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.

Recover $L_1$ from the given angular momentum raising operator and angular momentum lowering operator:This is the generator of rotations about the first coordinate axis. A natural candidate is the angular function pointing along that axis, $f_x=\sin \theta \cos \varphi$. Direct substitution givesAlso $f_x=(f_{+1}+f_{-1})/2$, so linearity and part (ii) give $L^2{f}_x=2{\hslash }^2{f}_x$. Thus an answer is $\sin \theta \cos \varphi$. Its squared norm on the sphere isso the normalized quantum state isThe supplied starting state gives an alternative derivation with a changed quantization axis. Put $g=\cos \theta +i\sin \theta \sin \varphi$ and $K_{+}=L_2+i{L}_3$. The angular momentum commutation relations give $[L_1,K_{+}]=\hslash K_{+}$ and $[L^2,K_{+}]=0$, so $K_{+}$ raises the $L_1$ projection from $-\hslash$ to zero. Direct differentiation giveswhich is nonzero and proportional to the same ${\psi }_x$. This also connects the construction directly to the given state.

Starting from $f_{-1}=\sin \theta e^{-i\varphi }$, part (i) gives $L^2{f}_{-1}=2{\hslash }^2{f}_{-1}$ and $L_3{f}_{-1}=-\hslash f_{-1}$. Apply the angular momentum raising operator explicitly:A second application givesSince $L_{+}$ commutes with $L^2$, both raised functions retain the eigenvalue $2{\hslash }^2$. Their $L_3$ eigenvalues follow either by differentiating their $\varphi$ dependence or from $[L_3,L_{+}]=\hslash L_{+}$. Thus three valid unnormalized functions, in the requested order, areFor later Born rule calculations, normalize with the solid angle measure $d\Omega =\sin \theta d\theta d\varphi$. Their squared norms are $8\pi /3$, $4\pi /3$, $8\pi /3$, respectively. For example ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. A convenient orthonormal basis is thereforeOrthogonality follows from the distinct Fourier factors in $\varphi$. An overall phase of each basis function is arbitrary; these choices need not impose the usual phase convention for the spherical harmonics.

Put $f_l(\theta ,\varphi )={\sin }^l\theta e^{-il\varphi }$. For the regular, single-valued orbital angular momentum states, $l$ is a nonnegative integer. Differentiation immediately givesFor $0<\theta <\pi$, the derivatives needed for $L^2$ areConsequently the angular differential expression inside the parentheses of $-L^2/{\hslash }^2$ equalsusing ${\csc }^2\theta -{\cot }^2\theta =1$. ThereforeThe expressions extend regularly through the polar coordinate singularities: $f_l$ is the restriction of the polynomial $(x-iy{)}^l$ to the unit sphere. It is proportional to the lowest-projection spherical harmonic of degree $l$; for $l=0$ it is the constant state.

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.

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.

The spatial quantum phase has unit modulus. Thus the probability density and total probability areThe exterior contributes zero. This establishes wavefunction normalization, although the spatial quantum phase means the state is not the ground energy eigenstate.

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.

Inside the infinite square well, the Schrödinger equation and Dirichlet boundary conditions areThe wavefunction vanishes outside the box. These walls constrain its values, not its first derivatives: requiring both to vanish at an endpoint would eliminate all the nonzero stationary solutions.

For a nonzero separated solution, separation of variables givesThe last equation is the time-independent Schrödinger equation. Its energy eigenvalue is real. Multiplication by ${\psi }^{\ast }$ and integration by parts, with the endpoint values zero, givesEquality would force $\psi$ to be constant and hence zero, so a nonzero energy eigenstate has $E>0$. Writing $k=\sqrt{2mE}/\hslash$, the general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$; the right gives $\sin ka=0$. Negative integers reproduce the same functions up to a constant, and $n=0$ gives the zero function, leaving $k=n\pi /a$ for $n=1,2,\dots$.

Finally ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. Thus wavefunction normalization with positive coefficient givesAn arbitrary constant global phase has no effect on the quantum state.

Twice applying the creation operator or twice applying the annihilation operator changes the energy eigenvalue by $2\hslash \omega$, unless they give zero. The same orthogonality argument as in part (i) therefore givesAlso the factorization in part (a) givesNow $P=(a_{+}+a_{-})/2$ and $X=(a_{+}-a_{-})/(2im\omega )$. Squaring in the correct operator order yieldsBy part (i), the means are zero, so these are the variances. Taking positive square roots gives the requested standard deviations:These harmonic-oscillator number-state second moments also show equal mean kinetic energy and potential energy, each $E/2$. Their uncertainty product is $E/\omega =(n+\frac{1}{2})\hslash$, so the ground state saturates the Heisenberg uncertainty relation.

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.

Let $\psi$ have energy $E_n$ and put $\gamma =2m\hslash \omega$. The lowering norm formula gives, inductively,In particular $(a_{-}{)}^n\psi$ is a nonzero ground state. By the assumed nondegeneracy, it equals $c{\psi }_0$ for some $c\ne 0$.

It remains to show that raising this vector recovers $\psi$, rather than merely another state at the same energy. On an energy eigenstate at $E_n$, factorization gives $a_{+}{a}_{-}\psi =\gamma n\psi$. Iterating the factorization down the ladder yieldsApplying $(a_{+}{)}^n$ to $(a_{-}{)}^n\psi =c{\psi }_0$ therefore provesFor $n=0$ this is the original ground-state assumption. This proves nondegeneracy propagated by an oscillator ladder: every energy eigenspace is one-dimensional if the ground state is nondegenerate.

Write $E_0=\hslash \omega /2$. Repeated application of the annihilation operator lowers the energy eigenvalue in steps of $\hslash \omega$. The lower bound from part (ii) forces this process to stop. To see why it also quantizes the starting energy, writeThe first $n$ lowering steps have positive squared norms. If $\delta >0$, the next step is also nonzero by part (ii), and its energy would be $E_0+(\delta -1)\hslash \omega <E_0$, contradicting the bound. Thus $\delta =0$, andExistence of the ground state can also be checked directly in the position representation. The equation $a_{-}{\psi }_0=0$ iswhose solution is proportional to $e^{-m\omega x^2/(2\hslash )}$. This Gaussian function is square-integrable, so it gives an actual ground state. Its normalization can be chosen using the Gaussian integral.

All the displayed levels occur: for a nonzero state at $E_n$,Thus repeated raising creates a nonzero energy eigenstate at every $E_n$.

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 $E=\hslash \omega /2$.

From $H=a_{+}{a}_{-}/(2m)+\hslash \omega /2$ and the commutator product rule,Consequently, if $H\psi =E\psi$,Thus the creation operator raises the energy eigenvalue by $\hslash \omega$, and the annihilation operator lowers it by $\hslash \omega$, provided the resulting vector is nonzero. The zero vector is not an eigenstate; in particular the ground state will be annihilated by $a_{-}$. The displayed operator identity holds even in that case.

Write $\rho =a\sqrt{2m{V}_0}/\hslash >0$. On the first tangent branch, $0<\beta <\pi /2$, the matching relations are equivalent toThe function $g(\beta )=\beta /\cos \beta$ is continuous and strictly increasing there:The intermediate value theorem therefore gives a unique first-branch root for every $\rho >0$. At this root $\beta <\rho$ and $\alpha =\beta \tan \beta >0$, so the energy eigenvalueobeys $-V_0<E<0$. This proves the even bound state in every attractive square well result.

To verify all the requested properties, take the resulting even wavefunction with exterior tails written in terms of their boundary values:The matching equation makes both the values and first derivatives agree at the endpoints. The exterior tails are square-integrable because $k>0$, and the interior interval is finite. More explicitly,so choosing $B$ as the positive inverse square root of the bracket gives wavefunction normalization. Thus an even, continuously differentiable, normalizable stationary solution always exists, however small the positive depth.

At a finite jump in potential energy, the wavefunction and its first derivative must be continuous. A jump in the derivative would create a Dirac delta function in ${\psi }^{\prime \prime }$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ givesA nonzero solution cannot have $B=0$. It also cannot have $\cos \beta =0$: the first condition would give $D=0$, while the second would force $B\kappa \sin \beta =0$, a contradiction. Division is therefore valid and yieldsParity makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants giveThe circle relation fixes the well strength; the tangent relation selects its even bound states.

For an even function, the interior sine term must vanish, because sine is odd and cosine is even. Comparing reflected exterior tails at any $x>a$ gives $D{e}^{-kx}=A{e}^{-kx}$. Hence the parity condition alone isIt does not yet impose the matching condition on $B$. Continuity at $x=a$, when imposed in the next part, additionally gives $A=D=B{e}^{ka}\cos (\kappa a)$.

In either exterior region of the finite square well, the time-independent Schrödinger equation reduces to ${\psi }^{\prime \prime }=k^2\psi$, since $E<0$. Its general solution is a linear combination of $e^{kx}$ and $e^{-kx}$. On the left, $e^{-kx}$ grows as $x\to -\infty$ and cannot be a square-integrable function, so its coefficient must be zero. On the right, the excluded growing term is $e^{kx}$. Hence the surviving tails are $A{e}^{kx}$ for $x\le -a$ and $D{e}^{-kx}$ for $x\ge a$.

Inside the well, $E+V_0>0$ and the equation becomes ${\psi }^{\prime \prime }=-{\kappa }^2\psi$, whose general solution is $B\cos \kappa x+C\sin \kappa x$. ThusThe coefficients are initially independent; wavefunction matching at a finite potential step imposes the remaining relations. Here the paper's $k$ is the exterior decay constant and $\kappa$ is the interior oscillation constant.

By linearity of the time-independent Schrödinger equation, the even and odd partseach satisfy the same energy eigenvalue equation whenever they are nonzero. They obey $R{\psi }_{+}={\psi }_{+}$ and $R{\psi }_{-}=-{\psi }_{-}$, so they have definite parity. At least one is nonzero, and $\psi ={\psi }_{+}+{\psi }_{-}$. Each nonzero part can be normalized separately.

Therefore a basis of stationary solutions can be chosen to have definite parity, and solving the even and odd sectors loses no energy eigenvalues. This does not assert that every possible quantum superposition has definite parity; a superposition of the two sectors need not. For one-dimensional bound states in the usual regular setting, nondegeneracy of one-dimensional bound states further implies that each individual energy eigenstate has definite parity.