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A pure state is a unit vector in a complex Hilbert space, with vectors differing by a global phase representing the same physical state. An observable is a self-adjoint operator, with measurement probabilities determined by its spectral projections. For vectors in the relevant operator domains,A squared norm vanishes exactly when its vector vanishes. Therefore zero dispersion is equivalent to $A\psi =E_{\psi }[A]\psi$.

In the position representation, $(X\psi )(x)=x\psi (x)$ and $(P\psi )(x)=-i\hslash {\psi }^{\prime }(x)$. On a common differentiable invariant domain,so $\boxed{[X,P]=i\hslash I}$.

For two observables set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=E_{\psi }[i[A,B]]$. For real $t$,If $b>0$, nonpositivity of its discriminant yields $\boxed{4ab\ge c^2}$. Equality holds exactly when the minimum squared norm vanishes, namelyThe degenerate case must also be included: if $B\psi =0$, then $b=c=0$ and equality holds for every $A\psi$. Thus the complete criterion is that $B\psi =0$, or that the displayed real-$t$ relation holds. This includes $A\psi =0$ by choosing $t=0$ when $b>0$.

Apply the inequality to centred operators $A=P-E_{\psi }[P]I$, $B=X-E_{\psi }[X]I$. Their squared norms are the variances and their commutator is $[P,X]=-i\hslash I$, givingAn eigenvector of either operator would have its corresponding dispersion zero, contradicting this positive lower bound if the other dispersion were finite and the stated domain conditions held. In the usual $L^2(\mathbb{R})$ representation there are in fact no normalizable exact position or momentum eigenvectors; their generalized eigenvectors are not counterexamples. The equality condition above is the uncentered second-moment commutator inequality.

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.

The position operator $X$ and momentum operator $P$ are self-adjoint operators, so the adjoint operator isUse the canonical commutation relation with the order of factors preserved:Subtracting the first product from the second provesThese creation and annihilation operators are unscaled and carry the paper's phase convention; their commutator is not the unit-normalized oscillator commutator.

If $a_{-}\psi \ne 0$, it has energy $E-\hslash \omega$, distinct from the energy of $\psi$. Eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal. Explicitly,so $⟨\psi ,a_{-}\psi ⟩=0$. If $a_{-}\psi =0$, the same conclusion is immediate.

Dividing by $\Vert \psi {\Vert }^2$ and using the definition of $a_{-}$ givesBoth expectation values are real because $P$ and $X$ are self-adjoint operators. Its real and imaginary parts therefore vanish separately:An oscillator energy eigenstate is centered in both position and momentum, despite having nonzero quantum uncertainties.

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.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.

The apparent interior identity $[H,P]=0$ cannot justify conservation here: the momentum operator does not preserve the Dirichlet boundary conditions of the Hamiltonian operator's operator domain. For sufficiently regular states, retaining the boundary terms in Ehrenfest theorem instead gives the boundary force in an infinite square well,In the strictly confined interval, $-i\hslash {\partial }_x$ with both endpoint values zero is symmetric but not itself a self-adjoint operator. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior commutator misses the wall force.

On a smooth wavefunction $f$, the position operator and momentum operator giveThus the canonical commutation relation is $\boxed{[X,P]=i\hslash I}$. Since $X$ and $P$ are self-adjoint operators and $m,\omega$ are real, the adjoint operator of $a_{-}$ isExpanding the products in the stated order is essential:Subtracting gives $\boxed{[a_{-},a_{+}]=2m\hslash \omega I}$. These identities initially hold on smooth rapidly decaying functions, which contain the quantum harmonic oscillator energy eigenstates used below.

The factorization also proves a lower bound: for a normalized energy eigenstate $H\psi =E\psi$,Therefore $E\ge E_0:=\hslash \omega /2$. The commutator product rule, applied to $H=a_{+}{a}_{-}/(2m)+E_0$, givesSo a nonzero application of the creation operator $a_{+}$ raises the energy eigenvalue by $\hslash \omega$, while the annihilation operator $a_{-}$ lowers it by the same amount.

Begin with any energy eigenstate, whose existence is assumed. Repeated lowering must terminate: otherwise it would eventually produce a nonzero energy eigenstate below $E_0$, contradicting the lower bound. More explicitly, if $(E-E_0)/(\hslash \omega )=N+r$ with integer $N\ge 0$ and $0\le r<1$, the first $N$ lowering steps are nonzero. If $r>0$, one further step is also nonzero by its positive squared norm and has energy $E_0+(r-1)\hslash \omega <E_0$, an impossibility. Thus $r=0$ and the last state satisfies $a_{-}{\psi }_0=0$. This provesEvery level exists as well: $\Vert a_{+}{\psi }_n{\Vert }^2=2m(E_n+\hslash \omega /2)=2m\hslash \omega (n+1)>0$, so repeated raising from ${\psi }_0$ never vanishes. The same factorization bounds the energy expectation of any normalized state in the one-dimensional harmonic oscillator form domain below by $E_0$, proving that the terminated state really is a ground state.

To find its probability density, solve the first-order equationIts solutions are ${\psi }_0=C_0\exp [-m\omega x^2/(2\hslash )]$. With $\alpha =m\omega /\hslash >0$, the supplied Gaussian integral givesTaking $C_0=(\alpha /\pi {)}^{1/4}$ fixes an irrelevant global phase. ConsequentlyThe ground-state energy is positive even though the potential minimum is zero: it is the oscillator's zero-point energy.

Put $\gamma =2m\hslash \omega$. If ${\psi }_n$ is a normalized energy eigenstate at $E_n$, the factorization from part (a) givesIt follows inductively that $v_n=(a_{+}{)}^n{\psi }_0$ has squared norm ${\gamma }^{n}n!$. The commutator with $H$ establishes its energy eigenvalue at each step. Henceis normalized and has $E_n=(n+\frac{1}{2})\hslash \omega$. Distinct levels are orthogonal because $H$ is a self-adjoint operator. In particular,

To decompose the specified initial wavefunction, retain this exact phase convention for the creation and annihilation operators. Their difference givesSince $a_{-}{\psi }_0=0$ and $a_{-}{a}_{+}{\psi }_0=\gamma {\psi }_0$,This quadratic excitation of an oscillator ground state, $\Psi (x,0)=C_0^{-1}{X}^2{\psi }_0$, contains only the $n=0$ and $n=2$ energy eigenstates. The printed initial function has an unspecified overall normalization, so the Born rule requires division by its squared norm. After wavefunction normalization the state isThe minus sign is consistent with the paper's convention: its $a_{+}$ is $i$ times the usual real-coordinate raising operator multiplied by $\sqrt{\gamma }$, so its ${\psi }_2$ has the opposite sign to the usual real second harmonic oscillator eigenfunction. The energy measurement probabilities are independent of this choice of phase.

Therefore the complete answer isEvery other energy has probability zero. At any later time the two coefficients only acquire their separate dynamical phases, so these probabilities remain the same. A direct check of the normalization isThis also confirms the factor of three in the squared norm of the two-component expansion.

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ is the identity operator. These are the relations for two independent pairs of creation and annihilation operators. The operator algebra in this question is understood on a common invariant operator domain, such as finite linear combinations of the number states constructed in part (c).

The angular momentum eigenstates have $J^2$ eigenvalue ${\hslash }^{2}j(j+1)$, where the angular momentum quantum number is $j=0,\frac{1}{2},1,\frac{3}{2},\dots$. For each $j$, the magnetic quantum number runs in unit steps from $-j$ to $j$. Here that magnetic quantum number is $+1/2$, so only half-integer $j$ are compatible:Integer $j$ cannot contain a $J_3$ eigenvalue $\hslash /2$. Applying the angular momentum lowering operator and using $[J_3,J_{-}]=-\hslash J_{-}$ givesThe lowered vector is nonzero, since the adjoint operator identity $J_{-}^{\ast }=J_{+}$ givesFinally, $J_3$ is a self-adjoint operator, so its eigenstates for distinct eigenvalues are orthogonal. Explicitly,and hence $\boxed{⟨\phi \mid \psi ⟩=0}$.

Since $X$ and $P$ are self-adjoint operators, the adjoint operator relation requires $c_{+}=c_{-}=c$ and $d_{+}=-d_{-}=-d$. Thus $a_{-}=cX+idP$ and $a_{+}=cX-idP$. The canonical commutation relation $[X,P]=i\hslash I$ givesMatching the $X^2$, $P^2$ and identity operator coefficients in $H=\hslash \omega (a_{+}{a}_{-}+\lambda I)$ yieldsConsequently all real solutions areThe conventional creation and annihilation operators use $\epsilon =1$. The other choice multiplies both by minus one and satisfies all the given equations. This harmless freedom is the oscillator ladder-operator sign convention; the subsequent formulas use the positive choice in both coordinates.

Recover the angular momentum operators from $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. The spin-one angular momentum matrices in the chosen descending-$J_3$ orthonormal basis areThey are self-adjoint operators, satisfy $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, and give $J^2=2{\hslash }^{2}I$, as required for $j=1$.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

For an isolated quantum system, pure states are normalized rays in a complex Hilbert space, observables are self-adjoint operators, and their probabilities obey the Born rule. A spectral projector $P_a$ has outcome probability $⟨\psi |P_a|\psi ⟩$; an ideal repeatable projective measurement has conditional state $P_a\psi /\sqrt{⟨\psi |P_a|\psi ⟩}$. The Schrödinger equation specifies continuous unitary evolution between such ideal measurements, given a Hamiltonian operator. General measurements need not have the projective state-update rule.

Composite systems use the tensor product of their state spaces. Identical particles additionally obey the bosonic or fermionic exchange symmetry appropriate to their species. The photoelectric effect, Davisson–Germer experiment and double-slit experiment motivate energy quanta, matter-wave phases and probability interference, respectively; these experiments constrain the framework but do not separately prove every abstract postulate or specify every system's Hamiltonian.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

For a normalized quantum state $\psi$ and a self-adjoint operator $H$ on a finite-dimensional Hilbert space, expand its unitary time evolution:Multiplying by its complex conjugate gives the survival probabilityThe function is even in $t$, so its odd coefficients vanish. The coefficient $(\Delta H{)}^2=⟨H^2⟩-⟨H{⟩}^2$ is the variance of the energy. This quadratic departure yields the quantum Zeno effect when ideal projective measurements repeatedly reset the quantum state.

For self-adjoint operators and an admissible pure state, the quantum covariance is the real, symmetrized second moment of two centered observables. The displayed expression uses their anticommutator; equivalently $C_{AB}=\mathrm{Re}⟨A^{\prime }\psi |B^{\prime }\psi ⟩$, which requires only the individual finite second moments. It reduces to ordinary covariance for commuting observables. For noncommuting observables, its definition does not assert that both have simultaneous sharp values drawn from a joint classical probability distribution.

Quantum mechanics describes a quantum system using a quantum state, with observables represented by self-adjoint operators. For a single particle in one dimension, a pure state can be described by a wavefunction $\psi (x,t)$: its squared modulus gives the probability density for position. The Schrödinger equation predicts how this probability amplitude evolves between interventions; measurement in quantum mechanics connects it to experimental outcomes through the Born rule. A wavefunction therefore predicts a distribution of possible positions rather than a single trajectory as in classical mechanics.

The infinite square well makes three central ideas visible with elementary sine functions. Confinement selects discrete energy eigenvalues; wavefunction normalization converts a shape into quantitative probabilities; and a quantum superposition can have a changing probability density even when each constituent energy eigenstate has a stationary one. This combination explains why quantum mechanics uses both linear algebra and differential equations: eigenfunctions describe the building blocks, while quantum phases determine how they interfere. The same tools help explain the discrete energies of chemical atoms and molecules, although their Quantum Hamiltonians require different potentials and often many particles.

In his first 1926 paper, Erwin Schrödinger formulated quantization as an eigenvalue problem for a hydrogen atom in nonrelativistic quantum mechanics. On its opening page he compared the emergence of integers with the number of nodes of a standing wave on a vibrating string. The infinite square well below realizes that appealing idea especially transparently: the allowed integers follow from the boundary conditions. This historical analogy concerns the origin of discrete energies; the Born rule supplies the statistical interpretation of the modern wavefunction. ^[ref].

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.