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Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.

For a state in the domain of the operator, its expectation value isFor a normalized state the denominator is one. Use an orthonormal energy eigenstate basis, choosing orthogonal vectors within degenerate eigenspaces. Inserting an expansion into the time-dependent Schrödinger equation gives $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$. ConsequentlyThe phases preserve normalization. Projection onto a specified basis eigenstate gives probability $\boxed{|a_p{|}^2}$, independent of time. If the measurement distinguishes only the energy and $E_p$ is degenerate, sum $|a_n{|}^2$ over all $n$ with $E_n=E_p$.

Using orthonormality to eliminate all cross terms,This is time independent whenever the expectation is defined. The time-independent Hamiltonian changes only relative phases, not the energy probabilities.

Define the interaction picture byDifferentiating and using the Schrödinger equation cancels the $H_0$ terms, leavingIts integral equation, iterated once from the initial $|a⟩$, gives $|{\psi }_I(t)⟩=|a⟩-(i\lambda /\hslash ){\int }_0^t{V}_I(t^{\prime })|a⟩d{t}^{\prime }+O({\lambda }^2)$.

Since $a,b$ are orthogonal energy eigenstates with distinct energies, the zeroth-order transition amplitude is zero, whileReturning to the Schrodinger picture changes this component only by a unit-modulus phase. Squaring the first-order amplitude therefore gives

For exponential switching, put $\Delta =E_b-E_a$. The integral isBecause $\mu >0$, its exponential vanishes at late time, yieldingThis is the leading weak-perturbation result, not an exact probability formula at arbitrary coupling.

Generalized position eigenstates satisfy $⟨x|x^{\prime }⟩=\delta (x-x^{\prime })$ and $\int |x⟩⟨x|dx=I$. Define the position-space wavefunction $\psi (x)=⟨x|\psi ⟩$. Completeness then gives $⟨\psi |\psi ⟩=\int |\psi (x){|}^{2}dx$. Position acts by multiplication, $⟨x|\hat{x}|\psi ⟩=x\psi (x)$. Using $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$ and momentum completeness, differentiating the Fourier integral gives $⟨x|\hat{p}|\psi ⟩=-i\hslash {\psi }^{\prime }(x)$ on the momentum operator's domain.

For the unit-mass, unit-frequency oscillator defineSince $[H,a]=-\hslash a$, a simultaneous energy eigenstate and $a$-eigenstate would satisfy $[H,a]|{\psi }_{\alpha }⟩=0=-\hslash \alpha |{\psi }_{\alpha }⟩$, forcing $\alpha =0$. For $\alpha =0$, the energy is $\hslash /2$, the smallest possible because $⟨a^{†}a⟩=\Vert a\psi {\Vert }^2\ge 0$.

The position-space equation is $(x+\hslash {\partial }_x){\psi }_{\alpha }=\sqrt{2\hslash }\alpha {\psi }_{\alpha }$. Solving and normalizing, with $\alpha =u+iv$, gives up to a constant phaseIts squared modulus is a normalized Gaussian, so every complex $\alpha$ gives a normalizable coherent state.

For a creation eigenstate, $(x-\hslash {\partial }_x)\psi =\sqrt{2\hslash }\beta \psi$ instead gives $\psi =C\exp [x^2/(2\hslash )-\sqrt{2/\hslash }\beta x]$. Every nonzero such function grows quadratically exponentially at infinity and fails square integrability. Hence there are no nonzero normalizable eigenstates of $a^{†}$. This last symbol is $a^{†}$ in the original PDF, although the converted TeX drops its dagger.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

Use units $\hslash =1$. The fluctuations about a classical path satisfy $f(0)=f(T)=0$. Expanding the action functional and integrating the term $m{\dot{q}}_c\dot{f}$ by parts givesThe linear term vanishes by the Euler-Lagrange equation and the endpoint conditions. Integrating the quadratic kinetic term by parts yields the Dirichlet fluctuation determinant of a semiclassical propagator:A finite time-slicing first turns this into a product of ordinary Gaussian integrals, whose limit is proportional to $(\det {\mathcal{O}}_c{)}^{-1/2}$. A useful normalization relative to the free operator ${\mathcal{O}}_0=-m{d}^2/d{t}^2$ isThe oscillatory square root is fixed by the short-time limit and the Feynman i-epsilon prescription. Away from zero modes this gives the semiclassical propagator; if several classical paths contribute, their saddle contributions are summed with their corresponding determinant phases.

For a general potential the prefactor depends on the endpoints through $q_c$; writing only $D(T)$ suppresses that dependence. If $V(q)=a{q}^2/2+bq+c$, the Taylor expansion has no terms beyond quadratic and $V^{\prime \prime }=a$ is independent of $q_c$. The Gaussian functional integral is then exact, and its prefactor genuinely depends only on $T$ and the potential parameters. Singular focusing times are understood by continuation of the kernel, rather than by assigning a finite value to a zero-mode determinant.

For the free-particle propagator, $q_c(t)=q_0+(q_1-q_0)t/T$ and $S_c=m(q_1-q_0{)}^2/(2T)$. The free determinant supplies the factor $\sqrt{m/(2\pi iT)}$. Its normalization can also be checked by the momentum-space representationwhere completing the square evaluates the Gaussian integral and the short-time limit gives the position-space delta function.

For the gravitational potential, the original PDF contains $-mgq$; the TeX aid drops the final $q$. The Euler-Lagrange equation is ${\ddot{q}}_c=g$, and its endpoint solution isSubstitution into the kinetic and potential terms givesSince $V^{\prime \prime }=0$, the fluctuation operator is the free one. Thus the quantum-mechanical propagator in a constant force is exactly

The quantum-mechanical propagator is $⟨q_1|e^{-iHT}|q_0⟩$, with Hamiltonian operator $H=-{\partial }_q^2/(2m)+V(q)$. For imaginary time $-i\tau$, insert an orthonormal complete set of energy eigenstates to obtainHere ${\psi }_n(q)=⟨q|n⟩$. This discrete sum assumes a discrete complete spectrum; continuous spectral components require the corresponding spectral integral. The heat-semigroup expression is well defined for an appropriate self-adjoint Hamiltonian bounded below. It is this spectral condition, not an unrestricted formal rotation for every conceivable potential, that justifies the imaginary-time kernel.

A path on the circle lifts to a path on the real line ending at $q_1+2\pi w$, where the integer $w$ is its winding number. The periodic quantum theory sums all these winding sectors with equal weight, giving the real-time winding representation of a quantum rotor kernel. For imaginary time, put $x=q_1-q_0$ and $y=\tau /m$ in the supplied Gaussian summation identity:Comparison with the spectral expansion determinesThe normalization uses $0\le q<2\pi$. The ground state is nondegenerate; levels with $n\ne 0$ have the degeneracy $n\leftrightarrow -n$. Returning to real time replaces $e^{-\tau E_n}$ by $e^{-iT{E}_n}$, with the usual limiting prescription. No twisted boundary condition or flux phase is present in the stated image sum.

For a normalized trial function in the quadratic-form domain of the Schrödinger operator, integration by parts gives $⟨E⟩=⟨\psi ,H\psi ⟩$. Expand in the complete orthonormal energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$ with ${\sum }_n{c}_n^2=1$. ThenMinimizing over a trial family therefore gives the best upper bound within that family, the quantum variational principle.

For $V=|x|$, choose ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ with $\alpha >0$. The Gaussian integrals giveThus $E(\alpha )=\alpha /2+(\pi \alpha {)}^{-1/2}$, whose derivative vanishes at $\alpha ={\pi }^{-1/3}$. It tends to infinity at both parameter endpoints and has positive second derivative, so this is its minimum. $\boxed{E_0\le \frac{3}{2}{\pi }^{-1/3}\simeq 1.0242}$ is the resulting estimate.

The potential is even; the unique nodeless ground state is even and the first excited state is odd, by the one-dimensional Sturm oscillation theorem. To estimate $E_1$, minimize the same Rayleigh quotient over odd trial functions, for example normalized $x{e}^{-\alpha x^2/2}$, or the compactly supported odd family $C_{a}x(1-|x|/a)$ for $|x|\le a$ and zero otherwise. Oddness gives exact orthogonality to the true ground state, not merely to an approximate trial ground state. The quantum variational principle restricted to the odd sector supplies an upper bound for $E_1$. No further integration is needed.

Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector $Q=0$ and in the one-kink topological sector $Q=1$. In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals ${\Psi }_Q$. Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with $\varphi (+\infty )-\varphi (-\infty )=2\pi Q$. A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time $\tau$, the energy eigenstate expansion gives $K_Q^E(\tau )\sim C_Q{e}^{-E_Q\tau }$. Thus the exact vacuum-subtracted soliton mass isEquivalently it is $i{T}^{-1}\log (K_1/K_0)$ at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.

With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, ${\varphi }_K^{\prime }=2\mathrm{sech}x$ and $1-\cos {\varphi }_K=2{\mathrm{sech}}^{2}x$, so $M_{\mathrm{cl}}=8m/{\beta }^2$. Write $\eta =\delta \varphi$. Expanding and integrating by parts givesThe first variation is ${\beta }^{-2}\int ({\varphi }_K^{\prime \prime }-\sin {\varphi }_K)\eta$, plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about $\eta$.

The printed expansion omits ${\beta }^{-2}$ despite the stated Lagrangian density. Its displayed form is the expansion of ${\beta }^{2}S$; it is not the physical $S$ at arbitrary coupling. Alternatively, writing $\eta =\beta \chi$ puts the quadratic term in canonical normalization, while leaving the classical term $-8mT/{\beta }^2$ unchanged. This normalization repair does not change ${\Delta }_x$ or the physical fluctuation frequencies.

The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and $A{\psi }_0=0$ gives the normalized translational zero mode of a sine-Gordon kink:A displacement $a$ changes ${\varphi }_K(x-a)$ by $-a{\varphi }_K^{\prime }(x)$. Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.

In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.

For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by $2\pi /L$, replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.

There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is $f_k\propto (\tanh x-ik)e^{ikx}$. Its transmission phase obeyswith the odd phase branch that tends to zero at large $|k|$. For $k>0$, the continuum roots have labels $n=1,2,\dots$; there is no periodic continuum root at $k=0$, since the limiting eigenfunction $\tanh x$ has opposite signs at the two ends. The bound state at ${\omega }_0=0$ replaces the free $k=0$ oscillator with ${\omega }_0^{(0)}=1$. Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite $-m/2$ under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large $L$. Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.

At high momentum, $\delta (k)=2/k+O(k^{-3})$, so both expressions have divergent part $-(m/\pi )\log \Lambda$. The canonical field $\phi =\varphi /\beta$ has a quartic interaction with coupling $-m^2{\beta }^2$. Its vacuum tadpole diagram shifts the squared mass by $-m^2{\beta }^{2}J(\Lambda )/4$, whereThe Sine-Gordon vacuum tadpole counterterm has $\delta m^2=+m^2{\beta }^{2}J/4$ and adds potential energy density $\delta m^2(1-\cos \varphi )/{\beta }^2$. Since $\int (1-\cos {\varphi }_K)dx=4$, its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then $8m/{\beta }^2-m/\pi +O(m{\beta }^2)$ in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.

Each bond of the classical Heisenberg antiferromagnet is minimized when its two spins are antiparallel. Since the periodic chain has an even number of sites, all bonds can be minimized simultaneously:The ground-state degeneracy is a continuous sphere of orientations $\hat{\mathbf{n}}\in S^2$. Choose the Néel state with even sites in $|S,S⟩$ and odd sites in $|S,-S⟩$.

This product state is not an energy eigenstate. The spin ladder operators inproduce states with both neighboring magnetic quantum numbers changed. On an up-down bond, $S_n^{-}{S}_{n+1}^{+}$ has a nonzero matrix element $2S$, hence coefficient $JS$ in the Hamiltonian. These orthogonal spin configurations show explicitly why the classical minimum is not an exact quantum eigenstate for $S>0$.

For the Hamiltonian $p^2/(2m)+V(q)$, the real-time configuration-space path integral iswith its measure defined by time slicing. Wick rotation gives the Euclidean path integral and the thermal trace.

For the infinite square well on $(0,L)$, the Dirichlet boundary conditions select the normalized energy eigenstatesThere is no $n=0$ state: the corresponding sine is identically zero. Taking the trace in this orthonormal basis gives the canonical partition function

Set $a=2m{L}^2/({\hslash }^2\beta )$ and $C=\sqrt{m/(2\pi {\hslash }^2\beta )}$. For the reflected images, the intervals $rL-q$, $0\le q\le L$, tile the real line once. A Gaussian integral givesConsequently $Z=CL{\sum }_r{e}^{-a{r}^2}-1/2$. The Poisson summation formula applied to the Gaussian yieldsThe zero dual mode cancels the reflected contribution, while the positive and negative modes pair:This proves equality with the energy eigenstate calculation. The image expansion converges rapidly at small $\beta$; the spectral expansion converges rapidly at large $\beta$.

Each energy eigenstate acquires the phase $e^{-i{E}_{n}t/\hslash }$ under unitary time evolution. Consequently at the requested time $t=1$,and the wavefunction is zero outside the well. Equivalently, in elementary functions,The distinct phases retain their relative sign from the initial expansion; an overall global phase would not change measurement probabilities.

Inside the infinite square well, the time-independent Schrödinger equation is $-({\hslash }^2/(2m)){\psi }^{\prime \prime }=E\psi$. The infinite exterior potential imposes $\psi (0)=\psi (a)=0$. For $E>0$, the first condition leaves a sine and the second requires $ka=n\pi$. A zero or negative energy gives only the zero solution under these conditions. The normalized energy eigenstates and eigenvalues areThe wavefunctions vanish outside the well, and a constant phase is immaterial. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$.

The Born rule gives the probability of $E_n$ as the squared coefficient of the initial state in the energy eigenstate basis. Direct integration givesThe strict energy cutoff corresponds to $n^2<5$, so only $n=1,2$ contribute. Both coefficients equal $2/\pi$, and thereforeThe discontinuous initial wavefunction is a legitimate normalized $L^2$ state for this spectral probability calculation, although it is not in the Hamiltonian's differentiability domain; finite mean energy is not required to compute this probability.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrödinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.