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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 Schrodinger 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 Schrodinger 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.

For a normalized trial state $\psi$, expand it in an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$. The Rayleigh quotient is thenThis is the quantum variational principle: minimizing the expectation of the Hamiltonian operator over any trial family gives an upper bound on the ground-state energy.

Let $\psi$ be the normalized ground-state wavefunction and form the normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{1/2}\psi (\lambda x)$. Its kinetic and potential expectations areThe true ground state makes the energy stationary at $\lambda =1$, soThus the power-law quantum virial theorem gives

If $n\le -3$ and $A>0$, the two sides have opposite signs, because $T>0$ and $V>0$. If $A<0$, the scaled energy ${\lambda }^{2}T+{\lambda }^{-n}V$ tends to $-\infty$ as $\lambda \to \infty$, since $-n>2$, so the Hamiltonian is not bounded below. If $A=0$, the free particle has no normalizable ground state. Hence there is no normalizable ground state for $n\le -3$.

For $n=1$ take the normalized Gaussian functionThe Gaussian integral givesand thereforeIts unique minimum satisfiesAt this point $⟨V⟩=2⟨T⟩$, consistently with the virial theorem, and the resulting variational estimate is

The states ${\chi }_1$ and ${\chi }_2$ are orthonormal energy eigenstates with hydrogenic energiesThe Born rule therefore gives

The particle on a circle has the complete orthonormal basis of energy eigenstates $\{|n⟩:n\in \mathbb{Z}\}$. Its spectral representation givesThe integer $n$ is the quantized angular momentum in units of $\hslash$.

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.

Long imaginary-time evolution multiplies an energy eigenstate by $e^{-T{E}_n}$. After normalization it projects onto the lowest energy component with nonzero initial overlap, or onto the lowest-energy subspace if the vacuum is degenerate. This provides vacuum boundary conditions for a path integral and is related to the Feynman i-epsilon prescription.

Wavefunction normalization makes the total Born rule probability equal to one. In an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\varphi }_n$ is normalized precisely when ${\sum }_n|c_n{|}^2=1$.