A quantum state is described using vectors, but a vector need not be an arrow in physical space. It can be a list of amplitudes or a function. The common structure is the ability to add vectors and multiply them by complex numbers. An inner product then supplies lengths, orthogonality, and transition probabilities.
This chapter starts with no assumed linear algebra. We use elementary real analysis, Euclidean metric spaces and topology, and complex arithmetic. Write \(\mathbb C\) for the complex numbers, \(\overline z\) for complex conjugation, and \(|z|=\sqrt{\overline z z}\). Every vector space below is over \(\mathbb C\), unless explicitly stated otherwise. Every sum is finite unless stated otherwise. Our inner products are conjugate-linear in the first slot and linear in the second, a common quantum-mechanics convention. The live Alphabeta pages instead use a pairing linear in the first slot. To translate, write \( (u,v)_{\mathrm{AB}}=\langle v,u\rangle \). Thus their rank-one formula \(T_{u,v}x=(x,v)_{\mathrm{AB}}u\) equals our \(u\langle v,x\rangle\). Lengths, orthogonality, adjoints, projections, and probabilities agree under this translation.
The finite-dimensional proofs here supply concrete models for framework and quantities, state geometry, and observables and Born probabilities. The functional-analysis chapter explains the additional hypotheses required in infinite dimensions. The choice of quantum states and the Born rule are physical postulates; the probability identities following those choices are mathematical consequences.
Six-chapter reading order
2.1. Vectors, spans, and coordinates
Definition 2.1. Complex vector space
A complex vector space is a set \(V\) with addition \(u+v\) and scalar multiplication \(av\), for \(u,v\in V\) and \(a\in\mathbb C\). Addition is associative and commutative, has a zero vector \(0\), and each \(v\) has an additive inverse \(-v\). The remaining rules are
\[a(u+v)=au+av,\quad (a+b)v=av+bv,\quad a(bv)=(ab)v,\quad 1v=v.\]These rules ensure that rearranging a finite weighted sum has its usual meaning. They imply \(0v=0\), \(a0=0\), and \((-1)v=-v\): for example \(0v=(0+0)v=0v+0v\), so adding the inverse of \(0v\) gives \(0v=0\); the other identities follow in the same way from \(a(0+0)\) and \((1-1)v\).
Example. The set \(\mathbb C^n\) consists of lists \((z_1,\ldots,z_n)\); add and scale each coordinate. The axioms hold because they hold for complex numbers in every coordinate. Functions \(f:X\to\mathbb C\) on any set \(X\) form a vector space with \((f+g)(x)=f(x)+g(x)\) and \((af)(x)=af(x)\), for the same reason. On a nonempty domain, nonnegative real-valued functions do not form a complex vector space: multiplying the constant function \(1\) by \(-1\) leaves that set.
Definition 2.2. Subspaces, span, and independence
A subspace \(W\subset V\) contains \(0\) and is closed under addition and complex scalar multiplication. The inherited operations then satisfy the vector-space axioms. A linear combination of \(v_1,\ldots,v_k\) is \(\sum_{j=1}^k a_jv_j\). Their span is the set of all such combinations. The span of an arbitrary set means all its finite linear combinations; the span of the empty set is \(\{0\}\).
A list is linearly independent if \(\sum_j a_jv_j=0\) forces every \(a_j=0\). Otherwise it is dependent. A basis is an independent spanning list. A space is finite-dimensional if it has a finite spanning list.
Proof.
A span contains zero by choosing zero coefficients. Adding two combinations and multiplying a combination by a scalar gives combinations again, so it is a subspace. Any subspace containing the original vectors contains their combinations; the span is therefore the smallest such subspace. A basis gives unique coordinates: spanning gives an expression \(v=\sum_j a_je_j\); subtracting two such expressions and using independence makes their coefficients equal.
□Example. In \(\mathbb C^2\), \(e_1=(1,0)\), \(e_2=(0,1)\) form a basis because \((a,b)=ae_1+be_2\) uniquely. The vectors \((1,1)\), \((2,2)\) are dependent and span the line \(\{(z,z):z\in\mathbb C\}\). The set \(\{(z,1):z\in\mathbb C\}\) is not a subspace because it omits zero. Real-coordinate vectors in \(\mathbb C^2\) form a real subspace but not a complex subspace: \(i(1,0)\) has a nonreal coordinate.
Proposition 2.3. Why dimension is well-defined
If \(V\) is spanned by \(n\) vectors, every independent list has at most \(n\) members. Every finite-dimensional space has a basis; all its bases have the same number of members. That number is its dimension, written \(\dim V\). Every subspace has a basis, and an independent list can be extended to a basis.
Proof: why dimension is well-defined
Proof.
Let \(w_1,\ldots,w_n\) span and let \(v_1,\ldots,v_k\) be independent. Express \(v_1\) in the spanning list. Some coefficient is nonzero; solving for its \(w_j\) lets us replace that vector by \(v_1\) and still span. After \(r\) replacements, express \(v_{r+1}\) using \(v_1,\ldots,v_r\) and the unreplaced \(w\)'s. At least one unreplaced coefficient is nonzero, or independence would fail. Replace that \(w\). There can be at most \(n\) replacements, proving \(k\le n\).
For existence, remove a member of a dependent finite spanning list: a dependence with a nonzero coefficient expresses that member using the others, so spanning survives. Continue until the list is independent. To extend an independent list, append a vector outside its span whenever possible. The bound just proved makes this stop with a basis. The same procedure within a subspace stops in at most \(n\) steps and gives a basis for that subspace. Comparing two bases by the independent-list bound in both directions proves equal sizes. The zero space has the empty basis and dimension zero.
□Example. The vectors \((1,1)\) and \((1,-1)\) form another basis of \(\mathbb C^2\): \((x,y)=\frac{x+y}{2}(1,1)+\frac{x-y}{2}(1,-1)\). Dimension counts independent coordinates, not a preferred basis. The polynomial functions \(1,t,t^2,\ldots\) on \(\mathbb R\) are independent: a finite polynomial that vanishes identically has all coefficients zero, as its successive derivatives at zero show. Their span therefore has no finite dimension.
2.2. Linear maps and matrices
Definition 2.4. Linear map, kernel, image, and rank
A map \(T:V\to W\) is linear if \(T(au+bv)=aT(u)+bT(v)\). Its kernel is \(\ker T=\{v:T(v)=0\}\), and its image is \(\operatorname{im}T=\{T(v):v\in V\}\). Both are subspaces: linearity preserves zero, sums, and scalar multiples. Its rank is \(\dim(\operatorname{im}T)\). A linear map is an isomorphism if it is bijective, meaning one-to-one and onto; its inverse is linear because applying \(T\) to the proposed linearity identity gives an equality and injectivity permits cancellation.
Example. The map \(T:\mathbb C^2\to\mathbb C\), \(T(x,y)=x+y\), has kernel \(\{(z,-z)\}\) and image \(\mathbb C\), so its rank is one. Translation \(v\mapsto v+b\) with \(b\ne0\) is not linear because it sends zero to \(b\). Conjugation \(z\mapsto\overline z\) is real-linear but not complex-linear: \(\overline{i\cdot1}=-i\ne i\overline1\).
Proposition 2.5. Rank and nullity
For a linear map with finite-dimensional domain,
\[\dim V=\dim\ker T+\operatorname{rank}T.\]Consequently an endomorphism \(T:V\to V\), meaning a linear map from a space to itself, is injective if and only if it is surjective, and then it has an inverse.
Proof.
Extend a basis \(k_1,\ldots,k_r\) of \(\ker T\) to a basis \(k_1,\ldots,k_r,v_1,\ldots,v_s\) of \(V\). The vectors \(Tv_j\) span the image because the \(k_i\) are killed. If \(\sum_j a_jTv_j=0\), then \(\sum_j a_jv_j\) lies in the kernel and is a combination of the \(k_i\). Independence of the extended basis forces all \(a_j=0\). Thus the image has dimension \(s\). A map is injective exactly when its kernel is zero: \(Tu=Tv\) is equivalent to \(u-v\in\ker T\). For an endomorphism, rank-nullity makes zero kernel equivalent to full image dimension. A subspace of full dimension equals the space by basis extension.
□Definition 2.6. The matrix of a map
Choose domain basis \(e_1,\ldots,e_n\) and target basis \(f_1,\ldots,f_m\). The matrix \(A\) of \(T\) is the rectangular array of coefficients
\[Te_j=\sum_{i=1}^m A_{ij}f_i.\]If \(v=\sum_j x_je_j\), its coordinate column is \(x=(x_1,\ldots,x_n)^T\), where \(T\) as a superscript means transpose of a row, not the linear map. Then \((Ax)_i=\sum_j A_{ij}x_j\). Conversely any array defines a linear map through this formula. The identity map \(I\) sends each vector to itself; its matrix has entries \(\delta_{ij}\), which equal one when \(i=j\) and zero otherwise.
Proof.
Expand \(T(\sum_jx_je_j)=\sum_jx_jTe_j\) and collect each \(f_i\)-coefficient to obtain the matrix formula. If \(S\) has matrix \(B\), then \((ST)e_j=\sum_i A_{ij}Sf_i\), so the matrix of the composite \(ST=S\circ T\) is \(BA\), with entries \((BA)_{kj}=\sum_iB_{ki}A_{ij}\). Composition is associative, hence matrix multiplication is associative; it also distributes over addition by expanding these finite sums.
If a new basis has columns \(e'_j=\sum_i S_{ij}e_i\), its coordinate columns satisfy \(x=Sx'\). The basis property makes \(S\) invertible. For an endomorphism, \(Ax=ASx'=S(A'x')\) gives \(A'=S^{-1}AS\). A matrix therefore depends on the basis, although the underlying map does not.
□Example. The matrix \(\begin{pmatrix}1&2\\0&3\end{pmatrix}\) sends \((x,y)\) to \((x+2y,3y)\). Its columns are the images of the standard basis. Matrix multiplication need not commute: for \(A=\begin{pmatrix}0&1\\0&0\end{pmatrix}\), \(B=\begin{pmatrix}0&0\\1&0\end{pmatrix}\), direct multiplication gives \(AB=\operatorname{diag}(1,0)\) and \(BA=\operatorname{diag}(0,1)\). Here \(\operatorname{diag}(a,b)\) denotes a diagonal matrix with diagonal entries \(a,b\) and all other entries zero.
Definition 2.7. Eigenvectors and eigenspaces
For an endomorphism \(A\), an eigenvector is a nonzero vector \(v\) satisfying \(Av=\lambda v\). The scalar \(\lambda\) is its eigenvalue, and the eigenspace is \(\ker(A-\lambda I)\), including zero. A map is diagonalizable if it has a basis of eigenvectors. Its matrix in that basis is diagonal by the definition of matrix columns. In finite dimensions, \(A-\lambda I\) fails to be invertible exactly when \(\lambda\) is an eigenvalue, by rank-nullity.
Example. For \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), the vectors \((1,1)\) and \((1,-1)\) have eigenvalues \(1\) and \(-1\), giving a diagonalizing basis. In contrast \(N=\begin{pmatrix}0&1\\0&0\end{pmatrix}\) satisfies \(N^2=0\). If \(Nv=\lambda v\), then \(0=N^2v=\lambda^2v\), hence \(\lambda=0\). Its eigenvectors all lie on the first coordinate line, so it cannot have an eigenbasis. An arbitrary matrix need not be diagonalizable.
2.3. Inner products and orthogonality
Definition 2.8. Inner product and its norm
An inner product is a function \(\langle\cdot,\cdot\rangle:V\times V\to\mathbb C\) satisfying
\[\langle u,av+bw\rangle=a\langle u,v\rangle+b\langle u,w\rangle,\quad \langle u,v\rangle=\overline{\langle v,u\rangle},\quad \langle v,v\rangle\ge0,\]with \(\langle v,v\rangle=0\) exactly when \(v=0\). The inequality refers to a real number, as conjugate symmetry makes \(\langle v,v\rangle\) real. The first slot consequently obeys \(\langle au+bv,w\rangle=\overline a\langle u,w\rangle+\overline b\langle v,w\rangle\). Define \(\|v\|=\sqrt{\langle v,v\rangle}\). For nonzero \(v\), its normalization is \(v/\|v\|\); the scaling rule below gives this vector norm one. Vectors are orthogonal when \(\langle u,v\rangle=0\); a vector is unit or normalized when its norm is one. A list is orthonormal if \(\langle e_i,e_j\rangle=\delta_{ij}\).
Conjugation is needed for complex lengths: it gives \(\|av\|=|a|\|v\|\), since \(\langle av,av\rangle=\overline aa\langle v,v\rangle\). A complex bilinear form instead has \(B(iv,iv)=-B(v,v)\), so it cannot be positive on both nonzero \(v\) and \(iv\).
Example. On \(\mathbb C^n\), \(\langle x,y\rangle=\sum_j\overline{x_j}y_j\). The axioms follow from conjugation and \(\sum_j|x_j|^2\ge0\), with equality only when every coordinate is zero. Positive weights \(w_j>0\) give another inner product \(\sum_jw_j\overline{x_j}y_j\). Negative weights fail positivity. For \(u=(1,i)/\sqrt2\) and \(v=(1,-i)/\sqrt2\), both norms are one and \(\langle u,v\rangle=(1+(-i)(-i))/2=0\).
Proposition 2.9. Cauchy–Schwarz and the triangle inequality
For all \(u,v\),
\[|\langle u,v\rangle|\le\|u\|\|v\|.\]Equality holds exactly when \(u,v\) are linearly dependent, including the case that one is zero. Also \(\|u+v\|\le\|u\|+\|v\|\). Thus this norm defines a metric \(d(u,v)=\|u-v\|\): it is nonnegative, vanishes exactly for equal vectors, is symmetric, and satisfies the triangle inequality.
Proof.
If \(v\ne0\), put \(a=\langle v,u\rangle/\|v\|^2\). Expanding using our slot convention gives
\[0\le\|u-av\|^2=\|u\|^2-\frac{|\langle v,u\rangle|^2}{\|v\|^2}.\]This proves the inequality; equality means \(u=av\) by positivity. If \(v=0\), both sides are zero and the list is dependent. Conversely proportional vectors attain equality by the norm scaling rule. Finally
\[\|u+v\|^2=\|u\|^2+2\operatorname{Re}\langle u,v\rangle+\|v\|^2 \le(\|u\|+\|v\|)^2.\]Taking nonnegative square roots proves the triangle inequality. Symmetry of the metric follows from \(\|-w\|=\|w\|\); its other properties follow from the norm properties already proved.
□Example. For \(u=(1,0)\), \(v=(1,1)\), the inequality reads \(1\le\sqrt2\), strictly. With \(v=iu\), it reads \(1=1\): equality permits any complex proportionality, not only real multiples. For unit vectors, \(|\langle u,v\rangle|^2\in[0,1]\), the range needed for a transition probability.
Proposition 2.10. Gram–Schmidt and orthonormal coordinates
Every finite-dimensional inner-product space has an orthonormal basis. For one such basis \(e_1,\ldots,e_n\),
\[v=\sum_j e_j\langle e_j,v\rangle,\qquad \|v\|^2=\sum_j|\langle e_j,v\rangle|^2.\]The second identity is the finite Parseval identity.
Proof.
Start with a basis \(v_1,\ldots,v_n\). Normalize \(v_1\). Given orthonormal \(e_1,\ldots,e_{k-1}\), set
\[w_k=v_k-\sum_{j<k}e_j\langle e_j,v_k\rangle,\qquad e_k=w_k/\|w_k\|.\]Taking the inner product with each previous \(e_i\) makes the two terms cancel, so \(w_k\) is orthogonal to them. It is nonzero because otherwise \(v_k\) would lie in the span of its predecessors. At each step the old and new lists span the same subspace. This gives an orthonormal basis. If \(v=\sum_jc_je_j\), taking \(\langle e_i,v\rangle\) gives \(c_i\). Expanding \(\langle v,v\rangle\) then leaves only the diagonal terms \(|c_j|^2\).
□Example. For \(v_1=(1,1)\), \(v_2=(1,0)\) in the standard inner product, Gram–Schmidt gives \(e_1=(1,1)/\sqrt2\), \(w_2=(1,-1)/2\), and \(e_2=(1,-1)/\sqrt2\). These two directions resolve any vector into perpendicular components.
A Hilbert space is an inner-product space complete in this metric, meaning every Cauchy sequence converges in the space. In finite dimension, the coordinate map above preserves distances and identifies the space with \(\mathbb C^n\). A Cauchy sequence has Cauchy real and imaginary parts in each coordinate, hence converges by completeness of \(\mathbb R\); the finite sum of coordinate errors tends to zero. Thus every finite-dimensional inner-product space is a Hilbert space. Completeness becomes an extra requirement in infinite dimensions.
Definition 2.11. Orthogonal complement and projection
For a subspace \(W\), its orthogonal complement is \(W^\perp=\{v:\langle w,v\rangle=0\text{ for every }w\in W\}\). It is a subspace by linearity in the second slot. If \(e_1,\ldots,e_r\) is an orthonormal basis of \(W\), define
\[P_Wv=\sum_{j=1}^r e_j\langle e_j,v\rangle.\]This is the orthogonal projection onto \(W\). It keeps precisely the components lying in \(W\).
Proposition 2.12. The projection is the closest vector
Every \(v\) has a unique decomposition \(v=w+z\), with \(w\in W\), \(z\in W^\perp\). Here \(w=P_Wv\). The map \(P_W\) is linear, independent of the chosen orthonormal basis, and satisfies \(P_W^2=P_W\). For every \(w'\in W\),
\[\|v-w'\|^2=\|v-P_Wv\|^2+\|P_Wv-w'\|^2.\]Thus \(P_Wv\) is the unique vector in \(W\) minimizing distance to \(v\), and \(\|P_Wv\|\le\|v\|\).
Proof.
The projection formula lies in \(W\), and \(\langle e_i,v-P_Wv\rangle=0\), hence the remainder lies in \(W^\perp\). The intersection \(W\cap W^\perp\) is zero because a vector in it is orthogonal to itself. Subtracting two decompositions therefore proves uniqueness and basis independence. Linearity follows from the formula; applying it to a vector already in \(W\) returns that vector, proving idempotence. Orthogonal vectors satisfy \(\|a+b\|^2=\|a\|^2+\|b\|^2\) by expansion, the Pythagorean identity. Apply this with \(a=v-P_Wv\), \(b=P_Wv-w'\). The second term is nonnegative and vanishes only at \(w'=P_Wv\). Applying Pythagoras to \(v=P_Wv+(v-P_Wv)\) proves the norm bound.
□Example. For the line spanned by \((1,1)\), \(P(x,y)=((x+y)/2,(x+y)/2)\), with matrix \(\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\). A map satisfying \(P^2=P\) alone need not be orthogonal: \(Q=\begin{pmatrix}1&1\\0&0\end{pmatrix}\) projects along a slanted direction onto the first coordinate line. It sends \((0,1)\) to \((1,0)\), which is not the closest point on that line.
2.4. Adjoints, positivity, and unitary maps
Definition 2.13. Adjoint
For a linear map \(A:V\to W\) between finite-dimensional inner-product spaces, its adjoint \(A^*:W\to V\) is the map satisfying
\[\langle u,Av\rangle_W=\langle A^*u,v\rangle_V\quad(u\in W,\ v\in V).\]In orthonormal bases, its matrix is the conjugate transpose, \((A^*)_{ji}=\overline{A_{ij}}\). The notation \(A^\dagger\) used in physics means the same adjoint. The adjoint transfers an operator across an inner product without changing the scalar.
Proof.
Expand the left side as \(\sum_{i,j}\overline{u_i}A_{ij}v_j\). Using the displayed conjugate-transpose formula makes the right side the same sum, proving existence. If \(B,C\) both satisfy the identity, then \(\langle(B-C)u,v\rangle=0\) for every \(v\); choose \(v=(B-C)u\) to obtain \(B=C\). Thus the construction does not depend on the orthonormal bases.
The defining identity applied twice, or conjugate-transpose multiplication, gives
\[(AB)^*=B^*A^*,\quad (A^*)^*=A,\quad (aA+bB)^*=\overline aA^*+\overline bB^*,\quad I^*=I.\]For example, \(\langle u,ABv\rangle=\langle A^*u,Bv\rangle=\langle B^*A^*u,v\rangle\), and uniqueness proves the first rule. Conjugation of matrix entries proves the other three. In particular \(\langle v,A^*Av\rangle=\|Av\|^2\).
□Example. If \(A=\begin{pmatrix}1&i\\2&0\end{pmatrix}\), then \(A^*=\begin{pmatrix}1&2\\-i&0\end{pmatrix}\). The orthonormal-basis hypothesis matters: in arbitrary bases \(e_i\) and \(f_j\), set \((G_V)_{ik}=\langle e_i,e_k\rangle\) and \((G_W)_{j\ell}=\langle f_j,f_\ell\rangle\). These are the Gram matrices. With these matrices, the defining identity is \(\overline u^{\,T}G_WAv=\overline{A^*u}^{\,T}G_Vv\). It gives \(A^*=G_V^{-1}\overline A^{\,T}G_W\), rather than just conjugate transpose. These matrices are invertible because \(Gx=0\) implies \(\overline x^{\,T}Gx=\|x\|^2=0\).
Definition 2.14. Self-adjoint, nonnegative, positive definite, and unitary
An endomorphism is self-adjoint or Hermitian if \(A^*=A\). It is nonnegative or positive semidefinite, written \(A\ge0\), if it is self-adjoint and \(\langle v,Av\rangle\ge0\) for all \(v\). It is positive definite if that quantity is strictly positive for every nonzero \(v\). Some sources use “positive” to mean nonnegative; we always distinguish the two.
A map \(U:V\to V\) is unitary if \(U^*U=UU^*=I\), equivalently \(U^{-1}=U^*\). The condition says that \(U\) preserves the inner product and has an inverse, so it preserves lengths and orthogonality.
Worked example for Definition 2.14
Counterexample: Positive eigenvalues do not replace self-adjointness
Proof.
For self-adjoint \(A\), \(\langle v,Av\rangle=\langle Av,v\rangle=\overline{\langle v,Av\rangle}\), so the quadratic value is real. The map \(A^*A\) is nonnegative by the adjoint rules and \(\langle v,A^*Av\rangle=\|Av\|^2\). It is positive definite exactly when \(A\) is injective. For unitary \(U\),
\[\langle Uu,Uv\rangle=\langle u,U^*Uv\rangle=\langle u,v\rangle.\]Conversely an inner-product-preserving endomorphism satisfies \(U^*U=I\), because \(\langle u,(U^*U-I)v\rangle=0\) for all \(u,v\). This gives injectivity; finite rank-nullity gives invertibility and then \(U^*=U^{-1}\), hence \(UU^*=I\). Also \(P_W^*=P_W\): expand the projection formula on both sides of the adjoint identity. Conversely a self-adjoint idempotent \(P\) is an orthogonal projection: \(v=Pv+(I-P)v\), and \(\langle Px,(I-P)y\rangle=\langle x,P(I-P)y\rangle=0\).
□Example. The matrices \(X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\) and \(Z=\operatorname{diag}(1,-1)\) are self-adjoint and unitary, since each is its adjoint and squares to \(I\). Neither is nonnegative because each has a unit vector with quadratic value \(-1\). The matrix \(\operatorname{diag}(1,0)\) is nonnegative but not positive definite. The matrix \(U=\operatorname{diag}(1,i)\) is unitary but not self-adjoint. Positive definite does not mean that every matrix entry is positive: \(\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\) has quadratic form \(|x-y|^2+|x|^2+|y|^2>0\) for nonzero \((x,y)\).
2.5. The finite spectral theorem
Why a continuous quadratic form attains a maximum on the unit sphere
A set is compact if every cover by open sets has a finite subcover. We prove the Euclidean fact needed here. A closed bounded box in \(\mathbb R^d\) is compact. Otherwise choose an open cover with no finite subcover. Bisect every coordinate interval. At least one of the finitely many smaller closed boxes has no finite subcover, because finite covers of all smaller boxes would together cover the original box. Repeat inside such a box. This produces nested boxes with side lengths tending to zero. The left endpoints in each coordinate increase, the right endpoints decrease, and their difference tends to zero; completeness of \(\mathbb R\) gives a common limit in every coordinate. The resulting point lies in every box. An open member of the cover containing that point contains a ball around it. Every sufficiently small nested box lies in that ball, contradicting its failure to have a finite subcover. Boxes with a zero-length side satisfy the same argument using the other coordinates; a single point is compact directly.
A closed subset \(F\) of a compact set \(K\) is compact: add the open complement of \(F\) to an open cover of \(F\), take a finite cover of \(K\), and discard that complement. The unit sphere in \(\mathbb C^n\cong\mathbb R^{2n}\) is a closed subset of a bounded box: its defining equation \(\sum_j|z_j|^2=1\) is continuous, so its complement is open. It is therefore compact.
The same bisection gives the Bolzano–Weierstrass property: every bounded sequence in \(\mathbb R^d\) has a convergent subsequence. Start with a box containing all its terms and repeatedly choose a smaller box containing infinitely many terms. Choose increasing sequence indices with the selected term in the corresponding nested box. The shrinking boxes have a common point as above, and the selected terms converge to it.
A compact Euclidean set \(K\) is bounded: the balls of positive integer radius centered at zero cover it, and a finite subcover bounds it. It is also closed. For \(x\notin K\), cover \(K\) by the balls centered at \(y\in K\) of radius \(|x-y|/3\). A finite subcover exists; the minimum \(\delta>0\) of these finitely many radii gives a ball of radius \(\delta\) about \(x\) disjoint from all the covering balls, by the triangle inequality. Thus the complement of \(K\) is open. Bolzano–Weierstrass therefore gives a convergent subsequence within \(K\) of every sequence in \(K\): boundedness supplies the subsequence, and closedness keeps its limit in \(K\).
A map \(f:K\to Y\), where \(Y\) is a metric space with distance \(d_Y\), is uniformly continuous if for every \(\varepsilon>0\) some \(\delta>0\) makes \(d_Y(f(x),f(y))<\varepsilon\) whenever \(x,y\in K\) and \(|x-y|<\delta\). Every continuous map on a compact Euclidean set has this property. If it failed, some fixed \(\varepsilon>0\) would allow choices \(x_n,y_n\in K\) with \(|x_n-y_n|<1/n\) and \(d_Y(f(x_n),f(y_n))\ge\varepsilon\). A subsequence of \(x_n\) tends to \(x\in K\), and the corresponding \(y_n\) tends to the same \(x\) by the triangle inequality. Continuity of \(f\) and the triangle inequality in \(Y\) make the displayed image distance tend to zero, a contradiction. This argument also covers vector-valued functions.
Finally a continuous real-valued \(q\) on a nonempty compact \(K\) attains a maximum. The sets \(q^{-1}((-N,N))\), for positive integers \(N\), give an open cover. A finite subcover bounds \(q\), so \(s=\sup q(K)\) is finite. Each closed set \(F_k=\{x\in K:q(x)\ge s-1/k\}\) is nonempty by the definition of supremum, and these sets are nested. Their intersection is nonempty: if it were empty, their open complements would cover \(K\); a finite subcover and nesting would make one \(F_k\) empty. A point in the intersection has \(q(x)=s\). This proves exactly the compactness and maximum facts used below.
Theorem 2.15. Orthonormal eigenvectors for self-adjoint maps
A self-adjoint map on a finite-dimensional complex inner-product space has an orthonormal basis of eigenvectors, and all its eigenvalues are real. Eigenspaces for distinct eigenvalues are orthogonal. If \(\lambda\) runs over its distinct eigenvalues and \(P_\lambda\) is the orthogonal projection onto its eigenspace, then
\[A=\sum_\lambda\lambda P_\lambda,\qquad \sum_\lambda P_\lambda=I,\qquad P_\lambda P_\mu=0\quad(\lambda\ne\mu).\]Worked example for Theorem 2.15
Counterexample: Positive eigenvalues do not replace self-adjointness
Proof: orthonormal eigenvectors for self-adjoint maps
Proof.
In dimension zero there is nothing to prove, so assume positive dimension. For a unit eigenvector \(e\), \(\lambda=\langle e,Ae\rangle\) is real. If \(Av=\lambda v\) and \(Aw=\mu w\), the adjoint identity gives \(\mu\langle v,w\rangle=\overline\lambda\langle v,w\rangle\), proving orthogonality when \(\lambda\ne\mu\).
We prove existence of an eigenvector without assuming polynomial root theorems. In an orthonormal basis, the unit sphere is the usual closed bounded sphere in \(\mathbb R^{2n}\), hence compact. The real-valued continuous function \(q(v)=\langle v,Av\rangle\) attains its maximum there, say at \(e\), with value \(\lambda\). For any \(w\perp e\), the unit vector \((e+tw)/\sqrt{1+t^2\|w\|^2}\), for real \(t\), has quadratic value
\[\frac{\lambda+2t\operatorname{Re}\langle w,Ae\rangle+t^2\langle w,Aw\rangle}{1+t^2\|w\|^2}.\]Its derivative at \(t=0\) is zero because it has a maximum there. Thus \(\operatorname{Re}\langle w,Ae\rangle=0\). Replacing \(w\) by \(iw\) gives \(\operatorname{Im}\langle w,Ae\rangle=0\). Hence \(Ae\) is orthogonal to every \(w\perp e\). The projection decomposition gives \(Ae=\langle e,Ae\rangle e=\lambda e\).
The complement \(e^\perp\) is preserved by \(A\): if \(v\perp e\), then \(\langle e,Av\rangle=\langle Ae,v\rangle=\lambda\langle e,v\rangle=0\). The restriction is self-adjoint on that complement. Induct on dimension, with the zero-dimensional case vacuous, to get an orthonormal eigenbasis. In that basis each displayed operator identity is checked on each basis vector; linearity makes it hold on all vectors.
□Example. For \(X\), put \(e_+=(1,1)/\sqrt2\), \(e_-=(1,-1)/\sqrt2\). Then \(P_+=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\), \(P_-=\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\), and \(X=P_+-P_-\). Repeated eigenvalues cause no difficulty: for \(A=2I\), the sole eigenspace is the whole space and its projection is \(I\). The nilpotent matrix from Section 2.2 fails the self-adjoint hypothesis and has no eigenbasis.
Proposition 2.16. Positivity, functions of matrices, and commuting observables
A self-adjoint map is nonnegative exactly when its eigenvalues are nonnegative, and positive definite exactly when they are positive. For any function \(f\) on the finite set of eigenvalues, define
\[f(A)=\sum_\lambda f(\lambda)P_\lambda.\]Then \((fg)(A)=f(A)g(A)\). A nonnegative \(A\) has the nonnegative square root \(\sqrt A=\sum_\lambda\sqrt\lambda P_\lambda\). Two commuting self-adjoint maps have a common orthonormal eigenbasis. A unitary map also has an orthonormal eigenbasis, with eigenvalues of modulus one.
Worked example for Proposition 2.16
Counterexample: Positive eigenvalues do not replace self-adjointness
Proof.
For \(v=\sum_jc_je_j\) in an eigenbasis, \(\langle v,Av\rangle=\sum_j\lambda_j|c_j|^2\). This proves both directions of the positivity characterizations, testing a single eigenvector for necessity. Products of spectral projections prove the multiplication rule and \((\sqrt A)^2=A\); the same quadratic formula proves nonnegativity of the root.
If \(AB=BA\), then \(Av=\lambda v\) implies \(A(Bv)=B(Av)=\lambda Bv\). Thus \(B\) preserves each eigenspace of \(A\), and its restriction there is self-adjoint. Apply the spectral theorem on each such space to obtain a common eigenbasis.
For unitary \(U\), set \(H=(U+U^*)/2\) and \(K=(U-U^*)/(2i)\). The adjoint rules show that both are self-adjoint. Since \(UU^*=U^*U\), expanding products shows \(HK=KH\). Their common eigenbasis also diagonalizes \(U=H+iK\). If \(Ue=ze\) and \(\|e\|=1\), preservation of norm gives \(|z|=1\).
□Example. For \(A=\operatorname{diag}(4,9)\), \(\sqrt A=\operatorname{diag}(2,3)\). If \(A\) is self-adjoint, the function \(f(\lambda)=e^{-it\lambda}\), with real \(t\), gives a unitary \(f(A)\): its adjoint has conjugate eigenvalues, so \(f(A)^*f(A)=I\). The complex exponential and the identity \(|e^{is}|=1\) are established in the complex-analysis chapter. The self-adjoint matrices \(X,Z\) above do not commute: \(XZ=-ZX\). They have no common eigenbasis, because matrices diagonal in the same basis commute.
2.6. Rank-one maps, rays, and density matrices
Definition 2.17. Rank-one operator and bra–ket notation
For vectors \(u,v\), define
\[R_{u,v}x=u\langle v,x\rangle.\]This is linear in \(x\) because our second slot is linear. Physicists write it \(|u\rangle\langle v|\): the ket \(|u\rangle\) means the vector \(u\), and the bra \(\langle v|\) means the linear functional \(x\mapsto\langle v,x\rangle\). A linear functional is a linear map to \(\mathbb C\). If \(u,v\ne0\), the image is exactly the line spanned by \(u\), so the rank is one: it lies in that line by the formula, and evaluating at \(v/\|v\|^2\) returns \(u\).
Proof.
Direct substitution and the adjoint identity give
\[R_{u,v}^*=R_{v,u},\qquad R_{u,v}R_{a,b}=\langle v,a\rangle R_{u,b}.\]Indeed the composite sends \(x\) to \(u\langle v,a\rangle\langle b,x\rangle\). Also \(\langle y,R_{u,v}x\rangle=\langle y,u\rangle\langle v,x\rangle=\langle R_{v,u}y,x\rangle\), proving the adjoint rule. For a unit \(u\), \(P_u=R_{u,u}\) is its line's orthogonal projection and has quadratic form \(\langle x,P_ux\rangle=|\langle u,x\rangle|^2\).
□Definition 2.18. Ray and phase
Here a ray is represented by a one-dimensional subspace \(\mathbb C u\) with \(u\ne0\). Complex projective space \(\mathbb P(V)\) is the set of these rays. Alphabeta calls the set \(\{au:a\ne0\}\), with zero removed, a ray; adjoining or removing zero gives a bijection between the two conventions. We require \(V\ne\{0\}\) when discussing normalized states. If two unit vectors represent the same ray, they differ by a scalar \(a\) with \(|a|=1\), called a phase. The map \(u\mapsto P_u\), on unit vectors, removes that choice of representative.
Proof.
If \(v=au\) and both are unit, the scaling rule gives \(|a|=1\). Then \(P_vx=au\langle au,x\rangle=a\overline aP_ux=P_ux\). Conversely equality of the projections gives equality of their images, hence equality of their rays. The number \(|\langle u,v\rangle|^2\) is unchanged by either unit representative's phase because conjugation and linearity multiply the inner product by a scalar of modulus one.
□Example. The vectors \((1,0)\), \((-1,0)\), and \((i,0)\) give the same ray and the same matrix \(P=\operatorname{diag}(1,0)\). The relative phase between coordinates survives: \((1,1)/\sqrt2\) and \((1,-1)/\sqrt2\) have different rays and are orthogonal. Multiplying an entire vector by a phase differs from changing only one coordinate's phase.
Definition 2.19. Trace and density matrix
In an orthonormal basis, the trace of an endomorphism is the sum of its diagonal entries:
\[\operatorname{tr}A=\sum_j\langle e_j,Ae_j\rangle.\]A density matrix is a nonnegative endomorphism \(\rho\) with \(\operatorname{tr}\rho=1\). Nonnegativity and this normalization will make measurement weights nonnegative and make their total equal one.
Proposition 2.20. Trace identities and mixtures
Trace is basis-independent and linear. For two endomorphisms, \(\operatorname{tr}(AB)=\operatorname{tr}(BA)\). Moreover
\[\operatorname{tr}R_{u,v}=\langle v,u\rangle,\qquad \operatorname{tr}(P_uA)=\langle u,Au\rangle\quad(\|u\|=1).\]Every density matrix has a decomposition \(\rho=\sum_j p_jP_{e_j}\), with orthonormal \(e_j\), \(p_j\ge0\), and \(\sum_jp_j=1\). More generally any finite weighted mixture of unit-vector projections, with nonnegative weights summing to one, is a density matrix.
Proof.
In coordinates, \(\operatorname{tr}(AB)=\sum_{i,j}A_{ij}B_{ji}=\sum_{j,i}B_{ji}A_{ij}=\operatorname{tr}(BA)\). A change of basis sends \(A\) to \(S^{-1}AS\), whose trace is \(\operatorname{tr}(SS^{-1}A)=\operatorname{tr}A\). Addition and scalar multiplication of diagonal entries prove linearity.
For a rank-one map,
\[\operatorname{tr}R_{u,v}=\sum_j\langle e_j,u\rangle\langle v,e_j\rangle =\left\langle v,\sum_je_j\langle e_j,u\rangle\right\rangle=\langle v,u\rangle.\]Also \(P_uA=R_{u,A^*u}\), so its trace is \(\langle A^*u,u\rangle=\langle u,Au\rangle\). The spectral theorem diagonalizes \(\rho\); positivity makes the eigenvalues \(p_j\) nonnegative and trace one makes their sum one. Conversely \(\sum_jp_jP_{u_j}\) is self-adjoint and has quadratic form \(\sum_jp_j|\langle u_j,x\rangle|^2\ge0\), and its trace is \(\sum_jp_j=1\).
□Example. For \(0\le p\le1\), \(\rho=\operatorname{diag}(p,1-p)\) is a density matrix. The state \(I/2\) has two different mixtures: \(\frac12(P_{e_1}+P_{e_2})=\frac12(P_{e_+}+P_{e_-})\), since either orthonormal basis's projections sum to \(I\). A mixture is not uniquely determined by \(\rho\). Trace one alone is insufficient: \(\operatorname{diag}(2,-1)\) has trace one but a negative quadratic value on \(e_2\).
A density matrix of the form \(P_u\) is called pure; all other density matrices are called mixed. Purity is equivalent to rank one: a rank-one density matrix has just one nonzero eigenvalue, which must equal one. It is also equivalent to \(\operatorname{tr}(\rho^2)=1\): writing eigenvalues \(p_j\in[0,1]\), the equality \(\sum_jp_j^2=\sum_jp_j=1\) forces every \(p_j(1-p_j)=0\), hence exactly one eigenvalue equals one.
Proposition 2.21. Finite Born probabilities and expectation
Let \(A=\sum_\lambda\lambda P_\lambda\) be self-adjoint and \(\rho\) a density matrix. The numbers
\[p(\lambda)=\operatorname{tr}(\rho P_\lambda)\]are nonnegative and sum to one. For a pure state \(\rho=P_u\),
\[p(\lambda)=\|P_\lambda u\|^2,\qquad \sum_\lambda\lambda p(\lambda)=\operatorname{tr}(\rho A).\]The latter number is the expectation, the probability-weighted average of possible values. The variance is their mean squared deviation from that average \(m\):
\[\sum_\lambda(\lambda-m)^2p(\lambda)=\operatorname{tr}(\rho(A-mI)^2)\ge0.\]Proof.
Write \(\rho=\sum_jp_jP_{e_j}\). The trace formulas and projection decomposition give \(\operatorname{tr}(\rho P_\lambda)=\sum_jp_j\langle e_j,P_\lambda e_j\rangle=\sum_jp_j\|P_\lambda e_j\|^2\ge0\). Summing over \(\lambda\), using \(\sum_\lambda P_\lambda=I\), gives \(\operatorname{tr}\rho=1\). For pure states the same trace identity gives the squared projection norm. Linearity and the spectral formula prove the expectation identity; multiplication of spectral projections gives \((A-mI)^2=\sum_\lambda(\lambda-m)^2P_\lambda\), proving the variance identity.
□Example. Measure \(Z=\operatorname{diag}(1,-1)\) in the unit state \(u=(\sqrt3/2,i/2)\). Its projections are onto the coordinate lines, so the probabilities are \(3/4\) and \(1/4\). The expectation is \(1/2\); the variance is \(1-(1/2)^2=3/4\). For \(I/2\), either result has probability \(1/2\). These calculations implement the Born postulate; linear algebra alone does not select that postulate.
Proposition 2.22. Convex mixtures and extreme states
A set is convex if it contains every combination \(t x+(1-t)y\) of its members, for \(0\le t\le1\). The density matrices form a convex set. A member is an extreme point if it cannot be written in that form with \(0<t<1\) and two distinct members. Its extreme points are exactly its pure states.
Proof: convex mixtures and extreme states
Proof.
Nonnegative weighted quadratic forms and linearity of trace prove convexity. If a density matrix has two positive eigenvalues, perturb one upward by \(\varepsilon\) and the other downward by \(\varepsilon\), with \(0<\varepsilon\) smaller than both. The two opposite perturbations are distinct density matrices whose average is the original, so it is not extreme.
If \(P_u=t\sigma+(1-t)\tau\), where \(0<t<1\), then for \(x\perp u\) the quadratic value is zero. Nonnegativity makes both \(\langle x,\sigma x\rangle\) and \(\langle x,\tau x\rangle\) zero. For any nonnegative map \(B\), the spectral expansion shows that \(\langle x,Bx\rangle=0\) implies \(Bx=0\): each positive-eigenvalue coordinate must vanish. Hence both maps kill \(u^\perp\). Self-adjointness then makes their images lie in \(\mathbb C u\). Trace one forces their only diagonal entry on that line to be one, so \(\sigma=\tau=P_u\).
□Example. The center \(I/2\) of the two-level state space is not extreme because it is the average of \(P_{e_1}\) and \(P_{e_2}\). A pure projection cannot be a nontrivial mixture of two different density matrices.
The framework also postulates an ideal conditional measurement. For an orthogonal projection \(P\), let \(p=\operatorname{tr}(\rho P)\). If \(p>0\), its conditional state is
\[\rho'=\frac{P\rho P}{p}.\]This is a density matrix: the numerator is self-adjoint, has quadratic form \(\langle Px,\rho Px\rangle\ge0\), and has trace \(\operatorname{tr}(\rho P^2)=p\). Dividing by \(p\) gives trace one. For \(\rho=P_u\), the numerator sends \(x\) to \(Pu\langle Pu,x\rangle\), so \(\rho'=P_{Pu/\|Pu\|}\), since \(p=\|Pu\|^2\). When \(p=0\), this normalization is undefined, reflecting an outcome of zero probability in the model. This formula is the ideal conditional-measurement postulate, not a model of an apparatus.
2.7. Combining finite systems: tensor products
Definition 2.23. Finite tensor product
Let \(V,W\) have bases \(e_1,\ldots,e_n\) and \(f_1,\ldots,f_m\). Construct \(V\otimes W\) as a vector space with formal basis symbols \(e_i\otimes f_j\), one for each ordered pair \((i,j)\). For \(v=\sum_ix_ie_i\) and \(w=\sum_jy_jf_j\), set
\[v\otimes w=\sum_{i,j}x_iy_j(e_i\otimes f_j).\]This operation is bilinear, meaning linear in each slot separately. A vector of this form is an elementary tensor. General tensors are finite sums of elementary tensors. The space has dimension \(nm\) by its constructed basis.
Proof: finite tensor product
Proof.
The coordinate formula proves bilinearity by distributing the coordinate products. To justify that the construction is independent of the chosen bases, let \(B:V\times W\to Z\) be any bilinear map. There is a unique linear map \(\widetilde B:V\otimes W\to Z\) with \(\widetilde B(e_i\otimes f_j)=B(e_i,f_j)\), because these symbols form a basis. Expanding coordinates gives \(\widetilde B(v\otimes w)=B(v,w)\). This is the universal property of the tensor product: bilinear maps correspond uniquely to linear maps on it. A construction with different bases satisfies the same property. Applying it to the elementary-tensor map into each construction gives mutually inverse linear maps, because their composites fix every elementary tensor, which spans the spaces.
□Proposition 2.24. Inner product and product operators
If the chosen bases are orthonormal, declare \(e_i\otimes f_j\) orthonormal. This gives a basis-independent inner product characterized by
\[\langle v\otimes w,v'\otimes w'\rangle=\langle v,v'\rangle\langle w,w'\rangle.\]In particular \(\|v\otimes w\|=\|v\|\|w\|\). For endomorphisms \(A\) of \(V\) and \(B\) of \(W\), there is a unique linear map \(A\otimes B\) satisfying \((A\otimes B)(v\otimes w)=Av\otimes Bw\). It obeys
\[(A\otimes B)^*=A^*\otimes B^*,\qquad (A\otimes B)(C\otimes D)=AC\otimes BD,\qquad \operatorname{tr}(A\otimes B)=\operatorname{tr}A\operatorname{tr}B.\]Tensor products of unitary maps are unitary. If \(\rho,\sigma\) are density matrices, \(\rho\otimes\sigma\) is a density matrix.
Proof: inner product and product operators
Proof.
The coordinate inner product is positive definite because it sums squared moduli of the \(nm\) coefficients. Expanding elementary tensors separates that finite sum into the product of two inner products. As elementary tensors span, the displayed formula and conjugate-linearity/linearity determine the inner product uniquely, so it is preserved by the basis-change isomorphism.
The map \((v,w)\mapsto Av\otimes Bw\) is bilinear, so the universal property defines \(A\otimes B\). Its adjoint rule follows by testing the adjoint identity on elementary tensors and then expanding sums. Its composition rule follows by acting on elementary tensors, and the trace rule follows by summing diagonal entries \(A_{ii}B_{jj}\). The adjoint and composition rules give unitarity when both factors are unitary. For nonnegative \(\rho,\sigma\), tensor their orthonormal eigenbases: their product eigenvalues are nonnegative. The trace product equals one, giving the density-matrix claim.
□Example. For two copies of \(\mathbb C^2\), the tensor basis is \(e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2\). The vector \(\psi=(e_1\otimes e_1+e_2\otimes e_2)/\sqrt2\) has norm one but is not an elementary tensor. If \(\psi=(ae_1+be_2)\otimes(ce_1+de_2)\), the coefficients would satisfy \(ac=bd=1/\sqrt2\) and \(ad=bc=0\). The first two equalities make all four scalars nonzero, contradicting the last two. Such a vector is called entangled. This algebraic definition of entanglement requires a specified division into tensor factors.
A useful way to describe one factor of a composite state is the partial trace. For an endomorphism \(R\) on \(V\otimes W\), define \(\rho_V\) by its entries in orthonormal bases:
\[(\rho_V)_{ik}=\sum_{j=1}^m\langle e_i\otimes f_j,R(e_k\otimes f_j)\rangle.\]This is characterized without coordinates by \(\operatorname{tr}(\rho_V A)=\operatorname{tr}(R(A\otimes I))\) for every endomorphism \(A\) of \(V\). To verify it, expand both traces: both equal \(\sum_{i,k,j}R_{(i,j),(k,j)}A_{ki}\). The characterization is unique, because testing \(A\) with a single nonzero matrix entry reads each entry of \(\rho_V\); it also proves independence of the chosen bases. If \(R\) is a density matrix, its partial trace is one too: it is self-adjoint by conjugating entries, its quadratic form is \(\sum_j\langle v\otimes f_j,R(v\otimes f_j)\rangle\ge0\), and its trace is \(\operatorname{tr}R=1\).
Example. For the entangled \(\psi\) above, \(R=P_\psi\) is pure, but its partial trace is \(I/2\): the entry formula gives two diagonal entries \(1/2\) and zero off-diagonal entries. A subsystem can therefore be mixed while the composite state is pure. For a product state \(R=\rho\otimes\sigma\), the entry sum instead gives \(\rho_V=\rho\operatorname{tr}\sigma=\rho\).
2.8. Finite autonomous dynamics
The autonomous-dynamics page uses unitary evolution generated by a self-adjoint energy operator. Here is its finite-dimensional version. Fix a self-adjoint \(H\) and a positive constant \(\hbar\). Define
\[U(t)=\sum_\lambda e^{-it\lambda/\hbar}P_\lambda\quad(t\in\mathbb R).\]The scalar exponential's product rule and modulus, proved in the complex-analysis chapter, give \(U(0)=I\), \(U(t+s)=U(t)U(s)\), and \(U(t)^*=U(-t)\). Products of the mutually orthogonal spectral projections prove these identities directly. They imply that each \(U(t)\) is unitary. Differentiating the finite sum gives
\[U'(t)=-\frac{i}{\hbar}HU(t),\qquad i\hbar\psi'(t)=H\psi(t)\quad\text{for }\psi(t)=U(t)\psi(0).\]The evolution preserves norm, and an eigenvector of energy \(\lambda\) evolves by the phase \(e^{-it\lambda/\hbar}\), so its ray is fixed. Conversely if a unit vector's ray is fixed for every \(t\), then its projection is fixed. The rank-one formula gives \(P_{U(t)u}=U(t)P_uU(t)^*\). Differentiating this identity at zero yields \(HP_u=P_uH\); applying this to \(u\) gives \(Hu=P_uHu\in\mathbb C u\), so the ray is an eigenray.
A density matrix evolves by \(\rho(t)=U(t)\rho(0)U(t)^*\). Conjugation preserves self-adjointness and positivity because \(\langle x,\rho(t)x\rangle=\langle U(t)^*x,\rho(0)U(t)^*x\rangle\ge0\), and preserves trace by cyclicity. Entrywise differentiation and the scalar product rule give
\[i\hbar\rho'(t)=[H,\rho(t)],\qquad [A,B]:=AB-BA.\]The brackets define the commutator. For an observable \(A\), define its Heisenberg evolution \(A(t)=U(t)^*AU(t)\). The same product rule gives \(A'(t)=(i/\hbar)[H,A(t)]\). Cyclicity gives \(\operatorname{tr}(\rho(t)A)=\operatorname{tr}(\rho(0)A(t))\); these are two ways to calculate the same expectation. If \(AH=HA\), it preserves every energy eigenspace and therefore commutes with all \(P_\lambda\) and \(U(t)\), so \(A(t)=A\) and its expectation is constant.
Example. For \(H=\operatorname{diag}(0,E)\) and \(u=(1,1)/\sqrt2\), the evolved state vector is \((1,e^{-itE/\hbar})/\sqrt2\). For \(X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), direct multiplication gives expectation \(\operatorname{Re}(e^{-itE/\hbar})=\cos(tE/\hbar)\). The diagonal energy probabilities stay \(1/2,1/2\), while the relative phase changes the expectation of \(X\). If the initial vector is \(e_2\), only an overall phase changes and all its Born probabilities stay fixed.
2.9. What changes in infinite dimensions?
All the spectral and trace sums in this chapter are finite. For wave functions, an orthonormal expansion may require an infinite convergent sum, projections require closed subspaces, and an unbounded operator includes its domain as part of its definition. A self-adjoint operator can have spectral values without eigenvectors, and a positive operator need not have a finite trace. The measure-theory chapter supplies integrals, the probability chapter supplies distributions, and the functional-analysis chapter proves the relevant extensions and explains where a full spectral theorem is needed.
For the finite setting, the definitions and proofs above are enough to read amplitudes, rays, density matrices, tensor composition, and discrete Born probabilities. Use the overview to choose the additional route for continuous position and momentum.
Verified Alphabeta reading links
The following pages supply the corresponding library material. The proofs in this chapter are given locally so no result depends on a reader having already studied these pages.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. A complex basis and a proper subspace
Applies to: Definition 2.1, Definition 2.2, Proposition 2.3.
Let \(V=\mathbb C^2\) with coordinatewise operations. Every vector-space axiom reduces to the corresponding complex-number identity in each coordinate, including additive inverses and distributivity. The vectors \(v_1=(1,i),v_2=(0,1)\) are independent: \(av_1+bv_2=(a,ia+b)=0\) forces \(a=b=0\). They span because \((z,w)=zv_1+(w-iz)v_2\). Thus they form a basis and \(\dim V=2\). The line \(W=\{(a,ia):a\in\mathbb C\}\) is a subspace, closed under sums and scalar multiples, with basis \(v_1\). It is proper since \((0,1)\notin W\); adjoining that vector extends its independent basis to one of \(V\). The list \(v_1,2v_1\) is dependent via \(2v_1-2v_1=0\) with coefficients 2 and \(-1\). This concretely checks the span, independence, dimension, and extension definitions.
Worked example 2. A matrix, its rank, and its eigenvectors
Applies to: Definition 2.4, Proposition 2.5, Definition 2.6, Definition 2.7.
On \(\mathbb C^2\) take \(T(z,w)=(z+w,z+w)\). Its standard matrix is \(A=\begin{pmatrix}1&1\\1&1\end{pmatrix}\), since its columns are \(Te_1,Te_2\). Distributivity proves complex linearity. Its kernel is \(\mathbb C(1,-1)\) and image \(\mathbb C(1,1)\); hence nullity plus rank is \(1+1=2\). It is neither injective nor onto. The nonzero vectors \((1,1),(1,-1)\) have eigenvalues 2 and 0 and form a basis, so it is diagonalizable. For comparison \(T+I\) has eigenvalues 3 and 1, is bijective, and has inverse \(\tfrac13\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\). Direct multiplication gives the identity, illustrating the finite endomorphism assertion in rank-nullity.
Worked example 3. Orthogonalization and the nearest vector
Applies to: Definition 2.8, Proposition 2.9, Proposition 2.10, Definition 2.11, Proposition 2.12.
Use \(\langle u,v\rangle=\bar u_1v_1+\bar u_2v_2\), linear in its second slot. Positive definiteness is the sum of squared moduli. Gram–Schmidt applied to \((1,1),(1,0)\) gives \(e_+=(1,1)/\sqrt2\), \(e_-=(1,-1)/\sqrt2\): subtracting the projection of the second vector leaves \((1/2,-1/2)\), of norm \(1/\sqrt2\). For \(x=(1,i)\), coordinates are \((1+i)/\sqrt2,(1-i)/\sqrt2\); their squared moduli sum to 2, exactly \(\|x\|^2\).
For \(W=\mathbb Ce_+\), \(P_Wx=((1+i)/2,(1+i)/2)\), while \(r=x-P_Wx=((1-i)/2,(-1+i)/2)\) lies in \(W^\perp=\mathbb Ce_-\). Thus \(P_W^2=P_W\) and for \(w=ae_+\), \(\|x-w\|^2=1+|(1+i)/\sqrt2-a|^2\), uniquely minimized by \(P_Wx\), with distance 1. Also \(|\langle e_+,x\rangle|=1<\sqrt2=\|e_+\|\|x\|\), a strict Cauchy–Schwarz case. For \(x=2ie_+\) equality holds. Finally \(\|e_++e_-\|=\sqrt2\le2\) checks the triangle inequality and its induced distance in this orthonormal example.
Worked example 4. Spectral projections, square roots, and adjoints
Applies to: Definition 2.13, Definition 2.14, Theorem 2.15, Proposition 2.16.
Let \(A=\begin{pmatrix}1&1\\1&1\end{pmatrix}\) as above. Its conjugate transpose equals itself, and \(\langle x,Ax\rangle=|x_1+x_2|^2\ge0\); it is nonnegative but not positive definite since it kills \(e_-\). The projections \(P_+=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\), \(P_-=\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\) obey \(P_++P_-=I\), \(P_+P_-=0\), and \(A=2P_+\). Therefore \(\sqrt A=\sqrt2P_+\), whose square is \(A\). For any functions \(f,g\) on \(\{0,2\}\), multiplying their projection sums gives \(f(A)g(A)=f(2)g(2)P_++f(0)g(0)P_-\). The operator \(A+I\) is positive definite (eigenvalues 3 and 1) and commutes with \(A\), sharing this orthonormal eigenbasis.
A non-self-adjoint example is \(C=\begin{pmatrix}0&i\\0&0\end{pmatrix}\), whose adjoint is \(\begin{pmatrix}0&0\\-i&0\end{pmatrix}\); inserting arbitrary coordinate vectors checks \(\langle u,Cv\rangle=\langle C^*u,v\rangle\). The map \(U=P_++iP_-\) has \(U^*=P_+-iP_-\), so \(U^*U=UU^*=I\) and eigenvalues \(1,i\) of modulus one. This checks the adjoint and unitary assertions with the declared inner-product convention.
Worked example 5. Rays, mixtures, and a measurement calculation
Applies to: Definition 2.17, Definition 2.18, Definition 2.19, Proposition 2.20, Proposition 2.21, Proposition 2.22.
For \(u=(1,i)/\sqrt2\), its rank-one projection is \(P_u=\tfrac12\begin{pmatrix}1&-i\\i&1\end{pmatrix}\): applying it to \(x\) gives \(u\langle u,x\rangle\). Its trace is \(\langle u,u\rangle=1\), its quadratic form is \(|\langle u,x\rangle|^2\), and it is unchanged by \(u\mapsto e^{i\theta}u\), so it represents the ray \(\mathbb Cu\). For \(\rho=\operatorname{diag}(1/4,3/4)=\tfrac14P_{e_1}+\tfrac34P_{e_2}\), positivity and trace one are immediate. For \(A=\operatorname{diag}(0,2)\), Born probabilities are \(1/4,3/4\), mean \(3/2\), second moment 3, and variance \(3-9/4=3/4\). They agree with \(\operatorname{tr}(\rho A)\) and \(\operatorname{tr}(\rho(A-\tfrac32I)^2)\).
The displayed mixture proves \(\rho\) is not extreme. To check extremality of \(P_u\), if \(P_u=t\sigma+(1-t)\eta\), \(0<t<1\), then for every \(w\perp u\), positivity forces \(\langle w,\sigma w\rangle=\langle w,\eta w\rangle=0\). The nonnegative spectral decompositions force \(\sigma w=\eta w=0\); trace one then forces both to equal \(P_u\). For the projection \(P_{e_2}\), the declared conditional update of \(\rho\) is \(P_{e_2}\rho P_{e_2}/(3/4)=P_{e_2}\). This also checks the unboxed update formula and its positive-probability hypothesis.
As a trace-cyclicity check, take \(B=\begin{pmatrix}0&1\\0&0\end{pmatrix}\), \(C=\begin{pmatrix}0&0\\1&0\end{pmatrix}\). Then \(BC=P_{e_1},CB=P_{e_2}\), both with trace 1 although the products differ. Under an orthonormal basis change \(V\), cyclicity gives \(\operatorname{tr}(V^*\rho V)=\operatorname{tr}(\rho VV^*)=1\), verifying the basis-independent trace in this calculation.
Further proof details
The positivity step in the extreme-point argument has an elementary proof. If \(S\ge0\) and \(\langle w,Sw\rangle=0\), then for every \(z\) and complex \(t\),
\[0\le\langle w+tz,S(w+tz)\rangle=2\operatorname{Re}(t\langle w,Sz\rangle)+|t|^2\langle z,Sz\rangle.\]If \(\langle w,Sz\rangle\ne0\), choose the phase of a sufficiently small \(t\) to make its linear term strictly negative; it dominates the quadratic term, a contradiction. Thus \(\langle w,Sz\rangle=0\) for all \(z\), and self-adjointness gives \(Sw=0\). Applying this to both terms of a positive mixture shows they kill \(u^\perp\) and send \(u\) to a scalar multiple of \(u\); trace one makes that scalar 1. This directly proves extremality of \(P_u\), without leaving the decisive implication to a named spectral theorem.
Worked example 6. Product operators and an entangled vector
Applies to: Definition 2.23, Proposition 2.24.
Take \(V=W=\mathbb C^2\) with standard orthonormal bases. The four tensors \(e_i\otimes e_j\) are an orthonormal basis, so the tensor space has dimension 4. For \(v=(1,i),w=(2,0)\), the coordinates of \(v\otimes w\) are \((2,0,2i,0)\), with norm \(2\sqrt2=\|v\|\|w\|\). Put \(Z=\operatorname{diag}(1,-1)\). Then \(Z\otimes Z=\operatorname{diag}(1,-1,-1,1)\), self-adjoint and unitary, with trace \(0=\operatorname{tr}Z\operatorname{tr}Z\); squaring it gives \(I\otimes I\), checking both product and adjoint rules. The product density \((I/2)\otimes P_{e_1}=\operatorname{diag}(1/2,0,1/2,0)\) is nonnegative with trace one.
The vector \(\Psi=(e_1\otimes e_1+e_2\otimes e_2)/\sqrt2\) is normalized. Its coefficient matrix is \(I/\sqrt2\), of determinant \(1/2\), whereas an elementary tensor has coefficient matrix \(vw^T\) of rank at most one. Thus \(\Psi\) is entangled. Its projection has entries \(R_{11,11}=R_{11,22}=R_{22,11}=R_{22,22}=1/2\); summing \(R_{ij,kj}\) gives partial trace \(I/2\), a mixed density matrix. For arbitrary \(A\), both \(\operatorname{tr}((I/2)A)\) and \(\langle\Psi,(A\otimes I)\Psi\rangle\) equal \((A_{11}+A_{22})/2\), checking the unboxed characterization of the partial trace.
Worked example 7. Schrödinger and Heisenberg calculations agree
Set \(\hbar=1,H=\operatorname{diag}(0,2)\), \(u=(1,1)/\sqrt2\), and \(X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\). Then \(U(t)=\operatorname{diag}(1,e^{-2it})\), \(\rho(t)=\tfrac12\begin{pmatrix}1&e^{2it}\\e^{-2it}&1\end{pmatrix}\). Differentiation gives \(i\rho'=[H,\rho]\) entry by entry. Also \(X(t)=U(t)^*XU(t)=\begin{pmatrix}0&e^{-2it}\\e^{2it}&0\end{pmatrix}\) and \(X'=(i)[H,X(t)]\). Both pictures give expectation \(\cos2t\). The energy expectation stays 1 because \(H\) commutes with \(U\). The eigenvector \(e_2\) evolves only by a phase, while the ray of \(u\) changes (at \(t=\pi/2\) it is orthogonal to its initial ray). This illustrates all the unboxed finite-dynamics conclusions.
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. Positive eigenvalues do not replace self-adjointness
Related statements: Definition 2.14, Theorem 2.15, Proposition 2.16.
Claim refuted. A diagonalizable complex matrix with positive real eigenvalues is automatically a nonnegative observable.
Proof.
Take \(A=\begin{pmatrix}1&1\\0&2\end{pmatrix}\). The independent vectors \(e_1\) and \((1,1)\) are eigenvectors of eigenvalues 1 and 2, so it is diagonalizable with positive real spectrum. But \(A^*=\begin{pmatrix}1&0\\1&2\end{pmatrix}\ne A\), and its eigenvectors have inner product 1 rather than zero. For \(v=(1,i)\), \(Av=(1+i,2i)\) and \(\langle v,Av\rangle=3+i\), which is not even real. Hence \(A\) is neither self-adjoint nor nonnegative under the chapter’s definitions. The eigenvalue criterion for positivity explicitly assumes self-adjointness, and a change to a nonorthonormal eigenbasis does not preserve the usual conjugate-transpose formula for the adjoint.
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