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The finite corners have interior angle $3\pi /2$, hence exterior-angle parameters $-1/2$. The channel end at $Z=0$ is logarithmic. Their Schwarz-Christoffel mapping is consequentlyChoose $R(Z)=\sqrt{Z^2-1}$ analytic in the upper half-plane with $R(Z)\sim Z$ at upper infinity. In particular $R(iy)=i\sqrt{1+y^2}$ for $y>0$. A direct differentiation givesHence $f=A+CH$ with $H=R+{\tan }^{-1}(1/R)$.

The inverse trigonometric functions here is the branch obtained by analytic continuation from $Z>1$ through the upper half-plane, not an independent Cauchy principal value evaluation at each point. Its endpoint values give $H(1)=\pi /2$, $H(-1)=-\pi /2$. Imposing $f(1)=1$ and $f(-1)=-1$ therefore gives $A=0$, $C=2/\pi$:Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to 0$; at upper infinity $H\sim Z$. The real intervals $(0,1)$ and $(-1,0)$ trace the channel walls $x=1$ and $x=-1$ downward, while the outer intervals trace the horizontal boundary. This verifies the stated end assignments as well as the constants. The map is the Schwarz-Christoffel map of a half-plane joined to a strip.

The two holomorphic function functions agree throughout their overlap region strip, so they define a single holomorphic function on the union of their domains, which is the whole complex plane. The resulting entire function is bounded outside a sufficiently large disk by the assumptions at infinity. It is bounded on the disk by continuity and compactness. Liouville theorem therefore makes it constant:Agreement on an open overlap region is what permits analytic continuation; equality of unspecified boundary values alone would need additional hypotheses.

The positive denominators in this part are Euclidean. Starting with the Minkowski spacetime Feynman rules, first perform a Wick rotation; a bare Minkowski metric integral without a pole prescription would not equal the displayed positive expression. Work initially with $m^2>0$ and $\Re n>D/2$, where the Euclidean massive loop integral converges.

Use Schwinger parameterization followed by the Gaussian integral:The remaining Gamma function integral givesThe factorial form is for positive integral $n$. Dimensional regularization defines the result elsewhere by analytic continuation.

For the scalar bubble integral, combine the two denominators with a Feynman parameter $x$ and shift the integration variable to $\ell =k-(1-x)p_E$. ThenFor $D=4-ϵ$ this isThe Gamma function recurrence gives $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral is $1+O(ϵ)$. ConsequentlyThis ultraviolet divergence is independent of the external momentum. The assumption $m^2>0$ avoids an infrared divergence. In particular the completely massless zero-momentum integral is a scaleless integral and vanishes in dimensional regularization; its ultraviolet and infrared poles must be separated before assigning it the displayed ultraviolet residue.

It is convenient to evaluate the one-particle irreducible vertices using the Euclidean action and then translate the counterterms back to the Minkowski spacetime Lagrangian density. For the tadpole integral, analytic continuation of solution givesIndeed $\Gamma (-1+ϵ/2)=\Gamma (ϵ/2)/(ϵ/2-1)=-2/ϵ+O(1)$. The negative pole is a property of analytic continuation, not the value of a convergent positive integral.

The Euclidean space inverse two-point vertex has the one-loop correctionThe factor $1/2$ is the symmetry factor weight of the tadpole diagram. A mass counterterm $\delta m^2={\lambda }_0{m}^2/(16{\pi }^2ϵ)$ cancels this pole. There is no momentum-dependent pole at this order, and hence no one-loop wave-function renormalization.

For the four-point vertex, the three bubble diagrams giveOne can check the relative minus sign independently from the one-loop effective action: its $\frac{1}{2}\mathrm{Tr}\log (1+G{\lambda }_0{\varphi }^2/2)$ term has a positive term linear in ${\lambda }_0{\varphi }^2$ and a negative term quadratic in it. Thus $\delta \lambda =3{\lambda }_0^2/(16{\pi }^2ϵ)$ cancels the quartic pole.

In the Euclidean action, these two counterterms are $+\delta m^2{\varphi }^2/2+\delta \lambda {\varphi }^4/4!$. Returning to the Minkowski spacetime Lagrangian density givesFor a direct sign check with Feynman rules, the two amputated loop pole insertions are $+i{\lambda }_0{m}^2/(16{\pi }^2ϵ)$ and $+i3{\lambda }_0^2/(16{\pi }^2ϵ)$, while the counterterm insertions are $-i\delta m^2$ and $-i\delta \lambda$.

As usual, the displayed pole residues suppress renormalization scale powers: $[{\lambda }_0]=ϵ$ away from four dimensions. The next part restores these powers explicitly with a dimensionless renormalized coupling. Replacing the tree coupling by its renormalized counterpart inside a one-loop residue changes only higher-loop terms.

Source correction: equation 11 on the original page has $\Gamma (\alpha -1/2)$. The dimensionally regulated integral instead contains $\Gamma (\alpha -d/2)$. This is not an interchangeable convention: for $\alpha =2$ and $d=4-ϵ$, the correct factor $\Gamma (ϵ/2)$ has a logarithmic ultraviolet pole, whereas the printed $\Gamma (3/2)$ is finite.

To obtain the correct expression, shift the loop momentum and perform a Wick rotation. For positive Euclidean ${\mathcal{M}}^2$, introduce a Schwinger parameterization:The derivation first holds in the convergent range and defines the result elsewhere by analytic continuation. For integer $\alpha$, rotating back giveswith the branch inherited from $i0$. This is the Lorentzian continuation of the Euclidean massive loop integral.

Set $p=0$, $\alpha =1$, $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,Consequently the directed tadpole in part d givesThe counterterm is $-i{\delta }_{m^2}$, so cancelling only the pole in the minimal subtraction scheme requiresThis is the quartic complex-scalar mass renormalization result. The scale factors and logarithms contribute finite terms at this order and do not change the displayed pole coefficient. A real-field check gives a tadpole coefficient $\frac{1}{2}(3g/2+g/2)=g$, in agreement with the oriented complex-field computation.

For a normalized state the position probability density is $\boxed{f=|\psi {|}^2}$. With a real potential, the Time-dependent Schrödinger equation and its conjugate giveThis is the continuity equation $\boxed{f_t+j_x=0}$. Integrating over the line yields $d\Vert \psi {\Vert }^2/dt=-j(\infty )+j(-\infty )=0$ under the stated decay. A real-valued wave function gives $\overline{\psi }{\psi }_x-\psi \overline{{\psi }_x}=0$, so its instantaneous current vanishes; the same is true up to a spatially constant phase.

Put $\alpha =m\omega /\hslash$. Complete the square in the overlap:The Gaussian integral is $\sqrt{\pi /\alpha }$, also for complex displacement by analytic continuation. HenceThe product of square roots fixes the branch convention implicit in the printed square root of $\overline{u}s$. At $u=s\ne 0$ the norm squared is $|N{|}^2|s|\sqrt{\pi /\alpha }{e}^{2\alpha |s{|}^2}$, which becomes one for the stated $N$. In particularThe centre is $-2\mathrm{Re}s$ and its position variance is $1/(2\alpha )$.

When differentiating in $s$, keep $N$ fixed on the chosen constant-radius orbit. ThenSubstitution gives $HG=\hslash \omega s{G}_s$. Since $\dot{s}=-i\omega s$, the chain rule gives $\boxed{i\hslash G_t=HG}$. Continue $\sqrt{s(t)}$ continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point phase. The case of zero displacement is obtained as a limit, since the separate printed $N$ and $\sqrt{s}$ expressions are singular at $s=0$.

Finally $j=(\hslash /m)\mathrm{Im}(\overline{G}{G}_x)=-2\omega \mathrm{Im}s|G{|}^2$. For real nonzero $a$ this isThe printed final current omits the Gaussian normalization factor $\sqrt{m\omega /(\pi \hslash )}$. Retaining that factor is required both by the normalized state and by the continuity equation. This current is the Gaussian probability density times the velocity of its moving centre, $2a\omega \sin \omega t$.

A branch point obstructs a single-valued analytic continuation around a closed circuit. Continuing $\log z=\log |z|+i\theta$ once around zero increases its value by $2\pi i$. A branch cut removes a curve joining branch points or extending to infinity so that a continuous argument can be chosen on the remainder. Zero and infinity are logarithmic branch points; the positive-axis cut used here selects $0<\arg z<2\pi$. The cut is a choice of representation, not a line of poles of the multivalued logarithm.

On the physical sheet the bound-state amplitude poles lie at $k=i{\kappa }_j$, ${\kappa }_j>0$, and map to real negative $E_j=-{\hslash }^2{\kappa }_j^2/(2m)$. The square-root map has two sheets meeting at the continuum threshold. Analytic continuation can also have virtual-state poles on the negative imaginary $k$ axis and resonance poles in the lower half-plane; these are not normalizable bound states. In the energy plane the resonances belong to the nonphysical sheet, rather than additional negative-axis bound levels.

For the algebraically continued probability, include the mirror poles contributed by $D(-k)$ as explained in part d; the two $k$ signs map to the same numerical energy but need not impose the same boundary condition. A sketch must label the sheet or the amplitude being plotted, rather than treating every singularity as a bound state.

A branch point is a point around which analytic continuation changes the value of a multivalued function. For $z^{1/2}=r^{1/2}{e}^{i\theta /2}$, one circuit around zero changes the sign. A branch cut removes paths producing this ambiguity; for example $-\pi <\arg z<\pi$ gives a single holomorphic square root off the nonpositive real axis. The position of a cut is a convention, while the branch point is intrinsic.

Set $q(z)=(z-1)/(z+1)$. For $z$ outside the closed segment joining $-1$ to $1$, $q(z)$ avoids the nonpositive real axis. Therefore the principal square root of $q$ defines a holomorphic function there, with the normalization $f(\infty )=1$. The endpoints are branch points. Expanding at infinity,For a positively oriented contour surrounding the cut, its integral equals the integral on a sufficiently large positively oriented circle, since the intervening annulus is holomorphic. ThusOn the upper and lower banks, $f_{+}(x)=+i\sqrt{(1-x)/(1+x)}$ and $f_{-}(x)=-i\sqrt{(1-x)/(1+x)}$. The upper bank is traversed from $1$ to $-1$, and the lower bank from $-1$ to $1$. Their sum is $-2i$ times the required real integral. Small endpoint arcs vanish because their singularities have exponent less than one. ConsequentlyFor $\alpha =1/n$, use $f_{\alpha }=q^{\alpha }$ with the same normalization. Its expansion is $1-2\alpha /z+O(z^{-2})$, and its bank values are $e^{\pm i\pi \alpha }[(1-x)/(1+x){]}^{\alpha }$. Equating the contour integral $-4\pi i\alpha$ with $-2i\sin (\pi \alpha )$ times the real integral givesThe condition $n\ge 2$ ensures endpoint integrability. Choosing the other square-root branch reverses $I_{\Gamma }$, but also reverses both bank values, leaving the real integral unchanged.

Write $\nu =m\gamma /{\hslash }^2$, so $t(k)\simeq k/(k+i\nu )$. Its nontrivial pole isFor negative $\gamma$ this is on the positive imaginary $k$ axis, the physical bound-state sheet. For positive $\gamma$ it lies on the negative imaginary axis and is a virtual-state pole reached by analytic continuation through the continuum cut. These different sheets can have the same negative numerical energy; specifying only $E_{\ast }$ would conceal the distinction. As a function of $E$, the momentum square root also introduces the continuum threshold branch point at zero. If $\gamma =0$, numerator and denominator cancel at this order and no pole is predicted by this leading approximation.

The delta function in the specified convolution kernel collapses the ${\mathbf{k}}^{\prime }$ integral to $-i\gamma G$. Set $\lambda =e\gamma$ and ${\omega }_0^2=k^2+m^2$. ThenThere is a change of assumptions here: the specified kernel is the transform of $V=-i\gamma$, which is not real. These later parts define a formal analytic continuation to an effective damped equation, $[{\partial }_{\tau }^2+2\lambda {\partial }_{\tau }+{\omega }_0^2]G=\delta (\tau )$; they cannot simultaneously describe the earlier real external potential. The following pole analysis uses that continued operator.

Completing the square gives $D(\omega )=-(\omega +i\lambda {)}^2+({\omega }_0^2-{\lambda }^2)$, soFor $e>0$, hence $\lambda >0$, the underdamped regime has two poles at $\pm \Omega -i\lambda$, with $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. Both lie below the real axis. In the overdamped regime they lie on the imaginary axis at $-i(\lambda \mp \kappa )$, where $\kappa =\sqrt{{\lambda }^2-k^2-m^2}<\lambda$ for $k^2+m^2>0$; both are again below the real axis. At equality they coalesce into a double pole at $-i\lambda$. If $m<|\lambda |$, the crossover occurs at $k_c=\sqrt{{\lambda }^2-m^2}$; increasing $k$ moves the two imaginary poles together and then apart horizontally. If $m>|\lambda |$, every $k$ is underdamped; if equal, only $k=0$ is critical.

The sign of $e$ is not specified: for $e<0$ all nonzero-frequency poles are instead in the upper half-plane, describing antidamping; for $e=0$ they lie on the real axis and a causal prescription is required. When $m=k=0$, one pole is at zero even for nonzero $\lambda$ and a real-axis prescription is again needed. The figure uses illustrative $\lambda =1$, not supplied numerical data.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

A free acoustic membrane wave is a zero of the outgoing dispersion relation $D$. At this zero, the impedance seen from one side equals the negative of that side's outgoing normal acoustic impedance. Thus the continued by analytic continuation reflection coefficient has a pole. A real guided mode is evanescent normal to the sheet; a complex root can describe a leaky mode. The pole describes nontrivial homogeneous motion with no incoming wave, rather than unbounded passive reflection at an ordinary real incidence angle.

The set of complex numbers $s$ for which the displayed integral is finite is the region of absolute convergence of a Laplace transform. It depends only on the real part $\sigma =\mathrm{Re}s$. Once the condition holds for one $\sigma$, it holds for every larger real part. For the exponential function $e^{-2t}$ it is exactly $\mathrm{Re}s>-2$; in general, convergence on the boundary must be checked separately. An analytic continuation outside this set need not equal a convergent defining integral.

For positive $Z$, the displayed identity follows by differentiating $Z^n=e^{n\log Z}$ at zero.

Integer powers of a partition function represent independent copies with the same disorder. Integer moments alone do not justify analytic continuation or interchanging disorder averages and thermodynamic limits.^[ref]

On a simply connected domain $D$, a germ of a holomorphic function admitting analytic continuation along every path from its base point determines a unique holomorphic function $F:D\to \mathbb{C}$ with that initial germ of a holomorphic function. ^[ref]

Indeed, any two paths with the same endpoints are homotopic with endpoints fixed. Define $F(b)$ using the terminal germ of a holomorphic function. The monodromy theorem makes this independent of the path. In a small open disc about $b$, extend the chosen route by a short segment; the same local representative supplies $F$ throughout the open disc, proving holomorphy. The identity theorem proves uniqueness. The existence hypothesis of the monodromy theorem must still be checked.

Zeta function regularization assigns a finite value to a divergent spectral sum by first forming a convergent complex-power series and then using its analytic continuation. For positive frequencies ${\omega }_n$, continue ${\sum }_n{\omega }_n^{-s}$ to $s=-1$ to define a regulated frequency sum, when that continuation is regular there. This does not turn the divergent positive-term sum into an ordinary convergent sum. It is a useful subtraction convention for the normal-ordering constant of a string and for vacuum energies.