Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.

With the specified ladder-operator normalization, $A_{+}^{†}=A_{-}$ andTherefore $\Vert A_{-}\psi {\Vert }^2=2m(E-\hslash \omega /2)\Vert \psi {\Vert }^2\ge 0$. This proves $\boxed{E\ge \hslash \omega /2}$, with equality exactly when $A_{-}\psi =0$. In position representation this equation is $\hslash {\psi }^{\prime }+m\omega x\psi =0$, whose general solution isFor a unit vector $|C|=(m\omega /\pi \hslash {)}^{1/4}$; its phase is arbitrary. The commutators $[H,A_{\pm }]=\pm \hslash \omega A_{\pm }$ make $A_{+}$ raise and $A_{-}$ lower energies by $\hslash \omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hslash \omega$. Conversely lowering any nonzero energy eigenvector must eventually terminate, since otherwise it would produce an energy below the lower bound. A terminating vector is annihilated by $A_{-}$ and hence has energy $\hslash \omega /2$. Thus no other eigenvalues occur.

Bilinearity of the two-dimensional commutators giveswhile all momentum-momentum and position-position commutators remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_{+}=(X_1+X_2)/\sqrt{2}$, $X_{-}=(X_1-X_2)/\sqrt{2}$ preserve the kinetic sum, andHence $\boxed{{\omega }_{+}=\omega ,{\omega }_{-}=3\omega }$ and $H=H_{+}+H_{-}$. Every cross-mode position/momentum pair commutes, so $[H_{+},H_{-}]=0$. On the tensor-product oscillator basis,For each $N\ge 0$, allowed pairs have $n_{-}=0,1,\dots ,\lfloor N/3\rfloor$ and $n_{+}=N-3n_{-}$. ThusEach pair contributes one product eigenvector; counting them accounts for the energy-level degeneracy.

Let $u=rF$. Multiplication of the radial equation by $-2mr/{\hslash }^2$ gives $u^{\prime \prime }+[A/r-\lambda (\lambda +1)/r^2-{\alpha }^2]u=0$. With $u=e^{-\alpha r}f$ the constant terms cancel, leavingThe indicial equation is $c(c-1)=\lambda (\lambda +1)$, with roots $c=\lambda +1$ and $c=-\lambda$. The usual physical Coulomb radial origin condition selects $c=\lambda +1$: it imposes the regular reduced radial boundary condition $u(0)=0$. Normalizability alone at the origin is insufficient to make this selection when an irregular branch is locally square-integrable, notably at $\lambda =0$.

Write $f=r^{\lambda +1}{\sum }_{j\ge 0}{a}_j{r}^j$. Coefficient comparison givesA nonterminating regular series has the exponentially growing asymptotic branch, suggested by $a_{j+1}/a_j\sim 2\alpha /j$: it supplies an $e^{2\alpha r}$ component, which the prefactor $e^{-\alpha r}$ cannot normalize. Equivalently the regular confluent-hypergeometric solution grows unless its first parameter is a nonpositive integer. Thus a bound state requires the numerator to vanish at some $j=k\ge 0$, givingThis is series-termination quantization of the Coulomb radial equation. It requires $A>0$ for the assumed $\alpha >0$; at $A=0$ there are no such negative-energy Coulomb bound states.

For $k=0$, $F=N{r}^{\lambda }{e}^{-\alpha r}$. The angular integral cancels when computing $⟨1/r⟩$, while the radial volume element contributes $r^{2}dr$. ConsequentlyThe ratio follows by one integration by parts and remains valid for real $\lambda \ge 0$; no integer-only factorial identity is needed. Since now $\alpha =A/[2(\lambda +1)]$,For a classical particle with that negative total energy, nonnegative kinetic energy requires $E-V={\hslash }^2(A/r-{\alpha }^2)/(2m)\ge 0$, hence $\boxed{r\le A/{\alpha }^2}$. This is a necessary radial bound; extra angular momentum can impose a stronger restriction.

At $k=\lambda =0$, $\alpha =A/2$ and the classical boundary is $r=2/\alpha$. The normalized radial probability density is $4{\alpha }^3{r}^2{e}^{-2\alpha r}$. With $z=2\alpha r$, the exterior quantum probability isThe final integration gives $(4^2+2\cdot 4+2)e^{-4}/2$. This is the nonzero quantum tail beyond the classical turning region.

Inside the infinite well, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$ with Dirichlet endpoints. Negative or zero energy gives no nonzero solution satisfying both endpoints. For $E>0$, the sine solution requires $ka=n\pi$. ThusThe normalization follows from ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. These functions form an orthonormal basis. Substitute the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take the inner product with each basis vector: orthonormality gives $i\hslash {\dot{c}}_n=E_n{c}_n$, equivalentlyFor a general infinite expansion this is interpreted in the usual Hamiltonian domain or its unitary evolution; the particular state below has only two terms.

Put $\theta =\pi x/a$. The identity $\sin \theta (16{\cos }^2\theta -1)=4\sin 3\theta +3\sin \theta$ gives $c_1=3/5$, $c_3=4/5$, all others zero. Their squared magnitudes sum to one. HenceThe evolved state is $(3e^{-i{E}_{1}t/\hslash }{\Psi }_1+4e^{-i{E}_{3}t/\hslash }{\Psi }_3)/5$. Its position probability includes an interference term with frequency $\Omega =(E_3-E_1)/\hslash =4{\pi }^2\hslash /(m{a}^2)$. Specifically,The three required integrals are $\pi /8-1/4$, $\pi /8+1/12$ and $1/4$, respectively. ThereforeThe energy expectation is constant while the spatial probability oscillates, because relative energy phases affect interference but not the energy probabilities.

Put $d=2^N$ and $\lambda =e^{2\pi i/d}$, with nonnegative integer $N$ so the register exists. The geometric sum satisfiesConsequently the matrix $F_{+}$ with entries $d^{-1/2}{\lambda }^{jk}$ has $F_{+}^{\ast }{F}_{+}=I$. This proves unitarity andFor a binary output string $j={\sum }_{r=1}^N{j}_r{2}^{N-r}$, expansion of the printed tensor product gives coefficientThus the printed product is exactly $F_{+}|k⟩$. It is a correct product decomposition of a Fourier phase state, but the printed recovery identity has inconsistent signs:by the square of the quantum Fourier transform. To recover $f(j)$ instead, use $F_{-}=F_{+}^{\ast }$, replacing every product phase by its complex conjugate. Equivalently one may reverse the sign in the definition of $\hat{f}$. The discrepancy is visible for $d\ge 4$ by taking $f$ concentrated at $1$: the printed expression returns the basis vector at $d-1$. The recovery display also uses $k$ as the summation variable on its right while its summand uses $j$; that dummy variable should be $j$.

For factorization first check $\gcd (3,91)=1$. Successive residues areThe first return is $\boxed{r=6}$. Its half power is $3^3=27$, which is neither $1$ nor $-1$ modulo $91$. Therefore

In Shor's algorithm prepare a uniform exponent register of size $Q=2^n$, conveniently $91^2\le Q<2\cdot 91^2$, and reversibly compute $3^x\mathrm{mod}91$ into a second register. Measuring that second register, or tracing it out, leaves periodic exponent correlations of period $r$. Conditional on one measured residue, the first register is a superposition over an arithmetic progression of exponents separated by $r$. The quantum Fourier transform converts this periodic structure into peaks near $k/Q=s/r$. Continued fractions recover $s/r$; when $s$ is coprime to $r$, its denominator is $r$. Verify by modular exponentiation and repeat when the reduced denominator is insufficient. For this example $Q=16384$ is suitable, and useful peaks include those near $Q/6$ and $5Q/6$. Either transform sign gives the same period information, so the source's inverse-sign error does not invalidate the order-finding mechanism.

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.

Choose an orthonormal eigenbasis $H_0|n⟩=E_n|n⟩$ and the phase convention $⟨0|{\psi }^{\prime }⟩=0$. Comparing coefficients of the eigenvalue equation at first order yieldsProjecting onto $|0⟩$ and the other eigenvectors givesProjection of the next-order equation onto $|0⟩$ givesThese are coefficients of the printed series, so $E^{\prime \prime }$ denotes the quadratic coefficient rather than the second derivative itself. Without a phase convention an arbitrary purely imaginary multiple of ${\psi }_0$ can be added to the first vector correction. Continuous spectrum, if present, requires the corresponding spectral integral; the oscillator has a discrete complete eigenbasis.

Put $K=m{\omega }^2/2$ and let $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ denote the ground-state density. Then $H^{\prime }=K(a^2-x^2)1_{[-a,a]}$ and $p={\int }_{-a}^{a}g(x)dx$. Because $g^{\prime }=-xg/{\sigma }^2$, integration by parts yieldsThus the two first-order shifts areThe first is the supplied ground-state expression; the second follows because the excited-state density is $x^{2}g/{\sigma }^2$, so $E_1^{\prime }=K(a^2{I}_2-I_4)/{\sigma }^2$.

Subtracting, and using $K{\sigma }^2=\hslash \omega /4$, givesThe bracket is zero at $a=0$, tends to one at infinity, and has derivative $2a^{2}g(a)/{\sigma }^2\ge 0$. Therefore the first-order predicted level gap isFor every $\lambda >2$ it becomes negative for all sufficiently large $a$, crossing when $p-2ag=2/\lambda$. It is still positive at $a=0$: the stated conclusion must be read with the large-$a$ regime, rather than for every interval width.

For $\lambda >1$ the inner potential curves downward, placing a central barrier between two symmetric minima at $\pm a$. The even unperturbed ground state has its greatest density at the central barrier, while the odd first excited state has a node there and is less penalized. This explains the relative first-order shifts. A negative predicted gap signals failure of that perturbation approximation, not inversion of the exact lowest two states: the true one-dimensional ground state remains nodeless and even, with the odd state above it. At a high barrier the actual low levels instead form a close double-well pair. The apparent level crossing from a finite-interval oscillator perturbation is useful precisely as a diagnostic of the approximation's range.

Let $U(t)=e^{-iHt/\hslash }$ and write ${\psi }_S(t)=U(t){\psi }_0$ in the Schrödinger picture, with observables $A_S$ constant unless they have an explicit time dependence. In the Heisenberg picture the state is ${\psi }_H={\psi }_0$ andDifferentiate the two unitary factors to obtain the equation. Unitarity gives $⟨{\psi }_S,A_S{\psi }_S⟩=⟨{\psi }_0,A_H{\psi }_0⟩$.

For the interaction picture choose a reference split $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and define ${\psi }_I=U_0^{\ast }{\psi }_S$, $A_I=U_0^{\ast }{A}_S{U}_0$. Then $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$, $V_I=U_0^{\ast }V{U}_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). AgainThese are different unitary allocations of the same dynamics between states and observables, so measured expectations agree.

The canonical commutation relation now givesMultiplication of the last equation by $e^{i\omega t}$ and integration givesSince $A_t^{\ast }-A_t=2im\omega X_t$ and $A_0{\psi }_0=0$, both initial expectations of $A_0,A_0^{\ast }$ vanish. ThereforeThe printed positive displacement endpoint has the wrong sign for the Hamiltonian's positive $\mathcal{E}X$ term. The two turning values are $0$ and $-2\mathcal{E}/(m{\omega }^2)$; their ordering depends on the sign of $\mathcal{E}$. The result oscillates about the displaced equilibrium $-\mathcal{E}/(m{\omega }^2)$, exactly as the force $-\mathcal{E}$ predicts.

Take a normalized vector in the required common operator domains. Expanding the squared norm and using self-adjointness givesThus its final term is $\hslash t⟨C⟩$. Set $a=\Vert A\psi {\Vert }^2$, $b=\Vert B\psi {\Vert }^2$, $c=⟨C⟩$. If $b>0$, minimizing the nonnegative quadratic at $t_{\ast }=-\hslash c/(2b)$ givesEquality holds precisely when the minimum norm vanishes, that is $(A-i{t}_{\ast }B)\psi =0$. If $b=0$, however, then $B\psi =0$ and $c=0$, so equality is automatic even when $A\psi \ne 0$. The printed equality criterion needs this zero-vector exception. For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is linear dependence of $A\psi ,B\psi$ together with vanishing real part of their inner product, as in the uncentered second-moment commutator inequality.

Apply the same calculation to $A-⟨A⟩I$, $B-⟨B⟩I$, whose commutator is unchanged. This gives the Robertson uncertainty principleIf $\psi$ is an eigenvector of $A$, its dispersion is zero; finite dispersion of $B$ then forces $\boxed{⟨C⟩=0}$.

In the position representation, $X\psi =x\psi$ and $P\psi =-i\hslash {\psi }^{\prime }$. The product rule gives $[P,X]\psi =-i\hslash \psi$, hence $\boxed{[P,X]=-i\hslash I}$. Since $V(X)$ commutes with $X$,For an energy eigenstate, $⟨[H,X]⟩=E⟨X⟩-E⟨X⟩=0$, so $\boxed{⟨P⟩=0}$. For the oscillator $[H,P]=i\hslash m{\omega }^{2}X$, and the same argument gives $\boxed{⟨X⟩=0}$. These expectation identities require the stated finite moments and commutator domains; formal position eigenvectors are not normalizable exceptions.

Apply the Robertson uncertainty principle to each angular-momentum pair:If $\psi$ is an eigenvector of $L_3$, then $\Delta L_3=0$. The inequalities for $(L_2,L_3)$ and $(L_3,L_1)$ force $\boxed{⟨L_1⟩=⟨L_2⟩=0}$, assuming the relevant moments exist.

Write $b=(\ell +1)a$, $R(r)=r^{\ell }{e}^{-r/b}$ and $Y(\theta ,\varphi )={\sin }^{\ell }\theta e^{i\ell \varphi }$, for nonnegative integer $\ell$. The angular Laplacian satisfies ${\Delta }_{S^2}Y=-\ell (\ell +1)Y$. Direct radial differentiation givesBecause $(\ell +1)/b=1/a$, the $1/r$ contribution cancels the Coulomb potential in the Hamiltonian. ThereforeThe volume element contributes $r^2\sin \theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular normalization cancels in every radial expectation. With $I_j={\int }_0^{\infty }{r}^j{e}^{-2r/b}dr$,The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)b{I}_j/2$. HenceIn particular $\mathrm{Var}(r)/{\mu }^2=(2\ell +3{)}^{-1}$. Put $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality gives

For the polar variable set $J_j={\int }_0^{\pi }{\sin }^j\theta d\theta$. The polar density is proportional to ${\sin }^{2\ell +1}\theta$, soAn exact expression for the requested tail, as well as its useful estimate, isThe inequality is Markov's inequality applied to ${\cos }^2\theta$. If a finite expression is desired, expand $(1-z^2{)}^{\ell }$: the numerator is ${\sum }_{j=0}^{\ell }(-1{)}^j(\genfrac{}{}{0ex}{}{\ell }{j})(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.

By the union bound, with probability at least $1-2h$ both $|r/\mu -1|\le h$ and $|\cos \theta |\le h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with probability tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The probability concentration of a circular Coulomb eigenstate is a precise version of the large-angular-momentum orbit picture.

For a normalized state the position probability density is $\boxed{f=|\psi {|}^2}$. With a real potential, the Time-dependent Schrödinger equation and its conjugate giveThis is the continuity equation $\boxed{f_t+j_x=0}$. Integrating over the line yields $d\Vert \psi {\Vert }^2/dt=-j(\infty )+j(-\infty )=0$ under the stated decay. A real-valued wave function gives $\overline{\psi }{\psi }_x-\psi \overline{{\psi }_x}=0$, so its instantaneous current vanishes; the same is true up to a spatially constant phase.

Put $\alpha =m\omega /\hslash$. Complete the square in the overlap:The Gaussian integral is $\sqrt{\pi /\alpha }$, also for complex displacement by analytic continuation. HenceThe product of square roots fixes the branch convention implicit in the printed square root of $\overline{u}s$. At $u=s\ne 0$ the norm squared is $|N{|}^2|s|\sqrt{\pi /\alpha }{e}^{2\alpha |s{|}^2}$, which becomes one for the stated $N$. In particularThe centre is $-2\mathrm{Re}s$ and its position variance is $1/(2\alpha )$.

When differentiating in $s$, keep $N$ fixed on the chosen constant-radius orbit. ThenSubstitution gives $HG=\hslash \omega s{G}_s$. Since $\dot{s}=-i\omega s$, the chain rule gives $\boxed{i\hslash G_t=HG}$. Continue $\sqrt{s(t)}$ continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point phase. The case of zero displacement is obtained as a limit, since the separate printed $N$ and $\sqrt{s}$ expressions are singular at $s=0$.

Finally $j=(\hslash /m)\mathrm{Im}(\overline{G}{G}_x)=-2\omega \mathrm{Im}s|G{|}^2$. For real nonzero $a$ this isThe printed final current omits the Gaussian normalization factor $\sqrt{m\omega /(\pi \hslash )}$. Retaining that factor is required both by the normalized state and by the continuity equation. This current is the Gaussian probability density times the velocity of its moving centre, $2a\omega \sin \omega t$.

Use periodic plane waves on the ring. For $n\in \mathbb{Z}$,They are orthonormal on one circumference and complete by Fourier series. Thus $\boxed{E_k={\hslash }^2{k}^2/(2m{a}^2)}$, with degeneracy $\boxed{1\text{ for }k=0,2\text{ for }k\ge 1}$: the two signs of nonzero angular momentum have the same energy. Equivalently one may use normalized sine and cosine eigenfunctions for each positive $k$.

For the position measurement put $\theta =x/a$. Direct integration givesThe energy measurement requires projection onto whole degenerate eigenspaces, rather than just onto one traveling wave. The expansion coefficients areFor instance the imaginary integral vanishes by reflection about $\theta =\pi$, and the product-to-sum identity evaluates the real integral as $4/(1-4n^2)$. ConsequentlyEvery numerator is positive and no integer $k$ makes a denominator zero, so every energy has nonzero probability. Parseval's identity guarantees that these probabilities sum to one. The supplied state is a valid normalized periodic $L^2$ state; its derivative need not obey the operator-domain periodic condition for the spectral probabilities to be defined.