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For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

For a normalized state and a self-adjoint operator $Q$ with finite second moment, the quantum uncertainty is defined by $\Delta Q=\Vert (Q-⟨Q⟩)\psi \Vert$. Since $⟨Q⟩$ is real,In an energy eigenstate of the quantum harmonic oscillator,Discarding the nonnegative squared means proves the stated lower bound. The Heisenberg uncertainty principle is $\Delta x\Delta p\ge \hslash /2$, following from $[x,p]=i\hslash$. Applying the arithmetic-geometric mean inequality to the two variance contributions givesHere $\omega >0$ is the oscillator frequency. The bound is the ground-state energy; equality requires zero means and the appropriate minimum-uncertainty width.

Let $(f_n)$ be a singular Weyl sequence for $L$ at $\lambda \in {\Sigma }_{\mathrm{e}}(L)$. Since $K$ is compact and $f_n\rightharpoonup 0$, the preceding result gives $K{f}_n\to 0$. ThereforeThe norms remain one and the weak limit remains zero. The singular Weyl sequence criterion yields $\lambda \in {\Sigma }_{\mathrm{e}}(L+K)$. Apply the same argument to $L+K$ and the compact self-adjoint operator $-K$ for the reverse inclusion. Both operators are bounded and self-adjoint. Thus the Weyl theorem for compact self-adjoint perturbations isThe corrected shifted-range definition is necessary. For the diagonal example in the preceding solutions, a rank-one perturbation changing the entry $2$ to $0$ removes $2$ from the spectrum. The unshifted printed definition had classified $2$ as essential merely because the original range was not closed, so it would make this invariance false.

No. On the complex l2 sequence space, let $L(x_n)=(x_n/n)$. This is a bounded self-adjoint operator. At $\lambda =0$, $\ker L=\{0\}$, so zero is not an eigenvalue (an eigenvector must be nonzero). But $y=(1/n)\in {\ell }^2$ has no preimage in ${\ell }^2$: its only formal preimage is the constant-one sequence. Thus $L$ is not surjective andAlternatively $\Vert L{e}_n\Vert =1/n\to 0$ excludes a bounded inverse. This example separates the spectrum of a bounded operator from its point spectrum.

The intended self-adjoint operator identity is $⟨Lf,g⟩=⟨f,Lg⟩$, with the inner product linear in its first argument. The printed second $Lf$ must be $Lg$. Taken literally, the printed identity forces $L=0$: put $g=0$ and then vary $g$. Its spectral conclusion is then trivial, but it does not define self-adjointness.

For the corrected identity, $⟨Lv,v⟩$ is real. If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality givesThus $L-\lambda I$ is injective and bounded below, so its range is closed: a convergent image sequence has a Cauchy sequence of preimages. Its range is dense because its orthogonal complement is $\ker (L-\bar{\lambda }I)$, also zero by the same lower bound. Hence it is surjective with inverse norm at most $1/|\mathrm{Im}\lambda |$. Therefore

For a bounded self-adjoint operator, $g$ is orthogonal to its range exactly when $⟨Lf,g⟩=⟨f,Lg⟩=0$ for every $f$, or equivalently $Lg=0$. ConsequentlyThis is image-kernel orthogonality for an adjoint. Taking the double orthogonal complement gives the closure; it does not assert that the range itself is closed.

If unit vectors $f_n$ satisfy $(L-\lambda I)f_n\to 0$, a bounded inverse would give $1\le \Vert (L-\lambda I{)}^{-1}\Vert \Vert (L-\lambda I)f_n\Vert \to 0$. Hence $\lambda$ belongs to the spectrum of a bounded operator.

Conversely a spectral point is real by the preceding argument. Put $A=L-\lambda I$, again a self-adjoint operator. If ${\inf }_{\Vert f\Vert =1}\Vert Af\Vert >0$, then $A$ is injective with closed range. Its range is dense by image-kernel orthogonality for an adjoint, so it is bijective with a bounded inverse, a contradiction. Thus this infimum is zero; choose unit $f_n$ with $\Vert A{f}_n\Vert <1/n$. We have provedSuch a sequence is a spectral Weyl sequence. No weak convergence is required here; nonreal $\lambda$ are ruled out by the lower bound in the preceding solution.

The real and imaginary parts of the operator are uniquelyThey are self-adjoint operators, and $\alpha ={\alpha }_1+i{\alpha }_2$. Conversely such a decomposition has ${\alpha }^{\ast }={\alpha }_1-i{\alpha }_2$. Expanding both products givesConsequently $\alpha$ is a normal operator exactly when its self-adjoint real and imaginary parts commute.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner-product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

Take the real inner productThe stationary distribution is strictly positive for an irreducible Markov chain. Detailed balance gives $⟨Pf,g{⟩}_{\pi }=⟨f,Pg{⟩}_{\pi }$, so $P$ is a self-adjoint operator. Choose a real orthonormal eigenbasis $f_1,\dots ,f_m$, where $m=|S|$, $P{f}_j={\lambda }_j{f}_j$, $f_1\equiv 1$, ${\lambda }_1=1$, and $\pi (f_j)=0$ for $j\ge 2$.

For fixed $y$, apply the spectral decomposition to $g_y(z)=1_{\{y\}}(z)/\pi (y)$. Its coefficient against $f_j$ is $f_j(y)$, whereas $(P^t{g}_y)(x)=P^t(x,y)/\pi (y)$. Thus the spectral kernel expansion of a reversible chain isThis holds for all integers $t\ge 0$; at $t=0$ each multiplier is one, including when ${\lambda }_j=0$. Negative eigenvalues are allowed, and retain their signs for odd $t$.

On a finite state space with at least two states, as in the surrounding questions, for a real function write $\pi (f)={\sum }_x\pi (x)f(x)$ and ${\mathrm{Var}}_{\pi }(f)={\sum }_x\pi (x)(f(x)-\pi (f){)}^2$. The Poincare inequality for a reversible Markov chain with constant $C$ iswhere the Dirichlet form of a Markov chain isThe associated spectral gap is the ordinary gap $\gamma =1-{\lambda }_2$, with variational formulaEquivalently minimize $\mathcal{E}(f,f)$ subject to $\pi (f)=0$ and $⟨f,f{⟩}_{\pi }=1$. Thus the Poincare inequality for a reversible Markov chain holds precisely for $C\ge 1/\gamma$, and the optimal constant is $1/\gamma$. It controls the ordinary rather than absolute spectral gap, so negative eigenvalues do not invalidate it. If the unspecified state space in this part is infinite, the Poincare inequality for a reversible Markov chain and the variational infimum still make sense on nonconstant functions in $L^2(\pi )$, using the self-adjoint operator on that Hilbert space; the resulting spectral gap need not be represented by an actual second eigenvalue. The finite-state formula above is the intended specialization.

Use $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. The Dirichlet form of a Markov chain isThe equality follows by expanding the square and using row sums one and the stationary distribution identity. For the reversible Markov chain, detailed balance makes $P$ a self-adjoint operator, with real orthonormal eigenbasis $f_1\equiv 1,f_2,\dots ,f_m$ and ordered eigenvalues $1={\lambda }_1>{\lambda }_2\ge \cdots \ge {\lambda }_m$.

Adding a constant changes neither the variance nor the Dirichlet form of a Markov chain. Expand the centered function as $f-\pi (f)={\sum }_{j\ge 2}{a}_j{f}_j$. ThenFor every nonconstant $f$ the quotient is at least $1-{\lambda }_2$, and equality is attained at $f=f_2$. This provesEquivalently minimize the Dirichlet form of a Markov chain over centered unit-norm functions. This is the variational characterization underlying the Poincare inequality for a reversible Markov chain, and does not require the transition matrix to have nonnegative eigenvalues. For a one-state chain there is no nonconstant test function; the displayed minimization concerns $m\ge 2$.

A self-adjoint operator $T$ satisfies $⟨Tx,y⟩=⟨x,Ty⟩$. If $Tx=\lambda x$, then $\lambda ⟨x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=\overline{\lambda }⟨x,x⟩$, so $\lambda \in \mathbb{R}$. An eigenvector exists over $\mathbb{C}$; its orthogonal complement is $T$-invariant. Induction on dimension produces an orthonormal basis of eigenvectors, proving the finite-dimensional spectral theorem.

A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $⟨Tx,x⟩={\sum }_i{\lambda }_i|x_i{|}^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.

The same eigenbasis shows that $\mathrm{spec}T\subset [a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda <a$ and negative definite for every $\lambda >b$. Finally, if $v$ is a unit eigenvector of $\alpha +\beta$ with eigenvalue $\gamma$, the Rayleigh quotient bounds giveThus every eigenvalue of $\alpha +\beta$ lies in $[a+c,b+d]$.

First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse $S=A_D^{-1}$ exists from $L^2$ into $H_0^1$. Elliptic regularity gives $Sf\in H^2\cap H_0^1$, and the Rellich-Kondrachov compactness theorem makes $S$ a compact operator on $L^2$. Symmetry givesso $S$ is a self-adjoint operator. It is injective, and $(Sf,f)=B[Sf,Sf]>0$ for $f\ne 0$. The spectral theorem for compact Hermitian operators supplies an orthonormal basis $(w_m)$ with $S{w}_m={\mu }_m{w}_m$, ${\mu }_m>0$, and ${\mu }_m\to 0$. Put ${\lambda }_m={\mu }_m^{-1}$. ThenRepeated elliptic regularity makes each $w_m$ smooth up to the boundary.

As with part (ii), the stated characterization omits boundary compatibility. At $k=0$, its right-hand side is finite for every $u\in L^2$ by Parseval identity, whereas its left-hand side would require $u\in H_0^1$. At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on $(0,\pi )$,Although this function belongs to every ordinary $H^k$, the weighted sum diverges at $k=3$.

The correct spectral characterization of elliptic Dirichlet domains iswith $X_k$ defined in part (ii). For finite eigenfunction sums the squared norm $((u,u){)}_k$ is exactly the weighted sum: even orders follow from $L^l{w}_m={\lambda }_m^l{w}_m$, and odd orders also use $B[w_m,w_j]={\lambda }_m{\delta }_{mj}$. These sums are dense in $X_1$: if $v$ is $B$-orthogonal to every $w_m$, then $0=B[v,w_m]={\lambda }_m(v,w_m)$, so $v=0$. They are dense in $X_0$ by construction, and the isomorphisms $A_D:X_{k+2}\to X_k$ propagate density to every $X_k$. Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for $k=1,2$.

Put $B=\Vert f{\Vert }_{\infty }$, $v={\mathrm{Var}}_{\mu }(f)$, and $N_n=n-m_n$. We first obtain a uniform-in-$f$ stationary bound, then compare the burned-in trajectory with a stationary one.

For a reversible Markov chain, geometric ergodicity gives an absolute L2 spectral gap of a reversible Markov chain. Thus, for some $r<1$, the Markov kernel viewed as an operator satisfiesOne can use the rate in the total-variation estimate. To see why, reversibility makes $P$ a self-adjoint operator. For bounded mean-zero $h$ supported on a set where $M$ is bounded,The spectral theorem for normal operators writes the left side as $\int t^{2k}d{\nu }_h(t)$. Any positive spectral mass where $|t|>r$ would contradict the inequality as $k\to \infty$. These test functions are dense in $L_0^2(\mu )$: truncate a function and restrict it to $\{M\le j\}$, then correct its mean using a bounded function supported on a fixed positive-measure set where $M$ is bounded. Hence the spectral restriction extends to the entire mean-zero space.

Let $(W_i)$ be the stationary chain. For $h=f-\mu f$, the stationary covariance bound for reversible Markov chains givesConsequently

The initial law of the burned-in trajectory is $P^{m_n}(x,\cdot )$. Its total variation distance from $\mu$ is at most ${\epsilon }_n=M(x)r^{m_n}$. Attaching the same conditional future evolution cannot increase that distance, so the entire length-$N_n$ path laws differ by at most ${\epsilon }_n$. For either path the sum $S$ has $|S|\le B{N}_n$. The dual bound for total variation distance gives a difference of at most $2B^2{N}_n^2{\epsilon }_n$ in the second moments, and at most $4B^2{N}_n^2{\epsilon }_n$ in the squares of the means. This burn-in total variation comparison yieldsSince $m_n=\lfloor n^{1/3}\rfloor$, $N_n{r}^{m_n}\to 0$. Dividing by $N_n$ givesThe constant depends on the chain, not on $f$.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

For an oriented smooth surface $S$, the Gauss map sends $p$ to the chosen unit normal vector $N(p)\in S^2$. Since $|N{|}^2=1$, differentiation shows that $D{N}_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_{p}S$.

In a local parametrization $\varphi (u,v)$ with $n=N\circ \varphi$, differentiatinggivesThus the bilinear form $(X,Y)\mapsto D{N}_p(X)\cdot Y$ is symmetric, so $D{N}_p$ is self-adjoint. The Gaussian curvature isWriting the coefficients of the first fundamental form asand those of the second fundamental form asone obtains

At an umbilic point, the self-adjoint map $D{N}_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ withEquality of mixed partial derivatives givesThe two tangent vectors are linearly independent, hence ${\lambda }_u={\lambda }_v=0$. Since ${\mathbb{R}}^2$ is connected, $\lambda$ is constant.

If $\lambda =0$, then $n$ is constant andso $S$ lies in a plane. If $\lambda \ne 0$, thenThus $n-\lambda \varphi =c$ is constant, andTherefore $S$ lies in a sphere of radius $1/|\lambda |$. This proves that the surface is part of a plane or part of a sphere.

Write $Z=X-\mu$, so the covariance operator isFor $g,h\in L^2[0,1]$,so $C_X$ is self-adjoint. Moreover,so it is a positive operator.

Let $(e_j)$ be any orthonormal basis. Tonelli theorem and Parseval identity giveA positive operator with finite trace is a trace-class operator, completing the proof.

Write $H={\partial }_x^2-V(x)$. The real potential and periodic boundary condition make $H$ a self-adjoint operator on the periodic domain. Since $u_t=-iHu$,because the inner product $⟨u,Hu⟩$ is real. Thus the Schrodinger equation preserves the $L^2$ norm.