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A branch cut is a curve removed from the complex plane so that a multivalued function can be treated as single-valued on the remaining domain. The cut is a convention; the branch points and the sheet structure are the underlying analytic facts.

Branch cuts matter in quantum mechanics because square roots, logarithms, resolvents, Green functions, scattering amplitudes, and special functions often become multivalued after analytic continuation.

Write a nonzero complex number as

z=reiθ,r>0.z=re^{i\theta}, \qquad r>0.

The angle is not unique:

θ∼θ+2πn,n∈Z.\theta \sim \theta+2\pi n, \qquad n\in\mathbb Z.

Therefore the logarithm has many possible values:

log⁡z=ln⁡r+i(θ+2πn).\log z = \ln r+i(\theta+2\pi n).

Similarly,

z=r ei(θ+2πn)/2\sqrt z = \sqrt r\,e^{i(\theta+2\pi n)/2}

has two values, differing by a sign. Going once around the origin changes the square root to its negative.

A branch is a single-valued choice of a multivalued function on a chosen domain. A branch cut is a curve removed from the plane to make such a choice possible.

For the principal logarithm, one common choice is

−π<Arg⁡z<π,-\pi<\operatorname{Arg}z<\pi,

with a cut along the negative real axis. Then

Log⁡z=ln⁡∣z∣+iArg⁡z\operatorname{Log}z = \ln\lvert z\rvert +i\operatorname{Arg}z

is single-valued on the cut plane.

The cut itself is not unique. One could instead cut along the positive real axis or along another ray from the origin. Different choices can be equally valid if used consistently.

A branch point is a point around which analytic continuation changes the value of the function. For log⁡z\log z and z\sqrt z, the origin is a branch point. Infinity is also part of the branch structure for these functions.

The diagnostic is monodromy: continue the function around a closed loop. If the value changes after returning to the starting point, the loop enclosed a branch point.

For example,

z1/2⟶−z1/2z^{1/2} \longrightarrow -z^{1/2}

after one loop around z=0z=0. A second loop returns to the original value.

For the principal logarithm with the cut on the negative real axis, take r>0r>0. Approaching the cut from above gives

Log⁡(−r+i0)=ln⁡r+iπ,\operatorname{Log}(-r+i0) = \ln r+i\pi,

while approaching from below gives

Log⁡(−r−i0)=ln⁡r−iπ.\operatorname{Log}(-r-i0) = \ln r-i\pi.

The discontinuity is

Disc⁡Log⁡(−r)=2πi.\operatorname{Disc}\operatorname{Log}(-r) = 2\pi i.

For the principal square root,

−r+i0=ir,−r−i0=−ir.\sqrt{-r+i0} = i\sqrt r, \qquad \sqrt{-r-i0} = -i\sqrt r.

The two boundary values are different sheets meeting at the cut.

A branch cut is not an isolated singularity. It restricts the domain so a chosen branch cannot change value around a closed loop. The branch point is the obstruction.

This distinction matters for contour integration:

  • an isolated pole contributes a residue;
  • a branch cut contributes an integral over the discontinuity across the cut;
  • a branch point may require a small contour around it;
  • a contour cannot be freely pushed across a cut without accounting for the jump.

This is why Residue Theorem does not by itself evaluate branch-cut integrals.

Worked example: integrating along both sides of a cut

Section titled “Worked example: integrating along both sides of a cut”

The discontinuity becomes useful when the two sides of a contour trace the same real interval. Consider the dimensionless integral

I(a)=∫0∞xa−11+x dx,0<a<1.I(a)=\int_0^\infty\frac{x^{a-1}}{1+x}\,dx, \qquad 0<a<1.

The condition a>0a>0 ensures integrability at zero; a<1a<1 ensures integrability at infinity. Choose

f(z)=za−11+z,za−1=exp⁡[(a−1)(ln⁡∣z∣+iarg⁡z)],0<arg⁡z<2π.f(z)=\frac{z^{a-1}}{1+z}, \qquad z^{a-1}=\exp[(a-1)(\ln|z|+i\arg z)], \qquad 0<\arg z<2\pi.

This branch has its cut on the positive real axis. Use a positively oriented keyhole contour: travel outward just above the cut from ε\varepsilon to RR, around the outer circle counterclockwise, inward just below the cut, then around the inner circle clockwise. Take 0<ε<1<R0<\varepsilon<1<R; the only enclosed pole is z=−1z=-1, with argument π\pi.

The boundary values and the opposite directions of the two straight pieces give

f(x+i0)=xa−11+x,f(x−i0)=e2πiaxa−11+x,∫upperf dz+∫lowerf dz=(1−e2πia)∫εRxa−11+x dx.\begin{aligned} f(x+i0)&=\frac{x^{a-1}}{1+x}, &f(x-i0)&=e^{2\pi ia}\frac{x^{a-1}}{1+x},\\ \int_{\text{upper}}f\,dz+\int_{\text{lower}}f\,dz &=(1-e^{2\pi ia})\int_\varepsilon^R\frac{x^{a-1}}{1+x}\,dx. \end{aligned}

The lower contribution carries both a branch phase and a minus sign from its direction. Meanwhile, the arc-length estimate gives

∣∫∣z∣=εf dz∣≤2πεa1−ε,∣∫∣z∣=Rf dz∣≤2πRaR−1.\left|\int_{|z|=\varepsilon}f\,dz\right| \le\frac{2\pi\varepsilon^a}{1-\varepsilon}, \qquad \left|\int_{|z|=R}f\,dz\right| \le\frac{2\pi R^a}{R-1}.

Both bounds vanish in the required limits precisely in the stated range. The residue is eiπ(a−1)e^{i\pi(a-1)}, so the residue theorem yields

(1−e2πia)I(a)=2πieiπ(a−1),I(a)=πsin⁡πa.(1-e^{2\pi ia})I(a)=2\pi i e^{i\pi(a-1)}, \qquad I(a)=\frac{\pi}{\sin\pi a}.

The result is positive, as the original integrand requires. Independently, the substitution t=x/(1+x)t=x/(1+x) gives I(a)=B(a,1−a)I(a)=B(a,1-a), agreeing with the beta integral and gamma reflection formula in Gamma and Beta Functions. The calculation uses a residue at the ordinary pole and the jump along the cut; it assigns no residue to the excluded branch point. For another worked contour of this type, see Dunkel, §11.4, Example 11.8.

A multivalued function can be represented as a single-valued function on a multi-sheeted surface. Each sheet corresponds to a consistent branch. Crossing a branch cut moves from one sheet to another.

For z\sqrt z, there are two sheets. For log⁡z\log z, there are infinitely many sheets. The branch cut is a way of drawing a boundary in one sheet; the full analytic continuation remembers the other sheets.

In scattering theory, the phrase “physical sheet” means the sheet reached from physical boundary values by the convention appropriate to the problem. Resonance poles often lie on an unphysical sheet reached by continuing through a branch cut.

For relative motion with reduced mass μ\mu,

E=ℏ2k22μ,E = \frac{\hbar^2k^2}{2\mu},

so

k(E)=2μEℏ.k(E) = \frac{\sqrt{2\mu E}}{\hbar}.

The square root makes E=0E=0 a branch point. The positive real energy axis is the physical continuum above threshold. Continuing around the threshold changes the sign of kk and therefore moves to a different sheet.

The cut need not be the principal square-root cut used earlier. For the outgoing free resolvent, a convenient physical-sheet convention is 0<arg⁡E<2π0<\arg E<2\pi, with the cut on the positive energy axis. It gives k(E+i0)=+2μE/ℏk(E+i0)=+\sqrt{2\mu E}/\hbar and k(E−i0)=−2μE/ℏk(E-i0)=-\sqrt{2\mu E}/\hbar for E>0E>0, while k=i2μ∣E∣/ℏk=i\sqrt{2\mu|E|}/\hbar below threshold. Thus Im⁡k>0\operatorname{Im}k>0 away from the cut on this sheet, selecting spatial decay. This is why the complex kk plane is often cleaner than the complex EE plane near a single nonrelativistic threshold. In the kk plane, the two energy sheets are unfolded.

For a self-adjoint Hamiltonian, the resolvent

R(z)=(z−H)−1R(z) = (z-H)^{-1}

is analytic away from the spectrum. For common Schrödinger operators, suitable matrix elements or weighted-space versions of the resolvent have boundary values on regular portions of the continuum, written

R(E+i0),R(E−i0).R(E+i0), \qquad R(E-i0).

These limits are not bounded-operator limits on the full Hilbert space at a spectral point. Singular thresholds may require separate analysis. Where the boundary values exist, the continuum often appears as a cut in the spectral parameter. The discontinuity across the cut is tied to the spectral density. This is the operator-level version of why branch cuts represent continua rather than isolated states.

Physical elastic scattering is measured on real positive energies, but poles associated with bound states, virtual states, and resonances are classified by analytic continuation.

For short-range one-channel scattering:

  • bound states appear as poles on the physical sheet below threshold;
  • virtual states appear on a different sheet near threshold;
  • resonances usually appear as poles at complex energy on an unphysical sheet.

The pole location alone is not enough unless the sheet is specified. This is why scattering pages carefully distinguish bound states, virtual states, and resonances rather than calling every nearby pole a resonance.

Special functions often inherit branch cuts from their defining expressions or from analytic continuation of differential-equation solutions. Hypergeometric functions, Bessel functions of noninteger order, Legendre functions, logarithms in Green functions, and square roots in WKB formulas all require branch choices.

The practical questions are:

  • where is the chosen cut;
  • which side of the cut defines the physical boundary value;
  • what is the discontinuity across the cut;
  • which sheet contains the pole or saddle being discussed.

For calculation-heavy special-function pages, the branch convention should be stated before using continuation formulas.

  • Treating the branch cut as unique rather than conventional.
  • Forgetting that the branch point, not the drawn cut, is the intrinsic obstruction.
  • Moving a contour across a branch cut without adding the discontinuity contribution.
  • Applying residue formulas to a branch point.
  • Referring to a pole without specifying the sheet when analytic continuation is involved.
  • Using a square root or logarithm from software without checking its branch convention.
  • Confusing an i0i0 boundary value with a small physical damping term in every context.
  1. Show that one loop around the origin changes the sign of z\sqrt z.
Solution

Write

z=reiθ.z=re^{i\theta}.

Then

z=r eiθ/2.\sqrt z = \sqrt r\,e^{i\theta/2}.

After one loop, θ↦θ+2π\theta\mapsto\theta+2\pi, so

r ei(θ+2π)/2=r eiθ/2eiπ=−r eiθ/2.\sqrt r\,e^{i(\theta+2\pi)/2} = \sqrt r\,e^{i\theta/2}e^{i\pi} = -\sqrt r\,e^{i\theta/2}.

The sign changes.

  1. Compute the jump of the principal logarithm across the negative real axis.
Solution

For r>0r>0,

Log⁡(−r+i0)=ln⁡r+iπ,\operatorname{Log}(-r+i0) = \ln r+i\pi,

and

Log⁡(−r−i0)=ln⁡r−iπ.\operatorname{Log}(-r-i0) = \ln r-i\pi.

Therefore

Disc⁡Log⁡(−r)=2πi.\operatorname{Disc}\operatorname{Log}(-r) = 2\pi i.
  1. Why is E=0E=0 a branch point for k(E)=2μE/ℏk(E)=\sqrt{2\mu E}/\hbar?
Solution

The square root changes sign after analytic continuation once around E=0E=0. Since k(E)k(E) is proportional to E\sqrt E, going around the threshold sends kk to −k-k. Thus E=0E=0 is a branch point connecting two sheets.

  1. Why can a branch cut represent a continuum in a Green function?
Solution

For a continuous spectrum, the resolvent has different boundary values above and below the spectral interval. The discontinuity across that interval is tied to the spectral density. Unlike an isolated eigenvalue, which produces a pole, a continuum produces extended nonanalytic structure, naturally represented as a cut in the spectral parameter.

  1. A resonance pole is reported at a complex energy, but no sheet is specified. What is missing?
Solution

The sheet of analytic continuation is essential. The same complex energy coordinate can represent different branches of the analytically continued amplitude. A resonance pole is usually on an unphysical sheet reached through a branch cut, while bound-state poles lie on the physical sheet below threshold. Without the sheet, the pole classification is incomplete.

  1. For dimensionless b>0b>0 and 0<a<10<a<1, evaluate J(a,b)=∫0∞xa−1/(x+b) dxJ(a,b)=\int_0^\infty x^{a-1}/(x+b)\,dx. Check a=1/2a=1/2, and explain why analytic continuation of the answer does not give the ordinary improper integral at a=3/2a=3/2.
Solution

Set x=bux=bu, including dx=b dudx=b\,du. The result is

J(a,b)=ba−1I(a)=πba−1sin⁡πa.J(a,b)=b^{a-1}I(a)=\frac{\pi b^{a-1}}{\sin\pi a}.

At a=1/2a=1/2, this is π/b\pi/\sqrt b. The independent substitution x=bt2x=bt^2 gives J(1/2,b)=2b−1/2∫0∞(1+t2)−1dtJ(1/2,b)=2b^{-1/2}\int_0^\infty(1+t^2)^{-1}dt, with the same value. At a=3/2a=3/2, the integrand behaves as x−1/2x^{-1/2} at infinity, so the integral diverges. The meromorphically continued expression would be −πb-\pi\sqrt b; it is not the value of this positive divergent integral. In the contour argument, the large-circle estimate no longer vanishes.

  • L. V. Ahlfors, Complex Analysis, 3rd ed., McGraw-Hill, 1979.
  • G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
  • J. W. Brown and R. V. Churchill, Complex Variables and Applications, 9th ed., McGraw-Hill, 2014.
  • J. Dunkel, 18.04 Complex Analysis with Applications, MIT lecture notes, Spring 2019, §11.4, Integrands with Branch Cuts, pp. 146–147; adapted from course materials by A. Nachbin and J. Orloff.
  • NIST, Digital Library of Mathematical Functions, §5.12: Beta Function, especially Eq. (5.12.3).
  • R. G. Newton, Scattering Theory of Waves and Particles, 2nd ed., Dover, 2002.
  • E. M. Stein and R. Shakarchi, Complex Analysis, Princeton University Press, 2003.
  • J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.