Orthogonal Projections and Spectral Resolution

Derivation

Orthogonal Projections and Spectral Resolution

D-089 Home PU-104 Threads symmetry · chance Depends on Spectral Theorem for Self-Adjoint Operators, Properties of Hermitian Operators

Statement

On a finite-dimensional complex inner-product space \(V\), an operator \(P\) is an orthogonal projector onto a subspace exactly when it is both idempotent, \(P^2=P\), and self-adjoint, \(P^\dagger=P\); such a \(P\) projects every vector onto the closed subspace \(M=\operatorname{ran}P\) along its orthogonal complement \(M^\perp=\ker P\). Consequently, for a self-adjoint operator \(A=A^\dagger\) with distinct eigenvalues \(\{\lambda_k\}\), the spectral theorem produces eigenprojectors \(\{P_k\}\) that are mutually orthogonal, \(P_jP_k=\delta_{jk}P_k\), resolve the identity, \(\sum_k P_k=\mathbf 1\), and resolve the operator itself, \(A=\sum_k\lambda_k P_k\).

Why it matters

The resolution of the identity is the algebraic backbone of the measurement postulate. Every observable is decomposed into a weighted sum of orthogonal projectors, one per possible outcome; the projectors are the objects the Born rule feeds on, turning an abstract state vector into a probability distribution over eigenvalues. This is where the symmetry thread (self-adjointness as a reality condition) and the chance thread (probabilities of outcomes) meet in a single equation.

Beyond quantum theory, \(A=\sum_k\lambda_kP_k\) is the cleanest form of a symmetric linear map: it diagonalizes \(A\) without choosing a basis and makes the functional calculus immediate, \(f(A)=\sum_k f(\lambda_k)P_k\) — which is how time-evolution operators \(e^{-iHt/\hbar}\), density matrices, and thermal averages are actually computed.

Assumptions

\(V\) is a complete inner-product space (finite-dimensional here) over \(\mathbb C\).Without an inner product the adjoint \(P^\dagger\) is undefined, so "self-adjoint" has no meaning; without completeness \(\operatorname{ran}P\) need not be closed and the orthogonal decomposition \(V=M\oplus M^\perp\) can fail. \(A\) is self-adjoint, \(A^\dagger=A\).Drop it and eigenvalues may be complex, eigenvectors of distinct eigenvalues need not be orthogonal, and the \(P_k\) fail to be orthogonal projectors — the resolution \(\sum_k P_k=\mathbf 1\) can collapse. \(A\) has purely discrete spectrum with a complete set of eigenvectors.For operators with continuous spectrum the sum \(\sum_k\lambda_kP_k\) must be replaced by a Stieltjes integral \(\int\lambda\,dE(\lambda)\) over a projection-valued measure; the finite resolution no longer holds verbatim.

Derivation

\[ P^2=P,\qquad P^\dagger=P \]

The two conditions we impose on \(P\): idempotent and self-adjoint. We show they force the range and kernel to be orthogonal complements. A

\[ v=Pv+(v-Pv),\qquad P(v-Pv)=Pv-P^2v=Pv-Pv=0 \]

Trivial add-and-subtract: the first term lies in \(\operatorname{ran}P\), and idempotency \(P^2=P\) puts the second in \(\ker P\). This gives the algebraic direct sum \(V=\operatorname{ran}P\oplus\ker P\) before any metric is used. A

\[ \langle Pu,\,(\mathbf 1-P)w\rangle=\langle u,\,P^\dagger(\mathbf 1-P)w\rangle=\langle u,\,P(\mathbf 1-P)w\rangle=\langle u,\,(P-P^2)w\rangle=0 \]

Now self-adjointness enters: move \(P\) across the inner product using \(P^\dagger=P\), then \(P-P^2=0\). Every vector of \(\operatorname{ran}P\) is orthogonal to every vector of \(\ker P\), upgrading the direct sum to an orthogonal decomposition. B

\[ \|Pv\|^2=\langle Pv,Pv\rangle=\langle v,P^\dagger Pv\rangle=\langle v,P^2v\rangle=\langle v,Pv\rangle\le\|v\|\,\|Pv\| \]

Combining \(P^\dagger=P\) and \(P^2=P\) gives \(P^\dagger P=P\); Cauchy–Schwarz then yields \(\|Pv\|\le\|v\|\). An orthogonal projector is a norm-one contraction — geometrically the "nearest point in \(M\)" map. C

\[ Av=\lambda v\ \Rightarrow\ \lambda\|v\|^2=\langle v,Av\rangle=\langle Av,v\rangle=\bar\lambda\|v\|^2\ \Rightarrow\ \lambda\in\mathbb R \]

From hermitian-operator-properties: self-adjointness forces every eigenvalue real, which is what lets \(\lambda_k\) label physical measurement values. B

\[ (\lambda_j-\lambda_k)\langle v_j,v_k\rangle=\langle Av_j,v_k\rangle-\langle v_j,Av_k\rangle=0 \]

Eigenvectors for distinct eigenvalues are orthogonal, again from \(A=A^\dagger\). Hence the eigenspaces \(\mathcal H_k=\ker(A-\lambda_k)\) are mutually orthogonal. B

\[ V=\bigoplus_k \mathcal H_k \]

The spectral theorem (spectral-theorem-self-adjoint) guarantees the eigenvectors span all of \(V\); orthogonality from step 6 makes the sum an orthogonal direct sum. C

\[ P_k:=\text{orthogonal projection onto }\mathcal H_k=\sum_{i}\big|e^{(k)}_i\big\rangle\big\langle e^{(k)}_i\big| \]

Define \(P_k\) via any orthonormal basis \(\{e^{(k)}_i\}\) of \(\mathcal H_k\). The outer-product form is manifestly idempotent and self-adjoint, so \(P_k\) is an orthogonal projector by steps 1–3. B

\[ P_jP_k=\delta_{jk}P_k \]

For \(j\ne k\), \(P_jP_k\) maps into \(\mathcal H_k\) then projects onto \(\mathcal H_j\perp\mathcal H_k\), giving \(0\); for \(j=k\), idempotency. The projectors are mutually orthogonal. A

\[ \sum_k P_k=\mathbf 1 \]

Because \(V=\bigoplus_k\mathcal H_k\) (step 7), summing the projections onto each summand reconstructs every vector, \(v=\sum_kP_kv\). This is the resolution of the identity. A

\[ A=A\Big(\sum_k P_k\Big)=\sum_k AP_k=\sum_k\lambda_k P_k \]

Insert the resolution of identity to the right of \(A\); on \(\mathcal H_k\) the operator acts as multiplication by \(\lambda_k\), so \(AP_k=\lambda_kP_k\). This is the spectral resolution of \(A\). B

\[ A^n=\sum_k\lambda_k^n P_k\quad\Rightarrow\quad f(A)=\sum_k f(\lambda_k)P_k \]

Because \(P_jP_k=\delta_{jk}P_k\), powers collapse term by term; extending linearly over polynomials and closing under uniform limits defines \(f(A)\) for continuous \(f\) by acting on eigenvalues alone — the functional calculus. C

Result

\[ P^\dagger=P,\ P^2=P\ \Rightarrow\ V=\operatorname{ran}P\overset{\perp}{\oplus}\ker P;\qquad A=\sum_k\lambda_k P_k,\quad \sum_k P_k=\mathbf 1,\quad P_jP_k=\delta_{jk}P_k \]

Reading. A self-adjoint idempotent is exactly an orthogonal projection: it splits the space into a subspace it keeps untouched and the perpendicular subspace it annihilates. The spectral theorem then writes any self-adjoint observable as a "menu" of its eigenvalues, each tagged with the projector onto the states that yield it, and those projectors tile the whole space with no overlap and no gaps. Functions of \(A\) act eigenvalue-by-eigenvalue: \(f(A)=\sum_k f(\lambda_k)P_k\).

Units check. Projectors and the identity are dimensionless, as they must be to be added to and equated with one another (\([P_k]=1\), consistent with \(\sum_k P_k=\mathbf 1\)). In \(A=\sum_k\lambda_kP_k\) each eigenvalue \(\lambda_k\) carries the physical units of the observable \(A\) (J for energy, \(\hbar\) for spin projection), so both sides share those units; the dimensionless \(P_k\) merely distributes the weight.

Limiting cases

Single eigenvalue (\(A=\lambda\mathbf 1\)): one projector \(P_1=\mathbf 1\), so the resolution reduces to the identity — a degenerate observable that distinguishes nothing.
Rank-one projector \(P=|\psi\rangle\langle\psi|\) with \(\|\psi\|=1\): idempotency and self-adjointness are immediate, and \(P\) selects the component along \(|\psi\rangle\).
Non-degenerate spectrum: every \(\mathcal H_k\) is one-dimensional, \(P_k=|k\rangle\langle k|\), and the resolution is the familiar \(\sum_k|k\rangle\langle k|=\mathbf 1\).
Two-outcome observable: \(A=\lambda_1P_1+\lambda_2P_2\) with \(P_2=\mathbf 1-P_1\); a projector \(A=P\) is its own resolution, \(P=1\cdot P+0\cdot(\mathbf 1-P)\).

Breaks when

Continuous spectrum. For position, momentum, or a free-particle Hamiltonian the eigenvalues form a continuum with no normalizable eigenvectors; the discrete sum \(\sum_k\lambda_kP_k\) is meaningless and must be replaced by the spectral integral \(\int\lambda\,dE(\lambda)\) over a projection-valued measure.
Non-normal / non-self-adjoint operators. A merely idempotent \(P^2=P\) that is not self-adjoint is an oblique projector: it splits the space along a slanted complement, so \(\operatorname{ran}P\not\perp\ker P\) and \(\|Pv\|\le\|v\|\) can fail. Defective (non-diagonalizable) operators admit no projector decomposition at all — they need the Jordan form.
Incomplete eigensystem / indefinite metric. If \(A\) is only symmetric but not essentially self-adjoint, or lives in a Krein/Minkowski indefinite inner product, the eigenvectors may fail to be complete or orthogonal, so \(\sum_k P_k\ne\mathbf 1\) and \(P_jP_k=\delta_{jk}P_k\) can break.

Failure modes

Confusing idempotent with unitary. Writing \(P^\dagger P=\mathbf 1\) for a projector; the correct relations are \(P^2=P\) and \(P^\dagger=P\), and \(P\) is not invertible unless \(P=\mathbf 1\).
Dropping self-adjointness and still assuming orthogonality. Concluding \(\operatorname{ran}P\perp\ker P\) from \(P^2=P\) alone; orthogonality requires \(P^\dagger=P\) (step 3), else the projection is oblique.
Forgetting normalization in \(|\psi\rangle\langle\psi|\). An unnormalized \(|\psi\rangle\) gives \(P^2=\|\psi\|^2P\ne P\), so it is not idempotent.
Not grouping degenerate eigenvalues. Treating each basis vector of a degenerate level as its own \(P_k\) makes the projectors basis-dependent; the spectral projector sums over the whole eigenspace, and Born weights must too: \(\sum_i|\langle e^{(k)}_i|\psi\rangle|^2\).
Treating \(\sum_k P_k=\mathbf 1\) as optional. Omitting an eigenvalue (truncating a basis) silently violates completeness and produces probabilities that do not sum to one.

Discussion

The equivalence "self-adjoint idempotent = orthogonal projection" makes measurement geometrically honest. An orthogonal projector maps a state to the closest point in a subspace (step 4), so collapse onto an eigenspace is the minimal, distance-minimizing update consistent with the measured value. Self-adjointness is precisely the metric statement that this nearest-point map is symmetric — it does not distort angles — whereas a bare idempotent (oblique projector) is a shear that keeps a subspace fixed but shadows it onto a tilted complement.

The resolution of the identity \(\sum_k P_k=\mathbf 1\) is the completeness relation dressed as an operator identity. Sandwiching it in a state gives \(\langle\psi|\psi\rangle=\sum_k\langle\psi|P_k|\psi\rangle=\sum_k p_k\), showing the Born probabilities \(p_k=\langle\psi|P_k|\psi\rangle\) sum to one for a normalized state. This is where the chance thread anchors: probability conservation is not an extra axiom but a corollary of the projectors tiling the space.

The functional calculus (step 12) is the practical payoff. Because commuting projectors turn operator functions into ordinary functions of eigenvalues, the Schrödinger propagator becomes \(e^{-iHt/\hbar}=\sum_k e^{-iE_kt/\hbar}P_k\) and the thermal density operator becomes \(\rho=Z^{-1}\sum_k e^{-\beta E_k}P_k\). The hard analytic content — existence of the eigenbasis — is imported once from the spectral theorem; everything downstream is bookkeeping with orthogonal projectors.

At the rigorous end, the leap to infinite dimensions is nontrivial: the sum \(\sum_k P_k\) converges only in the strong operator topology, not in norm, and for operators with continuous spectrum the eigenprojectors are replaced by a projection-valued measure \(E(\Omega)\) assigning an orthogonal projector to each Borel set of spectral values, with \(A=\int\lambda\,dE(\lambda)\) and \(E(\sigma(A))=\mathbf 1\). The discrete case here is the special instance where \(E\) is supported on isolated atoms and \(P_k=E(\{\lambda_k\})\). Relaxing \(P_jP_k=\delta_{jk}P_k\) to a set of positive operators \(\{E_k\}\) with \(\sum_k E_k=\mathbf 1\) that need not be projectors gives the POVMs of quantum information.

Common misconceptions. A projector is not a probability — it is an operator whose expectation in a state is a probability. "Orthogonal projector" refers to orthogonality of the geometry (\(\operatorname{ran}\perp\ker\)), not to the projector being an orthogonal/unitary matrix; nonzero projectors other than \(\mathbf 1\) are never invertible. And a degenerate self-adjoint operator still has a unique set of spectral projectors even though its eigenvectors are not unique.

Worked examples

Example 1 — Spin-\(\tfrac12\) along \(z\).

\[ S_z=\frac{\hbar}{2}\sigma_z=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=S_z^\dagger \]

Symbol first: self-adjoint observable with eigenvalues \(\lambda_\pm=\pm\hbar/2\) and eigenvectors \(|{\uparrow}\rangle,|{\downarrow}\rangle\). A

\[ P_+=|{\uparrow}\rangle\langle{\uparrow}|=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad P_-=|{\downarrow}\rangle\langle{\downarrow}|=\begin{pmatrix}0&0\\0&1\end{pmatrix} \]

Rank-one eigenprojectors. Check \(P_++P_-=\mathbf 1\), \(P_+P_-=0\), and \(\tfrac\hbar2 P_+-\tfrac\hbar2 P_-=S_z\). B

\[ |\psi\rangle=\alpha|{\uparrow}\rangle+\beta|{\downarrow}\rangle,\quad p_+=\langle\psi|P_+|\psi\rangle=|\alpha|^2,\quad \langle S_z\rangle=\tfrac\hbar2\big(|\alpha|^2-|\beta|^2\big) \]

Born rule from the projector; expectation from the spectral resolution. B

\[ \alpha=\tfrac{1}{\sqrt5},\ \beta=\tfrac{2}{\sqrt5}\ \Rightarrow\ |\alpha|^2=0.20,\ |\beta|^2=0.80,\ \langle S_z\rangle=\tfrac\hbar2(0.20-0.80)=-0.30\,\hbar \]

Numbers in only now; normalization \(|\alpha|^2+|\beta|^2=1\) checks, with \(\hbar=1.055\times10^{-34}\ \text{J·s}\). A

\[ p_+=0.20,\qquad \langle S_z\rangle=-0.30\,\hbar=-3.16\times10^{-35}\ \text{J·s} \]

Reading. The projector \(P_+\) assigns a 20% chance of measuring spin-up; the resolution \(S_z=\tfrac\hbar2 P_+-\tfrac\hbar2 P_-\) gives an expectation tilted toward spin-down. Units: \(p_+\) dimensionless; \(\langle S_z\rangle\) in J·s (action), correct for spin.

Example 2 — A real symmetric \(2\times2\), with a square root.

\[ A=\begin{pmatrix}3&1\\1&3\end{pmatrix}=A^\dagger,\qquad \det(A-\lambda\mathbf 1)=(3-\lambda)^2-1=0\ \Rightarrow\ \lambda=4,\,2 \]

Self-adjoint (real symmetric); the characteristic equation gives distinct eigenvalues. A

\[ e_1=\tfrac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\ (\lambda_1=4),\qquad e_2=\tfrac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}\ (\lambda_2=2) \]

Orthonormal eigenvectors, as guaranteed for a self-adjoint operator with distinct eigenvalues. B

\[ P_1=e_1e_1^\dagger=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix},\qquad P_2=e_2e_2^\dagger=\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix} \]

Outer products form the eigenprojectors. Check \(P_1+P_2=\mathbf 1\), \(P_1P_2=0\). B

\[ 4P_1+2P_2=\begin{pmatrix}2&2\\2&2\end{pmatrix}+\begin{pmatrix}1&-1\\-1&1\end{pmatrix}=\begin{pmatrix}3&1\\1&3\end{pmatrix}=A \]

Spectral resolution verified numerically. A

\[ \sqrt A=\sqrt4\,P_1+\sqrt2\,P_2=2P_1+1.414\,P_2\approx\begin{pmatrix}1.707&0.293\\0.293&1.707\end{pmatrix} \]

Functional calculus \(f(A)=\sum_k f(\lambda_k)P_k\) with \(f=\sqrt{\ \cdot\ }\); check \((\sqrt A)^2\approx A\). C

\[ A=4P_1+2P_2,\qquad \sqrt A=2P_1+\sqrt2\,P_2\approx\begin{pmatrix}1.707&0.293\\0.293&1.707\end{pmatrix} \]

Reading. The matrix is diagonal in the projector basis; taking a square root is done outcome-by-outcome on the eigenvalues, exactly what the functional calculus promises. Units: the entries of \(A\) and its eigenvalues share whatever unit \(A\) carries; \(P_k\) is dimensionless, so \(\sqrt A\) carries the square root of that unit.

Problems

Show that if \(P^2=P\) and \(P^\dagger=P\) then \(\mathbf 1-P\) is also an orthogonal projector, and identify its range and kernel.
Solution
Idempotent: \((\mathbf 1-P)^2=\mathbf 1-2P+P^2=\mathbf 1-2P+P=\mathbf 1-P\). Self-adjoint: \((\mathbf 1-P)^\dagger=\mathbf 1-P^\dagger=\mathbf 1-P\). Hence an orthogonal projector. Since \((\mathbf 1-P)v=0\iff v=Pv\iff v\in\operatorname{ran}P\), we get \(\ker(\mathbf 1-P)=\operatorname{ran}P\), and by complementarity \(\operatorname{ran}(\mathbf 1-P)=\ker P\). So \(\mathbf 1-P\) projects onto \(\ker P\) along \(\operatorname{ran}P\).
A qubit is prepared in \(|\psi\rangle=\cos\tfrac\theta2|{\uparrow}\rangle+\sin\tfrac\theta2|{\downarrow}\rangle\) with \(\theta=60^\circ\). Compute \(p_+\) for spin-up along \(z\) and the expectation \(\langle S_z\rangle\).
Solution
\(p_+=\langle\psi|P_+|\psi\rangle=\cos^2\tfrac\theta2=\cos^2 30^\circ=\big(\tfrac{\sqrt3}{2}\big)^2=0.75\); then \(p_-=\sin^2 30^\circ=0.25\). Expectation \(\langle S_z\rangle=\tfrac\hbar2(0.75-0.25)=0.25\,\hbar=2.64\times10^{-35}\ \text{J·s}\).
For \(A=\begin{pmatrix}2&0\\0&5\end{pmatrix}\), write the two eigenprojectors, verify the resolution of the identity, and compute \(e^{A}\) via the functional calculus.
Solution
Eigenvalues \(2,5\) with \(P_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}\), \(P_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}\). Check \(P_1+P_2=\mathbf 1\), \(2P_1+5P_2=A\). Then \(e^A=e^2P_1+e^5P_2=\begin{pmatrix}e^2&0\\0&e^5\end{pmatrix}\approx\begin{pmatrix}7.389&0\\0&148.41\end{pmatrix}\).
Give an explicit real \(2\times2\) matrix that is idempotent but not self-adjoint, and show its range and kernel are not orthogonal.
Solution
Take \(P=\begin{pmatrix}1&1\\0&0\end{pmatrix}\). Then \(P^2=\begin{pmatrix}1&1\\0&0\end{pmatrix}=P\), but \(P^\dagger=P^{\top}=\begin{pmatrix}1&0\\1&0\end{pmatrix}\ne P\). Range is \(\operatorname{span}(1,0)\); kernel solves \(x+y=0\), i.e. \(\operatorname{span}(1,-1)\). Their inner product \((1,0)\cdot(1,-1)=1\ne0\), so this is an oblique projector onto the \(x\)-axis along the line \(y=-x\).
A self-adjoint operator on \(\mathbb R^3\) has spectrum \(\{1,1,4\}\) (eigenvalue \(1\) doubly degenerate), with degenerate eigenspace \(\operatorname{span}\{e_1,e_2\}\) and \(\lambda=4\) eigenvector \(e_3\). Write the spectral resolution and compute \(p(\lambda=1)\) for \(|\psi\rangle=\tfrac1{\sqrt3}(1,1,1)\).
Solution
Degenerate projector \(P_1=e_1e_1^{\top}+e_2e_2^{\top}=\operatorname{diag}(1,1,0)\); \(P_2=\operatorname{diag}(0,0,1)\). Resolution \(A=1\cdot P_1+4\cdot P_2\), with \(P_1+P_2=\mathbf 1\). Born weight sums over the whole eigenspace: \(p(\lambda=1)=\langle\psi|P_1|\psi\rangle=|\langle e_1|\psi\rangle|^2+|\langle e_2|\psi\rangle|^2=\tfrac13+\tfrac13=\tfrac23\approx0.667\). (And \(p(\lambda=4)=\tfrac13\), summing to \(1\).)