix \(U_{ji}=\bra jU\ket i\); composition is the matrix product; probability is preserved exactly when \(U\) is unitary, and then \(U^{-1}=U^\dagger\).
  3. Postulates: unit vectors up to global phase; unitary closed evolution; measurement with probability \(|\braket\phi\psi|^2\) and state update.
  4. Composite systems live in the tensor product; states that do not factor are entangled.

Notes