Quantum States, Observables, and Measurement

A quantum prediction joins three logically different ingredients: a preparation, an evolution, and a question asked of a detector. The state does not by itself determine a laboratory probability, and a list of outcome probabilities does not by itself determine what state remains afterward. This separation is the cleanest way to read both elementary spin experiments and the more subtle measurements used in quantum fields on prescribed spacetimes.

Required background. Calculus and Linear Algebra supplies complex vector spaces, inner products, spectral decompositions, positivity, and traces; Ordinary Differential Equations supplies linear initial-value evolution and boundary-value reasoning.

Preparations, rays, and density operators

We keep $\hbar$ explicit. Begin in a finite-dimensional Hilbert space $\mathcal H$, where all operators are bounded; the domain issues needed in wave mechanics are introduced below.

A nonzero vector determines a ray: $|\psi\rangle$ and $c|\psi\rangle$ represent the same pure state for every nonzero complex $c$. After choosing a unit representative, the state is equivalently the rank-one projector

$$\rho_\psi=|\psi\rangle\langle\psi|, \qquad \rho_\psi^2=\rho_\psi, \qquad \operatorname{Tr}\rho_\psi=1.$$

The general state is a density operator $\rho$: it is self-adjoint, positive, and has unit trace. In an orthonormal eigenbasis, $\rho=\sum_j q_j|j\rangle\langle j|$ with $q_j\geq0$ and $\sum_jq_j=1$. A state is pure exactly when $\operatorname{Tr}\rho^2=1$; otherwise it is mixed. The same mixed $\rho$ can have many ensemble decompositions, so the coefficients in one chosen preparation story are not additional observable properties of the state. Tong 2017, §5.3, printed pp. 161–166 develops these density-operator properties and the nonuniqueness of ensemble descriptions.

Probabilities and state changes are separate

Closed-system evolution from $t_0$ to $t$ is a unitary operator $U(t,t_0)$. A measurement with outcomes $a$ supplies positive effects $E_a$ satisfying $\sum_aE_a=I$. If $\rho$ is the prepared state, then

$$\boxed{ p(a\mid\rho,U,E) =\operatorname{Tr}\!\left[E_aU(t,t_0)\rho U(t,t_0)^\dagger\right]. }$$

Positivity makes every probability nonnegative, and completeness gives $\sum_ap(a)=1$. The three factors answer different questions:

Physical role	Mathematical object	What it determines
Preparation	State $\rho$	Statistics available before the evolution
Closed evolution	Unitary $U$	How the preparation is transported
Outcome question	Effect $E_a$	Which event is counted as outcome $a$

Write $\rho_{\mathrm{pre}}=U(t,t_0)\rho U(t,t_0)^\dagger$ for the state just before the detector acts. This keeps the preceding evolution visible in the state-update formulas below.

A projection-valued measurement (PVM) is the sharp special case $E_a=P_a$, with $P_aP_b=\delta_{ab}P_a$. A positive-operator-valued measurement (POVM) permits effects that are not projectors. Neither object, by itself, specifies the conditional state after an outcome.

That extra information is a quantum instrument. In finite dimensions, one standard realization assigns measurement operators $M_{a\alpha}$ to each outcome and defines

$$\mathcal I_a(\rho_{\mathrm{pre}}) =\sum_\alpha M_{a\alpha}\rho_{\mathrm{pre}}M_{a\alpha}^\dagger, \qquad \sum_{a,\alpha}M_{a\alpha}^\dagger M_{a\alpha}=I.$$

Each map is completely positive and trace-nonincreasing, while their sum is trace preserving. Complete positivity means that the map remains positive when an untouched ancillary system is included. The outcome probability and normalized conditional state for $p(a)\gt0$ are

$$p(a)=\operatorname{Tr}[\mathcal I_a(\rho_{\mathrm{pre}})], \qquad \rho_a=\frac{\mathcal I_a(\rho_{\mathrm{pre}})}{p(a)}.$$

No normalized conditional state is assigned to an outcome whose probability is zero.

For one measurement operator $M_a$ per outcome, $\mathcal I_a(\rho_{\mathrm{pre}})=M_a\rho_{\mathrm{pre}}M_a^\dagger$ and $E_a=M_a^\dagger M_a$. Different $M_a$ can have the same $E_a$ and hence the same outcome probabilities while producing different conditional states. Tong makes this probability-versus-update distinction explicit in Tong 2017, §§5.4.2–5.4.3, printed pp. 175–180. For the ideal repeatable outcome $P_a$, the Lüders instrument is $\mathcal I_a(\rho)=P_a\rho P_a$; when the outcome has nonzero probability, its normalized conditional state is $P_a\rho P_a/\operatorname{Tr}(\rho P_a)$. This conditioning is a new preparation, not closed Schrödinger evolution. Preskill states the projective Born rule, conditional update, and unitary dynamics separately in Preskill 2015, §2.1, Eqs. (2.7)–(2.12), printed pp. 5–6.

A noisy spin detector

Let $0\leq\eta\leq1$ and define

$$E_\pm=\frac{I\pm\eta\sigma_z}{2}, \qquad \rho=\frac{I+\mathbf r\mathbin{\cdot}\boldsymbol\sigma}{2}.$$

Here $\sigma_z=\operatorname{diag}(1,-1)$, $\boldsymbol\sigma=(\sigma_x,\sigma_y,\sigma_z)$ is the Pauli triple, and the real Bloch vector obeys $\lVert\mathbf r\rVert\leq1$. The identities $\operatorname{Tr}\sigma_i=0$ and $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$ give

$$\operatorname{Tr}(\rho E_\pm) =\frac14\operatorname{Tr}\!\left[ (I+\mathbf r\mathbin{\cdot}\boldsymbol\sigma)(I\pm\eta\sigma_z) \right] =\frac{1\pm\eta r_z}{2}.$$

The eigenvalues of $E_\pm$ are $(1\pm\eta)/2$, so both effects are positive, and $E_++E_-=I$. Thus at $\eta=1$ this is a sharp $z$-spin measurement; at $\eta=0$ the outcome is a fair coin independent of the state. The effects interpolate between those probability laws, but they still do not specify the detector’s disturbance. The boundary $\lVert\mathbf r\rVert=1$ consists of pure qubit states; points inside the Bloch ball are mixed.

Observables include their domains

For a bounded discrete sharp observable $A=\sum_a aP_a$,

$$\langle A\rangle_\rho=\operatorname{Tr}(\rho A) =\sum_a a\,\operatorname{Tr}(\rho P_a).$$

For a continuous spectrum, the precise object is a projection-valued measure $P_A(\Delta)$, and the probability for a Borel set $\Delta$ is $\operatorname{Tr}[\rho P_A(\Delta)]$. Two sharp observables are compatible when all of their spectral projectors commute. This statement remains meaningful for unbounded observables even when a formal commutator has domain ambiguities.

An unbounded observable requires both an action—often, but not always, given by a differential expression—and a dense domain. Symmetry on $D(A)$,

$$\langle\phi,A\psi\rangle=\langle A\phi,\psi\rangle, \qquad \phi,\psi\in D(A),$$

does not imply self-adjointness. One needs $A=A^\dagger$ and $D(A)=D(A^\dagger)$. If $\mu_\psi$ is the spectral probability measure of a self-adjoint $A$, then an absolutely finite expectation requires $\int |\lambda|\,d\mu_\psi(\lambda)\lt\infty$. The stronger condition $\psi\in D(A)$ is equivalent to a finite second moment, $\int\lambda^2d\mu_\psi=\lVert A\psi\rVert^2\lt\infty$, and is required for the operator expression $\langle\psi,A\psi\rangle$.

For example, take $\mathcal H=L^2([0,L],dx)$ and the formal momentum $p=-i\hbar\,d/dx$. On absolutely continuous functions with square-integrable derivative, integration by parts gives the boundary form

$$\langle\phi,p\psi\rangle-\langle p\phi,\psi\rangle =-i\hbar[\phi^*(x)\psi(x)]_0^L.$$

For real $\alpha$ modulo $2\pi$, define

$$D(p_\alpha)=\left\{\psi\in H^1([0,L]): \psi(L)=e^{i\alpha}\psi(0)\right\},$$

where $H^1([0,L])$ consists of square-integrable functions with a square-integrable weak derivative. Imposing the same condition on both $\phi$ and $\psi$ makes the boundary form vanish and defines one self-adjoint realization. Other boundary data define different operators or fail to define a self-adjoint momentum at all. Bonneau, Faraut, and Valent distinguish symmetric and self-adjoint domains and classify these interval extensions in Bonneau, Faraut, and Valent 2001, §3, Eqs. (12)–(13), p. 5; §5.3, Eqs. (14)–(15), pp. 7–8; Appendix A, p. 18.

The boundary phase has a measurable spectral consequence. Solving $p_\alpha\psi=p_n\psi$ gives a plane wave, and its endpoint condition requires $e^{ip_nL/\hbar}=e^{i\alpha}$. Thus

$$p_n=\frac{\hbar}{L}(2\pi n+\alpha), \qquad \psi_n(x)=\frac{1}{\sqrt L}e^{i(2\pi n+\alpha)x/L}, \qquad n\in\mathbb Z.$$

The differential expression is unchanged as $\alpha$ varies, but its spectrum shifts. This models a periodic or twisted-periodic identification of the endpoints; it is not the momentum of a particle between two reflecting hard walls. The physical boundary problem selects the domain.

Worked example: a filter prepares a new state

Let $|\pm z\rangle$ and $|\pm x\rangle$ denote normalized eigenstates of $\sigma_z$ and $\sigma_x$, respectively. Prepare spin up along $z$:

$$|{+z}\rangle =\frac{|{+x}\rangle+|{-x}\rangle}{\sqrt2}.$$

An ideal $x$ measurement gives $p(+x)=p(-x)=1/2$. If the $+x$ output beam is retained, the conditional state is now $|+x\rangle$. Repeating the $x$ measurement yields $+x$ with probability one, whereas a subsequent $z$ measurement again gives equal probabilities. A negligible Hamiltonian is consistent with the whole sequence: the statistics changed because an outcome was selected and a new ensemble was prepared, not because measurement was secretly another interval of unitary motion.

Boundaries of the bridge to gravitation

Self-adjointness matters whenever a quantum problem has an origin, an asymptotic end, a material surface, or another place where flux can escape the chosen domain. The boundary condition must be derived from the physical problem; an event horizon is not automatically a reflecting wall, and an effective absorbing condition generally describes an open subsystem rather than unitary one-particle evolution.

This page does not supply local generally covariant observables, quantum field measurement theory, or particle creation. Those require the later field and curved-spacetime frameworks. The next page derives the exact equivalence of Unitary Dynamics and Pictures of Motion.

Exercises

Same probabilities, different disturbance

Let $E_\pm$ be the noisy-spin effects above, with $0\lt\eta\lt1$. The positive square root of an effect is obtained by taking the positive square root of each eigenvalue in its spectral decomposition. Compare $M_\pm=\sqrt{E_\pm}$ with $\widetilde M_\pm=\sigma_x\sqrt{E_\pm}$. Show that the two instruments have the same effects and outcome probabilities but generally different conditional states. Then calculate the two conditional states for the $+$ outcome when the input is $|+z\rangle$.

Solution

Because $\sigma_x$ is unitary,

$$\widetilde M_\pm^\dagger\widetilde M_\pm =\sqrt{E_\pm}\,\sigma_x^\dagger\sigma_x\sqrt{E_\pm} =E_\pm =M_\pm^\dagger M_\pm.$$

Both instruments therefore give $p(\pm)=\operatorname{Tr}(\rho E_\pm)$. Their unnormalized output states are $\sqrt{E_\pm}\rho\sqrt{E_\pm}$ and $\sigma_x\sqrt{E_\pm}\rho\sqrt{E_\pm}\sigma_x$, respectively. Unless the first output commutes with $\sigma_x$ or is invariant under that conjugation, the normalized conditional states differ.

For input $|+z\rangle$, the $+$ effect has eigenvalue $(1+\eta)/2$. Hence $M_+|+z\rangle$ is a positive multiple of $|+z\rangle$, so its normalized output is $|+z\rangle$. By contrast, $\widetilde M_+|+z\rangle$ is the same multiple of $\sigma_x|+z\rangle=|-z\rangle$, so the second normalized output is $|-z\rangle$. Both give $p(+)=(1+\eta)/2$. The POVM fixes the statistics, not the complete disturbance.

References

• Bonneau, Guy, Jacques Faraut, and Galliano Valent. “Self-adjoint Extensions of Operators and the Teaching of Quantum Mechanics.” American Journal of Physics 69 (2001): 322–331. doi:10.1119/1.1328351. Open PDF.
• Preskill, John. Quantum Information, Chapter 2, revised July 2015, §2.1, printed pp. 3–7. California Institute of Technology lecture notes. Official PDF.
• Tong, David. Topics in Quantum Mechanics, §§5.3–5.4, printed pp. 161–180. Cambridge lecture notes, 2017. Official PDF.