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

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

The $T$ matrix becomes

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

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

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

Then

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

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 a dimensionless bare contact interaction has produced the dimensionful 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 $p\ll\Lambda$,

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

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

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

The low-energy $T$ matrix is then

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

The corresponding $s$-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

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

Normalization alert

This is the fixed-center potential-scattering normalization. In nonrelativistic many-body EFT one often studies two particles with reduced mass and different state normalizations, so the same physical denominator may be accompanied by factors such as $4\pi/m$ rather than $2\pi/m$. The universal content is the low-energy inverse amplitude,

$$f_0(p)^{-1}=-a^{-1}-ip+O(p^2).$$

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.

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 $\Lambda$,

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

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; the matched low-energy amplitude remains fixed.

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 pure contact interaction captures $a$ but sets range corrections to zero at leading order. 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 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 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 bubble series as the potential problem. After doing the loop energy integral, the bubble reduces to the spatial integral $I_d(E,\Lambda)$. Thus the contact potential is not an unrelated quantum-mechanics curiosity; it is the simplest 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 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; in the field-theory formulation it is the spatial integral left after the bubble energy integral. The renormalized answer is not obtained by pretending the divergence was never there. It 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 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

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

from the matching condition

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

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

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

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

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

Therefore

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

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, 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, to first order in $\epsilon$,

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

Therefore

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

Matching gives

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

so

$$\boxed{ a={\pi\over2\Lambda\epsilon}+O(\epsilon^0). }$$

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

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.
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