Spectral Theorem for Self-Adjoint Operators

Derivation

Spectral Theorem for Self-Adjoint Operators

D-082 Home PU-104 Threads symmetry · matter · chance Depends on Properties of Hermitian Operators, Gram–Schmidt Orthonormalization, Diagonalizability and the Eigenbasis Criterion

Statement

Let \( \hat{A} \) be a self-adjoint operator on a finite-dimensional complex inner-product space \( V \) with \( \dim V = n \), so that \( \hat{A}=\hat{A}^{\dagger} \). Then \( \hat{A} \) has real eigenvalues, its eigenvectors belonging to distinct eigenvalues are orthogonal, and \( V \) possesses an orthonormal basis \( \{|e_1\rangle,\dots,|e_n\rangle\} \) consisting entirely of eigenvectors of \( \hat{A} \). Equivalently, the unitary \( \hat{U} \) whose columns are those eigenvectors diagonalizes \( \hat{A} \): \( \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n) \), and \( \hat{A}=\sum_{k}\lambda_k\,|e_k\rangle\langle e_k| \).

Why it matters

The spectral theorem is the mathematical backbone of quantum measurement: every observable is represented by a self-adjoint operator, and this theorem guarantees that its possible measured values (the eigenvalues) are real numbers and that the corresponding states (eigenvectors) form a complete orthonormal set. Completeness lets any state be expanded as a superposition of measurement outcomes; orthogonality makes those outcomes mutually exclusive. Without it the Born rule and the collapse postulate would have no rigorous footing.

Beyond quantum theory the result underwrites principal-axis analysis of the inertia and stress tensors, normal-mode decomposition of coupled oscillators, and the diagonalization of quadratic forms throughout classical mechanics. Wherever a Hermitian or real symmetric matrix appears, this theorem says it can be rotated into a diagonal, decoupled form.

Assumptions

The scalar field is \( \mathbb{C} \) (or \( \mathbb{R} \) for real symmetric matrices).Over a general field the characteristic polynomial need not split, so eigenvalues may fail to exist and the induction below cannot start.

The space is finite-dimensional.In infinite dimensions a self-adjoint operator may have purely continuous spectrum and no normalizable eigenvectors at all (e.g. \( \hat{x} \), \( \hat{p} \)); one then needs the projection-valued-measure form of the theorem.

Self-adjointness, not mere symmetry of the matrix, holds: \( \langle\hat{A}u|v\rangle=\langle u|\hat{A}v\rangle \) for all \( u,v \).A complex-symmetric matrix (\( A^{T}=A \) but \( A^{\dagger}\neq A \)) such as \( \left(\begin{smallmatrix}0&i\\ i&0\end{smallmatrix}\right) \) is not self-adjoint; its eigenbasis need not be orthogonal and unitary diagonalization fails.

The inner product is positive-definite.With an indefinite metric (as in Minkowski space or a Krein space) a "self-adjoint" operator can have complex-conjugate eigenvalue pairs or non-trivial Jordan blocks; the reality and orthogonality arguments both collapse.

Derivation

\[ \det(\hat{A}-\lambda\hat{I})=0 \;\Rightarrow\; \exists\,\lambda_1\in\mathbb{C},\ |v\rangle\neq 0:\ \hat{A}|v\rangle=\lambda_1|v\rangle \]

The characteristic polynomial has degree \( n\geq 1 \); by the fundamental theorem of algebra it has at least one complex root, hence at least one eigenpair. A

\[ \lambda_1\langle v|v\rangle=\langle v|\hat{A}v\rangle=\langle\hat{A}v|v\rangle=\overline{\lambda_1}\,\langle v|v\rangle \]

Insert \( \hat{A}|v\rangle=\lambda_1|v\rangle \), then move \( \hat{A} \) across the inner product using \( \hat{A}^{\dagger}=\hat{A} \) (hermitian-operator-properties); on the right it acts to the left as its adjoint. B

\[ (\lambda_1-\overline{\lambda_1})\langle v|v\rangle=0,\quad \langle v|v\rangle>0 \;\Rightarrow\; \lambda_1=\overline{\lambda_1}\in\mathbb{R} \]

Positive-definiteness gives \( \langle v|v\rangle\neq 0 \), so the scalar factor must vanish: every eigenvalue is real. A

\[ (\lambda-\mu)\langle\phi|\psi\rangle=\langle\hat{A}\phi|\psi\rangle-\langle\phi|\hat{A}\psi\rangle=0 \]

For eigenpairs \( \hat{A}|\phi\rangle=\lambda|\phi\rangle \), \( \hat{A}|\psi\rangle=\mu|\psi\rangle \) with \( \lambda,\mu \) real, self-adjointness gives this identity; if \( \lambda\neq\mu \) then \( \langle\phi|\psi\rangle=0 \): distinct-eigenvalue eigenvectors are orthogonal. B

\[ |e_1\rangle=\frac{|v\rangle}{\sqrt{\langle v|v\rangle}},\qquad W=\{|w\rangle\in V:\langle e_1|w\rangle=0\}=\big(\mathrm{span}\,|e_1\rangle\big)^{\perp} \]

Normalize the first eigenvector and take the orthogonal complement of the line it spans; the inner product gives \( V=\mathrm{span}\{|e_1\rangle\}\oplus W \) with \( \dim W=n-1 \). B

\[ |w\rangle\in W \;\Rightarrow\; \langle e_1|\hat{A}w\rangle=\langle\hat{A}e_1|w\rangle=\lambda_1\langle e_1|w\rangle=0 \;\Rightarrow\; \hat{A}|w\rangle\in W \]

Self-adjointness plus reality of \( \lambda_1 \) show \( W \) is \( \hat{A} \)-invariant: an eigenvector never leaks into its orthogonal complement. This invariance is the crux of the whole proof. C

\[ \hat{A}\big|_{W}:W\to W\ \text{ is self-adjoint on the }(n-1)\text{-dimensional }W \]

The inner product restricted to \( W \) is still positive-definite and \( \langle\hat{A}u|w\rangle=\langle u|\hat{A}w\rangle \) for \( u,w\in W \), so the restriction satisfies every hypothesis on a strictly smaller space. C

\[ \text{Induct on }n:\quad W \text{ yields } |e_2\rangle,\dots,|e_n\rangle\in W \;\Rightarrow\; \{|e_1\rangle,\dots,|e_n\rangle\}\ \text{orthonormal eigenbasis} \]

The base case \( n=1 \) is immediate; each step peels off one orthonormal eigenvector (automatically orthogonal to \( |e_1\rangle \) since it lies in \( W \)), terminating after \( n \) steps. B

\[ \text{Within a degenerate eigenspace } E_\lambda\ (\dim>1):\ \text{apply Gram–Schmidt to any basis} \]

If an eigenvalue repeats, gram-schmidt-orthonormalization produces an orthonormal basis of \( E_\lambda \); a subspace of one eigenvalue is closed under linear combination, so every produced vector remains an eigenvector. C

\[ \hat{U}=\big[\,|e_1\rangle\ |e_2\rangle\ \cdots\ |e_n\rangle\,\big],\qquad (\hat{U}^{\dagger}\hat{U})_{jk}=\langle e_j|e_k\rangle=\delta_{jk} \]

Stacking the orthonormal eigenvectors as columns, the orthonormality relations are exactly \( \hat{U}^{\dagger}\hat{U}=\hat{I} \): \( \hat{U} \) is unitary. A

\[ (\hat{U}^{\dagger}\hat{A}\hat{U})_{jk}=\langle e_j|\hat{A}|e_k\rangle=\lambda_k\langle e_j|e_k\rangle=\lambda_k\delta_{jk} \;\Rightarrow\; \hat{A}=\sum_{k}\lambda_k|e_k\rangle\langle e_k| \]

Apply \( \hat{A} \) to each eigen-column and use orthonormality; off-diagonal entries vanish, giving \( \hat{\Lambda} \) (diagonalizability-eigenbasis-criterion, now realized by a unitary) and the spectral decomposition. B

Result

\[ \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n),\quad \lambda_k\in\mathbb{R},\quad \hat{U}^{\dagger}\hat{U}=\hat{I},\quad \hat{A}=\sum_{k=1}^{n}\lambda_k\,|e_k\rangle\langle e_k| \]

Reading. A self-adjoint operator is nothing more than a set of real numbers \( \lambda_k \) attached to mutually orthogonal directions \( |e_k\rangle \); it stretches each direction by its eigenvalue and couples nothing. The unitary \( \hat{U} \) is the change of basis into those principal directions, and the spectral sum writes \( \hat{A} \) as a real-weighted sum of orthogonal projectors \( \hat{P}_k=|e_k\rangle\langle e_k| \) obeying \( \hat{P}_j\hat{P}_k=\delta_{jk}\hat{P}_k \) and \( \sum_k\hat{P}_k=\hat{I} \).

Units check. The projectors \( |e_k\rangle\langle e_k| \) are built from normalized (dimensionless) states, so each \( \lambda_k \) carries exactly the physical units of the observable \( \hat{A} \) represents — energy in joules for a Hamiltonian, \( \hbar \) for a spin component. \( \hat{U} \) is dimensionless (it preserves the dimensionless norm \( \langle\psi|\psi\rangle=1 \)), so both sides of \( \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda} \) carry the units of \( \hat{A} \), and \( \sum_k\hat{P}_k=\hat{I} \) is dimensionless as an identity operator must be.

Limiting cases

Non-degenerate spectrum (all \( \lambda_k \) distinct): each eigenspace is one-dimensional, the eigenbasis is unique up to phases, and no Gram–Schmidt is needed — step 9 is vacuous.
\( \hat{A}=c\hat{I} \) (multiple of the identity): every vector is an eigenvector with \( \lambda=c \); any orthonormal basis diagonalizes \( \hat{A} \), so \( \hat{U} \) is completely free.
\( \hat{A} \) a projector (\( \hat{A}^2=\hat{A} \)): eigenvalues collapse to \( \{0,1\} \), so \( \hat{A} \) is itself one of the spectral projectors.
Real symmetric matrix (\( V=\mathbb{R}^n \), \( \hat{A}=\hat{A}^{T} \)): the same proof runs, eigenvalues are automatically real, and \( \hat{U} \) becomes a real orthogonal matrix \( O \) with \( O^{T}\hat{A}O=\hat{\Lambda} \) — the principal-axis theorem.
\( \hat{A} \) already diagonal: the standard basis is the eigenbasis and \( \hat{U}=\hat{I} \); the theorem returns the input unchanged.

Breaks when

Non-normal operators. If \( \hat{A}\hat{A}^{\dagger}\neq\hat{A}^{\dagger}\hat{A} \), no orthonormal eigenbasis exists and \( \hat{A} \) may not be diagonalizable at all. The defective matrix \( \left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right) \) has only a one-dimensional eigenspace and reaches Jordan form, not diagonal form. Normality is the true necessary-and-sufficient condition for unitary diagonalizability; self-adjointness is a special case.
Infinite dimensions with continuous spectrum. For \( \hat{p}=-i\hbar\,d/dx \) on \( L^2(\mathbb{R}) \), the "eigenfunctions" \( e^{ikx} \) are not square-integrable and lie outside the Hilbert space. The sum \( \sum_k\lambda_k\hat{P}_k \) must be replaced by a spectral integral \( \int\lambda\,d\hat{E}(\lambda) \) over a projection-valued measure.
Merely symmetric but not self-adjoint operators. An operator can satisfy \( \langle\phi|\hat{A}\psi\rangle=\langle\hat{A}\phi|\psi\rangle \) on a domain yet have non-trivial deficiency indices; without a self-adjoint extension it possesses no complete real eigenbasis.
Indefinite inner product. Under a Minkowski metric the reality argument in step 3 uses \( \langle v|v\rangle>0 \), which no longer holds; pseudo-Hermitian operators can then have complex-conjugate eigenvalue pairs or non-trivial Jordan structure.

Failure modes

"Symmetric" \( \neq \) "Hermitian". Students check \( A_{ij}=A_{ji} \) on a complex matrix and conclude self-adjointness; the correct condition is \( A_{ij}=\overline{A_{ji}} \). A complex symmetric matrix like \( \left(\begin{smallmatrix}0&i\\ i&0\end{smallmatrix}\right) \) fails the theorem.
Forgetting to orthogonalize degenerate eigenvectors. Within a repeated eigenvalue any basis of the eigenspace is a valid set of eigenvectors, but two such vectors need not be orthogonal; skipping Gram–Schmidt yields a non-unitary \( \hat{U} \).
Normalizing with the wrong inner product. Using the real dot product \( u\cdot v \) instead of the Hermitian \( u^{\dagger}v \) when normalizing complex eigenvectors gives a norm that is complex or wrong, so \( \hat{U}^{\dagger}\hat{U}\neq\hat{I} \).
Assuming eigenvectors are orthogonal automatically. Orthogonality of distinct-eigenvalue eigenvectors is a theorem to be used (step 4), not an axiom; for degenerate eigenvalues it must be imposed by hand.
Confusing \( \hat{U}^{\dagger}\hat{A}\hat{U} \) with \( \hat{U}\hat{A}\hat{U}^{\dagger} \). The columns of \( \hat{U} \) are the eigenvectors, so the eigenvalues appear as \( \hat{U}^{\dagger}\hat{A}\hat{U} \); writing the conjugation backwards produces \( \hat{\Lambda} \) in the wrong basis.
Writing \( \hat{A}=\hat{S}\hat{\Lambda}\hat{S}^{-1} \) with non-unitary \( \hat{S} \). Any diagonalizable operator admits a general similarity diagonalization, but only a unitary \( \hat{U} \) preserves the inner product — and hence the probabilities — the physics relies on.

Discussion

The physical content of the theorem is that self-adjointness is precisely the condition under which an operator behaves like a collection of independent real-valued measurements along orthogonal axes. Reality of the spectrum makes the eigenvalues admissible as physical outcomes; orthogonality makes those outcomes distinguishable; completeness makes the set of outcomes exhaustive. These three facts are exactly what the Born rule requires, which is why observables in quantum mechanics are postulated to be self-adjoint rather than merely linear.

The invariant-subspace argument (steps 6–8) is the heart of the matter and generalizes far beyond finite dimensions. It says an eigenvector never leaks probability into its orthogonal complement, so the problem decouples into a one-dimensional piece and a smaller self-adjoint problem. Iterating this decoupling is what block-diagonalizes coupled systems into normal modes, and it is the finite-dimensional shadow of the spectral projections that appear in the general Hilbert-space theorem. The unitarity of \( \hat{U} \) is equally essential: unitary maps preserve inner products and therefore probabilities, so a change of basis that diagonalized \( \hat{A} \) but distorted lengths would misrepresent the physical amplitudes.

Unitary diagonalizability characterizes exactly the normal operators (those with \( \hat{A}\hat{A}^{\dagger}=\hat{A}^{\dagger}\hat{A} \)), of which self-adjoint operators are the case with real spectrum, unitary operators the case with unit-modulus spectrum, and skew-adjoint operators the case with imaginary spectrum. The spectral theorem for self-adjoint operators is therefore one facet of a single structural fact: commuting with one's own adjoint is equivalent to admitting an orthonormal eigenbasis. Commuting self-adjoint operators, \( [\hat{A},\hat{B}]=0 \), share a common eigenbasis and can be simultaneously diagonalized — the mathematical origin of compatible observables and complete sets of commuting observables. In infinite dimensions the clean eigenbasis statement dissolves into \( \hat{A}=\int_{\sigma(\hat{A})}\lambda\,d\hat{E}(\lambda) \) over a projection-valued measure on the spectrum \( \sigma(\hat{A})\subset\mathbb{R} \), and von Neumann's deficiency-index theory governs when a symmetric operator is genuinely self-adjoint.

Common misconceptions. The theorem does not say the eigenbasis is unique — within a degenerate eigenspace any orthonormal frame works, and even in the non-degenerate case each eigenvector carries an arbitrary phase; only the eigenspace projectors \( \hat{P}_\lambda \) are unique. Nor does diagonalizability alone imply self-adjointness: many non-normal matrices are diagonalizable, but only by a non-unitary similarity transform with no measurement interpretation. And real eigenvalues alone do not force self-adjointness — \( \left(\begin{smallmatrix}1&1\\0&2\end{smallmatrix}\right) \) has real eigenvalues but no orthogonal eigenbasis.

Worked examples

\[ \hat{A}=\begin{pmatrix}2&1-i\\ 1+i&3\end{pmatrix},\qquad \hat{A}=\hat{A}^{\dagger}\ \checkmark\quad(A_{21}=\overline{A_{12}}) \]

A genuinely complex Hermitian \( 2\times 2 \) matrix; confirm self-adjointness before applying the theorem. A

\[ \det(\hat{A}-\lambda\hat{I})=(2-\lambda)(3-\lambda)-|1-i|^2=\lambda^2-5\lambda+6-2=\lambda^2-5\lambda+4 \]

Form the characteristic polynomial, using \( |1-i|^2=2 \). A

\[ \lambda=\frac{5\pm\sqrt{25-16}}{2}=\frac{5\pm 3}{2}\;\Rightarrow\;\lambda_1=4,\ \lambda_2=1\quad(\text{both real }\checkmark) \]

Solve the quadratic; reality of both roots confirms self-adjointness numerically. A

\[ (\hat{A}-4\hat{I})|e_1\rangle=0:\ -2a+(1-i)b=0\;\Rightarrow\;b=(1+i)a,\quad |e_1\rangle=\tfrac{1}{\sqrt3}\begin{pmatrix}1\\1+i\end{pmatrix} \]

Solve for the \( \lambda_1=4 \) eigenvector (\( a=1 \)) and normalize with the Hermitian norm \( \|v\|^2=1+|1+i|^2=3 \). B

\[ |e_2\rangle=\tfrac{1}{\sqrt3}\begin{pmatrix}1-i\\-1\end{pmatrix},\qquad \langle e_1|e_2\rangle=\tfrac13\big(\overline{1}(1-i)+\overline{(1+i)}(-1)\big)=\tfrac13\big((1-i)-(1-i)\big)=0\ \checkmark \]

Repeat for \( \lambda_2=1 \), normalize, and verify orthogonality using conjugation on the first factor. B

\[ \hat{A}=4\,|e_1\rangle\langle e_1|+1\,|e_2\rangle\langle e_2|,\quad \hat{U}=\tfrac{1}{\sqrt3}\begin{pmatrix}1&1-i\\1+i&-1\end{pmatrix},\quad \hat{U}^{\dagger}\hat{A}\hat{U}=\begin{pmatrix}4&0\\0&1\end{pmatrix} \]

Reading. Despite complex entries, \( \hat{A} \) stretches by \( 4 \) along \( |e_1\rangle \) and by \( 1 \) along the orthogonal \( |e_2\rangle \); the eigenvalues are distinct so orthogonality was automatic. Units. If \( \hat{A} \) were an energy, \( \lambda_1=4\,\mathrm{J} \), \( \lambda_2=1\,\mathrm{J} \); \( \hat{U} \) and the projectors are dimensionless.

\[ \hat{H}=\varepsilon\,\hat{\sigma}_x=\varepsilon\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \hat{H}=\hat{H}^{\dagger},\ \varepsilon>0 \]

A two-level Hamiltonian (spin in a transverse field, or a symmetric double well) with \( \varepsilon \) carrying units of energy. A

\[ \det(\hat{H}-\lambda\hat{I})=\lambda^2-\varepsilon^2=0\;\Rightarrow\;\lambda_\pm=\pm\varepsilon \]

Diagonalize symbolically; eigenvalues are real and symmetric about zero, with level splitting \( 2\varepsilon \). A

\[ |+\rangle=\tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix},\quad |-\rangle=\tfrac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix},\quad \langle+|-\rangle=0 \]

Solve \( (\hat{H}\mp\varepsilon\hat{I})|\pm\rangle=0 \) and normalize; the eigenstates are the symmetric/antisymmetric combinations. A

\[ \hat{U}=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad \hat{U}^{\dagger}\hat{H}\hat{U}=\varepsilon\begin{pmatrix}1&0\\0&-1\end{pmatrix} \]

Assemble the (real orthogonal, hence unitary) eigenvector matrix; conjugation diagonalizes \( \hat{H} \). B

\[ \varepsilon=1.0\times10^{-3}\,\mathrm{eV}=1.60\times10^{-22}\,\mathrm{J}\;\Rightarrow\;\lambda_\pm=\pm1.60\times10^{-22}\,\mathrm{J},\quad 2\varepsilon=2.0\ \mathrm{meV} \]

Insert numbers at a typical tunneling-splitting scale. A

\[ \hat{H}=(+\varepsilon)\,|+\rangle\langle+|+(-\varepsilon)\,|-\rangle\langle-|,\qquad E_\pm=\pm1.60\times10^{-22}\,\mathrm{J} \]

Reading. The off-diagonal coupling that made the site basis non-diagonal is removed by rotating to the symmetric/antisymmetric eigenbasis, where the energies are simply \( \pm\varepsilon \). Units. \( \varepsilon \) carries energy (J); \( \hat{U} \) and the projectors are dimensionless, so both eigenvalues emerge in joules.

Problems

Diagonalize the real symmetric matrix \( \hat{A}=\begin{pmatrix}3&2\\2&3\end{pmatrix} \) by an orthogonal transformation.
Solution
\( \det(\hat{A}-\lambda\hat{I})=(3-\lambda)^2-4=0\Rightarrow\lambda=5,1 \). For \( \lambda=5 \): \( (3-5)x+2y=0\Rightarrow y=x\Rightarrow|e_1\rangle=\tfrac{1}{\sqrt2}(1,1)^{T} \). For \( \lambda=1 \): \( (3-1)x+2y=0\Rightarrow y=-x\Rightarrow|e_2\rangle=\tfrac{1}{\sqrt2}(1,-1)^{T} \). They are orthogonal, so \( O=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix} \) and \( O^{T}\hat{A}O=\mathrm{diag}(5,1) \).
Prove directly from \( \hat{A}^{\dagger}=\hat{A} \) that \( \langle\psi|\hat{A}|\psi\rangle \) is real for every \( |\psi\rangle \), and explain why this forces every eigenvalue to be real.
Solution
\( \overline{\langle\psi|\hat{A}|\psi\rangle}=\langle\hat{A}\psi|\psi\rangle=\langle\psi|\hat{A}^{\dagger}|\psi\rangle=\langle\psi|\hat{A}|\psi\rangle \) since \( \hat{A}^{\dagger}=\hat{A} \). A number equal to its own conjugate is real. Taking \( |\psi\rangle=|e\rangle \) a normalized eigenvector with \( \hat{A}|e\rangle=\lambda|e\rangle \) gives \( \langle e|\hat{A}|e\rangle=\lambda\langle e|e\rangle=\lambda \), so \( \lambda\in\mathbb{R} \).
Find the eigenvalues and an orthonormal eigenbasis of the Pauli matrix \( \hat{\sigma}_y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix} \), and write the unitary that diagonalizes it.
Solution
\( \det(\hat{\sigma}_y-\lambda\hat{I})=\lambda^2-1=0\Rightarrow\lambda=\pm1 \). For \( \lambda=+1 \): the row \( -x-iy=0\Rightarrow x=-iy \); take \( y=i \) so \( x=1 \), giving \( |+\rangle=\tfrac{1}{\sqrt2}(1,i)^{T} \); check \( \hat{\sigma}_y(1,i)^{T}=(-i\cdot i,\ i\cdot 1)^{T}=(1,i)^{T} \) \( \checkmark \). For \( \lambda=-1 \): \( |-\rangle=\tfrac{1}{\sqrt2}(1,-i)^{T} \). Orthogonality: \( \langle+|-\rangle=\tfrac12(\overline{1}\cdot1+\overline{i}\cdot(-i))=\tfrac12(1+(-i)(-i))=\tfrac12(1-1)=0 \) \( \checkmark \). \( \hat{U}=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\ i&-i\end{pmatrix} \), \( \hat{U}^{\dagger}\hat{\sigma}_y\hat{U}=\mathrm{diag}(1,-1) \).
The matrix \( \hat{A}=\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix} \) has a doubly-degenerate eigenvalue. Find all eigenvalues and construct an orthonormal eigenbasis, using Gram–Schmidt on the degenerate eigenspace.
Solution
Write \( \hat{A}=\hat{I}+J \) with \( J \) the all-ones matrix. \( J \) has eigenvalue \( 3 \) (eigenvector \( (1,1,1) \)) and \( 0 \) (multiplicity 2, the plane \( x+y+z=0 \)). Hence \( \hat{A} \) has \( \lambda=4 \), \( |e_1\rangle=\tfrac{1}{\sqrt3}(1,1,1)^{T} \), and \( \lambda=1 \) doubly degenerate. Pick \( a=(1,-1,0) \), \( b=(1,0,-1) \) in that plane. Gram–Schmidt: \( |e_2\rangle=\tfrac{1}{\sqrt2}(1,-1,0)^{T} \); then \( b-\langle e_2|b\rangle e_2=(1,0,-1)-\tfrac12(1,-1,0)=(\tfrac12,\tfrac12,-1) \), norm \( \sqrt{3/2} \), giving \( |e_3\rangle=\tfrac{1}{\sqrt6}(1,1,-2)^{T} \). Then \( O=[\,e_1\ e_2\ e_3\,] \) and \( O^{T}\hat{A}O=\mathrm{diag}(4,1,1) \).
For \( \hat{H}=\varepsilon\,\hat{\sigma}_x \) of Worked Example 2 with \( \varepsilon=1.0\times10^{-3}\,\mathrm{eV} \), a system starts in the site state \( |1\rangle=(1,0)^{T} \). Using the spectral decomposition, find the probability of measuring energy \( +\varepsilon \) and the time \( T \) for the state to first fully tunnel to \( |2\rangle=(0,1)^{T} \).
Solution
Expand \( |1\rangle=\tfrac{1}{\sqrt2}(|+\rangle+|-\rangle) \), so \( P(+\varepsilon)=|\langle+|1\rangle|^2=\tfrac12 \). Evolution: \( |\psi(t)\rangle=\tfrac{1}{\sqrt2}(e^{-i\varepsilon t/\hbar}|+\rangle+e^{+i\varepsilon t/\hbar}|-\rangle) \). With \( \langle2|+\rangle=\tfrac{1}{\sqrt2} \), \( \langle2|-\rangle=-\tfrac{1}{\sqrt2} \), the amplitude \( \langle2|\psi(t)\rangle=-i\sin(\varepsilon t/\hbar) \), so \( P_2(t)=\sin^2(\varepsilon t/\hbar) \). Full transfer first at \( \varepsilon T/\hbar=\pi/2 \), i.e. \( T=\dfrac{\pi\hbar}{2\varepsilon} \). Numerically \( \varepsilon=1.60\times10^{-22}\,\mathrm{J} \), \( \hbar=1.055\times10^{-34}\,\mathrm{J\,s} \): \( T=\dfrac{\pi(1.055\times10^{-34})}{2(1.60\times10^{-22})}\approx1.04\times10^{-12}\,\mathrm{s}\approx1.0\ \mathrm{ps} \).