A quantum prediction is a probability law for a declared measurement. To read that statement, we need to distinguish an event, a probability measure, a density, and an amplitude. This chapter builds those distinctions before connecting them to projections and the Born rule.
Use chapter 2 for complex vectors, orthogonal projections, the finite spectral theorem, and density matrices, and chapter 4 for measurable sets and Lebesgue integrals. Our inner product is conjugate-linear in the first slot. Alphabeta Math uses an inner product linear in the first slot: write \((u,v)_{\mathrm{AB}}=\langle v,u\rangle\) to translate its formulas into ours. Off-diagonal inner products therefore require swapping the arguments. For a self-adjoint operator \(F\), the expectation \(\langle\psi,F\psi\rangle\) is real, since its complex conjugate is \(\langle F\psi,\psi\rangle=\langle\psi,F\psi\rangle\); consequently either convention gives the same Born probabilities. Infinite-dimensional operator results are supplied in chapter 6; the probability arguments here state exactly where they are needed.
5.1. Events and probability spaces
An experiment can have several possible outcomes. We must specify which groups of outcomes count as events before assigning probabilities.
Definition 5.1. Probability space
A probability space is a triple \((\Omega,\mathcal F,\mathbb P)\). The set \(\Omega\) is the sample space of outcomes. The events are the members of the sigma-algebra \(\mathcal F\): it contains \(\Omega\), is closed under complements, and is closed under countable unions. The function \(\mathbb P:\mathcal F\to[0,1]\) satisfies \(\mathbb P(\Omega)=1\) and countable additivity: for pairwise disjoint events \(A_n\),
\[\mathbb P\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mathbb P(A_n).\]A statement holds almost surely if the event where it fails has probability zero.
Example. For a fair die, take \(\Omega=\{1,2,3,4,5,6\}\), all subsets as events, and \(\mathbb P(A)=\#A/6\), where \(\#A\) counts its members. Disjoint sets have disjoint members, so their counts add; only finitely many disjoint nonempty sets exist here. This verifies countable additivity. The event “even” is \(\{2,4,6\}\), with probability \(1/2\).
Proposition 5.2. Elementary event rules
We have \(\mathbb P(\varnothing)=0\), \(\mathbb P(A^c)=1-\mathbb P(A)\), and \(A\subset B\) implies \(\mathbb P(A)\le\mathbb P(B)\). For any events \(A_n\), \(\mathbb P(\bigcup_n A_n)\le\sum_n\mathbb P(A_n)\).
Proof.
Applying additivity to \(\Omega\) and empty sets gives \(1=1+\mathbb P(\varnothing)\). Next use the disjoint decompositions \(\Omega=A\cup A^c\) and \(B=A\cup(B\setminus A)\). For the last assertion replace \(A_n\) by \(C_n=A_n\setminus\bigcup_{k<n}A_k\). The \(C_n\) are disjoint, have the same union, and satisfy \(\mathbb P(C_n)\le\mathbb P(A_n)\). Countable additivity proves the bound.
□Example. On the die, \(A=\{2,4,6\}\) and \(B=\{4,5,6\}\) both have probability \(1/2\), but \(A\cup B\) has probability \(4/6\), not \(1\). Additivity requires disjointness.
5.2. Random variables and their laws
A recorded outcome is often a number, even if the underlying sample space is not numerical. A measurable map lets us transfer probabilities to those numbers.
Definition 5.3. Random variable and pushforward law
A real random variable is a measurable function \(X:\Omega\to\mathbb R\): for each Borel set \(B\subset\mathbb R\), its inverse image \(X^{-1}(B)=\{\omega:X(\omega)\in B\}\) belongs to \(\mathcal F\). The law of \(X\), also called its distribution or pushforward measure, is
\[\mu_X(B)=\mathbb P(X\in B):=\mathbb P(X^{-1}(B)).\]Here Borel sets are the sigma-algebra generated by open subsets of \(\mathbb R\), as in chapter 4. Vector-valued random variables use Borel sets of \(\mathbb R^d\) in the same way.
Why the law is a probability measure.
Inverse images preserve complements and countable unions, and the inverse images of disjoint sets are disjoint. Thus \(\mu_X(\mathbb R)=\mathbb P(\Omega)=1\), positivity is inherited, and countable additivity follows directly from that of \(\mathbb P\).
□Example. On the die let \(X(\omega)=1\) for an even face and \(X(\omega)=0\) for an odd face. Then \(\mu_X(\{0\})=\mu_X(\{1\})=1/2\). Many different outcomes of the original experiment have become the same recorded value.
Definition 5.4. Point masses and densities
A law supported on distinct values \(x_j\), in a finite or countable list, has point masses \(p_j=\mu(\{x_j\})\), with \(p_j\ge0\), \(\sum_jp_j=1\), and
\[\mu(B)=\sum_{j:x_j\in B}p_j.\]A point \(x\) with \(\mu(\{x\})>0\) is called an atom of the law. A law has a density with respect to Lebesgue measure if there is a measurable function \(f\ge0\) with \(\int_{\mathbb R}f(x)\,dx=1\) such that
\[\mu(B)=\int_B f(x)\,dx.\]“Discrete law” refers to the first case. Here we use “law with a density” for the second, rather than using “continuous” to imply a density without an additional hypothesis.
Verification.
For point masses, summing over a disjoint union partitions the nonnegative terms of the series. For a density, indicators of increasing finite unions of disjoint sets increase to the indicator of the whole union; monotone convergence from chapter 4 gives countable additivity. Both constructions are nonnegative and have total mass one.
□Example and warning. The uniform law on \([0,1/2]\) has density \(f=2\) there and zero elsewhere. A singleton has Lebesgue measure zero, so \(\mu(\{x\})=0\) for every \(x\). Yet \(\mu([0,1/4])=2(1/4)=1/2\). The density value \(2\) is not a probability. Changing a density at one point does not change the law because the integral over that point is zero.
Example. A law need not be wholly discrete or have a density: \(\mu(B)=\tfrac12\mathbf1_B(0)+\tfrac12\int_{B\cap[1,2]}dx\) has an atom of mass \(1/2\) at zero and mass \(1/2\) spread over an interval. It cannot have a Lebesgue density, since such a density gives zero mass to \(\{0\}\); it cannot be supported on a countable set, since every countable subset of \([1,2]\) has measure zero by countable additivity.
5.3. Expectations and finite moments
The mean is an integral of the outcome coordinate against its law. Writing an average symbol does not ensure that this integral is finite.
Definition 5.5. Expectation and variance
For a real random variable \(X\), a finite expectation exists when \(\int_\Omega|X|\,d\mathbb P<\infty\); then \(\mathbb E[X]=\int_\Omega X\,d\mathbb P\). A nonnegative random variable can instead have the extended expectation \(+\infty\). For a signed variable whose positive and negative parts both have infinite integrals, its expectation is undefined: \(\infty-\infty\) is not an integral.
The absolute moment of order \(k\) is \(\mathbb E[|X|^k]\), for a positive integer \(k\). If \(\mathbb E[X^2]<\infty\), set \(m=\mathbb E[X]\) and define the variance and standard deviation by
\[\operatorname{Var}(X)=\mathbb E[(X-m)^2],\qquad \sigma_X=\sqrt{\operatorname{Var}(X)}.\]We reserve a finite variance for this finite-second-moment case.
Alphabeta Math: moments and variance of the scalar spectral law
Proposition 5.6. Integrating the law and computing variance
For a nonnegative Borel function \(g\), and also for a signed Borel function with finite absolute integral,
\[\mathbb E[g(X)]=\int_{\mathbb R}g(x)\,d\mu_X(x).\]Thus discrete expectations are sums \(\sum_jg(x_j)p_j\), and expectations for a law with density \(f\) are integrals \(\int g(x)f(x)\,dx\). Finite second moment implies finite mean and
\[\operatorname{Var}(X)=\mathbb E[X^2]-(\mathbb E[X])^2\ge0.\]The variance vanishes exactly when \(X=m\) almost surely.
Proof.
The change-of-measure formula is the definition of \(\mu_X\) when \(g=\mathbf1_B\). Linearity proves it for nonnegative simple functions. Approximation from below and monotone convergence prove it for nonnegative \(g\); subtracting the integrals of positive and negative parts proves the integrable signed case. The formulas for sums and densities follow by the same indicator, simple-function, and convergence argument.
The inequality \(2|x|\le1+x^2\), obtained from \((|x|-1)^2\ge0\), proves finite absolute first moment. Also \((x-m)^2\le2x^2+2m^2\), so the variance is finite. Expand the square and use linearity to obtain the displayed identity. It is nonnegative because its defining integrand is nonnegative. If that integral is zero, then for every positive integer \(n\), the event \(\{|X-m|\ge1/n\}\) has probability zero: its indicator divided by \(n^2\) is bounded by \((X-m)^2\). Their union is \(\{X\ne m\}\), also of probability zero. The converse follows by integrating an almost surely zero function.
□Example. For the zero-or-one variable above, \(\mathbb E[X]=1/2\), \(\mathbb E[X^2]=1/2\), and \(\operatorname{Var}(X)=1/4\). For a uniform variable on \([0,1]\), direct integration gives mean \(\int_0^1x\,dx=1/2\), second moment \(1/3\), and variance \(1/12\).
Counterexample: an undefined mean. Set \(f(x)=1/[2(1+|x|)^2]\). On the positive half-line a primitive is \(-1/[2(1+x)]\), so \(\int_0^\infty f(x)dx=1/2\); symmetry gives total integral one. For \(x\ge1\), \(xf(x)\ge1/(8x)\), since \(1+x\le2x\). On every interval \([2^n,2^{n+1}]\), the integral of this lower bound is at least \(2^n/(8\,2^{n+1})=1/16\). Countably many disjoint such intervals therefore give an infinite positive-part integral. Symmetry gives an infinite negative-part integral too. Its mean is undefined, even though each symmetric truncated integral \(\int_{-R}^R xf(x)dx\) is zero. Symmetric cancellation is not expectation. Its second moment is infinite as well: for \(|x|\ge1\), \(x^2f(x)\ge1/8\).
The same failure for the Cauchy law. To avoid assuming an inverse trigonometric primitive, put \(h(x)=1/(1+x^2)\) and \(Z=\int_{\mathbb R}h(x)dx\). On \([-1,1]\), \(1/2\le h\le1\); outside it, \(h(x)\le x^{-2}\). The rational primitive gives \(\int_1^\infty x^{-2}dx=1\), so \(1\le Z\le4\). The normalized density \(h/Z\) is the standard Cauchy law. For \(x\ge1\), \(xh(x)/Z\ge1/(2Zx)\). Each dyadic interval contributes at least \(1/(4Z)\), proving an infinite positive-part integral; symmetry proves an infinite negative-part integral. Thus its mean is also undefined.
Counterexample: finite mean without finite variance. The density \(f(x)=2x^{-3}\) on \([1,\infty)\), zero elsewhere, has integral one. Its mean is \(\int_1^\infty2x^{-2}dx=2\), but its second moment is \(\int_1^\infty2x^{-1}dx=\infty\): on each interval \([2^n,2^{n+1}]\), the integrand is at least \(2/2^{n+1}\), so that interval contributes at least one. For \(x\ge4\), \((x-2)^2\ge x^2/4\), so its centered square also has infinite integral.
5.4. Conditioning and independence
Learning that an event occurred restricts the classical sample space. We first formulate this for events of positive probability; a division by zero cannot define a conditional law.
Definition 5.7. Conditional event probability and independence
If \(\mathbb P(B)>0\), define
\[\mathbb P(A\mid B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}.\]Events \(A,B\) are independent when \(\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)\). Real random variables \(X,Y\) are independent when \(\mathbb P(X\in C,Y\in D)=\mathbb P(X\in C)\mathbb P(Y\in D)\) for all Borel sets \(C,D\).
Worked example for Definition 5.7
Counterexample: Pairwise independence is weaker than mutual independence
Verification and consequences.
For a fixed positive-probability \(B\), intersections with \(B\) preserve disjoint unions; dividing additivity by \(\mathbb P(B)\) gives a probability measure, whose total mass is one. Its defining equation gives \(\mathbb P(A\cap B)=\mathbb P(A\mid B)\mathbb P(B)\). If \(A\) also has positive probability, reversing the roles gives Bayes’ identity \(\mathbb P(A\mid B)=\mathbb P(B\mid A)\mathbb P(A)/\mathbb P(B)\). Independence with \(\mathbb P(B)>0\) is equivalent to \(\mathbb P(A\mid B)=\mathbb P(A)\).
□Example. For a fair die, conditioned on an even result, the probability of \(\{4,6\}\) is \((2/6)/(3/6)=2/3\). For two fair coins, assign mass \(1/4\) to each of \(HH,HT,TH,TT\). The events “first is heads” and “second is heads” have intersection probability \(1/4=(1/2)(1/2)\), so they are independent. Distinct nonempty disjoint events of positive probability are not independent, since their intersection has probability zero but the product is positive.
Zero-probability conditioning. If \(X\) is uniform on \([0,1]\), then \(\mathbb P(X=1/2)=0\). The formula above cannot define \(\mathbb P(A\mid X=1/2)\): both numerator and denominator are zero. Conditioning on a random variable uses additional machinery, with values determined only almost surely. We do not introduce that machinery here or assign a meaning to this ratio.
5.5. Projection-valued measures produce probability laws
A sharp real-valued quantum measurement associates a projection with each Borel set of outcome values. A projection asks a yes-or-no question about a subspace; a projection-valued measure makes those questions consistent for all outcome sets.
Definition 5.8. Projection-valued measure
On a complex Hilbert space \(H\), a projection-valued measure (PVM) \(E\) on \(\mathbb R\) assigns a bounded orthogonal projection \(E(B)\) to every Borel set \(B\). Thus \(E(B)^*=E(B)=E(B)^2\). It satisfies \(E(\varnothing)=0\), \(E(\mathbb R)=I\),
\[E(B)E(C)=E(B\cap C),\]and strong countable additivity: for pairwise disjoint \(B_n\) and every \(v\in H\),
\[E\left(\bigcup_nB_n\right)v=\lim_{N\to\infty}\sum_{n=1}^NE(B_n)v\quad\text{in the norm of }H.\]“Strong” here means convergence after applying the operators to each vector. It does not demand convergence in operator norm. For this section, a Hilbert space means a complete inner-product space, developed in chapter 6; all arguments also apply to finite-dimensional spaces from chapter 2.
Finite example. Let \(A=\sum_{j=1}^m a_jP_j\) be the finite spectral decomposition of a self-adjoint matrix, with distinct real eigenvalues \(a_j\), mutually orthogonal projections \(P_j\), and \(\sum_jP_j=I\). Put \(E(B)=\sum_{a_j\in B}P_j\). Multiplying sums and using \(P_jP_k=0\) for \(j\ne k\) proves the projection and intersection identities. In a disjoint union each of the finitely many \(a_j\) occurs in at most one set, so the partial sums eventually equal \(E(\bigcup_nB_n)\). All PVM axioms hold.
Position example. On \(L^2(\mathbb R)\), define \((E(B)v)(x)=\mathbf1_B(x)v(x)\). It is a projection and is self-adjoint by the integral inner product; products of indicators give the intersection identity. For disjoint \(B_n\), the squared norm of the difference from the first \(N\) terms is \(\int_{(\cup_nB_n)\setminus(\cup_{n\le N}B_n)}|v|^2dx\), which tends to zero by monotone convergence of the integrals over the finite unions. The remaining axioms follow from the empty and full indicators.
Proposition 5.9. Scalar laws from a PVM
For a unit vector \(\psi\in H\),
\[\mu_\psi(B)=\langle\psi,E(B)\psi\rangle=\|E(B)\psi\|^2\]is a probability measure. If \(r_j\ge0\), \(\sum_jr_j=1\), and \(u_j\) are unit vectors in a finite or countable list, then \(\mu(B)=\sum_jr_j\mu_{u_j}(B)\) is also a probability measure. In finite dimension, for every density matrix \(\rho\),
\[\mu_\rho(B)=\operatorname{Tr}(\rho E(B))\]is such a mixture and is a probability measure.
Alphabeta Math: density operators and scalar spectral probability
Proof.
Self-adjointness and idempotence give \(\langle\psi,E(B)\psi\rangle=\langle E(B)\psi,E(B)\psi\rangle\ge0\). The full set has value \(\|\psi\|^2=1\). Strong additivity and Cauchy–Schwarz imply
\[\mu_\psi\left(\bigcup_nB_n\right)=\lim_{N\to\infty}\left\langle\psi,\sum_{n=1}^NE(B_n)\psi\right\rangle=\sum_n\mu_\psi(B_n).\]For the mixture, total mass is \(\sum_jr_j=1\). Countable additivity follows by interchanging two nonnegative sums. To justify the interchange, each iterated sum is the supremum of sums over finite rectangles of index pairs, so both equal the same supremum.
Finally diagonalize the positive matrix \(\rho=\sum_jr_j|u_j\rangle\langle u_j|\), with \(r_j\ge0\) and \(\sum_jr_j=\operatorname{Tr}\rho=1\), as in chapter 2. The rank-one map here sends \(v\) to \(u_j\langle u_j,v\rangle\). Taking the trace in an orthonormal basis gives \(\operatorname{Tr}(|u_j\rangle\langle u_j|E(B))=\langle u_j,E(B)u_j\rangle\). Linearity of the finite trace gives the claimed mixture.
□For an infinite-dimensional density operator on a separable Hilbert space, the missing ingredient is an operator theorem, not a new probability axiom. Chapter 6’s positive trace-class decomposition and trace formula supply \(\rho=\sum_jr_j|u_j\rangle\langle u_j|\), \(\sum_jr_j=1\), and \(\operatorname{Tr}(\rho B)=\sum_jr_j\langle u_j,Bu_j\rangle\) for bounded \(B\). Apply this with \(B=E(C)\) to reduce \(\mu_\rho(C)=\operatorname{Tr}(\rho E(C))\) to the proved countable-mixture case. We do not interchange an infinite operator trace and a strong limit without that theorem.
Example. For \(A=\operatorname{diag}(0,2)\) and \(\psi=(1,i)/\sqrt2\), the law has mass \(1/2\) at each of \(0,2\). Its mean is \(1\), second moment \(2\), and variance \(1\). The density matrix \(\rho=\operatorname{diag}(1/4,3/4)\) instead gives masses \(1/4,3/4\), mean \(3/2\), and variance \(3-9/4=3/4\).
Example. For the position PVM, \(\mu_\psi(B)=\int_B|\psi(x)|^2dx\). Unit norm gives \(\int|\psi|^2=1\), so \(|\psi|^2\) is a density. A probability belongs to a set \(B\), such as a detector bin, not to the number \(|\psi(x)|^2\) at one point. Changing an \(L^2\) representative on a null set changes neither the state nor this law.
Physical postulate. Born outcome law
For a declared sharp observable represented by a self-adjoint operator \(A\), use its spectral PVM \(E_A\). In a preparation represented by \(\rho\), the physical model assigns probability \(\operatorname{Tr}(\rho E_A(B))\) to an outcome in the Borel set \(B\). For a pure preparation \(\rho=|\psi\rangle\langle\psi|\), this is \(\|E_A(B)\psi\|^2\).
The assignment to an apparatus is a physical assumption. Proposition 5.9 proves that the assigned numbers satisfy probability axioms; it does not derive this physical assignment. Chapter 6 states the infinite-dimensional spectral theorem that supplies \(E_A\).
For the scalar law of a declared observable, finite mean requires \(\int|a|\,d\mu_\rho(a)<\infty\), and finite variance requires \(\int a^2\,d\mu_\rho(a)<\infty\). In finite dimension these conditions hold because the law has finite support. There \(\int a\,d\mu_\rho(a)=\operatorname{Tr}(\rho A)\) and \(\int a^2\,d\mu_\rho(a)=\operatorname{Tr}(\rho A^2)\): insert \(A=\sum_ja_jP_j\) and \(A^2=\sum_ja_j^2P_j\) and use trace linearity. For unbounded operators, the scalar integrals and operator domains must be checked before using an operator expression for a moment; see chapter 6.
Amplitudes are not probabilities. The components \(1/\sqrt2\) and \(i/\sqrt2\) above are complex amplitudes; their squared absolute values are the probabilities for the coordinate-basis measurement. Relative phase can affect another measurement. Set \(\psi_+=(e_1+e_2)/\sqrt2\), \(\psi_-=(e_1-e_2)/\sqrt2\), and \(f=(e_1+e_2)/\sqrt2\). Both states give coordinate probabilities \((1/2,1/2)\), but \(|\langle f,\psi_+\rangle|^2=1\) and \(|\langle f,\psi_-\rangle|^2=0\). In contrast, multiplying every component by the same phase \(e^{i\theta}\) changes no PVM probability, because \(\|E(B)e^{i\theta}\psi\|^2=\|E(B)\psi\|^2\).
5.6. Ideal projection updates and incompatible measurements
The outcome law specifies how often an outcome occurs. It does not specify what preparation follows it. A declared ideal projective measurement adopts an additional update rule.
Physical postulate. Finite Lüders update
Work in finite dimension. Let \(P_1,\ldots,P_m\) be mutually orthogonal projections with \(\sum_iP_i=I\). Outcome \(i\) has probability \(p_i=\operatorname{Tr}(\rho P_i)\). If \(p_i>0\), the declared selective update is
\[\rho_i=\frac{P_i\rho P_i}{p_i}.\]For a unit vector it is \(\psi_i=P_i\psi/\|P_i\psi\|\). If outcomes are discarded, the assigned density matrix is \(\rho'=\sum_iP_i\rho P_i\). These are the Lüders rules for this declared ideal apparatus, not consequences of the Born probabilities alone.
Normalization and repeatability.
For any \(v\), \(\langle v,P_i\rho P_iv\rangle=\langle P_iv,\rho P_iv\rangle\ge0\). Finite trace cyclicity from chapter 2 gives \(\operatorname{Tr}(P_i\rho P_i)=\operatorname{Tr}(\rho P_i^2)=p_i\), so \(\rho_i\) is positive with trace one. Likewise \(\rho'\) is positive and has trace \(\sum_ip_i=1\). Since \(P_jP_i=0\) for \(j\ne i\) and \(P_i^2=P_i\), a repeated measurement has probability \(\operatorname{Tr}(\rho_iP_j)=\delta_{ij}\). For pure states, \(p_i=\|P_i\psi\|^2\), which normalizes the displayed vector. If \(p_i=0\), neither selective formula is defined.
□For a subsequent yes-or-no projection \(Q\), the adopted sequential probability is
\[\mathbb P(Q\text{ yes after }i)=\operatorname{Tr}(\rho_iQ)=\frac{\operatorname{Tr}(\rho P_iQP_i)}{p_i}.\]This follows by substitution and finite trace cyclicity. If \(Q\) commutes with \(P_i\), then \(P_iQP_i=P_iQ\), a projection onto their common subspace, and the expression becomes the classical ratio for their joint sharp event.
Compatible events. For any two commuting projections \(P,Q\), the four operators \(PQ\), \(P(I-Q)\), \((I-P)Q\), and \((I-P)(I-Q)\) are projections, are mutually orthogonal, and sum to \(I\). These facts follow by multiplying the factors, using commutativity and \(P(I-P)=Q(I-Q)=0\). Their Born probabilities therefore define a four-outcome classical law. Within that law the two yes-events are independent exactly if \(\operatorname{Tr}(\rho PQ)=\operatorname{Tr}(\rho P)\operatorname{Tr}(\rho Q)\). Commutativity alone does not imply independence: take \(Q=P\) with probability \(1/2\), giving intersection probability \(1/2\), not \(1/4\).
Incompatible example. Let \(P=|e_1\rangle\langle e_1|\), \(Q=|f\rangle\langle f|\), with \(f=(e_1+e_2)/\sqrt2\), and prepare \(e_1\). Their matrices give \(PQ=\tfrac12\begin{pmatrix}1&1\\0&0\end{pmatrix}\) and \(QP=\tfrac12\begin{pmatrix}1&0\\1&0\end{pmatrix}\), so they do not commute and \(PQ\) is not an orthogonal projection. Measuring \(P\) first gives yes with certainty and then \(Q\)-yes with probability \(1/2\), so the yes-yes record has probability \(1/2\). Measuring \(Q\) first gives yes with probability \(1/2\), leaves \(f\), and gives \(P\)-yes with probability \(1/2\), so that yes-yes record has probability \(1/4\).
Classical independence concerns events within a specified joint probability law. Incompatible sharp measurements do not come with a joint sharp law formed by multiplying their projections. A sequential apparatus still has a classical law for its recorded outcomes, but the order and the adopted updates help determine it. The example proves an order dependence, not a general prohibition on constructing classical distributions with given marginal laws.
5.7. Finite effects and unsharp outcomes
A finite detector can have responses that are not projections. Probability normalization needs positivity and a sum equal to the identity; it does not require each response to be a yes-or-no subspace.
Definition 5.10. Effects and a finite POVM
In finite dimension, an effect is a self-adjoint operator \(F\) with \(0\le F\le I\), meaning \(\langle v,Fv\rangle\ge0\) and \(\langle v,(I-F)v\rangle\ge0\) for every \(v\). A finite positive-operator-valued measure (finite POVM) consists of positive operators \(F_1,\ldots,F_m\) with \(\sum_iF_i=I\). Each is automatically an effect because \(I-F_i=\sum_{j\ne i}F_j\ge0\). For outcome labels \(1,\ldots,m\), define \(F(S)=\sum_{i\in S}F_i\) for any subset \(S\).
The generalized Born assignment for a declared such measurement is \(p_i=\operatorname{Tr}(\rho F_i)\), or \(p_i=\langle\psi,F_i\psi\rangle\) for a pure preparation. Its physical application is part of the probability postulate.
Why the probabilities are normalized.
Diagonalize \(\rho=\sum_jr_j|u_j\rangle\langle u_j|\). Then \(p_i=\sum_jr_j\langle u_j,F_iu_j\rangle\ge0\) since each summand is nonnegative. Linearity gives \(\sum_ip_i=\operatorname{Tr}(\rho\sum_iF_i)=\operatorname{Tr}\rho=1\), so also \(p_i\le1\). Summing the \(p_i\) over a subset gives a probability law on the labels; disjoint subsets have additive sums. The pure-state formula is the rank-one trace identity already proved. For a countable trace-class decomposition in infinite dimension and finitely many bounded effects, chapter 6’s trace formula gives exactly the same argument.
□Example. The two effects \(F_1=\operatorname{diag}(4/5,1/5)\), \(F_2=I-F_1\) are positive and sum to \(I\). For \(e_1\) the probabilities are \(4/5,1/5\); for \(e_2\) they are \(1/5,4/5\). Since \(F_1^2\ne F_1\), this is not a projective measurement. The effect eigenvalues are allowed to lie between zero and one.
Same probabilities, different updates. Let both effects be \(I/2\). One possible model has outcome maps \(\rho\mapsto K_i\rho K_i^*\) with \(K_1=K_2=I/\sqrt2\). Another uses \(K_1=K_2=U/\sqrt2\), where \(U\) is a unitary matrix. In both models \(K_i^*K_i=I/2\); each output is positive and has trace \(1/2\), by trace cyclicity. Thus both outcomes have probability \(1/2\), but the normalized conditional states are respectively \(\rho\) and \(U\rho U^*\). If \(U\) exchanges \(e_1,e_2\) and the input is \(|e_1\rangle\langle e_1|\), these outputs differ. Effects specify outcome probabilities, not a unique state update.
You can now read the probability language in the framework and quantities, state geometry, and observables and Born probabilities. In particular, compare the Born outcome law’s properties and preparation moments with the distinction here between adopted physical assignments and proved measure-theoretic consequences. These links were checked against the live pages on October 15, 2026. Chapter 6 supplies the Hilbert-space and unbounded-operator machinery behind the same formulas.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Two fair bits and their laws
Applies to: Definition 5.1, Proposition 5.2, Definition 5.3, Definition 5.7.
Let \(\Omega=\{00,01,10,11\}\), with all subsets measurable and each point having probability \(1/4\). Additivity holds by counting disjoint points and the total is 1. The first bit \(X\) and second bit \(Y\) are measurable functions into \(\{0,1\}\). Each pushforward law assigns mass \(1/2\) to 0 and 1. For \(A=\{X=1\},B=\{Y=1\}\), \(\mathbb P(A\cap B)=1/4=\mathbb P(A)\mathbb P(B)\), and \(\mathbb P(A\mid B)=1/2\). Every joint choice of bit values has the product probability, so the random variables are independent. The union has probability \(3/4\le1/2+1/2\), the complement of \(A\) has probability \(1/2\), and \(A\cap B\subset A\) checks monotonicity. In contrast conditioning on \(A\) itself gives \(\mathbb P(A\mid A)=1\), so an event of probability \(1/2\) is not independent of itself.
Worked example 2. A uniform density and its moments
Applies to: Definition 5.4, Definition 5.5, Proposition 5.6.
On \(\Omega=[0,2]\) take \(\mathbb P(B)=\tfrac12\lambda(B)\), with Lebesgue measurable events, and \(X(\omega)=\omega\). Its law has density \(f(x)=\tfrac12\mathbf1_{[0,2]}(x)\), of integral one. Each singleton has probability zero, while \(\mathbb P(1\le X\le2)=1/2\). All absolute moments are finite because \(0\le X\le2\). In particular \(\mathbb E X=\tfrac12\int_0^2x\,dx=1\), \(\mathbb E X^2=\tfrac12\int_0^2x^2dx=4/3\), variance \(1/3\), and standard deviation \(1/\sqrt3\). Integrating \(g(X)=X^2\) on the original space or against its law gives the same \(4/3\), verifying the pushforward integral rule. A contrasting discrete variable taking 0 and 2 with equal probability has atoms, mean 1, and variance 1. A constant variable has variance zero because its deviation from its mean is zero everywhere.
Worked example 3. A finite spectral measure and a mixture
Applies to: Definition 5.8, Proposition 5.9, Physical postulate.
On \(\mathbb C^2\), let \(P_1,P_2\) project onto the standard coordinate lines and set \(E(B)=\mathbf1_B(0)P_1+\mathbf1_B(2)P_2\). It is a projection, \(E(B)E(C)=E(B\cap C)\), and \(E(\mathbb R)=I\). For disjoint Borel sets, each of the two eigenvalues occurs in at most one set, so the partial projection sums converge on every vector to the union projection. Thus it is a PVM, not just two isolated matrices.
For \(\psi=(\sqrt3/2,i/2)\), \(\|\psi\|=1\) and the scalar law has masses \(3/4,1/4\). The density matrix \(\rho=\tfrac12P_\psi+\tfrac12P_{e_2}\) has masses \(3/8,5/8\), both nonnegative with sum one; the trace formula gives the same values. If the physical model declares \(A=0P_1+2P_2\) to be the observable, these are its adopted Born probabilities. The calculation proves normalization of that assignment, while the assignment to an apparatus remains a postulate.
Worked example 4. An update and a later measurement
Applies to: Physical postulate.
With the preceding \(\psi\), a coordinate measurement has probabilities \(p_1=3/4,p_2=1/4\), both positive. The selective states are \(P_1P_\psi P_1/p_1=P_{e_1}\) and \(P_{e_2}\); ignoring the result gives \(\rho'=\operatorname{diag}(3/4,1/4)\). All have trace one and nonnegative quadratic forms. A subsequent test of \(Q=P_f\), \(f=(e_1+e_2)/\sqrt2\), gives probability \(1/2\) after either selective outcome, since \(|\langle f,e_j\rangle|^2=1/2\). The joint probability of first outcome 1 then \(Q\)-yes is therefore \(3/8=\operatorname{tr}(P_\psi P_1QP_1)\). Here \(P_1Q\ne QP_1\), so this is a sequential law determined by the declared update, rather than a product of compatible sharp events.
Worked example 5. Unsharp effects do not specify an update
Applies to: Definition 5.10.
Let \(F_1=\operatorname{diag}(3/4,1/4)\), \(F_2=I-F_1\). Both and their complements are nonnegative, so they are effects summing to \(I\). Since \(F_1^2=\operatorname{diag}(9/16,1/16)\ne F_1\), this POVM is not a PVM. For \(\rho=\operatorname{diag}(2/3,1/3)\), its outcome probabilities are \(7/12,5/12\), obtained by multiplying diagonal entries and taking the trace.
For a separate two-outcome POVM \(F_1=F_2=I/2\), the maps \(K_j=I/\sqrt2\) and \(L_j=X/\sqrt2\), with \(X\) the unitary coordinate swap, both satisfy \(K_j^*K_j=L_j^*L_j=I/2\). On input \(P_{e_1}\), each outcome has probability \(1/2\), but normalizing \(K_jP_{e_1}K_j^*\) gives \(P_{e_1}\), while normalizing \(L_jP_{e_1}L_j^*\) gives \(P_{e_2}\). This checks the unboxed claim that probabilities alone do not fix the conditional preparation.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Pairwise independence is weaker than mutual independence
Related statements: Definition 5.7.
Claim refuted. If every pair of random variables is independent, all the variables are jointly independent.
Proof.
On four equally likely outcomes \(00,01,10,11\), let \(X,Y\) be the first and second bits and \(Z=X+Y\pmod2\). The triples are \(000,011,101,110\), each with probability \(1/4\). For each of the pairs \((X,Y),(X,Z),(Y,Z)\), its four possible bit pairs occur once, so each joint probability is \(1/4=(1/2)(1/2)\). Since any Borel event for a bit reduces to a subset of \(\{0,1\}\), summing these four masses proves pairwise independence for all Borel choices. But \(\mathbb P(X=Y=Z=1)=0\), whereas the product of the three marginal probabilities is \(1/8\). Their joint law is not the product law.
Counterexample 2. Zero covariance does not imply independence
Related statements: Definition 5.5, Definition 5.7.
Claim refuted. If E[XY]=E[X]E[Y], then X and Y are independent.
Proof.
Take \(X\) uniform on \([-1,1]\), with density \(1/2\), and \(Y=X^2\). They are bounded measurable variables, so all moments below exist. Symmetry or direct integration gives \(\mathbb E X=0\), \(\mathbb E Y=\tfrac12\int_{-1}^1x^2dx=1/3\), and \(\mathbb E(XY)=\tfrac12\int_{-1}^1x^3dx=0\). Thus their covariance, defined as \(\mathbb E(XY)-\mathbb E X\mathbb E Y\), is zero. Nevertheless \(A=\{|X|\le1/2\}\) and \(B=\{Y\le1/4\}\) are the same event, of probability \(1/2\). Their intersection has probability \(1/2\), not the product \(1/4\); hence the variables are not independent. A single moment identity cannot replace all the event identities in the definition.
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