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For a normalized state and a self-adjoint operator, the expectation value and dispersion areUse the physics convention that the inner product is conjugate-linear in its first argument, and assume the displayed products and second moments are defined. Self-adjointness gives $\overline{⟨\psi |A\psi ⟩}=⟨A\psi |\psi ⟩=⟨\psi |A\psi ⟩$, so the expectation is real. MoreoverIt vanishes exactly when $\psi$ is an eigenstate, with eigenvalue its expectation.

The product rule gives $XPf=-i\hslash x{f}^{\prime }$ and $PXf=-i\hslash (f+x{f}^{\prime })$, hence $\boxed{[X,P]f=i\hslash f}$ on their common differentiable domain. For general operators, put $z=⟨A\psi |B\psi ⟩$. ThenThe Cauchy-Schwarz inequality yieldsApply this to centered position and momentum operators, whose commutator is unchanged. Their squared norms are the variances, givingEquality requires both proportionality in Cauchy-Schwarz and a purely imaginary inner product. Thus for real means $x_0,p_0$,with real $\gamma$. Solving this first-order equation givesNormalizability on the real line forces $\gamma >0$. Writing ${\sigma }^2=\hslash /(2\gamma )$, all normalized equality states are thereforeThey have $\Delta X=\sigma$ and $\Delta P=\hslash /(2\sigma )$. A nonconstant quadratic phase would increase the product, so the full family here has a real positive Gaussian width and only a linear spatial phase.

Use an expansion for the eigenstate continuing the nondegenerate level, with intermediate normalization $⟨{\psi }_0,\psi (ϵ)⟩=1$:Insert these expressions into the eigenvalue equation. At first order,Taking the inner product with ${\psi }_0$ eliminates the left side by self-adjointness. ThusFor $j\ne 0$, projection onto ${\psi }_j$ instead givesAt second order, projecting the eigenvalue equation onto ${\psi }_0$ gives $E^{(2)}=⟨{\psi }_0,H^{\prime }\chi ⟩$. Expand $\chi$ in the orthonormal energy basis and use self-adjointness of $H^{\prime }$:Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.

For the cosine perturbation of the ground level in a one-dimensional box, write $u=\pi x/a$. The product-to-sum identity givesso, with the given common normalization of the sine eigenfunctions,The diagonal matrix element is $-ϵ/2$, and the only nonzero off-diagonal matrix element out of the ground state is $⟨{\varphi }_3,V{\varphi }_1⟩=ϵ/2$. Let $E_{\mathrm{box}}={\pi }^2{\hslash }^2/(2m{a}^2)$. Since ${\mathcal{E}}_3=9E_{\mathrm{box}}$, the ground energy isThe second-order term is negative, as expected for a nondegenerate ground level: every excited-quantum state denominator is negative. The first-order shift is also negative for positive $ϵ$, because the ground-quantum state density is largest near the center, where this perturbing potential is negative.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.

A normalized quantum state satisfies $⟨\psi ,\psi ⟩=1$. Its expectation value is $\boxed{{\mathbb{E}}_{\psi }(A)=⟨\psi ,A\psi ⟩}$. The conjugate symmetry of the inner product and self-adjointness giveso the expectation value is real. If $A\varphi =\lambda \varphi$ with $\varphi \ne 0$, normalize $\varphi$ by dividing by its norm. ThenEvery eigenvalue of a self-adjoint operator is real.

The commutator of two linear operators is $\boxed{[A,B]=AB-BA}$. Write $T^{†}$ for the adjoint operator, using the inner product convention linear in its second argument. Since $A^{†}=A$ and $B^{†}=B$,Thus the commutator is anti-self-adjoint. With $C=-i[A,B]$, conjugation of the scalar givesAll products are understood on the stated common operator domain; for physical unbounded linear operators, the state must lie in the relevant product domains and have the finite moments used below.

The squared norm is obtained by moving the adjoint operator to the other side of the inner product:For an energy eigenstate, this isSince a squared norm is nonnegative, every nonzero energy eigenstate has $E\ge \hslash \omega /2$. Moreover $a_{-}\psi =0$ exactly when $E=\hslash \omega /2$.

The operator $R$ is the midpoint parity operator about $a/2$. The angle-subtraction identity givesConsequently $HR{\psi }_n=RH{\psi }_n=E_n(-1{)}^{n+1}{\psi }_n$. The sine functions form a complete orthonormal basis of $L^2(0,a)$, so the simultaneous diagonal action implies commutation on the Hamiltonian operator's operator domain. More directly, $R$ preserves the Dirichlet boundary conditions and $(R\psi {)}^{\prime \prime }(x)={\psi }^{\prime \prime }(a-x)$; hence $HR\psi =RH\psi$ whenever $H\psi$ is defined. This checks the domain as well as the differential expressions.

For arbitrary $f,g\in L^2(0,a)$, use $y=a-x$ in the inner product:Also $\Vert Rf\Vert =\Vert f\Vert$, so $R$ is bounded and defined everywhere. The identity therefore proves $\boxed{R^{\ast }=R}$, with no extra domain restriction. In particular $R$ is a self-adjoint operator; $R^2=I$ also makes it a unitary operator.

The requested energy eigenvalue is $E_2$. By the Born rule, its probability amplitude is the inner product $c_2=⟨{\psi }_2,\Psi (0)⟩$. Put $u=\pi x/a$ and use $e^{iu}\sin u=\frac{1}{2}\sin 2u+i{\sin }^{2}u$:The first integral is $\pi /2$; the second is zero under $u\mapsto \pi -u$. ThereforeFor the time-independent Hamiltonian operator, expand in its orthonormal basis of energy eigenstates:Unitary time evolution changes $c_2$ to $c_2{e}^{-i{E}_{2}t/\hslash }$, whose modulus is unchanged. Thus the same energy measurement probability holds at every time, even though relative quantum phases and other measurements can change.

Use the normalized $L_3$ eigenstates from part (ii). The quantum state found in part (iii) has the decompositionThere is no ${\chi }_0$ component. Equivalently, its inner product with ${\chi }_0$ contains ${\int }_0^{2\pi }\cos \varphi d\varphi =0$. The Born rule therefore givesFor completeness the only outcomes are $L_3=\pm \hslash$, each with probability $1/2$. Thus $⟨L_3⟩=0$ does not mean that zero is a possible result in this state: its quantum uncertainty in $L_3$ is $\hslash$, even though $L_1$ is exactly zero.

The common Fock vacuum obeys $N\psi ={\sum }_j{a}_j^{\ast }(a_j\psi )=0$. Put $B=(a_1^{\ast }{)}^{n_1}(a_2^{\ast }{)}^{n_2}$. By the commutator product rule and part (a),Moving $N$ through $B$ therefore givesDividing by $\sqrt{n_1!n_2!}$ provesThe constructed number state is nonzero: the relation $a_j(a_j^{\ast }{)}^r\psi =r(a_j^{\ast }{)}^{r-1}\psi$ follows from $[a_j,a_j^{\ast }]=I$ and $a_j\psi =0$. Commuting different modes and successively moving each creation operator across the inner product yieldsThus the given factorial denominator also gives unit norm, as expected for a number state.

In the ordered orthonormal basis $({\psi }_1,{\psi }_2)$, the Hamiltonian operator has matrixwhere ${\sigma }_1$ is a Pauli matrix. Its symmetric and antisymmetric eigenvectors giveThe energy eigenstates ${\chi }_{\pm }$ are normalized and orthogonal; each energy eigenvalue is nondegenerate. Their full stationary states are $e^{-i\omega t}{\chi }_{+}$ and $e^{i\omega t}{\chi }_{-}$. Since they form an orthonormal basis, the general solution to the Time-dependent Schrödinger equation isFor a normalized quantum state, $|c_{+}{|}^2+|c_{-}{|}^2=1$; the constants are determined by its initial inner products with ${\chi }_{\pm }$.

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are

Use metric signature $(+---)$ and four-velocities $U_i^a={\gamma }_{i}c(1,{\mathbf{\beta }}_i)$, with ${\gamma }_i=(1-{\beta }_i^2{)}^{-1/2}$. Their Lorentz invariant inner product isIn the rest frame of either particle this same scalar is the relative Lorentz factor ${\gamma }_{\mathrm{rel}}$. HenceLet $d={\mathbf{\beta }}_1\cdot {\mathbf{\beta }}_2$. The numerator after combining terms isUse $|{\mathbf{\beta }}_1\times {\mathbf{\beta }}_2{|}^2={\beta }_1^2{\beta }_2^2-d^2$ to obtainThe two rest frames agree on the relative speed, although their velocity vectors are expressed in different spatial frames. Checks include equal velocities giving zero, collinear velocities giving $|{\beta }_1-{\beta }_2|/|1-{\beta }_1{\beta }_2|$, and low speeds reducing to the Galilean relative speed. For timelike particles the four-velocity inner product also ensures $0\le {\beta }_{\mathrm{rel}}<1$.

Ordinary component differences at neighboring points depend on how the local bases vary. The covariant derivative corrects for that variation:Each upper tensor index adds a connection term and each lower index subtracts one. A scalar has ${\nabla }_{a}f={\partial }_{a}f$, while the derivative of a tensor transforms as a tensor with one additional covariant index. The Christoffel symbols themselves are not tensor components; their inhomogeneous transformation cancels that of the ordinary derivative.

For the Levi-Civita connection,These express metric compatibility and zero torsion. Along $x^a(\lambda )$, parallel transport is the initial value problemIt preserves inner products for the Levi-Civita connection. Parallel transport of the tangent itself gives the affinely parametrized geodesic equation, ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$.

Transport generally depends on the path. On a round sphere, transport a tangent vector around the triangle bounded by two meridians $90$ degrees apart and the intervening equatorial arc. All three sides are geodesics; the enclosed spherical area is $\pi a^2/2$. The returning vector is rotated by $90$ degrees, with sign set by traversal orientation. Locally this holonomy is controlled by curvature, even though the connection can be made zero at a single point. A vector can therefore have changing coordinate components while remaining parallel, and constant components do not generally mean parallel transport.

This construction compares directions at separate points through a chosen connection, rather than assuming that a curved manifold supplies a path-independent comparison.^[ref]

An energy measurement selects the ground state with probability equal to the squared inner product with ${\psi }_1$. Using $N=\sqrt{3}/L^{3/2}$ and integration by parts,ThusThe full Fourier sine series has $c_n=(-1{)}^{n+1}\sqrt{6}/(n\pi )$, so ${\sum }_n|c_n{|}^2=(6/{\pi }^2){\sum }_n{n}^{-2}=1$, an independent wavefunction normalization check. However $E_n|c_n{|}^2=3{\hslash }^2/(m{L}^2)$ for every $n$, hence ${\sum }_n{E}_n|c_n{|}^2$ diverges. This explicitly verifies the infinite-energy expectation qualification in part (b), without invalidating the individual projection probabilities.

For an active clockwise rotation of column vectors in the usual oriented $xy$ plane, takeDirect matrix multiplication givesThus the rotation matrix is an orthogonal matrix, preserving the Euclidean inner product, with determinant $+1$.

For a passive one-dimensional Lorentz boost, use rapidity $\eta$, $\beta =\tanh \eta$:This is not $I$ unless $\eta =0$. Hence a nontrivial Lorentz boost is not an orthogonal matrix for the Euclidean inner product. Instead it preserves the Minkowski metric:by ${\cosh }^2\eta -{\sinh }^2\eta =1$. Thus it is pseudo-orthogonal. The zero-rapidity identity is the explicit exception to the assertion that a Lorentz boost is not orthogonal.

As written, the assertion needs a nonzero-vector assumption: the singleton set containing the zero vector is mutually orthogonal but is linearly dependent. For the linear independence of nonzero orthogonal vectors, suppose ${\sum }_{i=1}^k{\alpha }_i{q}_i=0$. Taking the inner product with $q_j$ givesSince $q_j\ne 0$, positive definiteness gives $\Vert q_j{\Vert }^2>0$, hence ${\alpha }_j=0$. This holds for every $j$, proving linear independence under the necessary assumption.

In the column convention used for QR factorization, an $m\times k$ real matrix has orthonormal columns when $Q^{T}Q=I_k$. This says that each column has unit norm and distinct columns have zero inner product. If $Q$ is square, then $Q^{T}Q=I$ provides a left inverse; a square matrix with a left inverse is invertible, so $Q^{-1}=Q^T$, and also $Q{Q}^T=I$. Such a square matrix is an orthogonal matrix. A rectangular matrix with orthonormal columns generally has no two-sided inverse.

The first two columns are already orthogonal and have norms 15 and 5. Their normalized versions are $q_1=(3,4,0,0{)}^T/5$ and $q_2=(0,0,4,-3{)}^T/5$. Their inner products with the third column are 10 and 5. Subtracting these projections leaves $(20,-15,0,0{)}^T$, whose norm is 25. Thus an economy-size QR factorization isDirect multiplication gives $Q^{T}Q=I_3$ and $QR=A$. If a full square $Q$ is required, append $q_4=(0,0,3,4{)}^T/5$ and append a zero fourth row to $R$; the product is unchanged and the completed $Q$ is an orthogonal matrix.

Collect the orthonormal vectors as columns and compute $R=Q^{T}A$. This gives the thin QR decompositionThe inner products in part (e) prove $Q^{T}Q=I_3$; the column identities $a_1=2q_1$, $a_2=4q_1+2q_2$, $a_3=2q_1+2q_2+2q_3$ prove $A=QR$.