Inside the box, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$. The zero endpoint values eliminate the cosine term and require $ka=n\pi$. There is no nonzero zero-energy solution with both endpoints zero. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, soInsert the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take its inner product with ${\Psi }_m$. Orthonormality gives $i\hslash {\dot{c}}_m=E_m{c}_m$ for each $m$, and conversely these equations make every expansion coefficient of the residual zero. Thus $c_n(t)=c_n(0)e^{-i{E}_{n}t/\hslash }$ in the usual Hilbert-space or strong-solution class appropriate to the expansion.

For the specified state,The modes are real, so the interference term is a cosine of the energy difference. Writing $s_n=\sin (n\pi x/a)$,Its integral remains one because the cross term integrates to zero. The probabilities of the two energy outcomes are fixed at $9/25$ and $16/25$, even though the position probability density oscillates.

Differentiate the Lagrangian with respect to the generalized velocities:The Legendre transform cancels the vector-potential energy term, giving $H=p\cdot \dot{x}-L=m|\dot{x}{|}^2/2+V$. Eliminating velocity with $m\dot{x}=p-eA$ therefore yieldsFor the given gauge, set ${\pi }_1=p_1+eB{x}_2/2$, ${\pi }_2=p_2-eB{x}_1/2$, ${\pi }_3=p_3$. Hamilton's equations give ${\dot{x}}_j={\pi }_j/m$, ${\dot{p}}_1=(eB/2){\dot{x}}_2$, ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, ${\dot{p}}_3=0$. Hence the conserved quantities areThe two transverse combinations are components of magnetic pseudomomentum. The printed conserved combinations omit the electric charge factor $e$. With the stated vector potential and general electric charge they need the factors displayed above; as printed they hold only if $e=1$ or if electric charge is already absorbed into the field convention.

The initial data give $K_1=K_2=p_3=0$. Thus $p_1=eB{x}_2/2$, $p_2=-eB{x}_1/2$ throughout, andSolving with the initial point givesConsequently $x_1^2+x_2^2=a^2$: the guiding centre is the origin and the signed cyclotron frequency fixes the rotation direction. The energy is the constant $e^2{B}^2{a}^2/(2m)$.

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.

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.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

For a nonzero quantum state $\psi$ and a self-adjoint operator $A$, the expectation value and standard deviation of its measurement areThe squared deviation is also called the variance of the observable. All displayed products are understood on quantum states for which the relevant moments exist. Self-adjointness gives $\overline{⟨\psi ,A\psi ⟩}=⟨A\psi ,\psi ⟩=⟨\psi ,A\psi ⟩$, so $a$ is real. Expanding the squared norm gives the variance expression. A norm is nonnegative and is zero exactly for the zero vector, hence ${\Delta }_{\psi }(A)=0$ precisely when $A\psi =a\psi$. Thus zero dispersion is equivalent to being an eigenstate.

For bounded $H$, the operator exponential is the norm-convergent power seriesFor a general unbounded self-adjoint Hamiltonian operator, the spectral theorem defines the same notation byThis definition does not require the power series to converge on every quantum state. complex conjugation of the spectral multiplier gives $U_t^{\ast }=U_{-t}$, and multiplication of the phases gives $U_{-t}{U}_t=I$. Thus $U_t$ is unitary, preserves the quantum state norm, and has $U_0=I$.

For initial data in the domain of $H$, differentiating the spectral expression yields ${\partial }_t{U}_t{\psi }_0=-(i/\hslash )H{U}_t{\psi }_0$. Consequentlysolves the Schrödinger equation with the prescribed initial quantum state. Uniqueness follows by differentiating $U_{-t}{\psi }_t$: the two terms cancel, leaving the constant ${\psi }_0$.

In the two-dimensional energy basis, expand the initial quantum state as ${\psi }_0=c_1{\Psi }_1+c_2{\Psi }_2$. Each energy eigenvector receives its own phase under the operator exponential, givingThis is the general solution because every initial vector has a unique basis expansion.

In that orthonormal basis the other observable has the real symmetric matrixIts conjugate transpose equals itself, so it is self-adjoint. Also $A^2=I$ and $\mathrm{tr}A=0$, giving eigenvalues $+1$ and $-1$. Direct substitution gives the normalized orthogonal eigenvectorsAn irrelevant global phase may multiply either vector. With the chosen lower eigenstate initially,The Born rule says that the probability for the higher eigenstate is $|⟨v_{+},{\psi }_t⟩{|}^2$. Its amplitude is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. Using $|e^{-iu}-e^{-iv}{|}^2=4{\sin }^2((u-v)/2)$ givesThis is the transition probability for a rotated two-level observable. Its time dependence comes from the relative phase between two energy eigenstates. If their energies coincide, or if the measured observable has the energy eigenvectors as its eigenvectors, this transition probability is zero at all times.

Take $m,\omega >0$ and use the usual common invariant domain for the position operator, momentum operator and their products. Since $X$ and $P$ are self-adjoint,Expanding the commutator and using $[P,X]=-i\hslash I$ givesThe two products areConsequently the quantum harmonic oscillator Hamiltonian operator isThe identity $[AB,C]=A[B,C]+[A,C]B$ now givesThus $H(a_{\pm }\psi )=(E\pm \hslash \omega )a_{\pm }\psi$. Each ladder image is an eigenstate if it is nonzero; a zero vector is not an eigenstate. The distinction is needed for lowering the ground state. Their squared norms areThese follow by moving the operator to the other side of the inner product with its adjoint operator. The first inequality implies $E\ge \hslash \omega /2$ for every eigenstate.

For a lowest-energy quantum state, $a_{-}{\psi }_0$ must vanish, since otherwise it would be an eigenstate with lower energy. The norm formula then forces $E_0=\hslash \omega /2$. Raising never vanishes and producesTo see that there are no other eigenvalues, repeatedly lower any eigenstate. If $E/(\hslash \omega )-1/2$ were not a nonnegative integer, there would be a final nonzero lowered quantum state with energy strictly between $\hslash \omega /2$ and $3\hslash \omega /2$. Its lower image would have positive norm but energy below the lower bound, a contradiction. Thus every eigenstate reaches the ground eigenspace after an integer number of lowerings. On a positive level, lowering is injective by its strictly positive norm factor. The assumed one-dimensional ground eigenspace therefore makes every eigenspace one-dimensional, so the displayed raised quantum states give all eigenstates up to nonzero multiples. If ${\psi }_0$ is normalized, their norms are $\Vert {\psi }_n{\Vert }^2=n!$.

For $k\ge 1$, $(a_{+}{)}^k\psi$ has eigenvalue $E+k\hslash \omega$, distinct from $E$. eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal: applying self-adjointness to their inner product gives $(E^{\prime }-E)⟨\psi ,\varphi ⟩=0$. ThereforeThe expectation of $(a_{-}{)}^k$ also vanishes, either by lowering and orthogonality or by conjugating the adjoint relation.

Solving the ladder definitions for $X$ and $P$ givesso $T-V=\frac{1}{2}\hslash \omega ((a_{+}{)}^2+(a_{-}{)}^2)$. Both terms have zero expectation by the preceding orthogonality. Since $T+V=H$, the oscillator form of the quantum virial theorem isExpectations here are divided by $\Vert \psi {\Vert }^2$ if the eigenstate is not normalized; no normalization assumption was used in the ladder argument.

Use International System of Units, with $c^{-2}={\mu }_0{\epsilon }_0$. The Maxwell equations with charge density $\rho$ and current density $\mathbf{J}$ areLocally, or in a suitable contractible region, the zero divergence of $\mathbf{B}$ allows a magnetic vector potential $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law then makes $\mathbf{E}+{\partial }_t\mathbf{A}$ curl-free, so a electric potential can be chosen withThese formulas automatically satisfy the two homogeneous Maxwell equations. The local topology assumption rules out obstructions to expressing a closed field as a gradient or curl.

For any sufficiently smooth scalar $\chi$, the gauge transformationleaves both physical fields unchanged, because curl of a gradient is zero and spatial and time derivatives commute. In such a region this accounts for the freedom between potentials for the same fields.

The Lorenz gauge is $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\Phi =0$. Substitution into the sourced Maxwell equations, using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$, gives the decoupled wave equationsFor example, Gauss's law gives $-{\nabla }^2\Phi -{\partial }_t\nabla \cdot \mathbf{A}=\rho /{\epsilon }_0$; the gauge condition turns this into the scalar equation. The remaining gauge freedom preserving the condition has $({\nabla }^2-c^{-2}{\partial }_t^2)\chi =0$.

Put $\theta =\omega t-Kz$ for the given wave. Direct differentiation gives $\nabla \cdot \mathbf{E}=0$ and $\nabla \times \mathbf{E}=K\mathbf{E}$. Integrating Faraday's law in time yields the wave's magnetic fieldOne could superpose a time-independent divergence-free and curl-free magnetic field; the expression displayed chooses no such extra background. It has zero divergence and ${\partial }_t\mathbf{B}=-K\mathbf{E}$, so the magnetic Gauss and Faraday equations hold. MoreoverSince the time derivative is nonzero, the vacuum Ampere-Maxwell law holds precisely whenThis verifies all four source-free equations, including both divergence constraints. The sign of $\omega /K$ selects the direction of propagation.

A wave has circular polarization when, at a fixed spatial point, its transverse electric vector has constant nonzero magnitude and rotates uniformly through a circle. Here $E_x^2+E_y^2=E^2$ and the angle changes at rate $\omega$, so the wave is circularly polarized. Its magnetic vector is perpendicular to the electric vector, has magnitude $|E|/c$ under the dispersion relation, and rotates with it. A handedness label additionally requires a convention for the observer's viewing direction.

The Lorenz-gauge potentials for a circularly polarized plane wave give a convenient choice. Take $\Phi =0$ and $\mathbf{A}={\omega }^{-2}{\partial }_t\mathbf{E}$. Because ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$,Also $\nabla \cdot \mathbf{A}={\omega }^{-2}{\partial }_t\nabla \cdot \mathbf{E}=0$. Together with ${\partial }_t\Phi =0$, this proves these potentials satisfy the Lorenz gauge. They satisfy the source-free potential wave equations because every component has the same dispersion relation just obtained.