Quantum States

A quantum state is the mathematical object that encodes all probabilities for all measurements allowed on a declared physical system, conditional on a specified preparation. In the standard Hilbert-space formulation, a general state is represented by a positive trace-one operator $\rho$.

The definition has three essential qualifications:

• a state belongs to a specified system and choice of accessible degrees of freedom;
• it is assigned relative to a preparation, including any conditioning on recorded information;
• it determines probabilities only after a measurement is specified.

A state is therefore neither a single outcome nor one preferred wavefunction. It is the preparation-dependent prediction rule from which outcome laws are computed.

Helpful background. Minimal language introduces kets, operators, and the distinction between amplitudes and probabilities.

State as a Prediction Rule

In finite-dimensional quantum mechanics, a density operator satisfies

$$\rho=\rho^\dagger, \qquad \rho\ge0, \qquad \operatorname{Tr}\rho=1.$$

In infinite-dimensional Hilbert spaces, $\rho$ is additionally required to be trace class. Density operators describe the normal states used throughout ordinary wave mechanics and statistical mechanics. In a more general operator-algebraic formulation, a state is a normalized positive linear functional, and not every algebraic state need be represented by a density operator in every representation. This page stays with the density-operator sector and links to the algebraic extension below.

If a measurement has effects $\lbrace E_a\rbrace_a$, with

$$E_a\ge0, \qquad \sum_a E_a=I,$$

then the Born rule assigns

$$p(a\mid\rho,E) =\operatorname{Tr}(\rho E_a).$$

The density operator is not itself one probability distribution. It generates a different distribution for each allowed measurement. For a sharp observable $A=\sum_a aP_a$, the effects are its spectral projectors $P_a$; for a generalized measurement, the effects need not be orthogonal projectors.

Why positivity and unit trace appear

Positivity guarantees nonnegative probabilities. Every effect obeys $0\le E_a\le I$, so

$$0 \le\operatorname{Tr}(\rho E_a) \le\operatorname{Tr}\rho =1.$$

Unit trace and completeness of the effects give normalization:

$$\sum_a p(a\mid\rho,E) =\operatorname{Tr}\!\left( \rho\sum_a E_a \right) =1.$$

These conditions are therefore tied directly to the probability calculus, not added merely for mathematical elegance.

The full probability rule determines the state

Suppose two finite-dimensional density operators $\rho$ and $\sigma$ give the same probability for every effect:

$$\operatorname{Tr}(\rho E) =\operatorname{Tr}(\sigma E) \quad\text{for all }0\le E\le I.$$

Then $\rho=\sigma$. To see why, let $D=\rho-\sigma$. If $D\ne0$, the Hermitian operator $D$ has an eigenvector $|d\rangle$ with nonzero eigenvalue $d$. The rank-one effect $E_d=|d\rangle\langle d|$ would give

$$\operatorname{Tr}(DE_d)=d\ne0,$$

contradicting equality of all probabilities. A state can thus be identified with its complete measurement-probability assignment.

This is an operational statement. If a model restricts which measurements are allowed, two mathematical density operators can be indistinguishable within that restricted operational theory even though a larger theory distinguishes them.

Preparations and Operational Equivalence

A preparation procedure is a repeatable laboratory recipe: settings, source conditions, filters, control pulses, heralding outcomes, and selection rules used before the measurement under study. The state summarizes the predictive content of that preparation for the declared system.

Two preparation procedures are operationally equivalent on a system if every allowed measurement on that system gives the same outcome probabilities. Quantum mechanics represents operationally equivalent procedures by the same state.

This does not imply that their laboratory histories are identical. For example, a source may prepare $|0\rangle$ or $|1\rangle$ with equal probability, or instead prepare $|+\rangle$ or $|-\rangle$ with equal probability. If the classical preparation label is unavailable, both procedures give the qubit state

$$\rho=\frac{I}{2}.$$

No qubit-only measurement distinguishes the two unlabelled ensembles. If the choice is recorded in an accessible classical register, however, the larger system includes that register and the joint states need not be equivalent. The declared system and retained information are part of the state assignment.

Conditioning changes the assigned state

Suppose a heralding detector has outcomes $h$. Before reading the detector, the system may be assigned the average state

$$\rho=\sum_h p(h)\rho_h.$$

After learning $h$, the appropriate conditional state is $\rho_h$. This is not an inconsistency. The two states answer different conditional questions using different information. A careful calculation states whether it uses the unconditioned or conditioned preparation.

Ensembles and Preparation Procedures explains how preparation labels distinguish different decompositions.

Pure and Mixed States

The set of density operators is convex. If $\rho_1$ and $\rho_2$ are states and $0\le q\le1$, then

$$\rho=q\rho_1+(1-q)\rho_2$$

is also a state. Operationally, it describes a source that uses preparation $1$ with probability $q$ and preparation $2$ otherwise, when the choice is not retained as part of the measured system.

Pure states

A pure state is an extremal point of the convex state set: it cannot be written as a nontrivial convex combination of two distinct states. In the standard Hilbert-space formulation, a pure state has rank one,

$$\rho_\psi=|\psi\rangle\langle\psi|, \qquad \langle\psi|\psi\rangle=1.$$

All normalized vectors $e^{i\chi}|\psi\rangle$ on the same ray define the same projector. The physical pure state is therefore a ray, not one phase choice.

In finite dimensions, equivalent purity tests are

$$\rho^2=\rho, \qquad \operatorname{rank}\rho=1, \qquad \operatorname{Tr}(\rho^2)=1.$$

Mixed states

A state that is not pure is mixed. Its spectral decomposition is

$$\rho =\sum_j r_j|r_j\rangle\langle r_j|, \qquad r_j\ge0, \qquad \sum_j r_j=1.$$

For a $d$-dimensional state,

$$\frac1d \le\operatorname{Tr}(\rho^2) <1$$

when $\rho$ is mixed, with the lower bound attained only by $I/d$. Purity is a useful scalar diagnostic, but it does not specify the state or explain why it is mixed.

A mixed state can arise from an unrecorded classical choice, from discarding a part of an entangled system, from noise, or from coarse graining. These origins can matter for a larger physical description even when the reduced state of the declared system is the same.

Pure States and Pure vs Mixed States develop the geometry and diagnostic criteria without identifying purity with an interpretation.

State Vectors and Density Operators

A normalized ket is the standard representative of a pure state. In an orthonormal basis $\{|n\rangle\}$,

$$|\psi\rangle =\sum_n c_n|n\rangle, \qquad c_n=\langle n|\psi\rangle, \qquad \sum_n|c_n|^2=1.$$

The coefficients are probability amplitudes for the corresponding basis measurement. They are not basis-independent properties of the state.

Every pure-state prediction can be written either with the ket or with its density operator:

$$\langle\psi|E_a|\psi\rangle =\operatorname{Tr}(\rho_\psi E_a).$$

The density-operator language is more general because it also represents mixed states and reduced subsystem states. A generic mixed state cannot be represented by one ket in the system’s Hilbert space.

One can purify a mixed state by introducing a larger system, but the purifying ket belongs to the larger Hilbert space and is not a state vector of the original subsystem alone. Density Operators develops the general state formalism.

State Versus Representation

A state is not identical to one matrix or wavefunction. Those appear only after a representation is selected.

For a pure state,

$$\psi_n=\langle n|\psi\rangle$$

is its component column in a discrete basis, while

$$\psi(x)=\langle x|\psi\rangle$$

is its position-space wavefunction. The momentum-space wavefunction $\phi(p)=\langle p|\psi\rangle$ represents the same abstract state in another generalized basis.

A density operator likewise has basis-dependent matrix elements

$$\rho_{mn}=\langle m|\rho|n\rangle.$$

Under a passive unitary basis change, state and measurement matrices transform together, leaving $\operatorname{Tr}(\rho E_a)$ invariant. A matrix entry can change even though the physical state does not.

Mathematical Objects and Physical Meaning explains this distinction; Wavefunctions as Representations develops the continuous case.

State Versus Measurement Outcome

A state and an outcome are different kinds of object:

• the state is a prediction rule assigned before a measurement, possibly conditional on earlier information;
• an outcome is one recorded event, such as $S_z=+\hbar/2$;
• a post-measurement state is a new conditional prediction rule assigned after an outcome, using a specified measurement instrument.

For a projective measurement with projectors $P_a$, the outcome probability is

$$p(a)=\operatorname{Tr}(\rho P_a).$$

For an outcome with $p(a)>0$, the ideal Lüders update gives the conditional state

$$\rho_a =\frac{P_a\rho P_a} {\operatorname{Tr}(\rho P_a)}.$$

The label $a$ alone does not always specify $\rho_a$. If $P_a$ has rank larger than one, the outcome identifies an eigenspace, not a unique ray. More general instruments can produce different post-measurement states while having the same effects and outcome probabilities.

A state also need not be an eigenstate of the observable about to be measured. Most states assign nontrivial probabilities to several outcomes. The special case $P_a\rho P_a=\rho$ gives certainty for outcome $a$, but certainty for one observable does not imply definite values for all others.

Measurement in the Formalism develops measurement probabilities and state changes.

States of Composite Systems

For systems $A$ and $B$, a joint state $\rho_{AB}$ acts on $\mathcal H_A\otimes\mathcal H_B$. The state assigned to subsystem $A$ is the reduced density operator

$$\rho_A=\operatorname{Tr}_B\rho_{AB}.$$

It is characterized by the requirement that every local effect $E_A$ have the same probability whether computed jointly or locally:

$$\operatorname{Tr}_{AB} \!\left[ \rho_{AB}(E_A\otimes I_B) \right] =\operatorname{Tr}_A(\rho_AE_A).$$

A pure joint state can have mixed reduced states. For the Bell state

$$|\Phi^+\rangle =\frac{|00\rangle+|11\rangle}{\sqrt2},$$

one finds

$$\rho_A=\rho_B=\frac{I}{2}.$$

Purity is therefore relative to the system whose state is being described. “The state is pure” is incomplete unless the relevant system is clear. The calculation is explained in Reduced States.

States Change Under Physical Processes

A state may be indexed by time, control settings, or measurement records. For a closed system evolving unitarily,

$$\rho(t) =U(t,t_0)\rho(t_0)U^\dagger(t,t_0).$$

For a general input-output process represented by a quantum channel $\mathcal E$,

$$\rho_{\mathrm{out}} =\mathcal E(\rho_{\mathrm{in}}).$$

The state at time $t$ remains a prediction rule for measurements performed at that time. It should not automatically be read as a classical trajectory of simultaneously possessed observable values.

Unitary evolution preserves the eigenvalues and purity of $\rho$. A nonunitary channel can change them. Conditional measurement updates can also change the state stochastically, with the conditioning record specifying which branch is assigned.

Worked Example: Three Qubit States

Consider the pure coherent state

$$\rho_+ =|+\rangle\langle+| =\frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix},$$

the incoherent mixture

$$\rho_{\mathrm{mix}} =\frac12|0\rangle\langle0| +\frac12|1\rangle\langle1| =\frac I2,$$

and the definite computational-basis state

$$\rho_0 =|0\rangle\langle0| =\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}.$$

One measurement does not determine the state

For a computational-basis measurement,

State	$p(0)$	$p(1)$
$\rho_+$	$1/2$	$1/2$
$\rho_{\mathrm{mix}}$	$1/2$	$1/2$
$\rho_0$	$1$	$0$

Thus the $z$-basis measurement distinguishes $\rho_0$ from the other two but does not distinguish the coherent superposition from the mixture.

For the $x$-basis effects $P_+=|+\rangle\langle+|$ and $P_-=|-\rangle\langle-|$,

State	$p(+)$	$p(-)$
$\rho_+$	$1$	$0$
$\rho_{\mathrm{mix}}$	$1/2$	$1/2$
$\rho_0$	$1/2$	$1/2$

Now $\rho_+$ is distinguished, while $\rho_{\mathrm{mix}}$ and $\rho_0$ agree for this one measurement. A complete state assignment summarizes the probabilities for all measurements, not just one preferred basis.

Coherence appears relationally

The off-diagonal entries of $\rho_+$ in the $z$ basis encode the phase relation that produces a definite $x$ outcome. They vanish for $\rho_{\mathrm{mix}}$. Off-diagonal entries themselves are basis dependent, but whether two states give different interference statistics is physical.

The purities are

$$\operatorname{Tr}(\rho_+^2)=1, \qquad \operatorname{Tr}(\rho_0^2)=1, \qquad \operatorname{Tr}(\rho_{\mathrm{mix}}^2)=\frac12.$$

Both $\rho_+$ and $\rho_0$ are pure even though only one is an eigenstate of the computational-basis measurement.

Example: A Wave-Packet State

A normalized Gaussian wave packet on a line can be represented at one time by

$$\psi(x) =\frac{1}{(2\pi\sigma_x^2)^{1/4}} \exp\!\left[ -\frac{(x-x_0)^2}{4\sigma_x^2} +\frac{ip_0x}{\hbar} \right].$$

It satisfies

$$\int_{-\infty}^{\infty}|\psi(x)|^2\,dx=1,$$

with

$$\langle X\rangle=x_0, \qquad \operatorname{Var}(X)=\sigma_x^2, \qquad \langle P\rangle=p_0.$$

The function $\psi(x)$ is not an additional substance attached to the state; it is the position representation of the pure state $|\psi\rangle$. Its Fourier transform is the momentum representation of the same state. A position measurement samples the density $|\psi(x)|^2$, while another measurement uses a different state-dependent probability law.

The packet is not a position eigenstate. It assigns a distribution of positions and momenta, and its state evolves according to the Hamiltonian. Detailed packet structure and dynamics belong to Gaussian Wave Packets.

Example: An Energy Eigenstate

Let a time-independent Hamiltonian satisfy

$$H|n\rangle=E_n|n\rangle.$$

The state $\rho_n=|n\rangle\langle n|$ assigns energy $E_n$ with certainty. Its ket evolves by a phase,

$$|n,t\rangle =e^{-iE_n(t-t_0)/\hbar}|n\rangle,$$

but its density operator is stationary:

$$\rho_n(t)=\rho_n(t_0).$$

Stationarity means that the probabilities of every time-independent measurement stay fixed. It does not mean that every measurement has only one possible outcome. The state can be a superposition in the eigenbasis of an observable that does not commute with $H$, and that measurement can have several possible outcomes. Energy Eigenstates explains the spectral and dynamical details.

How a State Is Learned

The state is inferred from preparation records and measurement data; it is not read off from one individual system in a single trial. State tomography uses many similarly prepared systems and an informationally complete family of measurements to estimate $\rho$ with statistical uncertainty.

This distinction matters:

• one measurement outcome is a sample, not the state;
• repeated measurements in one basis generally reveal only part of the state;
• finite data produce an estimator and uncertainty region, not an exact matrix;
• drift or preparation dependence can invalidate the identical-preparation model used by the reconstruction.

State Tomography explains estimators, informational completeness, and scaling.

Operational Content and Interpretation

The formalism fixes how a declared state produces measurement probabilities and how states transform under specified physical processes. It does not by itself settle what kind of reality, knowledge, relation, or disposition a quantum state represents.

Different interpretations can agree on the same density operator and Born probabilities while disagreeing about ontology, single outcomes, or the status of the wavefunction. Conversely, experimental constraints on possible hidden- variable models do not license treating every interpretive gloss as part of the state definition.

This page uses the minimal operational claim: a state is the complete predictive object for measurements on the declared system within the quantum model. What the Postulates Do Not Say marks the boundary between formal rules and additional interpretation.

Common pitfalls

One outcome does not specify a state. A detector click is one draw from a probability distribution. Repeating one measurement estimates that distribution, but usually leaves other measurement probabilities undetermined.

A mixture is not a superposition. The two qubit tables above give a practical test: states that agree in the $z$ basis can disagree in the $x$ basis. A density operator records that distinction without choosing a preferred basis.

Purity depends on which system is described. A pure joint state can give a mixed state for one part. State the included degrees of freedom and which preparation labels are available before assigning a ket or density operator.

Connections

• State Vectors develops normalized ket representatives and basis expansions.
• Rays and Global Phase explains why normalized kets differing by a common phase represent one pure state.
• Density Operators develops positivity, trace rules, mixtures, and general states.
• Born Rule connects a state and specified measurement to probabilities.
• Measurement in the Formalism separates effects, outcomes, and state updates.
• Reduced States explains subsystem states and partial trace.
• States as Positive Linear Functionals gives the algebraic generalization in which a state is a normalized positive expectation-value functional on an observable algebra.

Exercises

1. A complete probability rule. Let

   $$\rho =\begin{pmatrix} 3/4&1/4\\ 1/4&1/4 \end{pmatrix}.$$

   Verify that $\rho$ is a density operator. Find the outcome probabilities for measurements in the $z$ and $x$ bases.

Solution

$\rho$ is Hermitian and has trace one. Its determinant is $3/16-1/16=1/8>0$, so both eigenvalues are positive.

The $z$-basis probabilities are the diagonal entries:

$$p(0)=\frac34, \qquad p(1)=\frac14.$$

Using $P_\pm=(I\pm\sigma_x)/2$ and $\operatorname{Tr}(\rho\sigma_x)=1/2$,

$$p(+)=\frac34, \qquad p(-)=\frac14.$$

Here the two measurements happen to give the same probability pair, even though their projectors differ. Equal probabilities in two bases still do not generally determine the state: an imaginary off-diagonal entry would change the $y$-basis statistics while leaving these two pairs unchanged.

2. Operational uniqueness. Let $D=\rho-\sigma$ for two distinct finite-dimensional density operators. Show explicitly that some two-outcome projective measurement distinguishes them.

Solution

Because $D$ is a nonzero Hermitian operator, it has a normalized eigenvector $|d\rangle$ with eigenvalue $d\ne0$. Choose

$$P=|d\rangle\langle d|, \qquad I-P.$$

For the first outcome, the probability difference is

$$\operatorname{Tr}(\rho P) -\operatorname{Tr}(\sigma P) =\operatorname{Tr}(DP) =d\ne0.$$

Thus the two states differ in the statistics of at least one projective measurement.

3. Bloch-vector positivity and purity. A qubit operator is written

   $$\rho =\frac12(I+\mathbf r\cdot\boldsymbol\sigma).$$

   Show that its eigenvalues are $(1\pm|\mathbf r|)/2$. Deduce the condition for $\rho$ to be a state and determine when it is pure.

Solution

The Pauli identity gives

$$(\mathbf r\cdot\boldsymbol\sigma)^2 =|\mathbf r|^2I,$$

so $\mathbf r\cdot\boldsymbol\sigma$ has eigenvalues $\pm|\mathbf r|$. Hence

$$\lambda_\pm =\frac{1\pm|\mathbf r|}{2}.$$

The trace is already one. Positivity requires $|\mathbf r|\le1$. Also,

$$\operatorname{Tr}(\rho^2) =\frac{1+|\mathbf r|^2}{2},$$

so the state is pure exactly when $|\mathbf r|=1$.

4. Superposition versus mixture. Compare $\rho_+=|+\rangle\langle+|$ with $\rho_{\mathrm{mix}}=I/2$. Show that they agree for every $z$-basis outcome but differ in the expectation of $\sigma_x$ and in purity.

Solution

Both matrices have diagonal entries $1/2$ in the $z$ basis, so both assign $p(0)=p(1)=1/2$. However,

$$\operatorname{Tr}(\rho_+\sigma_x)=1, \qquad \operatorname{Tr}(\rho_{\mathrm{mix}}\sigma_x)=0.$$

Furthermore,

$$\operatorname{Tr}(\rho_+^2)=1, \qquad \operatorname{Tr}(\rho_{\mathrm{mix}}^2)=\frac12.$$

Agreement for one measurement does not imply equality of states.

5. Nonunique ensembles. Verify

   $$\frac I2 =\frac12|0\rangle\langle0| +\frac12|1\rangle\langle1| =\frac12|+\rangle\langle+| +\frac12|-\rangle\langle-|.$$

   Why does this not imply that each individual system secretly belongs to all four pure states?

Solution

Expanding the $x$-basis projectors gives

$$|+\rangle\langle+| =\frac12 \begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad |-\rangle\langle-| =\frac12 \begin{pmatrix}1&-1\\-1&1\end{pmatrix}.$$

Their equal mixture is $I/2$, as is the equal mixture of the two computational basis projectors. Ensemble decompositions are preparation descriptions, not a unique decomposition into properties possessed by each member. Without an accessible preparation label, the density operator contains the complete qubit-only prediction rule.

6. Gaussian packet. Verify the normalization of the Gaussian wave packet on this page and compute $\langle X\rangle$ and $\operatorname{Var}(X)$.

Solution

Its probability density is

$$|\psi(x)|^2 =\frac{1}{\sqrt{2\pi\sigma_x^2}} \exp\!\left[ -\frac{(x-x_0)^2}{2\sigma_x^2} \right].$$

This is a normalized Gaussian distribution with mean $x_0$ and variance $\sigma_x^2$. Therefore

$$\int|\psi(x)|^2\,dx=1, \qquad \langle X\rangle=x_0, \qquad \langle(X-x_0)^2\rangle=\sigma_x^2.$$

The phase containing $p_0$ cancels from the position density but affects momentum statistics.

7. Energy eigenstate is not every eigenstate. Let $H=(\hbar\omega/2)\sigma_z$ and take $\rho=|0\rangle\langle0|$. Show that the state is stationary and has definite energy, then find the outcome probabilities for a $\sigma_x$ measurement.

Solution

$|0\rangle$ is an energy eigenstate, so its ket acquires only a phase and

$$\rho(t) =e^{-iHt/\hbar}\rho e^{iHt/\hbar} =\rho.$$

The energy outcome $+\hbar\omega/2$ has probability one. Since $|0\rangle=(|+\rangle+|-\rangle)/\sqrt2$, a $\sigma_x$ measurement gives

$$p(+)=p(-)=\frac12.$$

Definite energy does not imply definite values for incompatible observables.

8. Pure whole, mixed part. Compute the reduced state of qubit $A$ for

   $$|\Psi\rangle =\sqrt q\,|00\rangle +\sqrt{1-q}\,|11\rangle, \qquad 0\le q\le1.$$

   For which $q$ is the reduced state pure? Compare this with the purity of the joint state.

Solution

Tracing out $B$ removes the cross terms because $\langle0|1\rangle_B=0$:

$$\rho_A =q|0\rangle\langle0| +(1-q)|1\rangle\langle1|.$$

Its purity is

$$\operatorname{Tr}(\rho_A^2) =q^2+(1-q)^2.$$

This equals one only at $q=0$ or $q=1$. For $0<q<1$, the reduced state is mixed, while the joint state $|\Psi\rangle\langle\Psi|$ is pure for every $q$. Purity must be stated relative to the system.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014. Preparations, ensembles, and the statistical meaning of a state.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017. Convexity, purity, and the geometry of density operators.
• P. Busch, P. J. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016. Effects, probabilities, and measurement instruments.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, §§2.1–2.4. Finite-dimensional states, mixtures, measurements, and reduced states.