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The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrödinger equation. Just before switching the old ground state differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.

Let ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.

Use a real orthonormal basis of Fourier modes on the time circle:The eigenvalue of $L_{\omega }$ on the constant mode is ${\omega }^2$, and each sine/cosine pair has eigenvalue ${\omega }^2+{\nu }_r^2$. Retaining these modes giveswhere $C_N$ can depend on $N,\beta$ but is independent of $\omega$. Each real Gaussian integral contributes $\sqrt{2\pi /{\lambda }_r}$. Consequently the Fourier-cutoff oscillator functional determinant isA spectral cutoff between successive eigenvalue pairs selects precisely this finite subspace. In comparing two frequencies, the displayed regularization keeps the same number of modes $N$; holding an absolute spectral threshold fixed need not give the same $N$ at finite cutoff.

Use a genus-zero Polyakov path integral, target signature $(-,+,\dots ,+)$ and all four momenta incoming. Define the Mandelstam variables by $s=-(p_1+p_2{)}^2$, $t=-(p_1+p_3{)}^2$, $u=-(p_1+p_4{)}^2$. The string mass-shell condition is $p_i^2=4/{\alpha }^{\prime }$ and momentum conservation is ${\sum }_i{p}_i=0$, soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.

The free-boson worldsheet propagator is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give $g_s^4$ and the sphere contributes $g_s^{-2}$ through the string genus expansion, leaving $g_s^2$.

The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take $(z_1,z_2,z_3,z_4)=(z,0,1,\infty )$. The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply $|z_{23}{z}_{24}{z}_{34}{|}^2$. At finite $z_4$, this determinant grows as $|z_4{|}^4$, while the matter correlator decays as $|z_4{|}^{-4}$ by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight-$(1,1)$ matter operator at infinity is normalized with $|z_4{|}^4$, while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure $d^{2}z=idz\wedge d\bar{z}=2dxdy$. With $dxdy$ instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed $2\pi$ prefactor.

Set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$. Then $a+b+c=1$ and the integrand is $|z{|}^{2a-2}|1-z{|}^{2b-2}$. To evaluate the complex beta integral first work where $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c>0$. This ensures convergence near $0$, $1$ and infinity. Set $v=1-a$ and $w=1-b$. Schwinger parameterization givesCompleting the square, the $dxdy$ Gaussian integral is $\pi (\lambda +\rho {)}^{-1}\exp [-\lambda \rho /(\lambda +\rho )]$. Change variables to $q=\lambda +\rho$ and $x=\lambda /q$; their Jacobian determinant is $q$. Integrating $q$ gives $\Gamma (v+w-1)[x(1-x){]}^{-(v+w-1)}$. The remaining integral is the ordinary beta function $B(1-w,1-v)=B(b,a)$. ThusRestoring $d^{2}z=2dxdy$ and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.

The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal $1/\Gamma$ is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting $s,t,u$ leaves it unchanged. Its generic channel poles areand the same tower in $t$ and $u$. They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.

A precise check of factorization is the Virasoro–Shapiro amplitude pole residue. Near $a=-n$, set $c=1+n-b$ in the nonsingular factor. The gamma function recurrence givesCombining this with $\Gamma (a)\sim (-1{)}^n/[n!(a+n)]$ showsThe residue is a degree-$2n$ polynomial in $t$, consistent with exchange up to spin $2n$ and scattering-amplitude factorization. At exceptional kinematics reciprocal Gamma zeros can remove apparent poles. In particular, when both $a$ and $b$ approach nonpositive integers, the zero from $1/\Gamma (a+b)$ cancels the putative double pole, leaving channel simple-pole terms. Thus one must not count products of numerator poles without using $a+b+c=1$.

First make explicit the usual method of steepest descent hypotheses: the saddle point is simple, $G^{\prime }(z_0)=0$ and $g_2=G^{\prime \prime }(z_0)\ne 0$; a contour deformation leads to the two descent arms with one clockwise half-indentation around the pole; and endpoints or other contour portions contribute smaller terms. The sign of the exponential matters: $|e^{-i\lambda G}|=e^{\lambda \mathrm{Im}G}$, so decay is toward valleys of $\mathrm{Im}G$.

Use a local analytic function $w(z)$ as coordinate. Nondegeneracy allows an analytic square root after factoring the double zero of $G-G(z_0)$, so the coordinate can be chosen withThe sign of $c$ is selected by the orientation of the contour integral, with $w$ running from negative to positive real values. Matching cubic coefficients gives $d/c=-c{G}^{\prime \prime \prime }(z_0)/(6g_2)$. The transformed amplitude therefore has the Laurent seriesThe Cauchy principal value contribution of the pole on the two real arms is zero by oddness. Its clockwise semicircular indentation contributes $-i\pi F(z_0)$, since $\int dw/w=-i\pi$ there. The regular term contributes the Gaussian integral $B\sqrt{2\pi /\lambda }$. Consequently the simple saddle coinciding with a simple pole has the additive expansionThe remainder written here assumes the other contour contributions are negligible to that order. Otherwise the local result has to be supplemented by them. The next local correction is generically $O({\lambda }^{-1/2})$ times the displayed exponential. If $B=0$, it can vanish; odd Taylor terms integrate to zero between the symmetric arms. When $F(z_0)\ne 0$, the leading term gives the printed asymptotic equivalence. When $F(z_0)=0$, asymptotic equivalence to zero is not meaningful: the additive formula is still valid and a regular saddle-point approximation may lead instead.

Nondegeneracy is necessary and is not explicitly printed. For a counterexample to the unrestricted reading, take $F=1$, $z_0=0$ and $G(z)=-i{z}^4$. There is one critical point, but it is degenerate. Choose the contour from $+i\infty$ to $+\infty$, with a clockwise quarter-circle avoiding zero. Both rays are decaying valleys for $e^{-\lambda z^4}$. Their straight-ray contributions cancel, and the indentation gives $J=-i\pi /2$, not $-i\pi$. Thus the requested coefficient requires the simple saddle point and contour deformation assumptions above; the mere existence of one critical point is insufficient.

For a test function $f$, substitute $y=nx$ and use the Taylor expansion $f(y/n)=f(0)+(y/n)f^{\prime }(0)+O(n^{-2})$. Oddness removes the first term, while the Gaussian integral gives $\int y^2{e}^{-y^2}dy=\sqrt{\pi }/2$. HenceTherefore

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

For every interacting $n\ge 3$, the one-loop scalar effective action already contains counterexamples. On a constant background, with a positive auxiliary mass $\mu$ to control infrared divergences, its field-dependent part isThe term of order ${\lambda }^V$ has $E=(n-2)V$ external scalar fields and is a one-loop polygon one-particle-irreducible Feynman diagram. Choose $V>n/(n-2)$ and $2V>d$. Then $E>n$, its loop momentum integral is ultraviolet convergent, and its coefficient is nonzero. Removing the ultraviolet cutoff does not remove this finite higher-point scalar vertex. For $d>2$, generic nonexceptional external momenta give the same conclusion without retaining the auxiliary mass. In $d\le 2$, the massless theory still needs an infrared prescription; removing the ultraviolet cutoff does not remove that need or the induced vertices. Hence there are no interacting pairs $(n,d)$ under the literal wording. For $n=2$ the functional determinant is field independent; $n=1$ merely shifts a Gaussian integral when an infrared prescription exists.

There is a different conventional interpretation: absence of new independent divergent counterterms. The superficial degree of divergence of a connected diagram isThe standard perturbative counterterm closure criterion is $[\lambda ]\ge 0$, with the same pairs as in part ii, provided one includes all symmetry-allowed lower-degree potential terms, the kinetic term, and the vacuum energy. A pure monomial family need not itself have counterterm closure: a sextic interaction in three dimensions generates a quartic counterterm, and a quartic interaction generates a mass counterterm. This interpretation concerns the local divergent part or the continuum defining action, not the complete quantum effective action with its finite interaction vertices and derivative expansion.

For the massive four-dimensional quartic scalar field theory, write ${\lambda }_B$ for the bare quartic coupling, ${\lambda }_R$ for its renormalized value, and $m>0$ for the mass held fixed by a renormalization condition. Define the scalar bubble integralThe quartic one-particle-irreducible correlation function, defined as a derivative of the Euclidean quantum effective action, isThere are three bubble diagrams, one for each pairing of external momenta, and each has Feynman-diagram symmetry factor $2$. The minus sign follows equivalently from the quadratic term in the functional determinant $\frac{1}{2}\mathrm{Tr}\log [1+({\lambda }_B/2)(-{\partial }^2+m^2{)}^{-1}{\varphi }^2]$.

Radial integration at zero external momentum givesThe bare quartic vertex is not finite at fixed bare coupling. Its logarithmic ultraviolet divergence must be subtracted. Impose the momentum-subtraction scheme condition ${\Gamma }_R^{(4)}(0,0,0,0)={\lambda }_R$. To one loop this requiresFor the finite difference, a Feynman parameter and a shift of loop momentum giveThe boundary error due to shifting a sharp cutoff vanishes in this limit. Substitution yields the finite renormalized quartic scalar vertexThus finiteness requires holding the renormalized coupling fixed and allowing the bare coupling to depend on ${\Lambda }_0$. The unsubtracted assertion would be false.

The same one-loop scalar effective action explains the other effective interactions. For a constant background its expansion isThe $r=1$ tadpole diagram gives a mass correction $\delta m^2={\lambda }_R{I}_1/2$, withIt requires a mass counterterm. The field-independent vacuum energy also requires subtraction. The $r=2$ term is the quartic logarithm already treated. There is no one-loop wave-function renormalization from the quartic tadpole diagram.

For $r\ge 3$ the polygon diagrams generate finite higher-point scalar vertices, withFor example, the induced sextic term in the effective potential is $+{\lambda }_R^3{\varphi }^6/(768{\pi }^2{m}^2)$, or a six-point interaction vertex $15{\lambda }_R^3/(16{\pi }^2{m}^2)$ when normalized by $6!$. These finite terms survive the removal of the ultraviolet cutoff at fixed $m,{\lambda }_R$.

The external-momentum dependence of the bubble diagrams also generates a derivative expansion of quartic interactions. At small $P$, the subtracted scalar bubble integral is $-P^2/(96{\pi }^2{m}^2)+O(P^4/m^4)$, giving finite derivative couplings. At order ${\lambda }_R^2$ these are the new field-dependent interactions beyond the mass and quartic terms; the sextic and higher interaction vertices require higher powers of ${\lambda }_R$, though they still occur at one loop. Such coefficients are suppressed by powers of the physical mass or external momentum, not by powers of ${\Lambda }_0$. The calculation establishes an order-by-order perturbative limit; it does not establish a nonperturbative interacting continuum limit of a quantum field theory in four dimensions.

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

Rotate the contour for the Gaussian integral through angle $-\pi /4$; a small positive damping factor justifies the deformation and can then be removed. After scaling by $t^{-1/2}$, the oscillatory Gaussian integral with negative phase givesSimilarly, first scale $u=t^{-1/3}v$ and then rotate the positive ray through angle $-\pi /6$. Since ${\int }_0^{\infty }{e}^{-v^3}dv=\Gamma (4/3)$, the cubic oscillatory Gamma integral gives

For fixed $\alpha$, consider the complex phaseIts unique interior stationary point iswhereThe endpoint contributions are $O(x^{-1})$, while the one-dimensional stationary-phase formula has order $x^{-1/2}$. Taking its real part proves the stationary-phase asymptotic for the generalized Bessel cosine integral:

For $Q_x(x)$ the phase is $x(\sin \theta -\theta )$. Its stationary point has merged with the endpoint $\theta =0$, whereUse the cubic endpoint stationary-phase scaling $\theta =(6/x{)}^{1/3}u$. The other endpoint and the region away from zero contribute only $O(x^{-1})$, soTherefore

In the canonical ensemble, a one-particle velocity state has kinetic energy $E(\mathbf{v})=m|\mathbf{v}{|}^2/2$ and hence probability density proportional to $e^{-\beta E}$, where $\beta =(k_{B}T{)}^{-1}$. The three Cartesian Gaussian integrals normalize the density to the Maxwell-Boltzmann velocity distributionIntegrating over a spherical shell of radius $v$ and area $4\pi v^2$ gives the Maxwell speed distributionSinceits unique positive maximum is the most probable speed in the Maxwell distribution

The intended integral representation of the modified Bessel function isDifferentiation under the integral sign givesSinceintegration by parts and the vanishing endpoint term yield $x{I}_0^{\prime \prime }+I_0^{\prime }-x{I}_0=0$.

For large positive $x$, Laplace's method localizes the integral near the endpoint $\theta =0$. Put $\theta =u/\sqrt{x}$ and useExtending the exponentially small tail to infinity and evaluating the resulting Gaussian integrals givesand therefore

Now write $y=x^{-1/2}w$. Direct differentiation givesso substitution into the ordinary differential equation givesFor the Liouville–Green approximation, put $Q(x)=1-1/(4x^2)$. The two approximate solutions areso at leading order $y_{\pm }(x)\sim x^{-1/2}{e}^{\pm x}$. The growing solution agrees with the preceding asymptotic expansion, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.

For $0<a<1$, the phase has an interior minimum atThe inverse hyperbolic cosine selects the positive stationary point. At this point,Expanding the phase to second order and evaluating the resulting Gaussian integral by Laplace's method yieldsThe exponential grows because the minimum of $S$ is negative; the original integral remains convergent because $a\sinh t$ dominates $t$ at infinity.

Use the shape-rate convention for the Gamma distribution. DefineThe zero null space assumption makes $B$ a positive-definite matrix, and $u_{\ast }$ is the unique ordinary least squares solution. The Gaussian likelihood contributes ${\gamma }^{k/2}$, which must be retained because $\gamma$ is unknown. Multiplying it by the improper prior and the Gamma distribution gives the normal-gamma posterior with a flat prior:Here the constant density of the improper prior cancels on normalization. PutThe normal equation $A^{T}r=0$ makes the residual orthogonal to the range of $A$. The orthogonal decomposition by a closed subspace into that range and its orthogonal complement givesIntegration over $u$ by the Gaussian integral, followed by integration over $\gamma$ using the Gamma distribution, yieldsThus the posterior distribution is proper, despite the improper prior; here $k\ge d$ follows from the zero null space assumption.

Completing the square in $u$ and collecting the powers of $\gamma$ give the two conditional distributions:Explicitly their probability density functions areFor a joint maximum a posteriori estimate relative to $dud\gamma$, first fix $\gamma >0$. The unique maximizing $u$ is $u_{\ast }$. The remaining log density is $(a-1)\log \gamma -b_0\gamma$ plus a constant. Consequently, if $a>1$,The stated assumptions do not always imply $a>1$. If $a=1$, the supremum occurs as $\gamma \downarrow 0$ but is not attained on $\gamma >0$; if $a<1$, the density is unbounded there. In either case there is no joint maximum a posteriori estimate in the stated parameter space. Calling zero a boundary mode does not make it an admissible positive precision.

If “MAP estimators” instead means separate modes of the marginal distributions, integration givesThus the marginal maximum a posteriori estimate of $u$ is always $u_{\ast }$, and that of $\gamma$ is $(s-1)/b_0$ when $s>1$, with the same nonattainment at zero when $s\le 1$. Joint and marginal maximization are different operations; the usual joint interpretation gives the preceding boxed pair.

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