Contact Scattering and Renormalization in Quantum Mechanics

The previous page introduced effective actions as local long-distance descriptions. This page studies the smallest possible laboratory for that idea: a nonrelativistic particle scattering from a pointlike potential,

$$V(\boldsymbol r)=g_0\delta^{(d)}(\boldsymbol r).$$

This model is deliberately crude. A literal delta-function potential tries to replace all short-distance structure by one number. That replacement is harmless in one spatial dimension, logarithmically delicate in two, and power-divergent in three. The lesson is exactly the one QFT needs: a local interaction is not specified by its bare coefficient alone; it is specified by a prescription for how that coefficient is matched to physical observables as the cutoff is changed.

The calculation is also a perfect preview of loop diagrams. Repeated scattering from the fixed contact potential gives a geometric resolvent series. The divergent object is the free Green function evaluated at coincident points. In a nonrelativistic field theory the same algebra becomes a bubble sum after the loop energy is integrated out. Renormalization means replacing the singular bare strength $g_0$ by a measurable low-energy parameter such as a scattering length or a binding energy. The derivation keeps three objects distinct: the cutoff-dependent bare coupling $g_0(\Lambda)$, the coincident-point Green function $I_d(E,\Lambda)$, and cutoff-independent observables such as pole positions, the three-dimensional scattering length $a$, or the two-dimensional binding energy $B_2$.

The contact potential as a resolvent problem

Resolvent and scattering normalization. We use $d$ for the number of spatial dimensions. The nonrelativistic Hamiltonian is

$$H=H_0+V, \qquad H_0={\boldsymbol p^2\over 2m},$$

and momentum states obey

$$\langle \boldsymbol k|\boldsymbol k'\rangle=(2\pi)^d\delta^{(d)}(\boldsymbol k-\boldsymbol k').$$

For positive energy,

$$E={p^2\over 2m}, \qquad p>0,$$

we define the outgoing free resolvent by

$$G_0(E;\boldsymbol k)={1\over E-\boldsymbol k^2/(2m)+i0}.$$

The $T$ matrix is defined by the Lippmann–Schwinger equation $T=V+VG_0T$. Overall signs in the scattering amplitude differ across books because some define $f$ with an extra minus sign relative to $T$. The pole locations, cutoff dependence, and low-energy denominators below are convention-independent.

The exact scattering state with incoming momentum $\boldsymbol p$ satisfies

$$|\psi_{\boldsymbol p}^{(+)}\rangle =|\boldsymbol p\rangle+G_0(E+i0)V|\psi_{\boldsymbol p}^{(+)}\rangle.$$

Equivalently, the $T$ matrix obeys

$$T(E)=V+VG_0(E)T(E).$$

In momentum space this reads

$$T(\boldsymbol p',\boldsymbol p;E) = V(\boldsymbol p',\boldsymbol p) + \int^\Lambda {d^dk\over(2\pi)^d}\, V(\boldsymbol p',\boldsymbol k) {1\over E-\boldsymbol k^2/(2m)+i0} T(\boldsymbol k,\boldsymbol p;E),$$

where a cutoff $\Lambda$ has been inserted to make the short-distance question explicit.

For the contact potential,

$$V(\boldsymbol p',\boldsymbol p)=g_0,$$

so the solution is independent of the external momenta. We can write

$$T(\boldsymbol p',\boldsymbol p;E)=T(E).$$

The Lippmann–Schwinger equation collapses to the scalar equation

$$T(E)=g_0+g_0 I_d(E,\Lambda)T(E),$$

where

$$I_d(E,\Lambda) = \int^\Lambda {d^dk\over(2\pi)^d}\,{1\over E-\boldsymbol k^2/(2m)+i0}.$$

Thus

$$\boxed{ T(E)=\frac{g_0}{1-g_0I_d(E,\Lambda)} =\frac{1}{g_0^{-1}-I_d(E,\Lambda)}. }$$

The perturbation series is

$$T(E)=g_0+g_0I_d(E,\Lambda)g_0+g_0I_d(E,\Lambda)g_0I_d(E,\Lambda)g_0+\cdots.$$

Repeated scattering from a contact interaction gives a geometric series. The whole ultraviolet problem is contained in the coincident Green function $I_d(E)=G_0(E;\boldsymbol r=0)$.

In coordinate space the same object is even more transparent. Since

$$V(\boldsymbol r)=g_0\delta^{(d)}(\boldsymbol r),$$

the scattering wavefunction has the form

$$\psi^{(+)}_{\boldsymbol p}(\boldsymbol r) =e^{i\boldsymbol p\cdot\boldsymbol r} +g_0G_0(E;\boldsymbol r)\psi^{(+)}_{\boldsymbol p}(0),$$

where

$$G_0(E;\boldsymbol r) = \int^\Lambda {d^dk\over(2\pi)^d}\, {e^{i\boldsymbol k\cdot\boldsymbol r}\over E-\boldsymbol k^2/(2m)+i0}.$$

Setting $\boldsymbol r=0$ gives

$$\psi^{(+)}_{\boldsymbol p}(0) =1+g_0G_0(E;0)\psi^{(+)}_{\boldsymbol p}(0),$$

so the same denominator appears. The singularity of the zero-range potential is the singularity of the free Green function at coincident points.

Why the short-distance region appears

Write

$$E={p^2\over2m}.$$

Then

$$I_d(E,\Lambda) =2m\int_{|\boldsymbol k|<\Lambda}{d^dk\over(2\pi)^d}\,{1\over p^2-\boldsymbol k^2+i0}.$$

At large $k$, the integrand behaves as

$${1\over E-k^2/(2m)+i0}\sim -{2m\over k^2}.$$

The measure contributes $k^{d-1}dk$, so the high-momentum behavior is

$$I_d\sim -2m\int^\Lambda dk\,k^{d-3}.$$

Therefore:

$$\begin{array}{c|c|c} \text{spatial dimension} & \text{UV behavior of }I_d & \text{meaning} \\ \hline d=1 & \text{finite} & \text{a delta potential is a genuine potential} \\ d=2 & \log\Lambda & \text{a scale is generated by renormalization} \\ d=3 & \Lambda & \text{the bare coupling must be tuned} \\ d>3 & \Lambda^{d-2} & \text{more short-distance data are needed} \end{array}$$

This is the quantum-mechanical version of power counting. The contact interaction is more singular in higher dimension because the wavefunction has more angular phase space available near the origin.

There is also a useful relation to the relativistic propagator. A scalar Feynman propagator is

$$G_{\mathrm{rel}}(p)={i\over p_0^2-\boldsymbol p^2-m^2+i0}.$$

Near the positive-energy mass shell, write $p^0=m+E$ with $E\ll m$. Then

$$p_0^2-\boldsymbol p^2-m^2 =(m+E)^2-\boldsymbol p^2-m^2 \simeq 2m\left(E-{\boldsymbol p^2\over2m}\right),$$

so

$$G_{\mathrm{rel}}(p) \simeq {i\over 2m}\,{1\over E-\boldsymbol p^2/(2m)+i0}.$$

The nonrelativistic resolvent is the positive-energy, low-velocity limit of the relativistic propagator, up to the conventional normalization factor $2m$.

One spatial dimension: finite point scattering

In one dimension the integral can be evaluated with no UV cutoff:

$$I_1(E) =2m\int_{-\infty}^{\infty}{dk\over2\pi}\,{1\over p^2-k^2+i0}.$$

Using

$${1\over p^2-k^2+i0} =\operatorname{PV}{1\over p^2-k^2}-i\pi\delta(p^2-k^2),$$

the symmetric principal-value limit is zero, and

$$\delta(p^2-k^2)={1\over2p}\left[\delta(k-p)+\delta(k+p)\right].$$

Hence

$$I_1(E)=-{im\over p}.$$

The exact contact $T$ matrix is therefore

$$\boxed{ T_1(p)=\frac{1}{g_0^{-1}+im/p} =\frac{g_0}{1+img_0/p}. }$$

No cutoff dependence appears. The delta-function potential is a well-defined self-adjoint interaction in one dimension.

The pole structure is also simple. A bound state has energy

$$E_{\mathrm{bound}}=-{\kappa^2\over2m}, \qquad \kappa>0,$$

which corresponds to $p=i\kappa$. The pole condition is

$$g_0^{-1}+{m\over\kappa}=0,$$

so

$$\kappa=-mg_0.$$

Thus a bound state exists for attractive coupling $g_0<0$. Its binding-energy magnitude and energy are

$$B_1={\kappa^2\over2m}={mg_0^2\over2}, \qquad E_{\mathrm{bound}}=-B_1.$$

This is the only dimension in which the naive contact coupling is already physical without any renormalization.

Two spatial dimensions: logarithmic contact scattering

In two dimensions,

$$I_2(E,\Lambda) =2m\int_{|\boldsymbol k|<\Lambda}{d^2k\over(2\pi)^2}\,{1\over p^2-k^2+i0}.$$

Using polar coordinates,

$$I_2(E,\Lambda) ={m\over\pi}\int_0^\Lambda dk\,{k\over p^2-k^2+i0}.$$

The radial integral gives

$$\int_0^\Lambda dk\,{k\over p^2-k^2+i0} =-{1\over2}\left[\log{\Lambda^2\over p^2}+i\pi\right]+O\left({p^2\over\Lambda^2}\right),$$

so

$$\boxed{ I_2(E,\Lambda) =-{m\over2\pi}\left[\log{\Lambda^2\over p^2}+i\pi\right] +O\left({mp^2\over\Lambda^2}\right). }$$

The $T$ matrix becomes

$${1\over T_2(p)} ={1\over g_0} +{m\over2\pi}\left[\log{\Lambda^2\over p^2}+i\pi\right] +O\left({mp^2\over\Lambda^2}\right).$$

A finite bare coupling cannot survive the limit $\Lambda\to\infty$. Instead define a renormalized coupling at a subtraction scale $\mu$ by

$$\boxed{ {1\over g_R(\mu)} ={1\over g_0(\Lambda)}+{m\over2\pi}\log{\Lambda^2\over\mu^2}. }$$

After taking the zero-range limit at fixed $g_R(\mu)$,

$${1\over T_2(p)} ={1\over g_R(\mu)} +{m\over2\pi}\left[\log{\mu^2\over p^2}+i\pi\right].$$

The physical amplitude cannot depend on the arbitrary scale $\mu$. Therefore $g_R(\mu)$ must run. Differentiating the definition at fixed bare coupling gives

$$\mu{d\over d\mu}{1\over g_R(\mu)}=-{m\over\pi}.$$

Equivalently,

$$\boxed{ \mu{d g_R\over d\mu}={m\over\pi}g_R^2. }$$

This is the first beta function in the course. It is not yet a relativistic QFT beta function, but the mechanism is the same: logarithmic UV sensitivity is traded for scale dependence of a coupling.

A more physical parametrization uses the positive binding-energy magnitude. For an attractive interaction, the pole condition at $p=i\kappa$ defines

$$B_2={\kappa^2\over2m}>0, \qquad E_{\mathrm{bound}}=-B_2.$$

Analytic continuation gives

$${1\over g_R(\mu)}+{m\over2\pi}\log{\mu^2\over\kappa^2}=0.$$

Eliminating $g_R(\mu)$ yields

$$\boxed{ T_2(E)= {2\pi/m\over \log(B_2/E)+i\pi} }$$

for $E>0$. More generally, away from the positive-energy cut the denominator is $\log[B_2/(-E)]$; its upper-edge value is $\log(B_2/E)+i\pi$. Every attractive two-dimensional zero-range interaction therefore carries one bound-state scale. The essential fact is that the dimensionless coupling $mg_0$ has produced the energy scale $B_2$: dimensional transmutation in its simplest nonrelativistic form.

Three spatial dimensions: scattering length and resonance

In three dimensions,

$$I_3(E,\Lambda) =2m\int_{|\boldsymbol k|<\Lambda}{d^3k\over(2\pi)^3}\,{1\over p^2-k^2+i0}.$$

The angular integral gives

$$I_3(E,\Lambda) ={m\over\pi^2}\int_0^\Lambda dk\,{k^2\over p^2-k^2+i0}.$$

For $0<p<\Lambda$, the principal value and the outgoing boundary prescription give

$$I_3(E,\Lambda) ={m\over\pi^2}\left[-\Lambda+{p\over2}\log{\Lambda+p\over\Lambda-p}\right] -{imp\over2\pi}.$$

Thus, for $p\ll\Lambda$,

$$\boxed{ I_3(E,\Lambda) =-{m\Lambda\over\pi^2}-{imp\over2\pi}+O\left({mp^2\over\Lambda}\right). }$$

Thus

$${1\over T_3(p)} ={1\over g_0(\Lambda)}+{m\Lambda\over\pi^2}+{imp\over2\pi}+O\left({mp^2\over\Lambda}\right).$$

The linear divergence is absorbed by defining the physical scattering length $a$ through

$$\boxed{ {1\over g_0(\Lambda)}+{m\Lambda\over\pi^2}={m\over2\pi a}. }$$

At finite sharp cutoff the matched $T$ matrix is exactly

$$T_{3,\Lambda}(p) =\frac{2\pi/m}{a^{-1}+ip-\delta_\Lambda(p)}, \qquad \delta_\Lambda(p)={p\over\pi}\log{\Lambda+p\over\Lambda-p}.$$

The cutoff correction belongs in the inverse amplitude:

$${1\over T_{3,\Lambda}(p)} ={m\over2\pi}\left[a^{-1}+ip-{2p^2\over\pi\Lambda} +O\left({p^4\over\Lambda^3}\right)\right].$$

The zero-range limit at fixed $a$ and $p$ is therefore

$$\boxed{ T_3(p)=\frac{2\pi/m}{a^{-1}+ip}. }$$

An expansion of $T_{3,\Lambda}$ relative to this limit additionally requires $|\delta_\Lambda(p)|\ll|a^{-1}+ip|$. This condition must be reconsidered on analytic continuation near a pole; an additive $O(p^2/\Lambda)$ error on $T$ would have neither the correct dimensions nor a uniform pole estimate.

The corresponding zero-range $s$-wave scattering amplitude is

$$f_0(p)=-{m\over2\pi}T_3(p) =-{1\over a^{-1}+ip} ={1\over -a^{-1}-ip},$$

and the total low-energy cross-section is

$$\sigma(p)=4\pi |f_0(p)|^2 ={4\pi\over a^{-2}+p^2}.$$

From fixed-center to two-body scattering

Here $m$ is the kinetic mass in $H_0=p^2/(2m)$. For two equal-mass particles of constituent mass $M$, replace it by the reduced mass $m=M/2$, so the relative energy is $E=p^2/M$. The Lippmann–Schwinger matrix element then has numerator $4\pi/M$. The contact-interaction amplitude uses $\mathcal A=(4\pi/M)f_0=-T$: its denominator has the opposite sign. These are the same scattering amplitude with an explicit normalization change. In the finite-cutoff model,

$$f_{0,\Lambda}(p)^{-1}=-a^{-1}-ip+\delta_\Lambda(p).$$

The special point $a^{-1}=0$ is a zero-energy resonance, often called the unitary limit. At that point

$$f_0(p)={i\over p}, \qquad \sigma(p)={4\pi\over p^2},$$

which saturates the $s$-wave unitarity bound.

In the zero-range limit, the three-dimensional contact amplitude is governed by the denominator $a^{-1}+ip$. For $a>0$ the pole at $p=i/a$ is a bound state; for $a<0$ it is a virtual state. The resonance limit $|a|\to\infty$ places the pole at threshold.

Keeping $a$ finite while $\Lambda$ changes requires a cutoff-dependent bare coupling,

$$g_0(\Lambda) ={1\over -m\Lambda/\pi^2+m/(2\pi a)}.$$

For large $\Lambda$,

$$g_0(\Lambda) =-{\pi^2\over m\Lambda} \left[1+{\pi\over2a\Lambda}+O\left({1\over(a\Lambda)^2}\right)\right].$$

The bare coupling goes to zero, but it goes to zero in a very specific way. A finite physical scattering length is obtained only by tuning $g_0(\Lambda)$ close to the cutoff-dependent critical curve.

In three dimensions, fixed physical scattering length means $g_0^{-1}(\Lambda)+m\Lambda/\pi^2=m/(2\pi a)$. Changing the cutoff changes the bare parameter while preserving $a$. At nonzero momentum the finite sharp cutoff leaves the inverse-amplitude correction $\delta_\Lambda(p)$; the matched amplitude approaches its zero-range limit as the cutoff grows.

This is the key renormalization lesson. A divergent bare expression can still define finite physics if the bare parameters are regarded as cutoff-dependent coordinates on a space of effective theories.

The tuning becomes especially transparent at the resonance $a^{-1}=0$:

$$g_{0,c}(\Lambda)=-{\pi^2\over m\Lambda}.$$

A large but finite scattering length means that $g_0(\Lambda)$ is close to this cutoff-dependent critical curve. In other words, the low-energy resonance is not produced by a large bare coupling; it is produced by a precise cancellation in $g_0^{-1}-I_3(0,\Lambda)$.

From a singular potential to an effective field theory

A point interaction in three dimensions is not a normal function-valued potential. It is better understood as a boundary condition or as the leading term of a low-energy effective theory.

A finite-range potential of range $R$ has an $s$-wave amplitude that can be expanded at low momentum as

$$f_0(p)={1\over -a^{-1}+{1\over2}r_ep^2-ip+O(p^4R^3)},$$

where $a$ is the scattering length and $r_e$ is the effective range. The zero-range contact limit captures $a$ and has $r_e=0$; the finite sharp-cutoff approximation above instead has the regulator contribution $r_e=4/(\pi\Lambda)$. A more accurate effective interaction includes derivative terms,

$$V_{\mathrm{eff}}(\boldsymbol p',\boldsymbol p) =C_0(\Lambda) +C_2(\Lambda)(\boldsymbol p^2+\boldsymbol p'^2) +C_4(\Lambda)(\boldsymbol p^4+\cdots)+\cdots.$$

The scattering-length sign, effective-range expansion and equal-mass Born normalization are given in Braaten and Hammer 2006, § 2.1, pp. 11–13, PDF. Their $m$ is the constituent mass $M$ in the translation above; the sharp-cutoff correction displayed here follows from our explicit principal-value integral.

The coefficients $C_0,C_2,\ldots$ are not determined by the zero-range idealization. They must be matched to physical low-energy data: scattering length, effective range, and higher threshold parameters. This is the nonrelativistic version of the Wilsonian operator expansion.

The same structure appears in a nonrelativistic field theory with constituent mass $M$ and Lagrangian

$$\mathcal L =\psi^\dagger\left(i\partial_t+{\nabla^2\over2M}\right)\psi -{C_0\over2}(\psi^\dagger\psi)^2 +{C_2\over2}\left[\nabla(\psi^\dagger\psi)\right]^2+\cdots.$$

The quartic vertex $C_0$ generates the same geometric bubble structure as the potential problem. For two equal constituent masses $M$, the relative kinetic denominator after the loop energy integral is $E-k^2/M+i0$. Thus the spatial resolvent is $I_d$ with its fixed-center mass replaced by $M/2$; the vertex and external-state normalizations must also be matched. The contact potential supplies a simple exact example of a local interaction whose repeated short-distance fluctuations require renormalization.

Locality versus microscopic detail

The local contact theory does not claim that the true microscopic force literally has zero range. It claims that for $pR\ll1$ the details of that force can be replaced by a controlled sequence of local operators. The first coefficient is fixed by $a$; the next is fixed by $r_e$; and so on.

Example: unitarity from the denominator

In three dimensions, elastic unitarity for a single $s$-wave channel implies

$$\operatorname{Im}{1\over f_0(p)}=-p \qquad (p>0).$$

The zero-range contact result gives

$$f_0(p)={1\over -a^{-1}-ip},$$

so

$${1\over f_0(p)}=-a^{-1}-ip.$$

Therefore

$$\operatorname{Im}{1\over f_0(p)}=-p,$$

exactly as required. The real part $-a^{-1}$ is dynamical data. The imaginary part $-p$ is fixed by open phase space and probability conservation. This is why the low-energy denominator $a^{-1}+ip$ is more robust than any particular cutoff calculation.

Relativistic preview: a small coupling and a large logarithm

The manuscript next applies the same lesson to four-dimensional $\phi^4$ theory. A one-loop correction to the four-point vertex contains the schematic Euclidean integral

$$\delta\Gamma^{(4)}(q) \sim \lambda_0^2 \int^\Lambda {d^4k\over(2\pi)^4} {1\over k^2(k+q)^2} \sim {\lambda_0^2\over16\pi^2} \log{\Lambda^2\over q^2},$$

up to channel-dependent combinatorial coefficients and local terms. The perturbative parameter is therefore not just $\lambda_0$, but

$${\lambda_0\over16\pi^2}\log{\Lambda^2\over q^2}.$$

When this combination becomes order one, terms with the highest power of the logarithm at each loop order must be resummed even if $\lambda_0$ is small. The next lessons develop the scalar loop integral, the one-loop four-point function, leading logarithms, and the renormalization-group equation in detail.

Summary

A contact interaction is the smallest model in which renormalization is unavoidable but completely explicit.

In one spatial dimension, $G_0(E;0)$ is finite and the delta-function potential is an ordinary exactly solvable interaction. In two dimensions, the coincident Green function diverges logarithmically, producing a running coupling and a dynamically generated scale. In three dimensions, the divergence is linear; the bare coupling must be tuned with the cutoff so that the physical scattering length remains fixed.

The exact contact amplitude is a geometric series,

$$T(E)={1\over g_0^{-1}-I_d(E,\Lambda)}.$$

The integral $I_d$ is the coincident nonrelativistic resolvent; after replacing its kinetic mass by the reduced mass, the same spatial integral appears in the two-body bubble. The renormalized answer is obtained by replacing the unobservable bare coefficient $g_0(\Lambda)$ by physical low-energy data such as $a$ or $B_2$.

The three-dimensional zero-range result

$$T_3(p)=\frac{2\pi/m}{a^{-1}+ip}$$

is the prototype for much of what follows: a short-distance singularity is absorbed into a local parameter, while the remaining energy dependence is fixed by long-distance propagation and unitarity.

Common pitfalls

Treating $g_0$ as an observable. The bare strength of a point interaction depends on the cutoff and regulator. The observable is the scattering amplitude, or equivalently low-energy parameters such as $a$ and $r_e$.

Confusing the cutoff with the physical range. A cutoff $\Lambda$ is a calculation device. A real finite-range potential has a physical range $R$. A sensible EFT calculation takes $p\ll\Lambda$ while matching coefficients to physics at scales of order $R^{-1}$.

Dropping the $i0$. The imaginary part of $I_d$ is not a detail. It fixes outgoing boundary conditions and enforces elastic unitarity.

Expecting the same behavior in all dimensions. The contact interaction is finite in $d=1$, logarithmic in $d=2$, and linearly divergent in $d=3$. Dimensionality changes the physics.

Calling every divergence a failure. The divergence tells us that the zero-range interaction needs a matching condition. Once matched, it makes finite low-energy predictions.

Exercises

Exercise 1

Derive the geometric formula

$$T(E)={1\over g_0^{-1}-I_d(E,\Lambda)}$$

from the Lippmann–Schwinger equation for $V(\boldsymbol r)=g_0\delta^{(d)}(\boldsymbol r)$.

Solution

The momentum-space Lippmann–Schwinger equation is

$$T(\boldsymbol p',\boldsymbol p;E) = V(\boldsymbol p',\boldsymbol p) + \int^\Lambda {d^dk\over(2\pi)^d} V(\boldsymbol p',\boldsymbol k) {1\over E-k^2/(2m)+i0} T(\boldsymbol k,\boldsymbol p;E).$$

For the contact potential,

$$V(\boldsymbol p',\boldsymbol p)=g_0.$$

Because the kernel is independent of the external momenta, the solution is also independent of them:

$$T(\boldsymbol p',\boldsymbol p;E)=T(E).$$

Substituting gives

$$T(E)=g_0+g_0T(E)\int^\Lambda {d^dk\over(2\pi)^d}{1\over E-k^2/(2m)+i0}.$$

Define

$$I_d(E,\Lambda)=\int^\Lambda {d^dk\over(2\pi)^d}{1\over E-k^2/(2m)+i0}.$$

Then

$$T(E)=g_0+g_0I_d(E,\Lambda)T(E),$$

so

$$\left[1-g_0I_d(E,\Lambda)\right]T(E)=g_0.$$

Therefore

$$T(E)={g_0\over1-g_0I_d(E,\Lambda)}={1\over g_0^{-1}-I_d(E,\Lambda)}.$$

Exercise 2

Compute the one-dimensional integral

$$I_1(E)=2m\int_{-\infty}^{\infty}{dk\over2\pi}\,{1\over p^2-k^2+i0}, \qquad E={p^2\over2m},\quad p>0,$$

and find the binding-energy magnitude for an attractive delta-function potential.

Solution

Use the distribution identity

$${1\over x+i0}=\operatorname{PV}{1\over x}-i\pi\delta(x).$$

Then

$${1\over p^2-k^2+i0} =\operatorname{PV}{1\over p^2-k^2}-i\pi\delta(p^2-k^2).$$

The symmetric principal-value integral over the real line is zero. Also,

$$\delta(p^2-k^2)={1\over2p}\left[\delta(k-p)+\delta(k+p)\right].$$

Hence

$$\int_{-\infty}^{\infty}{dk\over2\pi}\delta(p^2-k^2) ={1\over2\pi p},$$

and

$$I_1(E)=2m\left(-i\pi\right){1\over2\pi p} =-{im\over p}.$$

Thus

$$T_1(p)={1\over g_0^{-1}+im/p}.$$

A bound state has $p=i\kappa$ with $\kappa>0$. The pole condition is

$$g_0^{-1}+{m\over\kappa}=0,$$

so

$$\kappa=-mg_0.$$

This requires $g_0<0$. The binding-energy magnitude and energy are

$$B_1={\kappa^2\over2m}={mg_0^2\over2}, \qquad E_{\mathrm{bound}}=-B_1.$$

Exercise 3

Show that in three dimensions

$$I_3(E,\Lambda) =-{m\Lambda\over\pi^2}-{imp\over2\pi}+O\left({mp^2\over\Lambda}\right).$$

Then derive the zero-range limit

$$T_3(p)=\frac{2\pi/m}{a^{-1}+ip}$$

from the matching condition

$${1\over g_0(\Lambda)}+{m\Lambda\over\pi^2}={m\over2\pi a}.$$

Also find the leading finite-cutoff correction to the inverse amplitude.

Solution

Start with

$$I_3(E,\Lambda)=2m\int_{|\boldsymbol k|<\Lambda}{d^3k\over(2\pi)^3}{1\over p^2-k^2+i0}.$$

The angular integral gives

$$I_3(E,\Lambda)={m\over\pi^2}\int_0^\Lambda dk\,{k^2\over p^2-k^2+i0}.$$

Write

$${k^2\over p^2-k^2+i0}=-1+{p^2\over p^2-k^2+i0}.$$

The first term gives

$$-{m\Lambda\over\pi^2}.$$

For the imaginary part, use

$$\operatorname{Im}{1\over p^2-k^2+i0}=-\pi\delta(p^2-k^2).$$

Thus

$$\operatorname{Im} I_3 ={m\over\pi^2}\int_0^\Lambda dk\,k^2\left[-\pi\delta(p^2-k^2)\right] =-{m\over\pi}\,{p\over2} =-{mp\over2\pi}.$$

For $0<p<\Lambda$, direct principal-value integration gives

$$\operatorname{Re} I_3 ={m\over\pi^2}\left[-\Lambda+{p\over2}\log{\Lambda+p\over\Lambda-p}\right].$$

The remaining real momentum-dependent part is $O(mp^2/\Lambda)$ for $p\ll\Lambda$. Therefore

$$I_3(E,\Lambda)=-{m\Lambda\over\pi^2}-{imp\over2\pi}+O\left({mp^2\over\Lambda}\right).$$

Now

$${1\over T_3(p)}={1\over g_0(\Lambda)}-I_3(E,\Lambda) ={1\over g_0(\Lambda)}+{m\Lambda\over\pi^2}+{imp\over2\pi}+O\left({mp^2\over\Lambda}\right).$$

Using the matching condition gives

$${1\over T_3(p)}={m\over2\pi a}+{imp\over2\pi}+O\left({mp^2\over\Lambda}\right) ={m\over2\pi}(a^{-1}+ip)+O\left({mp^2\over\Lambda}\right).$$

More precisely, expanding the logarithm in the exact principal value gives

$${1\over T_{3,\Lambda}(p)} ={m\over2\pi}\left[a^{-1}+ip-{2p^2\over\pi\Lambda} +O\left({p^4\over\Lambda^3}\right)\right].$$

Taking $\Lambda\to\infty$ at fixed $a,p$ gives $T_3(p)=(2\pi/m)/(a^{-1}+ip)$. Inverting the expansion at finite cutoff requires the denominator correction to be small compared with $a^{-1}+ip$; it is not a uniform approximation near a continued pole.

Exercise 4

In two dimensions, define $g_R(\mu)$ by

$${1\over g_R(\mu)}={1\over g_0(\Lambda)}+{m\over2\pi}\log{\Lambda^2\over\mu^2}.$$

Derive the beta function for $g_R(\mu)$ at fixed bare coupling.

Solution

Differentiate at fixed $g_0$ and $\Lambda$:

$$\mu{d\over d\mu}{1\over g_R(\mu)} ={m\over2\pi}\mu{d\over d\mu}\log{\Lambda^2\over\mu^2} =-{m\over\pi}.$$

But

$$\mu{d\over d\mu}{1\over g_R} =-{1\over g_R^2}\mu{dg_R\over d\mu}.$$

Therefore

$$-{1\over g_R^2}\mu{dg_R\over d\mu}=-{m\over\pi},$$

so

$$\boxed{\mu{dg_R\over d\mu}={m\over\pi}g_R^2.}$$

If one defines the dimensionless coupling $\lambda=mg_R/\pi$, then

$$\mu{d\lambda\over d\mu}=\lambda^2.$$

Exercise 5

For the three-dimensional contact amplitude in the zero-range limit, show that $a>0$ gives a bound-state pole and find its energy.

Solution

The amplitude has denominator

$$a^{-1}+ip.$$

A bound state corresponds to imaginary momentum

$$p=i\kappa, \qquad \kappa>0,$$

and bound-state energy

$$E_{\mathrm{bound}}=-{\kappa^2\over2m}.$$

The pole condition is

$$a^{-1}+i(i\kappa)=a^{-1}-\kappa=0.$$

Thus

$$\kappa={1\over a}.$$

This is positive only when $a>0$. The binding-energy magnitude and bound-state energy are

$$B_3={1\over2ma^2}, \qquad E_{\mathrm{bound}}=-B_3.$$

As $a\to\infty$, the pole approaches threshold and the scattering becomes resonant.

Exercise 6

At fixed cutoff $\Lambda$, define the critical bare coupling in three dimensions by the unitary-limit condition $a^{-1}=0$:

$$g_{0,c}(\Lambda)=-{\pi^2\over m\Lambda}.$$

Write $g_0=g_{0,c}(1+\epsilon)$ with $|\epsilon|\ll1$. Find the scattering length $a$ to leading order in $\epsilon$.

Solution

The matching condition is

$${1\over g_0}+{m\Lambda\over\pi^2}={m\over2\pi a}.$$

Since

$$g_0=g_{0,c}(1+\epsilon), \qquad {1\over g_{0,c}}=-{m\Lambda\over\pi^2},$$

we have exactly,

$${1\over g_0} ={1\over g_{0,c}}{1\over1+\epsilon} =-{m\Lambda\over\pi^2}{1\over1+\epsilon}.$$

Therefore

$${1\over g_0}+{m\Lambda\over\pi^2} ={m\Lambda\over\pi^2}{\epsilon\over1+\epsilon}.$$

Matching gives

$${m\over2\pi a}={m\Lambda\over\pi^2}{\epsilon\over1+\epsilon},$$

so

$$\boxed{ a={\pi(1+\epsilon)\over2\Lambda\epsilon} ={\pi\over2\Lambda\epsilon}[1+O(\epsilon)]. }$$

A large scattering length is therefore a near-critical tuning: $|a|\Lambda\gg1$ requires $|\epsilon|\ll1$.

References

• Braaten, Eric, and Hans-Werner Hammer. “Universality in Few-Body Systems with Large Scattering Length.” Physics Reports 428 (2006): 259–390. DOI. Open PDF, version 3, 18 August 2006; locators above use the printed preprint pages.

Further reading

• Hammer, Hans-Werner, and Richard J. Furnstahl. “Effective Field Theory for Dilute Fermi Systems.” Nuclear Physics A 678, nos. 3–4 (2000): 277–294.
• Kaplan, David B., Martin J. Savage, and Mark B. Wise. “A New Expansion for Nucleon–Nucleon Interactions.” Physics Letters B 424, nos. 3–4 (1998): 390–396.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge: Cambridge University Press, 2014, Chapters 22–23.
• Srednicki, Mark. Quantum Field Theory. Cambridge: Cambridge University Press, 2007, Sections 27–29.
• Weinberg, Steven. The Quantum Theory of Fields. Volume I: Foundations. Cambridge: Cambridge University Press, 1995, Chapter 3.
• Zee, A. Quantum Field Theory in a Nutshell. 2nd ed. Princeton: Princeton University Press, 2010.